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Today’s Outline - October 26, 2017
• The Stern-Gerlach experiment
• Total spin of hydrogen
• Adding angular momentum
• More chapter 4 problems
Reading Assignment: Chapter 4.4
Homework Assignment #07:Chapter 4: 1,2,4,5,7,8due Tuesday, October 24, 2017
Homework Assignment #08:Chapter 4: 10,13,14,15,16,38due Tuesday, October 31, 2017
Midterm Exam 2: Thursday, November 9, 2017covers through Chapter 4
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 1 / 17
Today’s Outline - October 26, 2017
• The Stern-Gerlach experiment
• Total spin of hydrogen
• Adding angular momentum
• More chapter 4 problems
Reading Assignment: Chapter 4.4
Homework Assignment #07:Chapter 4: 1,2,4,5,7,8due Tuesday, October 24, 2017
Homework Assignment #08:Chapter 4: 10,13,14,15,16,38due Tuesday, October 31, 2017
Midterm Exam 2: Thursday, November 9, 2017covers through Chapter 4
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 1 / 17
Today’s Outline - October 26, 2017
• The Stern-Gerlach experiment
• Total spin of hydrogen
• Adding angular momentum
• More chapter 4 problems
Reading Assignment: Chapter 4.4
Homework Assignment #07:Chapter 4: 1,2,4,5,7,8due Tuesday, October 24, 2017
Homework Assignment #08:Chapter 4: 10,13,14,15,16,38due Tuesday, October 31, 2017
Midterm Exam 2: Thursday, November 9, 2017covers through Chapter 4
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 1 / 17
Today’s Outline - October 26, 2017
• The Stern-Gerlach experiment
• Total spin of hydrogen
• Adding angular momentum
• More chapter 4 problems
Reading Assignment: Chapter 4.4
Homework Assignment #07:Chapter 4: 1,2,4,5,7,8due Tuesday, October 24, 2017
Homework Assignment #08:Chapter 4: 10,13,14,15,16,38due Tuesday, October 31, 2017
Midterm Exam 2: Thursday, November 9, 2017covers through Chapter 4
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 1 / 17
Today’s Outline - October 26, 2017
• The Stern-Gerlach experiment
• Total spin of hydrogen
• Adding angular momentum
• More chapter 4 problems
Reading Assignment: Chapter 4.4
Homework Assignment #07:Chapter 4: 1,2,4,5,7,8due Tuesday, October 24, 2017
Homework Assignment #08:Chapter 4: 10,13,14,15,16,38due Tuesday, October 31, 2017
Midterm Exam 2: Thursday, November 9, 2017covers through Chapter 4
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 1 / 17
Today’s Outline - October 26, 2017
• The Stern-Gerlach experiment
• Total spin of hydrogen
• Adding angular momentum
• More chapter 4 problems
Reading Assignment: Chapter 4.4
Homework Assignment #07:Chapter 4: 1,2,4,5,7,8due Tuesday, October 24, 2017
Homework Assignment #08:Chapter 4: 10,13,14,15,16,38due Tuesday, October 31, 2017
Midterm Exam 2: Thursday, November 9, 2017covers through Chapter 4
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 1 / 17
Today’s Outline - October 26, 2017
• The Stern-Gerlach experiment
• Total spin of hydrogen
• Adding angular momentum
• More chapter 4 problems
Reading Assignment: Chapter 4.4
Homework Assignment #07:Chapter 4: 1,2,4,5,7,8due Tuesday, October 24, 2017
Homework Assignment #08:Chapter 4: 10,13,14,15,16,38due Tuesday, October 31, 2017
Midterm Exam 2: Thursday, November 9, 2017covers through Chapter 4
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 1 / 17
Today’s Outline - October 26, 2017
• The Stern-Gerlach experiment
• Total spin of hydrogen
• Adding angular momentum
• More chapter 4 problems
Reading Assignment: Chapter 4.4
Homework Assignment #07:Chapter 4: 1,2,4,5,7,8due Tuesday, October 24, 2017
Homework Assignment #08:Chapter 4: 10,13,14,15,16,38due Tuesday, October 31, 2017
Midterm Exam 2: Thursday, November 9, 2017covers through Chapter 4
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 1 / 17
Today’s Outline - October 26, 2017
• The Stern-Gerlach experiment
• Total spin of hydrogen
• Adding angular momentum
• More chapter 4 problems
Reading Assignment: Chapter 4.4
Homework Assignment #07:Chapter 4: 1,2,4,5,7,8due Tuesday, October 24, 2017
Homework Assignment #08:Chapter 4: 10,13,14,15,16,38due Tuesday, October 31, 2017
Midterm Exam 2: Thursday, November 9, 2017covers through Chapter 4
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 1 / 17
Angular momentum in a non-uniform B-field
An inhomogeneous magnetic produces a force on a magnetic moment aswell as a torque
~F = ∇(~µ · ~B)
If a beam of heavy neutral atoms traveling in the y direction passesthrough a field
~B = −αxx + (B0 + αz)z
it will experience a force
~F = γα(−Sx x + Sz z)
since the moment precesses about ~B0, the x-component will average tozero leaving a net force Fz = γαSz
Suppose we use a beam of atoms with spin 12 . What do we expect to see
in an experiment?
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 2 / 17
Angular momentum in a non-uniform B-field
An inhomogeneous magnetic produces a force on a magnetic moment aswell as a torque
~F = ∇(~µ · ~B)
If a beam of heavy neutral atoms traveling in the y direction passesthrough a field
~B = −αxx + (B0 + αz)z
it will experience a force
~F = γα(−Sx x + Sz z)
since the moment precesses about ~B0, the x-component will average tozero leaving a net force Fz = γαSz
Suppose we use a beam of atoms with spin 12 . What do we expect to see
in an experiment?
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 2 / 17
Angular momentum in a non-uniform B-field
An inhomogeneous magnetic produces a force on a magnetic moment aswell as a torque
~F = ∇(~µ · ~B)
If a beam of heavy neutral atoms traveling in the y direction passesthrough a field
~B = −αxx + (B0 + αz)z
it will experience a force
~F = γα(−Sx x + Sz z)
since the moment precesses about ~B0, the x-component will average tozero leaving a net force Fz = γαSz
Suppose we use a beam of atoms with spin 12 . What do we expect to see
in an experiment?
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 2 / 17
Angular momentum in a non-uniform B-field
An inhomogeneous magnetic produces a force on a magnetic moment aswell as a torque
~F = ∇(~µ · ~B)
If a beam of heavy neutral atoms traveling in the y direction passesthrough a field
~B = −αxx + (B0 + αz)z
it will experience a force
~F = γα(−Sx x + Sz z)
since the moment precesses about ~B0, the x-component will average tozero leaving a net force Fz = γαSz
Suppose we use a beam of atoms with spin 12 . What do we expect to see
in an experiment?
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 2 / 17
Angular momentum in a non-uniform B-field
An inhomogeneous magnetic produces a force on a magnetic moment aswell as a torque
~F = ∇(~µ · ~B)
If a beam of heavy neutral atoms traveling in the y direction passesthrough a field
~B = −αxx + (B0 + αz)z
it will experience a force
~F = γα(−Sx x + Sz z)
since the moment precesses about ~B0, the x-component will average tozero leaving a net force Fz = γαSz
Suppose we use a beam of atoms with spin 12 . What do we expect to see
in an experiment?
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 2 / 17
Angular momentum in a non-uniform B-field
An inhomogeneous magnetic produces a force on a magnetic moment aswell as a torque
~F = ∇(~µ · ~B)
If a beam of heavy neutral atoms traveling in the y direction passesthrough a field
~B = −αxx + (B0 + αz)z
it will experience a force
~F = γα(−Sx x + Sz z)
since the moment precesses about ~B0, the x-component will average tozero leaving a net force Fz = γαSz
Suppose we use a beam of atoms with spin 12 . What do we expect to see
in an experiment?
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 2 / 17
The Stern-Gerlach experiment (SGE)
In 1922, Stern and Gerlach performed an experiment designed to test theBohr-Sommerfeld theory of the atom which predicted that the electronorbits had a property called “space quantization” in a magnetic field
In the classical regime we would ex-pect that the relative orientation of thespin with respect to the magnetic fieldwould determine where the atom is de-tected spatially.
Because of the Bohr quantized circularorbits, the direction of the orbital mo-tion with respect to the magnetic field,would cause the atoms to be deflectedin opposite directions
N
S
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 3 / 17
The Stern-Gerlach experiment (SGE)
In 1922, Stern and Gerlach performed an experiment designed to test theBohr-Sommerfeld theory of the atom which predicted that the electronorbits had a property called “space quantization” in a magnetic field
In the classical regime we would ex-pect that the relative orientation of thespin with respect to the magnetic fieldwould determine where the atom is de-tected spatially.
Because of the Bohr quantized circularorbits, the direction of the orbital mo-tion with respect to the magnetic field,would cause the atoms to be deflectedin opposite directions
N
S
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 3 / 17
The Stern-Gerlach experiment (SGE)
In 1922, Stern and Gerlach performed an experiment designed to test theBohr-Sommerfeld theory of the atom which predicted that the electronorbits had a property called “space quantization” in a magnetic field
In the classical regime we would ex-pect that the relative orientation of thespin with respect to the magnetic fieldwould determine where the atom is de-tected spatially.
Because of the Bohr quantized circularorbits, the direction of the orbital mo-tion with respect to the magnetic field,would cause the atoms to be deflectedin opposite directions
N
S
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 3 / 17
The SGE result
When the experiment was performed with neutral silver atoms two distinctspots on the screen were seen.
Gerlach sent the above picture on a postcard to announce this result.
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 4 / 17
The SGE result
When the experiment was performed with neutral silver atoms two distinctspots on the screen were seen.
Gerlach sent the above picture on a postcard to announce this result.
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 4 / 17
The SGE result
When the experiment was performed with neutral silver atoms two distinctspots on the screen were seen.
Gerlach sent the above picture on a postcard to announce this result.C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 4 / 17
Reaction to the SGE
Arnold Sommerfeld – Through clever experimental arrangement Stern andGerlach not only demostrated ad oculus the space quantization of atomsin a magnetic field, but they also proved the quantum origin of electricityand its connection with atomic structure
Albert Einstein – The most interesting achievement at this point is theexperiment of Stern and Gerlach. The alignment of the atoms withoutcollisions via radiative exchange is not comprehensible based on currenttheoretical methods.
James Franck – More important is whether this proves the existence ofspace quantization. Please add a few words of explanation to your puzzle,such as what’s really going on.
Niels Bohr – I would be very grateful if you or Stern could let me know, ina few lines, whether you interpret your experimental results in this waythat the atoms are oriented only parallel or opposed, but not normal to thefield, as one could provide theoretical reasons for the latter assertion.
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 5 / 17
Reaction to the SGE
Arnold Sommerfeld – Through clever experimental arrangement Stern andGerlach not only demostrated ad oculus the space quantization of atomsin a magnetic field, but they also proved the quantum origin of electricityand its connection with atomic structure
Albert Einstein – The most interesting achievement at this point is theexperiment of Stern and Gerlach. The alignment of the atoms withoutcollisions via radiative exchange is not comprehensible based on currenttheoretical methods.
James Franck – More important is whether this proves the existence ofspace quantization. Please add a few words of explanation to your puzzle,such as what’s really going on.
Niels Bohr – I would be very grateful if you or Stern could let me know, ina few lines, whether you interpret your experimental results in this waythat the atoms are oriented only parallel or opposed, but not normal to thefield, as one could provide theoretical reasons for the latter assertion.
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 5 / 17
Reaction to the SGE
Arnold Sommerfeld – Through clever experimental arrangement Stern andGerlach not only demostrated ad oculus the space quantization of atomsin a magnetic field, but they also proved the quantum origin of electricityand its connection with atomic structure
Albert Einstein – The most interesting achievement at this point is theexperiment of Stern and Gerlach. The alignment of the atoms withoutcollisions via radiative exchange is not comprehensible based on currenttheoretical methods.
James Franck – More important is whether this proves the existence ofspace quantization. Please add a few words of explanation to your puzzle,such as what’s really going on.
Niels Bohr – I would be very grateful if you or Stern could let me know, ina few lines, whether you interpret your experimental results in this waythat the atoms are oriented only parallel or opposed, but not normal to thefield, as one could provide theoretical reasons for the latter assertion.
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 5 / 17
Reaction to the SGE
Arnold Sommerfeld – Through clever experimental arrangement Stern andGerlach not only demostrated ad oculus the space quantization of atomsin a magnetic field, but they also proved the quantum origin of electricityand its connection with atomic structure
Albert Einstein – The most interesting achievement at this point is theexperiment of Stern and Gerlach. The alignment of the atoms withoutcollisions via radiative exchange is not comprehensible based on currenttheoretical methods.
James Franck – More important is whether this proves the existence ofspace quantization. Please add a few words of explanation to your puzzle,such as what’s really going on.
Niels Bohr – I would be very grateful if you or Stern could let me know, ina few lines, whether you interpret your experimental results in this waythat the atoms are oriented only parallel or opposed, but not normal to thefield, as one could provide theoretical reasons for the latter assertion.
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 5 / 17
Interesting SGE facts
Silver atoms, we now know, have orbital angular momentum l = 0, but asingle electron in the outermost orbital so there cannot be a third beam inthe center. We are seeing the result of the spin of the electron.
The splitting observed was exactly what was predicted by Bohr theory forl = 1. Why would that be when s = 1
2? As we will see later, thegyromagnetic ratio for the electron spin is twice that for its orbital angularmomentum
The Stern-Gerlach result is often cited as proving the existence of electronspin but it was conducted years before a true quantum theory wasdeveloped and before the existence of spin was postulated in 1925. Indeed,only in 1927 was this result understood when Fraser noted that the groundstate of the silver atom has l = 0
The concept of force is not really valid in a quantum context. A moreproper way to look at it is to allow the Hamiltonian to act on the spin 1
2spin wavefunction.
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 6 / 17
Interesting SGE facts
Silver atoms, we now know, have orbital angular momentum l = 0, but asingle electron in the outermost orbital so there cannot be a third beam inthe center. We are seeing the result of the spin of the electron.
The splitting observed was exactly what was predicted by Bohr theory forl = 1. Why would that be when s = 1
2?
As we will see later, thegyromagnetic ratio for the electron spin is twice that for its orbital angularmomentum
The Stern-Gerlach result is often cited as proving the existence of electronspin but it was conducted years before a true quantum theory wasdeveloped and before the existence of spin was postulated in 1925. Indeed,only in 1927 was this result understood when Fraser noted that the groundstate of the silver atom has l = 0
The concept of force is not really valid in a quantum context. A moreproper way to look at it is to allow the Hamiltonian to act on the spin 1
2spin wavefunction.
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 6 / 17
Interesting SGE facts
Silver atoms, we now know, have orbital angular momentum l = 0, but asingle electron in the outermost orbital so there cannot be a third beam inthe center. We are seeing the result of the spin of the electron.
The splitting observed was exactly what was predicted by Bohr theory forl = 1. Why would that be when s = 1
2? As we will see later, thegyromagnetic ratio for the electron spin is twice that for its orbital angularmomentum
The Stern-Gerlach result is often cited as proving the existence of electronspin but it was conducted years before a true quantum theory wasdeveloped and before the existence of spin was postulated in 1925. Indeed,only in 1927 was this result understood when Fraser noted that the groundstate of the silver atom has l = 0
The concept of force is not really valid in a quantum context. A moreproper way to look at it is to allow the Hamiltonian to act on the spin 1
2spin wavefunction.
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 6 / 17
Interesting SGE facts
Silver atoms, we now know, have orbital angular momentum l = 0, but asingle electron in the outermost orbital so there cannot be a third beam inthe center. We are seeing the result of the spin of the electron.
The splitting observed was exactly what was predicted by Bohr theory forl = 1. Why would that be when s = 1
2? As we will see later, thegyromagnetic ratio for the electron spin is twice that for its orbital angularmomentum
The Stern-Gerlach result is often cited as proving the existence of electronspin but it was conducted years before a true quantum theory wasdeveloped and before the existence of spin was postulated in 1925.
Indeed,only in 1927 was this result understood when Fraser noted that the groundstate of the silver atom has l = 0
The concept of force is not really valid in a quantum context. A moreproper way to look at it is to allow the Hamiltonian to act on the spin 1
2spin wavefunction.
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 6 / 17
Interesting SGE facts
Silver atoms, we now know, have orbital angular momentum l = 0, but asingle electron in the outermost orbital so there cannot be a third beam inthe center. We are seeing the result of the spin of the electron.
The splitting observed was exactly what was predicted by Bohr theory forl = 1. Why would that be when s = 1
2? As we will see later, thegyromagnetic ratio for the electron spin is twice that for its orbital angularmomentum
The Stern-Gerlach result is often cited as proving the existence of electronspin but it was conducted years before a true quantum theory wasdeveloped and before the existence of spin was postulated in 1925. Indeed,only in 1927 was this result understood when Fraser noted that the groundstate of the silver atom has l = 0
The concept of force is not really valid in a quantum context. A moreproper way to look at it is to allow the Hamiltonian to act on the spin 1
2spin wavefunction.
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 6 / 17
Interesting SGE facts
Silver atoms, we now know, have orbital angular momentum l = 0, but asingle electron in the outermost orbital so there cannot be a third beam inthe center. We are seeing the result of the spin of the electron.
The splitting observed was exactly what was predicted by Bohr theory forl = 1. Why would that be when s = 1
2? As we will see later, thegyromagnetic ratio for the electron spin is twice that for its orbital angularmomentum
The Stern-Gerlach result is often cited as proving the existence of electronspin but it was conducted years before a true quantum theory wasdeveloped and before the existence of spin was postulated in 1925. Indeed,only in 1927 was this result understood when Fraser noted that the groundstate of the silver atom has l = 0
The concept of force is not really valid in a quantum context. A moreproper way to look at it is to allow the Hamiltonian to act on the spin 1
2spin wavefunction.
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 6 / 17
Space quantization
Consider the frame of the par-ticles in the beam. At timet = 0, the beam begins the tran-sit through the magnetic fieldgradient for a time T . Ignor-ing the x-component, the timedependent Hamiltonian is
start with a beam of mixed spin12 with a wave function
H(t) =
0, t < 0
−γ(B0 + αz)Sz , 0 ≤ t ≤ T
0, t > T
χ(0) = aχ+ + bχ−
E± = ∓γ(B0 + αz)~2
χ(t) = aχ+e−iE+t/~ + bχ−e
−iE−t/~
χ(T ) =(ae iγTB0/2χ+
)e i(αγT/2)z +
(be−iγTB0/2χ−
)e−i(αγT/2)z
〈pz〉 = χ(T )∗~i
∂
∂zχ(T )
= |a|2αγT~2− |b|2αγT~
2
There is momentum in the ±z-direction imparted to the beam as ittransits the magnetic field
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 7 / 17
Space quantization
Consider the frame of the par-ticles in the beam.
At timet = 0, the beam begins the tran-sit through the magnetic fieldgradient for a time T . Ignor-ing the x-component, the timedependent Hamiltonian is
start with a beam of mixed spin12 with a wave function
H(t) =
0, t < 0
−γ(B0 + αz)Sz , 0 ≤ t ≤ T
0, t > T
χ(0) = aχ+ + bχ−
E± = ∓γ(B0 + αz)~2
χ(t) = aχ+e−iE+t/~ + bχ−e
−iE−t/~
χ(T ) =(ae iγTB0/2χ+
)e i(αγT/2)z +
(be−iγTB0/2χ−
)e−i(αγT/2)z
〈pz〉 = χ(T )∗~i
∂
∂zχ(T )
= |a|2αγT~2− |b|2αγT~
2
There is momentum in the ±z-direction imparted to the beam as ittransits the magnetic field
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 7 / 17
Space quantization
Consider the frame of the par-ticles in the beam. At timet = 0, the beam begins the tran-sit through the magnetic fieldgradient for a time T .
Ignor-ing the x-component, the timedependent Hamiltonian is
start with a beam of mixed spin12 with a wave function
H(t) =
0, t < 0
−γ(B0 + αz)Sz , 0 ≤ t ≤ T
0, t > T
χ(0) = aχ+ + bχ−
E± = ∓γ(B0 + αz)~2
χ(t) = aχ+e−iE+t/~ + bχ−e
−iE−t/~
χ(T ) =(ae iγTB0/2χ+
)e i(αγT/2)z +
(be−iγTB0/2χ−
)e−i(αγT/2)z
〈pz〉 = χ(T )∗~i
∂
∂zχ(T )
= |a|2αγT~2− |b|2αγT~
2
There is momentum in the ±z-direction imparted to the beam as ittransits the magnetic field
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 7 / 17
Space quantization
Consider the frame of the par-ticles in the beam. At timet = 0, the beam begins the tran-sit through the magnetic fieldgradient for a time T . Ignor-ing the x-component, the timedependent Hamiltonian is
start with a beam of mixed spin12 with a wave function
H(t) =
0, t < 0
−γ(B0 + αz)Sz , 0 ≤ t ≤ T
0, t > T
χ(0) = aχ+ + bχ−
E± = ∓γ(B0 + αz)~2
χ(t) = aχ+e−iE+t/~ + bχ−e
−iE−t/~
χ(T ) =(ae iγTB0/2χ+
)e i(αγT/2)z +
(be−iγTB0/2χ−
)e−i(αγT/2)z
〈pz〉 = χ(T )∗~i
∂
∂zχ(T )
= |a|2αγT~2− |b|2αγT~
2
There is momentum in the ±z-direction imparted to the beam as ittransits the magnetic field
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 7 / 17
Space quantization
Consider the frame of the par-ticles in the beam. At timet = 0, the beam begins the tran-sit through the magnetic fieldgradient for a time T . Ignor-ing the x-component, the timedependent Hamiltonian is
start with a beam of mixed spin12 with a wave function
H(t) =
0, t < 0
−γ(B0 + αz)Sz , 0 ≤ t ≤ T
0, t > T
χ(0) = aχ+ + bχ−
E± = ∓γ(B0 + αz)~2
χ(t) = aχ+e−iE+t/~ + bχ−e
−iE−t/~
χ(T ) =(ae iγTB0/2χ+
)e i(αγT/2)z +
(be−iγTB0/2χ−
)e−i(αγT/2)z
〈pz〉 = χ(T )∗~i
∂
∂zχ(T )
= |a|2αγT~2− |b|2αγT~
2
There is momentum in the ±z-direction imparted to the beam as ittransits the magnetic field
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 7 / 17
Space quantization
Consider the frame of the par-ticles in the beam. At timet = 0, the beam begins the tran-sit through the magnetic fieldgradient for a time T . Ignor-ing the x-component, the timedependent Hamiltonian is
start with a beam of mixed spin12 with a wave function
H(t) =
0, t < 0
−γ(B0 + αz)Sz , 0 ≤ t ≤ T
0, t > T
χ(0) = aχ+ + bχ−
E± = ∓γ(B0 + αz)~2
χ(t) = aχ+e−iE+t/~ + bχ−e
−iE−t/~
χ(T ) =(ae iγTB0/2χ+
)e i(αγT/2)z +
(be−iγTB0/2χ−
)e−i(αγT/2)z
〈pz〉 = χ(T )∗~i
∂
∂zχ(T )
= |a|2αγT~2− |b|2αγT~
2
There is momentum in the ±z-direction imparted to the beam as ittransits the magnetic field
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 7 / 17
Space quantization
Consider the frame of the par-ticles in the beam. At timet = 0, the beam begins the tran-sit through the magnetic fieldgradient for a time T . Ignor-ing the x-component, the timedependent Hamiltonian is
start with a beam of mixed spin12 with a wave function
H(t) =
0, t < 0
−γ(B0 + αz)Sz , 0 ≤ t ≤ T
0, t > T
χ(0) = aχ+ + bχ−
E± = ∓γ(B0 + αz)~2
χ(t) = aχ+e−iE+t/~ + bχ−e
−iE−t/~
χ(T ) =(ae iγTB0/2χ+
)e i(αγT/2)z +
(be−iγTB0/2χ−
)e−i(αγT/2)z
〈pz〉 = χ(T )∗~i
∂
∂zχ(T )
= |a|2αγT~2− |b|2αγT~
2
There is momentum in the ±z-direction imparted to the beam as ittransits the magnetic field
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 7 / 17
Space quantization
Consider the frame of the par-ticles in the beam. At timet = 0, the beam begins the tran-sit through the magnetic fieldgradient for a time T . Ignor-ing the x-component, the timedependent Hamiltonian is
start with a beam of mixed spin12 with a wave function
H(t) =
0, t < 0
−γ(B0 + αz)Sz , 0 ≤ t ≤ T
0, t > T
χ(0) = aχ+ + bχ−
E± = ∓γ(B0 + αz)~2
χ(t) = aχ+e−iE+t/~ + bχ−e
−iE−t/~
χ(T ) =(ae iγTB0/2χ+
)e i(αγT/2)z +
(be−iγTB0/2χ−
)e−i(αγT/2)z
〈pz〉 = χ(T )∗~i
∂
∂zχ(T )
= |a|2αγT~2− |b|2αγT~
2
There is momentum in the ±z-direction imparted to the beam as ittransits the magnetic field
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 7 / 17
Space quantization
Consider the frame of the par-ticles in the beam. At timet = 0, the beam begins the tran-sit through the magnetic fieldgradient for a time T . Ignor-ing the x-component, the timedependent Hamiltonian is
start with a beam of mixed spin12 with a wave function
H(t) =
0, t < 0
−γ(B0 + αz)Sz , 0 ≤ t ≤ T
0, t > T
χ(0) = aχ+ + bχ−
E± = ∓γ(B0 + αz)~2
χ(t) = aχ+e−iE+t/~ + bχ−e
−iE−t/~
χ(T ) =(ae iγTB0/2χ+
)e i(αγT/2)z +
(be−iγTB0/2χ−
)e−i(αγT/2)z
〈pz〉 = χ(T )∗~i
∂
∂zχ(T )
= |a|2αγT~2− |b|2αγT~
2
There is momentum in the ±z-direction imparted to the beam as ittransits the magnetic field
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 7 / 17
Space quantization
Consider the frame of the par-ticles in the beam. At timet = 0, the beam begins the tran-sit through the magnetic fieldgradient for a time T . Ignor-ing the x-component, the timedependent Hamiltonian is
start with a beam of mixed spin12 with a wave function
H(t) =
0, t < 0
−γ(B0 + αz)Sz , 0 ≤ t ≤ T
0, t > T
χ(0) = aχ+ + bχ−
E± = ∓γ(B0 + αz)~2
χ(t) = aχ+e−iE+t/~ + bχ−e
−iE−t/~
χ(T ) =(ae iγTB0/2χ+
)e i(αγT/2)z +
(be−iγTB0/2χ−
)e−i(αγT/2)z
〈pz〉 = χ(T )∗~i
∂
∂zχ(T )
= |a|2αγT~2− |b|2αγT~
2
There is momentum in the ±z-direction imparted to the beam as ittransits the magnetic field
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 7 / 17
Space quantization
Consider the frame of the par-ticles in the beam. At timet = 0, the beam begins the tran-sit through the magnetic fieldgradient for a time T . Ignor-ing the x-component, the timedependent Hamiltonian is
start with a beam of mixed spin12 with a wave function
H(t) =
0, t < 0
−γ(B0 + αz)Sz , 0 ≤ t ≤ T
0, t > T
χ(0) = aχ+ + bχ−
E± = ∓γ(B0 + αz)~2
χ(t) = aχ+e−iE+t/~ + bχ−e
−iE−t/~
χ(T ) =(ae iγTB0/2χ+
)e i(αγT/2)z +
(be−iγTB0/2χ−
)e−i(αγT/2)z
〈pz〉 = χ(T )∗~i
∂
∂zχ(T ) = |a|2αγT~
2− |b|2αγT~
2
There is momentum in the ±z-direction imparted to the beam as ittransits the magnetic field
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 7 / 17
Space quantization
Consider the frame of the par-ticles in the beam. At timet = 0, the beam begins the tran-sit through the magnetic fieldgradient for a time T . Ignor-ing the x-component, the timedependent Hamiltonian is
start with a beam of mixed spin12 with a wave function
H(t) =
0, t < 0
−γ(B0 + αz)Sz , 0 ≤ t ≤ T
0, t > T
χ(0) = aχ+ + bχ−
E± = ∓γ(B0 + αz)~2
χ(t) = aχ+e−iE+t/~ + bχ−e
−iE−t/~
χ(T ) =(ae iγTB0/2χ+
)e i(αγT/2)z +
(be−iγTB0/2χ−
)e−i(αγT/2)z
〈pz〉 = χ(T )∗~i
∂
∂zχ(T ) = |a|2αγT~
2− |b|2αγT~
2
There is momentum in the ±z-direction imparted to the beam as ittransits the magnetic field
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 7 / 17
Total spin of the hydrogen atom
Suppose we have a hydrogen atom in its ground state (L = 0).
Theelectron, and the proton, each have spin S = 1
2. What is the total angular
momentum of the atom?
Each particle can have spin up or down, for 4 combinations.
↑↑ ↑↓ ↓↑ ↓↓
The total angular momentum is just the sum
~S ≡ ~S (1) + ~S (2)
each of the four possible states is an eigenfunction of the z-component ofthe total angular momentum
Szχ1χ2 = (S(1)z + S
(2)z )χ1χ2 = (S
(1)z χ1)χ2 + χ1(S
(2)z χ2)
= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 8 / 17
Total spin of the hydrogen atom
Suppose we have a hydrogen atom in its ground state (L = 0). Theelectron, and the proton, each have spin S = 1
2.
What is the total angularmomentum of the atom?
Each particle can have spin up or down, for 4 combinations.
↑↑ ↑↓ ↓↑ ↓↓
The total angular momentum is just the sum
~S ≡ ~S (1) + ~S (2)
each of the four possible states is an eigenfunction of the z-component ofthe total angular momentum
Szχ1χ2 = (S(1)z + S
(2)z )χ1χ2 = (S
(1)z χ1)χ2 + χ1(S
(2)z χ2)
= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 8 / 17
Total spin of the hydrogen atom
Suppose we have a hydrogen atom in its ground state (L = 0). Theelectron, and the proton, each have spin S = 1
2. What is the total angular
momentum of the atom?
Each particle can have spin up or down, for 4 combinations.
↑↑ ↑↓ ↓↑ ↓↓
The total angular momentum is just the sum
~S ≡ ~S (1) + ~S (2)
each of the four possible states is an eigenfunction of the z-component ofthe total angular momentum
Szχ1χ2 = (S(1)z + S
(2)z )χ1χ2 = (S
(1)z χ1)χ2 + χ1(S
(2)z χ2)
= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 8 / 17
Total spin of the hydrogen atom
Suppose we have a hydrogen atom in its ground state (L = 0). Theelectron, and the proton, each have spin S = 1
2. What is the total angular
momentum of the atom?
Each particle can have spin up or down, for 4 combinations.
↑↑ ↑↓ ↓↑ ↓↓
The total angular momentum is just the sum
~S ≡ ~S (1) + ~S (2)
each of the four possible states is an eigenfunction of the z-component ofthe total angular momentum
Szχ1χ2 = (S(1)z + S
(2)z )χ1χ2 = (S
(1)z χ1)χ2 + χ1(S
(2)z χ2)
= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 8 / 17
Total spin of the hydrogen atom
Suppose we have a hydrogen atom in its ground state (L = 0). Theelectron, and the proton, each have spin S = 1
2. What is the total angular
momentum of the atom?
Each particle can have spin up or down, for 4 combinations.
↑↑
↑↓ ↓↑ ↓↓
The total angular momentum is just the sum
~S ≡ ~S (1) + ~S (2)
each of the four possible states is an eigenfunction of the z-component ofthe total angular momentum
Szχ1χ2 = (S(1)z + S
(2)z )χ1χ2 = (S
(1)z χ1)χ2 + χ1(S
(2)z χ2)
= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 8 / 17
Total spin of the hydrogen atom
Suppose we have a hydrogen atom in its ground state (L = 0). Theelectron, and the proton, each have spin S = 1
2. What is the total angular
momentum of the atom?
Each particle can have spin up or down, for 4 combinations.
↑↑ ↑↓
↓↑ ↓↓
The total angular momentum is just the sum
~S ≡ ~S (1) + ~S (2)
each of the four possible states is an eigenfunction of the z-component ofthe total angular momentum
Szχ1χ2 = (S(1)z + S
(2)z )χ1χ2 = (S
(1)z χ1)χ2 + χ1(S
(2)z χ2)
= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 8 / 17
Total spin of the hydrogen atom
Suppose we have a hydrogen atom in its ground state (L = 0). Theelectron, and the proton, each have spin S = 1
2. What is the total angular
momentum of the atom?
Each particle can have spin up or down, for 4 combinations.
↑↑ ↑↓ ↓↑
↓↓
The total angular momentum is just the sum
~S ≡ ~S (1) + ~S (2)
each of the four possible states is an eigenfunction of the z-component ofthe total angular momentum
Szχ1χ2 = (S(1)z + S
(2)z )χ1χ2 = (S
(1)z χ1)χ2 + χ1(S
(2)z χ2)
= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 8 / 17
Total spin of the hydrogen atom
Suppose we have a hydrogen atom in its ground state (L = 0). Theelectron, and the proton, each have spin S = 1
2. What is the total angular
momentum of the atom?
Each particle can have spin up or down, for 4 combinations.
↑↑ ↑↓ ↓↑ ↓↓
The total angular momentum is just the sum
~S ≡ ~S (1) + ~S (2)
each of the four possible states is an eigenfunction of the z-component ofthe total angular momentum
Szχ1χ2 = (S(1)z + S
(2)z )χ1χ2 = (S
(1)z χ1)χ2 + χ1(S
(2)z χ2)
= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 8 / 17
Total spin of the hydrogen atom
Suppose we have a hydrogen atom in its ground state (L = 0). Theelectron, and the proton, each have spin S = 1
2. What is the total angular
momentum of the atom?
Each particle can have spin up or down, for 4 combinations.
↑↑ ↑↓ ↓↑ ↓↓
The total angular momentum is just the sum
~S ≡ ~S (1) + ~S (2)
each of the four possible states is an eigenfunction of the z-component ofthe total angular momentum
Szχ1χ2 = (S(1)z + S
(2)z )χ1χ2 = (S
(1)z χ1)χ2 + χ1(S
(2)z χ2)
= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 8 / 17
Total spin of the hydrogen atom
Suppose we have a hydrogen atom in its ground state (L = 0). Theelectron, and the proton, each have spin S = 1
2. What is the total angular
momentum of the atom?
Each particle can have spin up or down, for 4 combinations.
↑↑ ↑↓ ↓↑ ↓↓
The total angular momentum is just the sum
~S ≡ ~S (1) + ~S (2)
each of the four possible states is an eigenfunction of the z-component ofthe total angular momentum
Szχ1χ2 = (S(1)z + S
(2)z )χ1χ2 = (S
(1)z χ1)χ2 + χ1(S
(2)z χ2)
= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 8 / 17
Total spin of the hydrogen atom
Suppose we have a hydrogen atom in its ground state (L = 0). Theelectron, and the proton, each have spin S = 1
2. What is the total angular
momentum of the atom?
Each particle can have spin up or down, for 4 combinations.
↑↑ ↑↓ ↓↑ ↓↓
The total angular momentum is just the sum
~S ≡ ~S (1) + ~S (2)
each of the four possible states is an eigenfunction of the z-component ofthe total angular momentum
Szχ1χ2 = (S(1)z + S
(2)z )χ1χ2
= (S(1)z χ1)χ2 + χ1(S
(2)z χ2)
= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 8 / 17
Total spin of the hydrogen atom
Suppose we have a hydrogen atom in its ground state (L = 0). Theelectron, and the proton, each have spin S = 1
2. What is the total angular
momentum of the atom?
Each particle can have spin up or down, for 4 combinations.
↑↑ ↑↓ ↓↑ ↓↓
The total angular momentum is just the sum
~S ≡ ~S (1) + ~S (2)
each of the four possible states is an eigenfunction of the z-component ofthe total angular momentum
Szχ1χ2 = (S(1)z + S
(2)z )χ1χ2 = (S
(1)z χ1)χ2 + χ1(S
(2)z χ2)
= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 8 / 17
Total spin of the hydrogen atom
Suppose we have a hydrogen atom in its ground state (L = 0). Theelectron, and the proton, each have spin S = 1
2. What is the total angular
momentum of the atom?
Each particle can have spin up or down, for 4 combinations.
↑↑ ↑↓ ↓↑ ↓↓
The total angular momentum is just the sum
~S ≡ ~S (1) + ~S (2)
each of the four possible states is an eigenfunction of the z-component ofthe total angular momentum
Szχ1χ2 = (S(1)z + S
(2)z )χ1χ2 = (S
(1)z χ1)χ2 + χ1(S
(2)z χ2)
= (~m1χ1)χ2 + χ1(~m2χ2)
= ~(m1 + m2)χ1χ2
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 8 / 17
Total spin of the hydrogen atom
Suppose we have a hydrogen atom in its ground state (L = 0). Theelectron, and the proton, each have spin S = 1
2. What is the total angular
momentum of the atom?
Each particle can have spin up or down, for 4 combinations.
↑↑ ↑↓ ↓↑ ↓↓
The total angular momentum is just the sum
~S ≡ ~S (1) + ~S (2)
each of the four possible states is an eigenfunction of the z-component ofthe total angular momentum
Szχ1χ2 = (S(1)z + S
(2)z )χ1χ2 = (S
(1)z χ1)χ2 + χ1(S
(2)z χ2)
= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 8 / 17
Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1.
Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ = ~2 12(32
)↑↑+ ~2 12
(32
)↑↑+ 2
(~24 ↓↓ −
~24 ↓↓+ ~2
4 ↑↑)
= 2~2↑↑
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 9 / 17
Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ = ~2 12(32
)↑↑+ ~2 12
(32
)↑↑+ 2
(~24 ↓↓ −
~24 ↓↓+ ~2
4 ↑↑)
= 2~2↑↑
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 9 / 17
Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2))
= (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ = ~2 12(32
)↑↑+ ~2 12
(32
)↑↑+ 2
(~24 ↓↓ −
~24 ↓↓+ ~2
4 ↑↑)
= 2~2↑↑
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 9 / 17
Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ = ~2 12(32
)↑↑+ ~2 12
(32
)↑↑+ 2
(~24 ↓↓ −
~24 ↓↓+ ~2
4 ↑↑)
= 2~2↑↑
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 9 / 17
Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ = ~2 12(32
)↑↑+ ~2 12
(32
)↑↑+ 2
(~24 ↓↓ −
~24 ↓↓+ ~2
4 ↑↑)
= 2~2↑↑
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 9 / 17
Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑
=~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ = ~2 12(32
)↑↑+ ~2 12
(32
)↑↑+ 2
(~24 ↓↓ −
~24 ↓↓+ ~2
4 ↑↑)
= 2~2↑↑
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 9 / 17
Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)
=~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ = ~2 12(32
)↑↑+ ~2 12
(32
)↑↑+ 2
(~24 ↓↓ −
~24 ↓↓+ ~2
4 ↑↑)
= 2~2↑↑
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 9 / 17
Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)
=~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ = ~2 12(32
)↑↑+ ~2 12
(32
)↑↑+ 2
(~24 ↓↓ −
~24 ↓↓+ ~2
4 ↑↑)
= 2~2↑↑
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 9 / 17
Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ = ~2 12(32
)↑↑+ ~2 12
(32
)↑↑+ 2
(~24 ↓↓ −
~24 ↓↓+ ~2
4 ↑↑)
= 2~2↑↑
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 9 / 17
Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑
=~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ = ~2 12(32
)↑↑+ ~2 12
(32
)↑↑+ 2
(~24 ↓↓ −
~24 ↓↓+ ~2
4 ↑↑)
= 2~2↑↑
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 9 / 17
Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)
= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ = ~2 12(32
)↑↑+ ~2 12
(32
)↑↑+ 2
(~24 ↓↓ −
~24 ↓↓+ ~2
4 ↑↑)
= 2~2↑↑
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 9 / 17
Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)
= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ = ~2 12(32
)↑↑+ ~2 12
(32
)↑↑+ 2
(~24 ↓↓ −
~24 ↓↓+ ~2
4 ↑↑)
= 2~2↑↑
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 9 / 17
Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ = ~2 12(32
)↑↑+ ~2 12
(32
)↑↑+ 2
(~24 ↓↓ −
~24 ↓↓+ ~2
4 ↑↑)
= 2~2↑↑
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 9 / 17
Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑
=~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ = ~2 12(32
)↑↑+ ~2 12
(32
)↑↑+ 2
(~24 ↓↓ −
~24 ↓↓+ ~2
4 ↑↑)
= 2~2↑↑
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 9 / 17
Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)
=~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ = ~2 12(32
)↑↑+ ~2 12
(32
)↑↑+ 2
(~24 ↓↓ −
~24 ↓↓+ ~2
4 ↑↑)
= 2~2↑↑
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 9 / 17
Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)
=~2
4↑↑
S2↑↑ = ~2 12(32
)↑↑+ ~2 12
(32
)↑↑+ 2
(~24 ↓↓ −
~24 ↓↓+ ~2
4 ↑↑)
= 2~2↑↑
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 9 / 17
Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ = ~2 12(32
)↑↑+ ~2 12
(32
)↑↑+ 2
(~24 ↓↓ −
~24 ↓↓+ ~2
4 ↑↑)
= 2~2↑↑
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 9 / 17
Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ =
~2 12(32
)↑↑+ ~2 12
(32
)↑↑+ 2
(~24 ↓↓ −
~24 ↓↓+ ~2
4 ↑↑)
= 2~2↑↑
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 9 / 17
Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ = ~2 12(32
)↑↑
+ ~2 12(32
)↑↑+ 2
(~24 ↓↓ −
~24 ↓↓+ ~2
4 ↑↑)
= 2~2↑↑
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 9 / 17
Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ = ~2 12(32
)↑↑+ ~2 12
(32
)↑↑
+ 2(~24 ↓↓ −
~24 ↓↓+ ~2
4 ↑↑)
= 2~2↑↑
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 9 / 17
Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ = ~2 12(32
)↑↑+ ~2 12
(32
)↑↑+ 2
(~24 ↓↓ −
~24 ↓↓+ ~2
4 ↑↑)
= 2~2↑↑
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 9 / 17
Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ = ~2 12(32
)↑↑+ ~2 12
(32
)↑↑+ 2
(~24 ↓↓ −
~24 ↓↓+ ~2
4 ↑↑)
= 2~2↑↑
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 9 / 17
Triplet state (cont.)
Use the lowering operator to generate the state with total s = 1 and m = 0
S−↑↑ = (S(1)− + S
(2)− )↑↑ = (S
(1)− ↑)↑+ ↑(S (2)
− ↑)
= ~
√1
2
(3
2
)− 1
2
(−1
2
)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =
√2~
1√2
(↓↑+ ↑↓)
Apply the lowering operator again to get s = 1, m = −1
S−1√2
(↓↑+ ↑↓) = (S(1)− + S
(2)− )
1√2
(↓↑+ ↑↓)
=1√2
[S(1)− (↓↑+ ↑↓) + S
(2)− (↓↑+ ↑↓)
]=
1√2
[~↓↓+ ~↓↓]
=√
2~↓↓
The triplet state is thus
|1 1〉 = ↑↑, |1 0〉 =1√2
(↓↑+ ↑↓), |1−1〉 = ↓↓
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 10 / 17
Triplet state (cont.)
Use the lowering operator to generate the state with total s = 1 and m = 0
S−↑↑ = (S(1)− + S
(2)− )↑↑
= (S(1)− ↑)↑+ ↑(S (2)
− ↑)
= ~
√1
2
(3
2
)− 1
2
(−1
2
)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =
√2~
1√2
(↓↑+ ↑↓)
Apply the lowering operator again to get s = 1, m = −1
S−1√2
(↓↑+ ↑↓) = (S(1)− + S
(2)− )
1√2
(↓↑+ ↑↓)
=1√2
[S(1)− (↓↑+ ↑↓) + S
(2)− (↓↑+ ↑↓)
]=
1√2
[~↓↓+ ~↓↓]
=√
2~↓↓
The triplet state is thus
|1 1〉 = ↑↑, |1 0〉 =1√2
(↓↑+ ↑↓), |1−1〉 = ↓↓
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 10 / 17
Triplet state (cont.)
Use the lowering operator to generate the state with total s = 1 and m = 0
S−↑↑ = (S(1)− + S
(2)− )↑↑ = (S
(1)− ↑)↑+ ↑(S (2)
− ↑)
= ~
√1
2
(3
2
)− 1
2
(−1
2
)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =
√2~
1√2
(↓↑+ ↑↓)
Apply the lowering operator again to get s = 1, m = −1
S−1√2
(↓↑+ ↑↓) = (S(1)− + S
(2)− )
1√2
(↓↑+ ↑↓)
=1√2
[S(1)− (↓↑+ ↑↓) + S
(2)− (↓↑+ ↑↓)
]=
1√2
[~↓↓+ ~↓↓]
=√
2~↓↓
The triplet state is thus
|1 1〉 = ↑↑, |1 0〉 =1√2
(↓↑+ ↑↓), |1−1〉 = ↓↓
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 10 / 17
Triplet state (cont.)
Use the lowering operator to generate the state with total s = 1 and m = 0
S−↑↑ = (S(1)− + S
(2)− )↑↑ = (S
(1)− ↑)↑+ ↑(S (2)
− ↑)
= ~
√1
2
(3
2
)− 1
2
(−1
2
)(↓↑+ ↑↓)
= ~(↓↑+ ↑↓) =√
2~1√2
(↓↑+ ↑↓)
Apply the lowering operator again to get s = 1, m = −1
S−1√2
(↓↑+ ↑↓) = (S(1)− + S
(2)− )
1√2
(↓↑+ ↑↓)
=1√2
[S(1)− (↓↑+ ↑↓) + S
(2)− (↓↑+ ↑↓)
]=
1√2
[~↓↓+ ~↓↓]
=√
2~↓↓
The triplet state is thus
|1 1〉 = ↑↑, |1 0〉 =1√2
(↓↑+ ↑↓), |1−1〉 = ↓↓
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 10 / 17
Triplet state (cont.)
Use the lowering operator to generate the state with total s = 1 and m = 0
S−↑↑ = (S(1)− + S
(2)− )↑↑ = (S
(1)− ↑)↑+ ↑(S (2)
− ↑)
= ~
√1
2
(3
2
)− 1
2
(−1
2
)(↓↑+ ↑↓) = ~(↓↑+ ↑↓)
=√
2~1√2
(↓↑+ ↑↓)
Apply the lowering operator again to get s = 1, m = −1
S−1√2
(↓↑+ ↑↓) = (S(1)− + S
(2)− )
1√2
(↓↑+ ↑↓)
=1√2
[S(1)− (↓↑+ ↑↓) + S
(2)− (↓↑+ ↑↓)
]=
1√2
[~↓↓+ ~↓↓]
=√
2~↓↓
The triplet state is thus
|1 1〉 = ↑↑, |1 0〉 =1√2
(↓↑+ ↑↓), |1−1〉 = ↓↓
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 10 / 17
Triplet state (cont.)
Use the lowering operator to generate the state with total s = 1 and m = 0
S−↑↑ = (S(1)− + S
(2)− )↑↑ = (S
(1)− ↑)↑+ ↑(S (2)
− ↑)
= ~
√1
2
(3
2
)− 1
2
(−1
2
)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =
√2~
1√2
(↓↑+ ↑↓)
Apply the lowering operator again to get s = 1, m = −1
S−1√2
(↓↑+ ↑↓) = (S(1)− + S
(2)− )
1√2
(↓↑+ ↑↓)
=1√2
[S(1)− (↓↑+ ↑↓) + S
(2)− (↓↑+ ↑↓)
]=
1√2
[~↓↓+ ~↓↓]
=√
2~↓↓
The triplet state is thus
|1 1〉 = ↑↑, |1 0〉 =1√2
(↓↑+ ↑↓), |1−1〉 = ↓↓
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 10 / 17
Triplet state (cont.)
Use the lowering operator to generate the state with total s = 1 and m = 0
S−↑↑ = (S(1)− + S
(2)− )↑↑ = (S
(1)− ↑)↑+ ↑(S (2)
− ↑)
= ~
√1
2
(3
2
)− 1
2
(−1
2
)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =
√2~
1√2
(↓↑+ ↑↓)
Apply the lowering operator again to get s = 1, m = −1
S−1√2
(↓↑+ ↑↓) = (S(1)− + S
(2)− )
1√2
(↓↑+ ↑↓)
=1√2
[S(1)− (↓↑+ ↑↓) + S
(2)− (↓↑+ ↑↓)
]=
1√2
[~↓↓+ ~↓↓]
=√
2~↓↓
The triplet state is thus
|1 1〉 = ↑↑, |1 0〉 =1√2
(↓↑+ ↑↓), |1−1〉 = ↓↓
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 10 / 17
Triplet state (cont.)
Use the lowering operator to generate the state with total s = 1 and m = 0
S−↑↑ = (S(1)− + S
(2)− )↑↑ = (S
(1)− ↑)↑+ ↑(S (2)
− ↑)
= ~
√1
2
(3
2
)− 1
2
(−1
2
)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =
√2~
1√2
(↓↑+ ↑↓)
Apply the lowering operator again to get s = 1, m = −1
S−1√2
(↓↑+ ↑↓) = (S(1)− + S
(2)− )
1√2
(↓↑+ ↑↓)
=1√2
[S(1)− (↓↑+ ↑↓) + S
(2)− (↓↑+ ↑↓)
]=
1√2
[~↓↓+ ~↓↓]
=√
2~↓↓
The triplet state is thus
|1 1〉 = ↑↑, |1 0〉 =1√2
(↓↑+ ↑↓), |1−1〉 = ↓↓
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 10 / 17
Triplet state (cont.)
Use the lowering operator to generate the state with total s = 1 and m = 0
S−↑↑ = (S(1)− + S
(2)− )↑↑ = (S
(1)− ↑)↑+ ↑(S (2)
− ↑)
= ~
√1
2
(3
2
)− 1
2
(−1
2
)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =
√2~
1√2
(↓↑+ ↑↓)
Apply the lowering operator again to get s = 1, m = −1
S−1√2
(↓↑+ ↑↓) = (S(1)− + S
(2)− )
1√2
(↓↑+ ↑↓)
=1√2
[S(1)− (↓↑+ ↑↓) + S
(2)− (↓↑+ ↑↓)
]
=1√2
[~↓↓+ ~↓↓]
=√
2~↓↓
The triplet state is thus
|1 1〉 = ↑↑, |1 0〉 =1√2
(↓↑+ ↑↓), |1−1〉 = ↓↓
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 10 / 17
Triplet state (cont.)
Use the lowering operator to generate the state with total s = 1 and m = 0
S−↑↑ = (S(1)− + S
(2)− )↑↑ = (S
(1)− ↑)↑+ ↑(S (2)
− ↑)
= ~
√1
2
(3
2
)− 1
2
(−1
2
)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =
√2~
1√2
(↓↑+ ↑↓)
Apply the lowering operator again to get s = 1, m = −1
S−1√2
(↓↑+ ↑↓) = (S(1)− + S
(2)− )
1√2
(↓↑+ ↑↓)
=1√2
[S(1)− (↓↑+ ↑↓) + S
(2)− (↓↑+ ↑↓)
]=
1√2
[~↓↓+ ~↓↓]
=√
2~↓↓
The triplet state is thus
|1 1〉 = ↑↑, |1 0〉 =1√2
(↓↑+ ↑↓), |1−1〉 = ↓↓
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 10 / 17
Triplet state (cont.)
Use the lowering operator to generate the state with total s = 1 and m = 0
S−↑↑ = (S(1)− + S
(2)− )↑↑ = (S
(1)− ↑)↑+ ↑(S (2)
− ↑)
= ~
√1
2
(3
2
)− 1
2
(−1
2
)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =
√2~
1√2
(↓↑+ ↑↓)
Apply the lowering operator again to get s = 1, m = −1
S−1√2
(↓↑+ ↑↓) = (S(1)− + S
(2)− )
1√2
(↓↑+ ↑↓)
=1√2
[S(1)− (↓↑+ ↑↓) + S
(2)− (↓↑+ ↑↓)
]=
1√2
[~↓↓+ ~↓↓]
=√
2~↓↓
The triplet state is thus
|1 1〉 = ↑↑, |1 0〉 =1√2
(↓↑+ ↑↓), |1−1〉 = ↓↓
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 10 / 17
Triplet state (cont.)
Use the lowering operator to generate the state with total s = 1 and m = 0
S−↑↑ = (S(1)− + S
(2)− )↑↑ = (S
(1)− ↑)↑+ ↑(S (2)
− ↑)
= ~
√1
2
(3
2
)− 1
2
(−1
2
)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =
√2~
1√2
(↓↑+ ↑↓)
Apply the lowering operator again to get s = 1, m = −1
S−1√2
(↓↑+ ↑↓) = (S(1)− + S
(2)− )
1√2
(↓↑+ ↑↓)
=1√2
[S(1)− (↓↑+ ↑↓) + S
(2)− (↓↑+ ↑↓)
]=
1√2
[~↓↓+ ~↓↓]
=√
2~↓↓
The triplet state is thus
|1 1〉 = ↑↑, |1 0〉 =1√2
(↓↑+ ↑↓), |1−1〉 = ↓↓
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 10 / 17
Triplet state (cont.)
Use the lowering operator to generate the state with total s = 1 and m = 0
S−↑↑ = (S(1)− + S
(2)− )↑↑ = (S
(1)− ↑)↑+ ↑(S (2)
− ↑)
= ~
√1
2
(3
2
)− 1
2
(−1
2
)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =
√2~
1√2
(↓↑+ ↑↓)
Apply the lowering operator again to get s = 1, m = −1
S−1√2
(↓↑+ ↑↓) = (S(1)− + S
(2)− )
1√2
(↓↑+ ↑↓)
=1√2
[S(1)− (↓↑+ ↑↓) + S
(2)− (↓↑+ ↑↓)
]=
1√2
[~↓↓+ ~↓↓]
=√
2~↓↓
The triplet state is thus
|1 1〉 = ↑↑, |1 0〉 =1√2
(↓↑+ ↑↓), |1−1〉 = ↓↓
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 10 / 17
Singlet state
Initially we had four possible combinations of the two spins, but we haveused all four to generate only three states so there is one combinationremaining and it must be
|0 0〉 = 1√2
(↑↓ − ↓↑)
S2 1√2
(↑↓ − ↓↑) = (S (1))2 1√2
(↑↓ − ↓↑) + (S (2))2 1√2
(↑↓ − ↓↑)
+ 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z ) 1√
2(↑↓ − ↓↑)
S (1)S (1) 1√2
(↑↓ − ↓↑) = 34~
2 1√2
(↑↓ − ↓↑) = S (2)S (2) 1√2
(↑↓ − ↓↑)
S(1)x S
(2)x
1√2
(↑↓ − ↓↑) = 1√2
(14~
2↓↑ − 14~
2↑↓)
= −14~
2 1√2
(↑↓ − ↓↑)
S(1)y S
(2)y
1√2
(↑↓ − ↓↑) = 1√2
(−i2 14~
2↓↑+ i2 14~2↑↓)
= −14~
2 1√2
(↑↓ − ↓↑)
S(1)z S
(2)z
1√2
(↑↓ − ↓↑) = 1√2
(−1
4~2↑↓+ 1
4~2↓↑)
= −14~
2 1√2
(↑↓ − ↓↑)
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 11 / 17
Singlet state
Initially we had four possible combinations of the two spins, but we haveused all four to generate only three states so there is one combinationremaining and it must be
|0 0〉 = 1√2
(↑↓ − ↓↑)
S2 1√2
(↑↓ − ↓↑) = (S (1))2 1√2
(↑↓ − ↓↑) + (S (2))2 1√2
(↑↓ − ↓↑)
+ 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z ) 1√
2(↑↓ − ↓↑)
S (1)S (1) 1√2
(↑↓ − ↓↑) = 34~
2 1√2
(↑↓ − ↓↑) = S (2)S (2) 1√2
(↑↓ − ↓↑)
S(1)x S
(2)x
1√2
(↑↓ − ↓↑) = 1√2
(14~
2↓↑ − 14~
2↑↓)
= −14~
2 1√2
(↑↓ − ↓↑)
S(1)y S
(2)y
1√2
(↑↓ − ↓↑) = 1√2
(−i2 14~
2↓↑+ i2 14~2↑↓)
= −14~
2 1√2
(↑↓ − ↓↑)
S(1)z S
(2)z
1√2
(↑↓ − ↓↑) = 1√2
(−1
4~2↑↓+ 1
4~2↓↑)
= −14~
2 1√2
(↑↓ − ↓↑)
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 11 / 17
Singlet state
Initially we had four possible combinations of the two spins, but we haveused all four to generate only three states so there is one combinationremaining and it must be
|0 0〉 = 1√2
(↑↓ − ↓↑)
S2 1√2
(↑↓ − ↓↑)
= (S (1))2 1√2
(↑↓ − ↓↑) + (S (2))2 1√2
(↑↓ − ↓↑)
+ 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z ) 1√
2(↑↓ − ↓↑)
S (1)S (1) 1√2
(↑↓ − ↓↑) = 34~
2 1√2
(↑↓ − ↓↑) = S (2)S (2) 1√2
(↑↓ − ↓↑)
S(1)x S
(2)x
1√2
(↑↓ − ↓↑) = 1√2
(14~
2↓↑ − 14~
2↑↓)
= −14~
2 1√2
(↑↓ − ↓↑)
S(1)y S
(2)y
1√2
(↑↓ − ↓↑) = 1√2
(−i2 14~
2↓↑+ i2 14~2↑↓)
= −14~
2 1√2
(↑↓ − ↓↑)
S(1)z S
(2)z
1√2
(↑↓ − ↓↑) = 1√2
(−1
4~2↑↓+ 1
4~2↓↑)
= −14~
2 1√2
(↑↓ − ↓↑)
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 11 / 17
Singlet state
Initially we had four possible combinations of the two spins, but we haveused all four to generate only three states so there is one combinationremaining and it must be
|0 0〉 = 1√2
(↑↓ − ↓↑)
S2 1√2
(↑↓ − ↓↑) = (S (1))2 1√2
(↑↓ − ↓↑) + (S (2))2 1√2
(↑↓ − ↓↑)
+ 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z ) 1√
2(↑↓ − ↓↑)
S (1)S (1) 1√2
(↑↓ − ↓↑) = 34~
2 1√2
(↑↓ − ↓↑) = S (2)S (2) 1√2
(↑↓ − ↓↑)
S(1)x S
(2)x
1√2
(↑↓ − ↓↑) = 1√2
(14~
2↓↑ − 14~
2↑↓)
= −14~
2 1√2
(↑↓ − ↓↑)
S(1)y S
(2)y
1√2
(↑↓ − ↓↑) = 1√2
(−i2 14~
2↓↑+ i2 14~2↑↓)
= −14~
2 1√2
(↑↓ − ↓↑)
S(1)z S
(2)z
1√2
(↑↓ − ↓↑) = 1√2
(−1
4~2↑↓+ 1
4~2↓↑)
= −14~
2 1√2
(↑↓ − ↓↑)
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 11 / 17
Singlet state
Initially we had four possible combinations of the two spins, but we haveused all four to generate only three states so there is one combinationremaining and it must be
|0 0〉 = 1√2
(↑↓ − ↓↑)
S2 1√2
(↑↓ − ↓↑) = (S (1))2 1√2
(↑↓ − ↓↑) + (S (2))2 1√2
(↑↓ − ↓↑)
+ 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z ) 1√
2(↑↓ − ↓↑)
S (1)S (1) 1√2
(↑↓ − ↓↑)
= 34~
2 1√2
(↑↓ − ↓↑) = S (2)S (2) 1√2
(↑↓ − ↓↑)
S(1)x S
(2)x
1√2
(↑↓ − ↓↑) = 1√2
(14~
2↓↑ − 14~
2↑↓)
= −14~
2 1√2
(↑↓ − ↓↑)
S(1)y S
(2)y
1√2
(↑↓ − ↓↑) = 1√2
(−i2 14~
2↓↑+ i2 14~2↑↓)
= −14~
2 1√2
(↑↓ − ↓↑)
S(1)z S
(2)z
1√2
(↑↓ − ↓↑) = 1√2
(−1
4~2↑↓+ 1
4~2↓↑)
= −14~
2 1√2
(↑↓ − ↓↑)
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 11 / 17
Singlet state
Initially we had four possible combinations of the two spins, but we haveused all four to generate only three states so there is one combinationremaining and it must be
|0 0〉 = 1√2
(↑↓ − ↓↑)
S2 1√2
(↑↓ − ↓↑) = (S (1))2 1√2
(↑↓ − ↓↑) + (S (2))2 1√2
(↑↓ − ↓↑)
+ 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z ) 1√
2(↑↓ − ↓↑)
S (1)S (1) 1√2
(↑↓ − ↓↑) = 34~
2 1√2
(↑↓ − ↓↑)
= S (2)S (2) 1√2
(↑↓ − ↓↑)
S(1)x S
(2)x
1√2
(↑↓ − ↓↑) = 1√2
(14~
2↓↑ − 14~
2↑↓)
= −14~
2 1√2
(↑↓ − ↓↑)
S(1)y S
(2)y
1√2
(↑↓ − ↓↑) = 1√2
(−i2 14~
2↓↑+ i2 14~2↑↓)
= −14~
2 1√2
(↑↓ − ↓↑)
S(1)z S
(2)z
1√2
(↑↓ − ↓↑) = 1√2
(−1
4~2↑↓+ 1
4~2↓↑)
= −14~
2 1√2
(↑↓ − ↓↑)
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 11 / 17
Singlet state
Initially we had four possible combinations of the two spins, but we haveused all four to generate only three states so there is one combinationremaining and it must be
|0 0〉 = 1√2
(↑↓ − ↓↑)
S2 1√2
(↑↓ − ↓↑) = (S (1))2 1√2
(↑↓ − ↓↑) + (S (2))2 1√2
(↑↓ − ↓↑)
+ 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z ) 1√
2(↑↓ − ↓↑)
S (1)S (1) 1√2
(↑↓ − ↓↑) = 34~
2 1√2
(↑↓ − ↓↑) = S (2)S (2) 1√2
(↑↓ − ↓↑)
S(1)x S
(2)x
1√2
(↑↓ − ↓↑) = 1√2
(14~
2↓↑ − 14~
2↑↓)
= −14~
2 1√2
(↑↓ − ↓↑)
S(1)y S
(2)y
1√2
(↑↓ − ↓↑) = 1√2
(−i2 14~
2↓↑+ i2 14~2↑↓)
= −14~
2 1√2
(↑↓ − ↓↑)
S(1)z S
(2)z
1√2
(↑↓ − ↓↑) = 1√2
(−1
4~2↑↓+ 1
4~2↓↑)
= −14~
2 1√2
(↑↓ − ↓↑)
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 11 / 17
Singlet state
Initially we had four possible combinations of the two spins, but we haveused all four to generate only three states so there is one combinationremaining and it must be
|0 0〉 = 1√2
(↑↓ − ↓↑)
S2 1√2
(↑↓ − ↓↑) = (S (1))2 1√2
(↑↓ − ↓↑) + (S (2))2 1√2
(↑↓ − ↓↑)
+ 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z ) 1√
2(↑↓ − ↓↑)
S (1)S (1) 1√2
(↑↓ − ↓↑) = 34~
2 1√2
(↑↓ − ↓↑) = S (2)S (2) 1√2
(↑↓ − ↓↑)
S(1)x S
(2)x
1√2
(↑↓ − ↓↑)
= 1√2
(14~
2↓↑ − 14~
2↑↓)
= −14~
2 1√2
(↑↓ − ↓↑)
S(1)y S
(2)y
1√2
(↑↓ − ↓↑) = 1√2
(−i2 14~
2↓↑+ i2 14~2↑↓)
= −14~
2 1√2
(↑↓ − ↓↑)
S(1)z S
(2)z
1√2
(↑↓ − ↓↑) = 1√2
(−1
4~2↑↓+ 1
4~2↓↑)
= −14~
2 1√2
(↑↓ − ↓↑)
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 11 / 17
Singlet state
Initially we had four possible combinations of the two spins, but we haveused all four to generate only three states so there is one combinationremaining and it must be
|0 0〉 = 1√2
(↑↓ − ↓↑)
S2 1√2
(↑↓ − ↓↑) = (S (1))2 1√2
(↑↓ − ↓↑) + (S (2))2 1√2
(↑↓ − ↓↑)
+ 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z ) 1√
2(↑↓ − ↓↑)
S (1)S (1) 1√2
(↑↓ − ↓↑) = 34~
2 1√2
(↑↓ − ↓↑) = S (2)S (2) 1√2
(↑↓ − ↓↑)
S(1)x S
(2)x
1√2
(↑↓ − ↓↑) = 1√2
(14~
2↓↑ − 14~
2↑↓)
= −14~
2 1√2
(↑↓ − ↓↑)
S(1)y S
(2)y
1√2
(↑↓ − ↓↑) = 1√2
(−i2 14~
2↓↑+ i2 14~2↑↓)
= −14~
2 1√2
(↑↓ − ↓↑)
S(1)z S
(2)z
1√2
(↑↓ − ↓↑) = 1√2
(−1
4~2↑↓+ 1
4~2↓↑)
= −14~
2 1√2
(↑↓ − ↓↑)
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 11 / 17
Singlet state
Initially we had four possible combinations of the two spins, but we haveused all four to generate only three states so there is one combinationremaining and it must be
|0 0〉 = 1√2
(↑↓ − ↓↑)
S2 1√2
(↑↓ − ↓↑) = (S (1))2 1√2
(↑↓ − ↓↑) + (S (2))2 1√2
(↑↓ − ↓↑)
+ 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z ) 1√
2(↑↓ − ↓↑)
S (1)S (1) 1√2
(↑↓ − ↓↑) = 34~
2 1√2
(↑↓ − ↓↑) = S (2)S (2) 1√2
(↑↓ − ↓↑)
S(1)x S
(2)x
1√2
(↑↓ − ↓↑) = 1√2
(14~
2↓↑ − 14~
2↑↓)
= −14~
2 1√2
(↑↓ − ↓↑)
S(1)y S
(2)y
1√2
(↑↓ − ↓↑) = 1√2
(−i2 14~
2↓↑+ i2 14~2↑↓)
= −14~
2 1√2
(↑↓ − ↓↑)
S(1)z S
(2)z
1√2
(↑↓ − ↓↑) = 1√2
(−1
4~2↑↓+ 1
4~2↓↑)
= −14~
2 1√2
(↑↓ − ↓↑)
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 11 / 17
Singlet state
Initially we had four possible combinations of the two spins, but we haveused all four to generate only three states so there is one combinationremaining and it must be
|0 0〉 = 1√2
(↑↓ − ↓↑)
S2 1√2
(↑↓ − ↓↑) = (S (1))2 1√2
(↑↓ − ↓↑) + (S (2))2 1√2
(↑↓ − ↓↑)
+ 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z ) 1√
2(↑↓ − ↓↑)
S (1)S (1) 1√2
(↑↓ − ↓↑) = 34~
2 1√2
(↑↓ − ↓↑) = S (2)S (2) 1√2
(↑↓ − ↓↑)
S(1)x S
(2)x
1√2
(↑↓ − ↓↑) = 1√2
(14~
2↓↑ − 14~
2↑↓)
= −14~
2 1√2
(↑↓ − ↓↑)
S(1)y S
(2)y
1√2
(↑↓ − ↓↑)
= 1√2
(−i2 14~
2↓↑+ i2 14~2↑↓)
= −14~
2 1√2
(↑↓ − ↓↑)
S(1)z S
(2)z
1√2
(↑↓ − ↓↑) = 1√2
(−1
4~2↑↓+ 1
4~2↓↑)
= −14~
2 1√2
(↑↓ − ↓↑)
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 11 / 17
Singlet state
Initially we had four possible combinations of the two spins, but we haveused all four to generate only three states so there is one combinationremaining and it must be
|0 0〉 = 1√2
(↑↓ − ↓↑)
S2 1√2
(↑↓ − ↓↑) = (S (1))2 1√2
(↑↓ − ↓↑) + (S (2))2 1√2
(↑↓ − ↓↑)
+ 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z ) 1√
2(↑↓ − ↓↑)
S (1)S (1) 1√2
(↑↓ − ↓↑) = 34~
2 1√2
(↑↓ − ↓↑) = S (2)S (2) 1√2
(↑↓ − ↓↑)
S(1)x S
(2)x
1√2
(↑↓ − ↓↑) = 1√2
(14~
2↓↑ − 14~
2↑↓)
= −14~
2 1√2
(↑↓ − ↓↑)
S(1)y S
(2)y
1√2
(↑↓ − ↓↑) = 1√2
(−i2 14~
2↓↑+ i2 14~2↑↓)
= −14~
2 1√2
(↑↓ − ↓↑)
S(1)z S
(2)z
1√2
(↑↓ − ↓↑) = 1√2
(−1
4~2↑↓+ 1
4~2↓↑)
= −14~
2 1√2
(↑↓ − ↓↑)
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 11 / 17
Singlet state
Initially we had four possible combinations of the two spins, but we haveused all four to generate only three states so there is one combinationremaining and it must be
|0 0〉 = 1√2
(↑↓ − ↓↑)
S2 1√2
(↑↓ − ↓↑) = (S (1))2 1√2
(↑↓ − ↓↑) + (S (2))2 1√2
(↑↓ − ↓↑)
+ 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z ) 1√
2(↑↓ − ↓↑)
S (1)S (1) 1√2
(↑↓ − ↓↑) = 34~
2 1√2
(↑↓ − ↓↑) = S (2)S (2) 1√2
(↑↓ − ↓↑)
S(1)x S
(2)x
1√2
(↑↓ − ↓↑) = 1√2
(14~
2↓↑ − 14~
2↑↓)
= −14~
2 1√2
(↑↓ − ↓↑)
S(1)y S
(2)y
1√2
(↑↓ − ↓↑) = 1√2
(−i2 14~
2↓↑+ i2 14~2↑↓)
= −14~
2 1√2
(↑↓ − ↓↑)
S(1)z S
(2)z
1√2
(↑↓ − ↓↑) = 1√2
(−1
4~2↑↓+ 1
4~2↓↑)
= −14~
2 1√2
(↑↓ − ↓↑)
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 11 / 17
Singlet state
Initially we had four possible combinations of the two spins, but we haveused all four to generate only three states so there is one combinationremaining and it must be
|0 0〉 = 1√2
(↑↓ − ↓↑)
S2 1√2
(↑↓ − ↓↑) = (S (1))2 1√2
(↑↓ − ↓↑) + (S (2))2 1√2
(↑↓ − ↓↑)
+ 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z ) 1√
2(↑↓ − ↓↑)
S (1)S (1) 1√2
(↑↓ − ↓↑) = 34~
2 1√2
(↑↓ − ↓↑) = S (2)S (2) 1√2
(↑↓ − ↓↑)
S(1)x S
(2)x
1√2
(↑↓ − ↓↑) = 1√2
(14~
2↓↑ − 14~
2↑↓)
= −14~
2 1√2
(↑↓ − ↓↑)
S(1)y S
(2)y
1√2
(↑↓ − ↓↑) = 1√2
(−i2 14~
2↓↑+ i2 14~2↑↓)
= −14~
2 1√2
(↑↓ − ↓↑)
S(1)z S
(2)z
1√2
(↑↓ − ↓↑)
= 1√2
(−1
4~2↑↓+ 1
4~2↓↑)
= −14~
2 1√2
(↑↓ − ↓↑)
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 11 / 17
Singlet state
Initially we had four possible combinations of the two spins, but we haveused all four to generate only three states so there is one combinationremaining and it must be
|0 0〉 = 1√2
(↑↓ − ↓↑)
S2 1√2
(↑↓ − ↓↑) = (S (1))2 1√2
(↑↓ − ↓↑) + (S (2))2 1√2
(↑↓ − ↓↑)
+ 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z ) 1√
2(↑↓ − ↓↑)
S (1)S (1) 1√2
(↑↓ − ↓↑) = 34~
2 1√2
(↑↓ − ↓↑) = S (2)S (2) 1√2
(↑↓ − ↓↑)
S(1)x S
(2)x
1√2
(↑↓ − ↓↑) = 1√2
(14~
2↓↑ − 14~
2↑↓)
= −14~
2 1√2
(↑↓ − ↓↑)
S(1)y S
(2)y
1√2
(↑↓ − ↓↑) = 1√2
(−i2 14~
2↓↑+ i2 14~2↑↓)
= −14~
2 1√2
(↑↓ − ↓↑)
S(1)z S
(2)z
1√2
(↑↓ − ↓↑) = 1√2
(−1
4~2↑↓+ 1
4~2↓↑)
= −14~
2 1√2
(↑↓ − ↓↑)
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 11 / 17
Singlet state
Initially we had four possible combinations of the two spins, but we haveused all four to generate only three states so there is one combinationremaining and it must be
|0 0〉 = 1√2
(↑↓ − ↓↑)
S2 1√2
(↑↓ − ↓↑) = (S (1))2 1√2
(↑↓ − ↓↑) + (S (2))2 1√2
(↑↓ − ↓↑)
+ 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z ) 1√
2(↑↓ − ↓↑)
S (1)S (1) 1√2
(↑↓ − ↓↑) = 34~
2 1√2
(↑↓ − ↓↑) = S (2)S (2) 1√2
(↑↓ − ↓↑)
S(1)x S
(2)x
1√2
(↑↓ − ↓↑) = 1√2
(14~
2↓↑ − 14~
2↑↓)
= −14~
2 1√2
(↑↓ − ↓↑)
S(1)y S
(2)y
1√2
(↑↓ − ↓↑) = 1√2
(−i2 14~
2↓↑+ i2 14~2↑↓)
= −14~
2 1√2
(↑↓ − ↓↑)
S(1)z S
(2)z
1√2
(↑↓ − ↓↑) = 1√2
(−1
4~2↑↓+ 1
4~2↓↑)
= −14~
2 1√2
(↑↓ − ↓↑)
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 11 / 17
Singlet state
S (1)S (1) 1√2
(↑↓ − ↓↑) = S (2)S (2) 1√2
(↑↓ − ↓↑) = 34~
2 1√2
(↑↓ − ↓↑)
S(1)x S
(2)x
1√2
(↑↓ − ↓↑) = S(1)y S
(2)y
1√2
(↑↓ − ↓↑) = S(1)z S
(2)z
1√2
(↑↓ − ↓↑)
= −14~
2 1√2
(↑↓ − ↓↑)
S2 1√2
(↑↓ − ↓↑) = (S (1))2 1√2
(↑↓ − ↓↑) + (S (2))2 1√2
(↑↓ − ↓↑)
+ 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z ) 1√
2(↑↓ − ↓↑)
=[
34~
2 + 34~
2 + 2(−1
4~2 − 1
4~2 − 1
4~2)
]1√2
(↑↓ − ↓↑)
= 0
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 12 / 17
Singlet state
S (1)S (1) 1√2
(↑↓ − ↓↑) = S (2)S (2) 1√2
(↑↓ − ↓↑) = 34~
2 1√2
(↑↓ − ↓↑)
S(1)x S
(2)x
1√2
(↑↓ − ↓↑) = S(1)y S
(2)y
1√2
(↑↓ − ↓↑) = S(1)z S
(2)z
1√2
(↑↓ − ↓↑)
= −14~
2 1√2
(↑↓ − ↓↑)
S2 1√2
(↑↓ − ↓↑) = (S (1))2 1√2
(↑↓ − ↓↑) + (S (2))2 1√2
(↑↓ − ↓↑)
+ 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z ) 1√
2(↑↓ − ↓↑)
=[
34~
2 + 34~
2 + 2(−1
4~2 − 1
4~2 − 1
4~2)
]1√2
(↑↓ − ↓↑)
= 0
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 12 / 17
Singlet state
S (1)S (1) 1√2
(↑↓ − ↓↑) = S (2)S (2) 1√2
(↑↓ − ↓↑) = 34~
2 1√2
(↑↓ − ↓↑)
S(1)x S
(2)x
1√2
(↑↓ − ↓↑) = S(1)y S
(2)y
1√2
(↑↓ − ↓↑) = S(1)z S
(2)z
1√2
(↑↓ − ↓↑)
= −14~
2 1√2
(↑↓ − ↓↑)
S2 1√2
(↑↓ − ↓↑) = (S (1))2 1√2
(↑↓ − ↓↑) + (S (2))2 1√2
(↑↓ − ↓↑)
+ 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z ) 1√
2(↑↓ − ↓↑)
=[
34~
2 + 34~
2 + 2(−1
4~2 − 1
4~2 − 1
4~2)
]1√2
(↑↓ − ↓↑)
= 0
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 12 / 17
Singlet state
S (1)S (1) 1√2
(↑↓ − ↓↑) = S (2)S (2) 1√2
(↑↓ − ↓↑) = 34~
2 1√2
(↑↓ − ↓↑)
S(1)x S
(2)x
1√2
(↑↓ − ↓↑) = S(1)y S
(2)y
1√2
(↑↓ − ↓↑) = S(1)z S
(2)z
1√2
(↑↓ − ↓↑)
= −14~
2 1√2
(↑↓ − ↓↑)
S2 1√2
(↑↓ − ↓↑) = (S (1))2 1√2
(↑↓ − ↓↑) + (S (2))2 1√2
(↑↓ − ↓↑)
+ 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z ) 1√
2(↑↓ − ↓↑)
=[34~
2
+ 34~
2 + 2(−1
4~2 − 1
4~2 − 1
4~2)
]1√2
(↑↓ − ↓↑)
= 0
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 12 / 17
Singlet state
S (1)S (1) 1√2
(↑↓ − ↓↑) = S (2)S (2) 1√2
(↑↓ − ↓↑) = 34~
2 1√2
(↑↓ − ↓↑)
S(1)x S
(2)x
1√2
(↑↓ − ↓↑) = S(1)y S
(2)y
1√2
(↑↓ − ↓↑) = S(1)z S
(2)z
1√2
(↑↓ − ↓↑)
= −14~
2 1√2
(↑↓ − ↓↑)
S2 1√2
(↑↓ − ↓↑) = (S (1))2 1√2
(↑↓ − ↓↑) + (S (2))2 1√2
(↑↓ − ↓↑)
+ 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z ) 1√
2(↑↓ − ↓↑)
=[34~
2 + 34~
2
+ 2(−1
4~2 − 1
4~2 − 1
4~2)
]1√2
(↑↓ − ↓↑)
= 0
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 12 / 17
Singlet state
S (1)S (1) 1√2
(↑↓ − ↓↑) = S (2)S (2) 1√2
(↑↓ − ↓↑) = 34~
2 1√2
(↑↓ − ↓↑)
S(1)x S
(2)x
1√2
(↑↓ − ↓↑) = S(1)y S
(2)y
1√2
(↑↓ − ↓↑) = S(1)z S
(2)z
1√2
(↑↓ − ↓↑)
= −14~
2 1√2
(↑↓ − ↓↑)
S2 1√2
(↑↓ − ↓↑) = (S (1))2 1√2
(↑↓ − ↓↑) + (S (2))2 1√2
(↑↓ − ↓↑)
+ 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z ) 1√
2(↑↓ − ↓↑)
=[34~
2 + 34~
2 + 2(−1
4~2 − 1
4~2 − 1
4~2) ]
1√2
(↑↓ − ↓↑)
= 0
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 12 / 17
Singlet state
S (1)S (1) 1√2
(↑↓ − ↓↑) = S (2)S (2) 1√2
(↑↓ − ↓↑) = 34~
2 1√2
(↑↓ − ↓↑)
S(1)x S
(2)x
1√2
(↑↓ − ↓↑) = S(1)y S
(2)y
1√2
(↑↓ − ↓↑) = S(1)z S
(2)z
1√2
(↑↓ − ↓↑)
= −14~
2 1√2
(↑↓ − ↓↑)
S2 1√2
(↑↓ − ↓↑) = (S (1))2 1√2
(↑↓ − ↓↑) + (S (2))2 1√2
(↑↓ − ↓↑)
+ 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z ) 1√
2(↑↓ − ↓↑)
=[34~
2 + 34~
2 + 2(−1
4~2 − 1
4~2 − 1
4~2) ]
1√2
(↑↓ − ↓↑)
= 0
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 12 / 17
Orthonormality
The three triplet states are orthonormal by construction because they weregenerated by the lowering operator for the total spin.
〈1 1|1 −1〉 ≡ 0 〈1 1|1 0〉 ≡ 0 〈1 0|1 −1〉 ≡ 0
What about the singlet state?
〈1 1|0 0〉
= (↑↑)† 1√2
(↑↓ − ↓↑) = 0
〈1 −1|0 0〉
= (↓↓)† 1√2
(↑↓ − ↓↑) = 0
〈1 0|0 0〉
= 1√2
(↑↓+ ↓↑)† 1√2
(↑↓ − ↓↑) = 12(1− 1) = 0
We have recovered all four orthonormal states.
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 13 / 17
Orthonormality
The three triplet states are orthonormal by construction because they weregenerated by the lowering operator for the total spin.
〈1 1|1 −1〉 ≡ 0
〈1 1|1 0〉 ≡ 0 〈1 0|1 −1〉 ≡ 0
What about the singlet state?
〈1 1|0 0〉
= (↑↑)† 1√2
(↑↓ − ↓↑) = 0
〈1 −1|0 0〉
= (↓↓)† 1√2
(↑↓ − ↓↑) = 0
〈1 0|0 0〉
= 1√2
(↑↓+ ↓↑)† 1√2
(↑↓ − ↓↑) = 12(1− 1) = 0
We have recovered all four orthonormal states.
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 13 / 17
Orthonormality
The three triplet states are orthonormal by construction because they weregenerated by the lowering operator for the total spin.
〈1 1|1 −1〉 ≡ 0 〈1 1|1 0〉 ≡ 0
〈1 0|1 −1〉 ≡ 0
What about the singlet state?
〈1 1|0 0〉
= (↑↑)† 1√2
(↑↓ − ↓↑) = 0
〈1 −1|0 0〉
= (↓↓)† 1√2
(↑↓ − ↓↑) = 0
〈1 0|0 0〉
= 1√2
(↑↓+ ↓↑)† 1√2
(↑↓ − ↓↑) = 12(1− 1) = 0
We have recovered all four orthonormal states.
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 13 / 17
Orthonormality
The three triplet states are orthonormal by construction because they weregenerated by the lowering operator for the total spin.
〈1 1|1 −1〉 ≡ 0 〈1 1|1 0〉 ≡ 0 〈1 0|1 −1〉 ≡ 0
What about the singlet state?
〈1 1|0 0〉
= (↑↑)† 1√2
(↑↓ − ↓↑) = 0
〈1 −1|0 0〉
= (↓↓)† 1√2
(↑↓ − ↓↑) = 0
〈1 0|0 0〉
= 1√2
(↑↓+ ↓↑)† 1√2
(↑↓ − ↓↑) = 12(1− 1) = 0
We have recovered all four orthonormal states.
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 13 / 17
Orthonormality
The three triplet states are orthonormal by construction because they weregenerated by the lowering operator for the total spin.
〈1 1|1 −1〉 ≡ 0 〈1 1|1 0〉 ≡ 0 〈1 0|1 −1〉 ≡ 0
What about the singlet state?
〈1 1|0 0〉
= (↑↑)† 1√2
(↑↓ − ↓↑) = 0
〈1 −1|0 0〉
= (↓↓)† 1√2
(↑↓ − ↓↑) = 0
〈1 0|0 0〉
= 1√2
(↑↓+ ↓↑)† 1√2
(↑↓ − ↓↑) = 12(1− 1) = 0
We have recovered all four orthonormal states.
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 13 / 17
Orthonormality
The three triplet states are orthonormal by construction because they weregenerated by the lowering operator for the total spin.
〈1 1|1 −1〉 ≡ 0 〈1 1|1 0〉 ≡ 0 〈1 0|1 −1〉 ≡ 0
What about the singlet state?
〈1 1|0 0〉
= (↑↑)† 1√2
(↑↓ − ↓↑) = 0
〈1 −1|0 0〉
= (↓↓)† 1√2
(↑↓ − ↓↑) = 0
〈1 0|0 0〉
= 1√2
(↑↓+ ↓↑)† 1√2
(↑↓ − ↓↑) = 12(1− 1) = 0
We have recovered all four orthonormal states.
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 13 / 17
Orthonormality
The three triplet states are orthonormal by construction because they weregenerated by the lowering operator for the total spin.
〈1 1|1 −1〉 ≡ 0 〈1 1|1 0〉 ≡ 0 〈1 0|1 −1〉 ≡ 0
What about the singlet state?
〈1 1|0 0〉 = (↑↑)†
1√2
(↑↓ − ↓↑) = 0
〈1 −1|0 0〉
= (↓↓)† 1√2
(↑↓ − ↓↑) = 0
〈1 0|0 0〉
= 1√2
(↑↓+ ↓↑)† 1√2
(↑↓ − ↓↑) = 12(1− 1) = 0
We have recovered all four orthonormal states.
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 13 / 17
Orthonormality
The three triplet states are orthonormal by construction because they weregenerated by the lowering operator for the total spin.
〈1 1|1 −1〉 ≡ 0 〈1 1|1 0〉 ≡ 0 〈1 0|1 −1〉 ≡ 0
What about the singlet state?
〈1 1|0 0〉 = (↑↑)† 1√2
(↑↓ − ↓↑)
= 0
〈1 −1|0 0〉
= (↓↓)† 1√2
(↑↓ − ↓↑) = 0
〈1 0|0 0〉
= 1√2
(↑↓+ ↓↑)† 1√2
(↑↓ − ↓↑) = 12(1− 1) = 0
We have recovered all four orthonormal states.
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 13 / 17
Orthonormality
The three triplet states are orthonormal by construction because they weregenerated by the lowering operator for the total spin.
〈1 1|1 −1〉 ≡ 0 〈1 1|1 0〉 ≡ 0 〈1 0|1 −1〉 ≡ 0
What about the singlet state?
〈1 1|0 0〉 = (↑↑)† 1√2
(↑↓ − ↓↑) = 0
〈1 −1|0 0〉
= (↓↓)† 1√2
(↑↓ − ↓↑) = 0
〈1 0|0 0〉
= 1√2
(↑↓+ ↓↑)† 1√2
(↑↓ − ↓↑) = 12(1− 1) = 0
We have recovered all four orthonormal states.
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 13 / 17
Orthonormality
The three triplet states are orthonormal by construction because they weregenerated by the lowering operator for the total spin.
〈1 1|1 −1〉 ≡ 0 〈1 1|1 0〉 ≡ 0 〈1 0|1 −1〉 ≡ 0
What about the singlet state?
〈1 1|0 0〉 = (↑↑)† 1√2
(↑↓ − ↓↑) = 0
〈1 −1|0 0〉
= (↓↓)† 1√2
(↑↓ − ↓↑) = 0
〈1 0|0 0〉
= 1√2
(↑↓+ ↓↑)† 1√2
(↑↓ − ↓↑) = 12(1− 1) = 0
We have recovered all four orthonormal states.
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 13 / 17
Orthonormality
The three triplet states are orthonormal by construction because they weregenerated by the lowering operator for the total spin.
〈1 1|1 −1〉 ≡ 0 〈1 1|1 0〉 ≡ 0 〈1 0|1 −1〉 ≡ 0
What about the singlet state?
〈1 1|0 0〉 = (↑↑)† 1√2
(↑↓ − ↓↑) = 0
〈1 −1|0 0〉 = (↓↓)†
1√2
(↑↓ − ↓↑) = 0
〈1 0|0 0〉
= 1√2
(↑↓+ ↓↑)† 1√2
(↑↓ − ↓↑) = 12(1− 1) = 0
We have recovered all four orthonormal states.
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 13 / 17
Orthonormality
The three triplet states are orthonormal by construction because they weregenerated by the lowering operator for the total spin.
〈1 1|1 −1〉 ≡ 0 〈1 1|1 0〉 ≡ 0 〈1 0|1 −1〉 ≡ 0
What about the singlet state?
〈1 1|0 0〉 = (↑↑)† 1√2
(↑↓ − ↓↑) = 0
〈1 −1|0 0〉 = (↓↓)† 1√2
(↑↓ − ↓↑)
= 0
〈1 0|0 0〉
= 1√2
(↑↓+ ↓↑)† 1√2
(↑↓ − ↓↑) = 12(1− 1) = 0
We have recovered all four orthonormal states.
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 13 / 17
Orthonormality
The three triplet states are orthonormal by construction because they weregenerated by the lowering operator for the total spin.
〈1 1|1 −1〉 ≡ 0 〈1 1|1 0〉 ≡ 0 〈1 0|1 −1〉 ≡ 0
What about the singlet state?
〈1 1|0 0〉 = (↑↑)† 1√2
(↑↓ − ↓↑) = 0
〈1 −1|0 0〉 = (↓↓)† 1√2
(↑↓ − ↓↑) = 0
〈1 0|0 0〉
= 1√2
(↑↓+ ↓↑)† 1√2
(↑↓ − ↓↑) = 12(1− 1) = 0
We have recovered all four orthonormal states.
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 13 / 17
Orthonormality
The three triplet states are orthonormal by construction because they weregenerated by the lowering operator for the total spin.
〈1 1|1 −1〉 ≡ 0 〈1 1|1 0〉 ≡ 0 〈1 0|1 −1〉 ≡ 0
What about the singlet state?
〈1 1|0 0〉 = (↑↑)† 1√2
(↑↓ − ↓↑) = 0
〈1 −1|0 0〉 = (↓↓)† 1√2
(↑↓ − ↓↑) = 0
〈1 0|0 0〉
= 1√2
(↑↓+ ↓↑)† 1√2
(↑↓ − ↓↑) = 12(1− 1) = 0
We have recovered all four orthonormal states.
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 13 / 17
Orthonormality
The three triplet states are orthonormal by construction because they weregenerated by the lowering operator for the total spin.
〈1 1|1 −1〉 ≡ 0 〈1 1|1 0〉 ≡ 0 〈1 0|1 −1〉 ≡ 0
What about the singlet state?
〈1 1|0 0〉 = (↑↑)† 1√2
(↑↓ − ↓↑) = 0
〈1 −1|0 0〉 = (↓↓)† 1√2
(↑↓ − ↓↑) = 0
〈1 0|0 0〉 = 1√2
(↑↓+ ↓↑)†
1√2
(↑↓ − ↓↑) = 12(1− 1) = 0
We have recovered all four orthonormal states.
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 13 / 17
Orthonormality
The three triplet states are orthonormal by construction because they weregenerated by the lowering operator for the total spin.
〈1 1|1 −1〉 ≡ 0 〈1 1|1 0〉 ≡ 0 〈1 0|1 −1〉 ≡ 0
What about the singlet state?
〈1 1|0 0〉 = (↑↑)† 1√2
(↑↓ − ↓↑) = 0
〈1 −1|0 0〉 = (↓↓)† 1√2
(↑↓ − ↓↑) = 0
〈1 0|0 0〉 = 1√2
(↑↓+ ↓↑)† 1√2
(↑↓ − ↓↑)
= 12(1− 1) = 0
We have recovered all four orthonormal states.
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 13 / 17
Orthonormality
The three triplet states are orthonormal by construction because they weregenerated by the lowering operator for the total spin.
〈1 1|1 −1〉 ≡ 0 〈1 1|1 0〉 ≡ 0 〈1 0|1 −1〉 ≡ 0
What about the singlet state?
〈1 1|0 0〉 = (↑↑)† 1√2
(↑↓ − ↓↑) = 0
〈1 −1|0 0〉 = (↓↓)† 1√2
(↑↓ − ↓↑) = 0
〈1 0|0 0〉 = 1√2
(↑↓+ ↓↑)† 1√2
(↑↓ − ↓↑) = 12(1− 1)
= 0
We have recovered all four orthonormal states.
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 13 / 17
Orthonormality
The three triplet states are orthonormal by construction because they weregenerated by the lowering operator for the total spin.
〈1 1|1 −1〉 ≡ 0 〈1 1|1 0〉 ≡ 0 〈1 0|1 −1〉 ≡ 0
What about the singlet state?
〈1 1|0 0〉 = (↑↑)† 1√2
(↑↓ − ↓↑) = 0
〈1 −1|0 0〉 = (↓↓)† 1√2
(↑↓ − ↓↑) = 0
〈1 0|0 0〉 = 1√2
(↑↓+ ↓↑)† 1√2
(↑↓ − ↓↑) = 12(1− 1) = 0
We have recovered all four orthonormal states.
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 13 / 17
Orthonormality
The three triplet states are orthonormal by construction because they weregenerated by the lowering operator for the total spin.
〈1 1|1 −1〉 ≡ 0 〈1 1|1 0〉 ≡ 0 〈1 0|1 −1〉 ≡ 0
What about the singlet state?
〈1 1|0 0〉 = (↑↑)† 1√2
(↑↓ − ↓↑) = 0
〈1 −1|0 0〉 = (↓↓)† 1√2
(↑↓ − ↓↑) = 0
〈1 0|0 0〉 = 1√2
(↑↓+ ↓↑)† 1√2
(↑↓ − ↓↑) = 12(1− 1) = 0
We have recovered all four orthonormal states.
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 13 / 17
Eigenvalues of Sz
We can also check that these are eigenfunctions of the Sz operator
Sz |0 0〉 = (S(1)z + S
(2)z ) 1√
2(↑↓ − ↓↑)
= 1√2
(S(1)z ↑↓+ S
(2)z ↑↓ − S
(1)z ↓↑ − S
(2)z ↓↑)
= 1√2~(12↑↓ −
12↑↓+ 1
2↓↑ −12↓↑)
= 0
Sz |1 1〉 = (S(1)z + S
(2)z )↑↑ = ~
(12 + 1
2
)↑↑ = +1~↑↑
Sz |1 0〉 = (S(1)z + S
(2)z ) 1√
2(↑↓+ ↓↑)
= 1√2~(12↑↓ −
12↑↓ −
12↓↑+ 1
2↓↑)
= 0
Sz |1−1〉 = (S(1)z + S
(2)z )↓↓ = ~
(−1
2 −12
)↓↓ = −1~↓↓
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 14 / 17
Eigenvalues of Sz
We can also check that these are eigenfunctions of the Sz operator
Sz |0 0〉 = (S(1)z + S
(2)z ) 1√
2(↑↓ − ↓↑)
= 1√2
(S(1)z ↑↓+ S
(2)z ↑↓ − S
(1)z ↓↑ − S
(2)z ↓↑)
= 1√2~(12↑↓ −
12↑↓+ 1
2↓↑ −12↓↑)
= 0
Sz |1 1〉 = (S(1)z + S
(2)z )↑↑ = ~
(12 + 1
2
)↑↑ = +1~↑↑
Sz |1 0〉 = (S(1)z + S
(2)z ) 1√
2(↑↓+ ↓↑)
= 1√2~(12↑↓ −
12↑↓ −
12↓↑+ 1
2↓↑)
= 0
Sz |1−1〉 = (S(1)z + S
(2)z )↓↓ = ~
(−1
2 −12
)↓↓ = −1~↓↓
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 14 / 17
Eigenvalues of Sz
We can also check that these are eigenfunctions of the Sz operator
Sz |0 0〉 = (S(1)z + S
(2)z ) 1√
2(↑↓ − ↓↑)
= 1√2
(S(1)z ↑↓+ S
(2)z ↑↓ − S
(1)z ↓↑ − S
(2)z ↓↑)
= 1√2~(12↑↓ −
12↑↓+ 1
2↓↑ −12↓↑)
= 0
Sz |1 1〉 = (S(1)z + S
(2)z )↑↑ = ~
(12 + 1
2
)↑↑ = +1~↑↑
Sz |1 0〉 = (S(1)z + S
(2)z ) 1√
2(↑↓+ ↓↑)
= 1√2~(12↑↓ −
12↑↓ −
12↓↑+ 1
2↓↑)
= 0
Sz |1−1〉 = (S(1)z + S
(2)z )↓↓ = ~
(−1
2 −12
)↓↓ = −1~↓↓
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 14 / 17
Eigenvalues of Sz
We can also check that these are eigenfunctions of the Sz operator
Sz |0 0〉 = (S(1)z + S
(2)z ) 1√
2(↑↓ − ↓↑)
= 1√2
(S(1)z ↑↓+ S
(2)z ↑↓ − S
(1)z ↓↑ − S
(2)z ↓↑)
= 1√2~(12↑↓ −
12↑↓+ 1
2↓↑ −12↓↑)
= 0
Sz |1 1〉 = (S(1)z + S
(2)z )↑↑ = ~
(12 + 1
2
)↑↑ = +1~↑↑
Sz |1 0〉 = (S(1)z + S
(2)z ) 1√
2(↑↓+ ↓↑)
= 1√2~(12↑↓ −
12↑↓ −
12↓↑+ 1
2↓↑)
= 0
Sz |1−1〉 = (S(1)z + S
(2)z )↓↓ = ~
(−1
2 −12
)↓↓ = −1~↓↓
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 14 / 17
Eigenvalues of Sz
We can also check that these are eigenfunctions of the Sz operator
Sz |0 0〉 = (S(1)z + S
(2)z ) 1√
2(↑↓ − ↓↑)
= 1√2
(S(1)z ↑↓+ S
(2)z ↑↓ − S
(1)z ↓↑ − S
(2)z ↓↑)
= 1√2~(12↑↓ −
12↑↓+ 1
2↓↑ −12↓↑)
= 0
Sz |1 1〉 = (S(1)z + S
(2)z )↑↑ = ~
(12 + 1
2
)↑↑ = +1~↑↑
Sz |1 0〉 = (S(1)z + S
(2)z ) 1√
2(↑↓+ ↓↑)
= 1√2~(12↑↓ −
12↑↓ −
12↓↑+ 1
2↓↑)
= 0
Sz |1−1〉 = (S(1)z + S
(2)z )↓↓ = ~
(−1
2 −12
)↓↓ = −1~↓↓
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 14 / 17
Eigenvalues of Sz
We can also check that these are eigenfunctions of the Sz operator
Sz |0 0〉 = (S(1)z + S
(2)z ) 1√
2(↑↓ − ↓↑)
= 1√2
(S(1)z ↑↓+ S
(2)z ↑↓ − S
(1)z ↓↑ − S
(2)z ↓↑)
= 1√2~(12↑↓ −
12↑↓+ 1
2↓↑ −12↓↑)
= 0
Sz |1 1〉 = (S(1)z + S
(2)z )↑↑
= ~(12 + 1
2
)↑↑ = +1~↑↑
Sz |1 0〉 = (S(1)z + S
(2)z ) 1√
2(↑↓+ ↓↑)
= 1√2~(12↑↓ −
12↑↓ −
12↓↑+ 1
2↓↑)
= 0
Sz |1−1〉 = (S(1)z + S
(2)z )↓↓ = ~
(−1
2 −12
)↓↓ = −1~↓↓
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 14 / 17
Eigenvalues of Sz
We can also check that these are eigenfunctions of the Sz operator
Sz |0 0〉 = (S(1)z + S
(2)z ) 1√
2(↑↓ − ↓↑)
= 1√2
(S(1)z ↑↓+ S
(2)z ↑↓ − S
(1)z ↓↑ − S
(2)z ↓↑)
= 1√2~(12↑↓ −
12↑↓+ 1
2↓↑ −12↓↑)
= 0
Sz |1 1〉 = (S(1)z + S
(2)z )↑↑ = ~
(12 + 1
2
)↑↑
= +1~↑↑
Sz |1 0〉 = (S(1)z + S
(2)z ) 1√
2(↑↓+ ↓↑)
= 1√2~(12↑↓ −
12↑↓ −
12↓↑+ 1
2↓↑)
= 0
Sz |1−1〉 = (S(1)z + S
(2)z )↓↓ = ~
(−1
2 −12
)↓↓ = −1~↓↓
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 14 / 17
Eigenvalues of Sz
We can also check that these are eigenfunctions of the Sz operator
Sz |0 0〉 = (S(1)z + S
(2)z ) 1√
2(↑↓ − ↓↑)
= 1√2
(S(1)z ↑↓+ S
(2)z ↑↓ − S
(1)z ↓↑ − S
(2)z ↓↑)
= 1√2~(12↑↓ −
12↑↓+ 1
2↓↑ −12↓↑)
= 0
Sz |1 1〉 = (S(1)z + S
(2)z )↑↑ = ~
(12 + 1
2
)↑↑ = +1~↑↑
Sz |1 0〉 = (S(1)z + S
(2)z ) 1√
2(↑↓+ ↓↑)
= 1√2~(12↑↓ −
12↑↓ −
12↓↑+ 1
2↓↑)
= 0
Sz |1−1〉 = (S(1)z + S
(2)z )↓↓ = ~
(−1
2 −12
)↓↓ = −1~↓↓
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 14 / 17
Eigenvalues of Sz
We can also check that these are eigenfunctions of the Sz operator
Sz |0 0〉 = (S(1)z + S
(2)z ) 1√
2(↑↓ − ↓↑)
= 1√2
(S(1)z ↑↓+ S
(2)z ↑↓ − S
(1)z ↓↑ − S
(2)z ↓↑)
= 1√2~(12↑↓ −
12↑↓+ 1
2↓↑ −12↓↑)
= 0
Sz |1 1〉 = (S(1)z + S
(2)z )↑↑ = ~
(12 + 1
2
)↑↑ = +1~↑↑
Sz |1 0〉 = (S(1)z + S
(2)z ) 1√
2(↑↓+ ↓↑)
= 1√2~(12↑↓ −
12↑↓ −
12↓↑+ 1
2↓↑)
= 0
Sz |1−1〉 = (S(1)z + S
(2)z )↓↓ = ~
(−1
2 −12
)↓↓ = −1~↓↓
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 14 / 17
Eigenvalues of Sz
We can also check that these are eigenfunctions of the Sz operator
Sz |0 0〉 = (S(1)z + S
(2)z ) 1√
2(↑↓ − ↓↑)
= 1√2
(S(1)z ↑↓+ S
(2)z ↑↓ − S
(1)z ↓↑ − S
(2)z ↓↑)
= 1√2~(12↑↓ −
12↑↓+ 1
2↓↑ −12↓↑)
= 0
Sz |1 1〉 = (S(1)z + S
(2)z )↑↑ = ~
(12 + 1
2
)↑↑ = +1~↑↑
Sz |1 0〉 = (S(1)z + S
(2)z ) 1√
2(↑↓+ ↓↑)
= 1√2~(12↑↓ −
12↑↓ −
12↓↑+ 1
2↓↑)
= 0
Sz |1−1〉 = (S(1)z + S
(2)z )↓↓ = ~
(−1
2 −12
)↓↓ = −1~↓↓
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 14 / 17
Eigenvalues of Sz
We can also check that these are eigenfunctions of the Sz operator
Sz |0 0〉 = (S(1)z + S
(2)z ) 1√
2(↑↓ − ↓↑)
= 1√2
(S(1)z ↑↓+ S
(2)z ↑↓ − S
(1)z ↓↑ − S
(2)z ↓↑)
= 1√2~(12↑↓ −
12↑↓+ 1
2↓↑ −12↓↑)
= 0
Sz |1 1〉 = (S(1)z + S
(2)z )↑↑ = ~
(12 + 1
2
)↑↑ = +1~↑↑
Sz |1 0〉 = (S(1)z + S
(2)z ) 1√
2(↑↓+ ↓↑)
= 1√2~(12↑↓ −
12↑↓ −
12↓↑+ 1
2↓↑)
= 0
Sz |1−1〉 = (S(1)z + S
(2)z )↓↓ = ~
(−1
2 −12
)↓↓ = −1~↓↓
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 14 / 17
Eigenvalues of Sz
We can also check that these are eigenfunctions of the Sz operator
Sz |0 0〉 = (S(1)z + S
(2)z ) 1√
2(↑↓ − ↓↑)
= 1√2
(S(1)z ↑↓+ S
(2)z ↑↓ − S
(1)z ↓↑ − S
(2)z ↓↑)
= 1√2~(12↑↓ −
12↑↓+ 1
2↓↑ −12↓↑)
= 0
Sz |1 1〉 = (S(1)z + S
(2)z )↑↑ = ~
(12 + 1
2
)↑↑ = +1~↑↑
Sz |1 0〉 = (S(1)z + S
(2)z ) 1√
2(↑↓+ ↓↑)
= 1√2~(12↑↓ −
12↑↓ −
12↓↑+ 1
2↓↑)
= 0
Sz |1−1〉 = (S(1)z + S
(2)z )↓↓
= ~(−1
2 −12
)↓↓ = −1~↓↓
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 14 / 17
Eigenvalues of Sz
We can also check that these are eigenfunctions of the Sz operator
Sz |0 0〉 = (S(1)z + S
(2)z ) 1√
2(↑↓ − ↓↑)
= 1√2
(S(1)z ↑↓+ S
(2)z ↑↓ − S
(1)z ↓↑ − S
(2)z ↓↑)
= 1√2~(12↑↓ −
12↑↓+ 1
2↓↑ −12↓↑)
= 0
Sz |1 1〉 = (S(1)z + S
(2)z )↑↑ = ~
(12 + 1
2
)↑↑ = +1~↑↑
Sz |1 0〉 = (S(1)z + S
(2)z ) 1√
2(↑↓+ ↓↑)
= 1√2~(12↑↓ −
12↑↓ −
12↓↑+ 1
2↓↑)
= 0
Sz |1−1〉 = (S(1)z + S
(2)z )↓↓ = ~
(−1
2 −12
)↓↓
= −1~↓↓
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 14 / 17
Eigenvalues of Sz
We can also check that these are eigenfunctions of the Sz operator
Sz |0 0〉 = (S(1)z + S
(2)z ) 1√
2(↑↓ − ↓↑)
= 1√2
(S(1)z ↑↓+ S
(2)z ↑↓ − S
(1)z ↓↑ − S
(2)z ↓↑)
= 1√2~(12↑↓ −
12↑↓+ 1
2↓↑ −12↓↑)
= 0
Sz |1 1〉 = (S(1)z + S
(2)z )↑↑ = ~
(12 + 1
2
)↑↑ = +1~↑↑
Sz |1 0〉 = (S(1)z + S
(2)z ) 1√
2(↑↓+ ↓↑)
= 1√2~(12↑↓ −
12↑↓ −
12↓↑+ 1
2↓↑)
= 0
Sz |1−1〉 = (S(1)z + S
(2)z )↓↓ = ~
(−1
2 −12
)↓↓ = −1~↓↓
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 14 / 17
“Good” quantum numbers
|0 0〉 = 1√2
(↑↓ − ↓↑)
|1 1〉 = ↑↑, |1 0〉 = 1√2
(↓↑+ ↑↓), |1−1〉 = ↓↓
Inspecting the four solutions, we see that these states are still eigenstatesof the individual spin operators, S2 and S2 but not of their z-components,
S(1)z and S
(2)z
Thus the valid quantum numbers for these states are s, m, s, s – a generalresult
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 15 / 17
“Good” quantum numbers
|0 0〉 = 1√2
(↑↓ − ↓↑)
|1 1〉 = ↑↑, |1 0〉 = 1√2
(↓↑+ ↑↓), |1−1〉 = ↓↓
Inspecting the four solutions, we see that these states are still eigenstatesof the individual spin operators, S2 and S2
but not of their z-components,
S(1)z and S
(2)z
Thus the valid quantum numbers for these states are s, m, s, s – a generalresult
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 15 / 17
“Good” quantum numbers
|0 0〉 = 1√2
(↑↓ − ↓↑)
|1 1〉 = ↑↑, |1 0〉 = 1√2
(↓↑+ ↑↓), |1−1〉 = ↓↓
Inspecting the four solutions, we see that these states are still eigenstatesof the individual spin operators, S2 and S2 but not of their z-components,
S(1)z and S
(2)z
Thus the valid quantum numbers for these states are s, m, s, s – a generalresult
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 15 / 17
“Good” quantum numbers
|0 0〉 = 1√2
(↑↓ − ↓↑)
|1 1〉 = ↑↑, |1 0〉 = 1√2
(↓↑+ ↑↓), |1−1〉 = ↓↓
Inspecting the four solutions, we see that these states are still eigenstatesof the individual spin operators, S2 and S2 but not of their z-components,
S(1)z and S
(2)z
Thus the valid quantum numbers for these states are s, m, s, s – a generalresult
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 15 / 17
Adding angular momentum
Addition of spin for hydrogen is the simplest case of a more general case,which can be to add any form of angular momentum from multiple sources
For two spins s1 and s2, their“sum” gives all possible values of the spinfrom (s1 + s2) down to |s1 − s2| in integer steps
s = (s1 + s2), (s1 + s2 − 1), (s1 + s2 − 2), . . . , |s1 − s2|
the combined eigenstate of the total spin, |s m〉 is given by a linearcombination of the composite states
|s m〉 =∑
m1+m2=m
C s1s2m1m2m|s1 m1〉|s2 m2〉
The probabilities are given by the Clebsch-Gordan coefficients, C s1s2m1m2m
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 16 / 17
Adding angular momentum
Addition of spin for hydrogen is the simplest case of a more general case,which can be to add any form of angular momentum from multiple sources
For two spins s1 and s2, their“sum” gives all possible values of the spinfrom (s1 + s2) down to |s1 − s2| in integer steps
s = (s1 + s2), (s1 + s2 − 1), (s1 + s2 − 2), . . . , |s1 − s2|
the combined eigenstate of the total spin, |s m〉 is given by a linearcombination of the composite states
|s m〉 =∑
m1+m2=m
C s1s2m1m2m|s1 m1〉|s2 m2〉
The probabilities are given by the Clebsch-Gordan coefficients, C s1s2m1m2m
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 16 / 17
Adding angular momentum
Addition of spin for hydrogen is the simplest case of a more general case,which can be to add any form of angular momentum from multiple sources
For two spins s1 and s2, their“sum” gives all possible values of the spinfrom (s1 + s2) down to |s1 − s2| in integer steps
s = (s1 + s2), (s1 + s2 − 1), (s1 + s2 − 2), . . . , |s1 − s2|
the combined eigenstate of the total spin, |s m〉 is given by a linearcombination of the composite states
|s m〉 =∑
m1+m2=m
C s1s2m1m2m|s1 m1〉|s2 m2〉
The probabilities are given by the Clebsch-Gordan coefficients, C s1s2m1m2m
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 16 / 17
Adding angular momentum
Addition of spin for hydrogen is the simplest case of a more general case,which can be to add any form of angular momentum from multiple sources
For two spins s1 and s2, their“sum” gives all possible values of the spinfrom (s1 + s2) down to |s1 − s2| in integer steps
s = (s1 + s2), (s1 + s2 − 1), (s1 + s2 − 2), . . . , |s1 − s2|
the combined eigenstate of the total spin, |s m〉 is given by a linearcombination of the composite states
|s m〉 =∑
m1+m2=m
C s1s2m1m2m|s1 m1〉|s2 m2〉
The probabilities are given by the Clebsch-Gordan coefficients, C s1s2m1m2m
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 16 / 17
Adding angular momentum
Addition of spin for hydrogen is the simplest case of a more general case,which can be to add any form of angular momentum from multiple sources
For two spins s1 and s2, their“sum” gives all possible values of the spinfrom (s1 + s2) down to |s1 − s2| in integer steps
s = (s1 + s2), (s1 + s2 − 1), (s1 + s2 − 2), . . . , |s1 − s2|
the combined eigenstate of the total spin, |s m〉 is given by a linearcombination of the composite states
|s m〉 =∑
m1+m2=m
C s1s2m1m2m|s1 m1〉|s2 m2〉
The probabilities are given by the Clebsch-Gordan coefficients, C s1s2m1m2m
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 16 / 17
Adding angular momentum
Addition of spin for hydrogen is the simplest case of a more general case,which can be to add any form of angular momentum from multiple sources
For two spins s1 and s2, their“sum” gives all possible values of the spinfrom (s1 + s2) down to |s1 − s2| in integer steps
s = (s1 + s2), (s1 + s2 − 1), (s1 + s2 − 2), . . . , |s1 − s2|
the combined eigenstate of the total spin, |s m〉 is given by a linearcombination of the composite states
|s m〉 =∑
m1+m2=m
C s1s2m1m2m|s1 m1〉|s2 m2〉
The probabilities are given by the Clebsch-Gordan coefficients, C s1s2m1m2m
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 16 / 17
Clebsch-Gordan coefficients
consider the 2× 1 table of Clebsch-Gordan coefficients
|3 0 > =√
15 |2 1 > |1−1 > +
√35 |2 0 > |1 0 > +
√15 |2−1 > |1 1 >
|2 0 > |1 0 > =√
35 |3 0 > −
√25 |1 0 >
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 17 / 17
Clebsch-Gordan coefficients
consider the 2× 1 table of Clebsch-Gordan coefficients
|3 0 > =√
15 |2 1 > |1−1 > +
√35 |2 0 > |1 0 > +
√15 |2−1 > |1 1 >
|2 0 > |1 0 > =√
35 |3 0 > −
√25 |1 0 >
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 17 / 17
Clebsch-Gordan coefficients
consider the 2× 1 table of Clebsch-Gordan coefficients
|3 0 >
=√
15 |2 1 > |1−1 > +
√35 |2 0 > |1 0 > +
√15 |2−1 > |1 1 >
|2 0 > |1 0 > =√
35 |3 0 > −
√25 |1 0 >
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 17 / 17
Clebsch-Gordan coefficients
consider the 2× 1 table of Clebsch-Gordan coefficients
|3 0 > =√
15 |2 1 > |1−1 >
+√
35 |2 0 > |1 0 > +
√15 |2−1 > |1 1 >
|2 0 > |1 0 > =√
35 |3 0 > −
√25 |1 0 >
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 17 / 17
Clebsch-Gordan coefficients
consider the 2× 1 table of Clebsch-Gordan coefficients
|3 0 > =√
15 |2 1 > |1−1 > +
√35 |2 0 > |1 0 >
+√
15 |2−1 > |1 1 >
|2 0 > |1 0 > =√
35 |3 0 > −
√25 |1 0 >
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 17 / 17
Clebsch-Gordan coefficients
consider the 2× 1 table of Clebsch-Gordan coefficients
|3 0 > =√
15 |2 1 > |1−1 > +
√35 |2 0 > |1 0 > +
√15 |2−1 > |1 1 >
|2 0 > |1 0 > =√
35 |3 0 > −
√25 |1 0 >
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 17 / 17
Clebsch-Gordan coefficients
consider the 2× 1 table of Clebsch-Gordan coefficients
|3 0 > =√
15 |2 1 > |1−1 > +
√35 |2 0 > |1 0 > +
√15 |2−1 > |1 1 >
|2 0 > |1 0 >
=√
35 |3 0 > −
√25 |1 0 >
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 17 / 17
Clebsch-Gordan coefficients
consider the 2× 1 table of Clebsch-Gordan coefficients
|3 0 > =√
15 |2 1 > |1−1 > +
√35 |2 0 > |1 0 > +
√15 |2−1 > |1 1 >
|2 0 > |1 0 > =√
35 |3 0 >
−√
25 |1 0 >
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 17 / 17
Clebsch-Gordan coefficients
consider the 2× 1 table of Clebsch-Gordan coefficients
|3 0 > =√
15 |2 1 > |1−1 > +
√35 |2 0 > |1 0 > +
√15 |2−1 > |1 1 >
|2 0 > |1 0 > =√
35 |3 0 > −
√25 |1 0 >
C. Segre (IIT) PHYS 405 - Fall 2017 October 19, 2017 17 / 17
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