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C H A P T E R
1
What is quantum mechanics anyway? • A quantum is a discontinuous step in any property (as opposed to a continuous
function) • Quantized properties were only discovered as scientists were able to measure smaller
and smaller particles (e.g. charge on an electron) • Quantum mechanics developed along two main tracks:
1. Matrix Mechanics: energy exchange between subatomic particles is quantized: blackbody radiation (Planck), hydrogen atom (Bohr), matrix version of QM (Heisenberg)
2. Wave Mechanics: wave particle duality of light (Einstein), matter (de Broglie), wave equation for matter waves (Schroedinger) and probability waves (Born)
• Paul Dirac showed that these were equivalent • There is a huge difference between this and classical mechanics, where everything is
continuous, a particle is a particle, a wave is a wave, and all parameters can be uniquely determined at any time, simultaneously.
Matrix Mechanics: Evidence for Quantum Phenomena
1) Max Planck: The UV Catastrophe
• From thermodynamics, the average energy per degree of freedom is kTE = where k is the Boltzmann constant, 1.381x10-23 J/K
• Rayleigh-Jeans law says that a wave of frequency, ν, has 3
28cπν degrees of freedom
• So the radiation energy density is kTc
Tu 3
28),( πνν =
• There are two problems with this:
1. The intensity of light predicted by this ( 2vu ∝ ) is much stronger than that which is actually observed
2. If we integrate the radiation density over all possible frequencies, the total energy is infinite! This is called the ultraviolet catastrophe.
How to fix this problem?
• Classically,
kTdEe
dEEeE
kTE
kTE
==
∫
∫∞
−
∞−
0
/
0
/
• Planck suggested energy is discrete, in quanta of ε, and not continuous. Then the integral becomes a sum and E=nε
• Math trick: let x=exp(-ε)/kT, then
• We know how to do sums like this (series progressions). Look in rubber bible.
∑∞
= −=++++=
0
32
111
n
n
xxxxx K and ∑
∞
= −=++++=
02
32
)1(320
n
n
xxxxxnx K
So we now have 111 //
/
−=
−=
−= −
−
kTkT
KT
eee
xxE εε
ε εεε
Planck defined ε=hν, where h=Planck’s constant=6.626x10-34 Js,
so for Planck, 1/ −
= kThehE ν
ν
And therefore the energy density is given by 1
18),( /3
3
−= kThec
hTu ν
νπν
If we integrate this function over all frequencies we now get a finite total energy:
∫∞
− −=
0/
3
3 18
kThTot ed
chU ν
ννπ
Change variables: let x=hν/kT, then dx=hdν/kT and ν=kTx/h
Then, ∫∫∫∞
−
−∞
−
−∞
−=
−
=
−
=
0
3
33
44
0
3
33
44
0
34
3 18
18
18
x
x
xx
x
xTot edxex
hcTk
edxx
ee
hcTk
edxx
hkT
chU πππ
∑
∑∞
=
−
∞
=
−
=
0
/
0
/
n
kTn
n
kTn
e
enE
ε
εε
∑
∑∞
=
∞
==
0
0
n
n
n
n
x
nxE
ε
Again, use a math trick, ∑∞
=
−− =
− 011
n
nxx e
e
So now ∑∫∫ ∑∞
=
∞+−
∞ ∞−− ==
0 0
)1(333
44
0 0
333
44 88n
xnnxxTot dxex
hcTkdxeex
hcTkU ππ
Change variables again: let y=(n+1)x, then dy=(n+1)dx, and
==
+=
+= ∑∑∑ ∫
∞
=
∞
=
∞
=
∞−
9048148
1!3
)1(18
)1(18 4
33
44
1433
44
04433
44
0 0
3433
44 πππππhc
Tknhc
Tknhc
Tkdyeynhc
TkUnnn
yTot
4
33
45
158 T
chkUTot
=
π , where [ ]=σ=Stefan-Boltzmann constant
At low frequency, steps (hν) are small, and it is likely that the classical, E=kT describes a given energy level. At high frequency, steps are large, and it is unlikely that a given level is occupied, more likely E=0, so the intensity è0
Inte
nsity
= u
(ν,T
) Classical
Planck
2) Nils Bohr:The Bohr atom After Rutherford’s alpha scattering experiment he proposed that the nucleus consisted of small negatively charged electrons orbiting a larger positively charged nucleus. è electrons moving in circles are accelerating è accelerating charged particles emit radiation; è this results in energy loss which reduces orbital speed è this will result in the electrons collapsing into the nucleus. But this does not happen—why not? Bohr proposed that a) The radius of an electron’s orbit is quantized, its angular momentum= hnmvr = where n=1,2,3, and within a given orbit no radiation is emitted b) The energy required to jump from one orbit to another is quantized: the electron emits or absorbs a photon of frequency ν, such that Ei−Ef=hν In order to move in a circular orbit, the centripetal force must be equal to the coulomb attraction of the electron to the nucleus (assume hydrogen, qnucleus=e)
So: )4
1 (where 0
2
22
πε== k
rke
rmv
h
h
h
h
nkev
mkenr
mnrmker
mrnv
mvker
kermv
2
2
22
22
222
2
2
22
(1)) (from
=⇒=⇒=
=
=
=
(in c.g.s, k=1) The energy for an electron in orbit n is the sum of its kinetic energy and its potential energy:
−=
−=−=−=
−
=−= 222
42
22
42
22
42
22
42
2
22
22222
n1eV6.131
222221
nemk
nemk
nemk
nemk
mkenke
nkem
rkemvEn
hhhhhh
Ground state, n=1, nmÅmke
ar 0529.0529.0radiusBohr 3
2
01 =====h , E1=−13.6 eV
First excited state, n = 2, r2 = 4a0, E2 = −13.6/4 = −3.4 eV, etc.
So what is the energy change between levels?
−
=
−
=⇒
−
==− 223
222
222
22
222
22 112112
112 fififi
fi nnhemk
nnhemk
nnemkhEE πνν
hh
This predicted frequency agrees well with the observed Balmer series for hydrogen with nf=2, and ni=3,4,5….
3) Stern-Gerlach Experiment 1922: Stern and Gerlach fired a beam of neutral silver atoms through a magnetic field into a detector èatoms separate out into two: one set deflected upward and the other deflected downward Why? Neutral atoms should feel no magnetic force, )( BvqF
rrr×= , however there can be a
force if the atom has a magnetic moment, µr . The energy of an magnetic moment is BE
rr⋅−= µ , so the resulting force is )( BF
rrrr⋅∇= µ .
In the Stern-Gerlach Sectup, kzBB ˆ)(=r
, so z
BF zzz ∂
∂= µ , and deflects the beam in the ±z
direction
Classicaly we envisage the electrons as orbiting in a loop, creating a current, that results in a magnetic moment µ=IA. If we consider the electron to have a charge q, speed v, and orbiting at radius r, then
Lmqqrvr
vrq
22)(
)/2(2 === π
πµ , in terms of the angular momentum L=mrv.
Classically, the angular momentum is continuous, and so we would expect a continous spread of particles in the detector. So if we get two discrete groups, then the momentum must be quantized into two discrete values. It turns out that these two values are due to the ‘spin’ angular momentum of the electron having two possible values, spin up or spin down. Silver has 50 electrons: why do we only see one? All but one electron are in closed shells, which have total momentum of zero. This angular momentum quantization will be examined in much more detail later in the course. 4) Heisenberg arranged the quantum steps into arrays, and thus this became known as the
matrix form of quantum mechanics. If there were two possible values, then you got a 2x2 array, if three values, then a 3x3 array, etc.
WAVE MECHANICS
1) Einstein Photons: (photoelectric effect) A photon is a particle of light, with E=hν=pc, and since ω=ck and ω=2πν, we get kp h=
ph
pcch
pch
=
=
=
λ
λ
ν
2) De Broglie said: if light waves can be moving particles, why can’t we treat moving particles as waves? Treated Bohr’s orbits as stationary (standing waves) and said, smallest radius=1λ, as as n increases, orbits become multiples of the wavelength. This was observed by Davisson-Germer: they shot an electron beam through a diffraction grating consisting of a single crystal. If they are particles, we would expect to just see a blob of incident electrons on the other side, instead they saw a diffraction pattern. So what is the wavelength of a particle? Well, in analogy with photons, let
mvh
ph
==λ (for a wave, λ=2π/k)
Eg: Derive Bohr’s angular momentum quantization for the Bohr atom from de Broglie’s relation. i.e. one circumference = integer number of wavelengths
formula sBohr'2
2
===
⋅=
mvn
mvnhr
phnr
h
π
π
e.g. 2: Scattering of high-energy protons is like diffraction of light waves from a slit. 200GeV protons are scattered by a hydrogen target at an angle given by psinθ=1.2 GeV/c. Estimate the radius of the proton. For diffraction, dsinθ=nλ, where we can take the proton size to be approximately d (a hydrogen nucleus is just a proton). So if we solve for d,
mmd
m
GeVJcGeV
sJph
rad
cGeVpGeVpcEp
cGeV
nd
1518
1810
34
1004.1006.01021.6 so
1021.6
1106.1/200
.10626.6,Broglie de from and
006.0200
2.1sin so
/200200 and
/2.1sin above, from
sin
−−
−−
−
×=×
=
×≈
×⋅
×==
==
=⇒==
=
=
λ
θ
θ
θλ
2) Schrödinger’s Equation for Matter Waves
The classical wave equation with which you are familiar has the form 2
2
2
2
dxdtΨ∂
∝Ψ∂
But Schrödinger’s equation has 2
2
dxdtΨ∂
∝Ψ∂
The one dimensional Schrödinger equation (1D S-E) for a particle in a potential V is:
Ψ+∂Ψ∂−
=∂
Ψ∂ Vx
txmt
txi 2
22 ),(2
),( hh (Gr. 1-1)
( π2/h=h ) Why doesn’t the Schrödinger equation have a second derivative with time? If it did, it would not reduce to the non-relativistic case, which is a requirement. Proof: Take a plane wave solution for a free particle (V=0) )( tkxie ω−=Ψ . For a de Broglie wave,
kp h= and ωh=E so we can rewrite the wave function as )( Etpxi
e−
=Ψ h . If we now plug this wavefunction into the 1D S-E, we get:
energy kinetic classical21)(
21
2
2222:
:
222
222)(2
2
22
)(
====⇒=
Ψ=Ψ
−
=
∂−
=Ψ∂−
Ψ=
−
=∂Ψ∂
−
−
mvm
mvm
pERHSLHS
mppi
mepi
dxmdxmRHS
EeEiit
iLHS
Etpxi
Etpxi
h
h
h
hh
hhh
h
h
If Schrödinger had used the second derivative with time, you would get the relativistic case, E=p2c2+m0
2c4 (check this yourself!) The fact that Plank’s constant now appears in the S-E is what imposes quantization on the wave and all the properties of matter. One problem is that the presence of the first derivative means that the wave function is going to be complex, not real, and this is what gives rise to the “probability” interpretation.
Complex Number Review Cartesian Form and Argand Diagram In Cartesian form, any complex number, z, can be written as the sum of its real and imaginary parts. If x=Re(z)=real part of z, and y=Im(z)=imaginary part of z, then we can write, z = x + iy, where ±i are the two possible square roots of −1. We can further interpret this in terms of an Argand diagram, as shown below, with the real component of z on the x axis, and the imaginary part on the y axis.
This further allows us to transform into polar coordinates, with 22 yxr += as the magnitude of the complex number, z, and
=
xyarctanθ as its argument or phase. If given polar
coordinates, it is simple trigonometry to convert back to Cartesian form. Euler’s Formula θθθ sincos iei += for any real angle expressed in radians. So we can also write z=reiθ
Adding and subtracting complex numbers: Combine real and imaginary parts separately.
( ) ( ) ( ) ( )2121221121 yyixxiyxiyxzz ±+±=+±+=± Multiplying and Dividing Complex numbers: Two methods, depending on whether complex or polar form:
Remember: 1))((
1))((1))((
+=−−=−−
−=
iiii
ii
In polar form: multiply/divide magnitudes and add/subtract arguments
( )( ) ( )2121212121
θθθθ +== iii errererzz ( ) ( ) ( ) ( )2121212121 /// θθθθ −== iii errererzz
In Cartesian form, multiply using the foil method
( )( ) )()( 1221212121122121221121 yxyxiyyxxyyyixyixxxiyxiyxzz ++−=−++=++= In Cartesian form, divide using the complex conjugate, z*=x−iy:
r
More on Complex conjugates:
( )( ) 2222 zyxrrerezz ii =+=== −∗ θθ Statistical Interpretation of Complex Wavefunctions So what is the meaning of a wavefunction for a particle? Classically, we can say where a particle is, and what its speed/momentum is at any time. How do we do this when we have a complex wavefunction describing the particle as a function of position and time? We use Born’s statistical interpretation, where the square of the magnitude of the wavefunction at (x,t), ),(),(),( 2 txtxtx ∗ΨΨ=Ψ indicates the probability of that particle being in location x at time t. The probability is always positive, by definition.
Note that: ∫ Ψb
a
dxtx 2),( is the probability of the particle lying between a and b at time t. [1-3]
We are thus more likely to find a particle at a location where 2),( txΨ is large, than at a
location where 2),( txΨ is small, but it is impossible to predict where the particle will be when you measure it. In 1964 John Bell showed that if you knew where the particle was before the measurement is done, you actually change the result of the measurement. The best viewpoint is the orthodox one, that before you do the measurement, the particle is not localized anywhere, and only by doing the measurement, do you force the particle to “choose” a location.
( )( )
( )( )
( )( ) 2
222
12212121
22
22
22
11
22
11
2
1 )()(yx
yxyxiyyxxiyxiyx
iyxiyx
iyxiyx
zz
++−++
=−−
++
=++
=
When you actually do the measurement, and find, for example, the particle at position C, the wavefunction collapses to a spike.
Probability Review Discrete Variables: If we have a set of N measurements of a discrete variable, xj (can only have integer values) where j=1 to N, we can define certain features: The number of values with a particular value of x, N(x)
This is of course, constrained such that the total number of values ∑∞
=
=0
)(x
xNN [1-4]
The probability of making a measurement x, is then N
xNxP )()( = [1-5]
**** Note that ∑∞
=
=0
1)(x
xP [1-6]
The most probable x is the x for which P(x) is greatest The median x is the value of x for which there are the same number of data points above and below.
The mean or average x is defined as ∑∑ ∞
=
==0
)()(
xxxP
NxxN
x [1-7]
The mean square is 2
0
22 )( xxPxxx
≠= ∑∞
=
in general [1-8]
The deviation is xxx −=∆ , which can be positive or negative, so 0=∆x [1-10] The variance σ2, is defined as 2222 )( xxx −=∆=σ
The standard deviation is: 222)( xxx −=∆=σ [1-11]
For a given function f(x), ∑∞
=
=0
)()()(x
xPxfxf [1-9]
Eg: If we have 14 people in a room, 1 aged 14 years, 1 age 15, 3 aged 16, 2 aged 22 and 2 aged 24., and 5 aged 25 Then P(16)=3/14, Most probable age is 25, Median age is 23 (7 above, 7 below), leave it to you to calculate Mean, Variance and Standard Deviation Continuous variables: eg, position, speed, etc we replace sums by integrals, and probability P(x), by probability density ρ(x):
The probability of having a value of x between a and b is ∫=b
aab dxxP )(ρ [1-15]
And following the derivation for discrete variables, we have:
∫∞
∞−
=1)( dxxρ [1-16]
POSITION OPERATOR ∫∞
∞−
= dxxxx )(ρ [1-17]
∫∞
∞−
= dxxxfxf )()()( ρ [1-18]
e.g. Drop a rock from rest at the top of a cliff of height h, and measure x(t). What is the average distance traveled? (i.e. take pictures at uniform time intervals, what is the most common x for these pictures)
By definition, ∫=h
dxxxx0
)(ρ
From continuity, dttdxx )()( ρρ = ρ(x) is not easy to define, but ρ(t) is just the probability of each time, which is constant, so ρ(t)=1/T, where T is the total time to fall.
First year kinematics: 221)( gttx = and gt
dtdxv ==
è gtdtdx =
gxt 2
= and ghT 2
=
So hx
dxgdx
xg
hg
gtdx
hgdttdxx
2222)()( ==== ρρ
332
21
21
21)(
0
2/3
000
hxh
dxxhx
dxxh
dxxxxhhhh
=
==== ∫∫∫ ρ
This makes sense, since we would expect the rock to move slower in the top part of its trajectory, and thus the average value should be less than h/2
Check: is ?1)(0
=∫h
dxxρ
Yes! ( ) 122
12
1)( 02/1
00
=== ∫∫hhh
xhx
dxh
dxxρ
Normalization and Probability Currents: The particle has to be SOMEWHERE. This means that we must be able to normalize the wavefunction.
i.e. ∞<Ψ∫∞
∞−
dxtx 2),( , or in other words, Ψ(x,t) is square-integrable
In order for this to occur, then δ+→Ψ∞→ 5.0x
1n faster tha 0),(, as txx
Normalization implies that 1)(),( 2 ==Ψ ∫∫∞
∞−
∞
∞−
dxxdxtx ρ [1-20]
What rationalization do we have for requiring this normalization? Well if Ψ(x,t) is a solution to the Schroedinger equation, then so is any complex number A times the wavefunction. So all
we have to do is pick A so that 1),( 2 =Ψ∫∞
∞−
dxtx
If we normalize the wavefunction at t=0, it remains normalized for all time.
For this to be true, we must have 0),(),( 22 =Ψ∂∂
=Ψ∂∂
∫∫∞
∞−
∞
∞−
dxtxt
dxtxt
[1-21]
Proof: ttt
tx∂Ψ∂
Ψ+Ψ∂Ψ∂
=∂
Ψ∂ **),( 2
[1-22]
From the one-dimensional Schrödinger equation:
Ψ+
∂
Ψ∂−=
Ψ−
∂
Ψ∂−−
=∂Ψ∂
Ψ−
∂Ψ∂
=
Ψ+
∂Ψ∂−
=∂Ψ∂
Vixm
iVxmit
Vixm
iVxmit
h
hh
h
h
hh
h
2
2
2
22
2
2
2
22
2**
21*
221
ΨΨ+
∂
Ψ∂Ψ
−+
ΨΨ−Ψ
∂
Ψ∂=
∂Ψ∂
Ψ+Ψ∂Ψ∂
=∂
Ψ∂**
2**
2**
),(2
2
2
22
Vixm
iVixm
ittt
txh
h
h
h
∂Ψ∂
Ψ−Ψ∂Ψ∂
∂∂
=
∂
Ψ∂Ψ−Ψ
∂
Ψ∂=
∂Ψ∂
xxmi
xxxmi
ttx **
2**
2),(
2
2
2
22hh
[1-25]
We define the probability current, j, as the negative of the quantity in brackets, above, so
Ψ
∂Ψ∂
−∂Ψ∂
Ψ= **2
),(xxm
itxj h (like the current density in electrodynamics)
And that long equation above becomes:
0),(),(=
∂∂
+∂
∂x
txjt
txρ (analog of ρ(x,t) is charge density in electrodynamics)
If we integrate this over space to solve for Gr [1-21], i.e., we get:
+∞
∞−
∞
∞−
∞
∞−∫∫
∂Ψ∂
Ψ−Ψ∂Ψ∂
=Ψ∂∂
=Ψ∂∂
=∂
∂xxm
idxtxt
dxtxtt
tx **2
),(),(),( 22 hρ [1-26]
However, since the wavefunctions are normalizable, both Ψ and Ψ∗→0 as x→∞
And so 0),(=
∂∂
ttxρ QED [1-27]
Also, this means that both the wavefunction and its position derivative must be continuous everywhere, otherwise j(x,t) would diverge (matter creation and destruction).
[1-23 and 1-24] Substitute into [1-22] above
Postion: We have defined position as ∫∞
∞−
Ψ= dxtxxx 2),( . Remember that this is NOT the
average position of a given particle, but rather the average you would get if you measured the position of a large number of identical particles all with the same wavefunction (ensemble). Momentum: How do we define the momentum in quantum mechanics?
Since dtdxmmvp == , we anticipate that
dtxd
mp =
As in the previous section,
dxxxx
xm
idxtxt
xdt
xd∫ ∫∞
∞−
∞
∞−
∂Ψ∂
Ψ−Ψ∂Ψ∂
∂∂
=Ψ∂∂
=**
2),( 2 h [1-29]
Remember for differentiation and integration by parts:
[ ]
∫∫ −=
+=
b
a
b
a
b
agdx
dxdffgdx
dxdgf
dxdfg
dxdgfxgxf
dxd )()(
So
∂Ψ∂
Ψ−Ψ∂Ψ∂
−
∂Ψ∂
Ψ−Ψ∂Ψ∂
=
∂Ψ∂
Ψ−Ψ∂Ψ∂
∂∂
∫∫∞
∞−
∞
∞−
∞
∞−
dxxx
xxxm
idxxxx
xm
i ****2
**2
hh
The first term on the RHS goes to zero since Ψ goes to zero at ±∞ So now we have
∫∞
∞−
∂Ψ∂
Ψ−Ψ∂Ψ∂−
= dxxxm
idt
xd **2
h [1-30]
Use the same trick again, and discard the boundary term.
∫
∫∞
∞−
∞
∞−
∂Ψ∂
Ψ−=
∂Ψ∂
Ψ−
=
dxx
ip
dxxm
idt
xd
*
*
h
h
[1-33]
Define operator O, such that (more in Chapter 3)
∫
∫
∫
∞
∞−
∞
∞−
∞
∞−
Ψ
∂∂
−Ψ=
ΨΨ=
ΨΨ=
dxx
ip
dxxx
dxOO
h*
*
*
[1-34], [1-35]
So the position operator is x, and the momentum operator is )/( xip ∂∂−= h For any given operator that is a function of x and p we can write
∫∞
∞−
Ψ∂∂
−Ψ= dxx
ixQQ ),(* h [1-36]
Example: Kinetic energy operator: T=p2/2m
∫∫∞
∞−
∞
∞− ∂Ψ∂
Ψ=Ψ
∂∂
−Ψ= dx
xmdx
mx
iT 2
22
2
*22
* hh
[1-37]
The Uncertainty Principle This arises due to the impossibility to state where exactly a wave is located AND what its speed is at the same time. Remember that for a wave, the momentum is defined by the wavelength.
λπ
λh2
==hp So in a wave with a well defined wavelength, you can measure the momentum
accurately. However, the location of the wave is hard to determine. If the position of the wave is well-defined (i.e. a spike) then the wavelength (i.e. speed) is poorly defined. Both can not be well defined at the same time. This leads to Heisenberg’s uncertainty principle, which we will discuss more in Chapter 3.
2h
≥pxσσ [1-40]
C H A P T E R
2
1
Chapter 2: The time independent Schrödinger equation For a given potential V(x,t) the 1D SE is:
V
x
tx
mt
txi
2
22 ),(
2
),( (TDSE) [2.1]
If the potential is time-independent, i.e. V=V(x), we can simplify things using separation of variables.
i.e. )()(),( txtx [2.2] This is not applicable to all problems, but we can use it to obtain separate solutions that can be patched together to give the correct result. With this assumption, we can substitute our new wavefunction into the TDSE. Note that now the partial derivatives turn into ordinary derivatives, i.e.
dt
tdx
t
tx )()(
),(
and )(
)(),(2
2
2
2
tdx
xd
x
tx
So, putting this in TDSE gives: (dropping the explicit position and time dependence for brevity)
Vdx
d
mdt
di
2
22
2
Now divide this by the wavefunction, i.e. .
Vdx
d
mdt
di
2
22 1
2
1
[2.3]
In this equation, the LHS depends only on TIME and the RHS depends only on POSITION. This equality, then, can only be true if each of the sides is equal to some constant (called a separation constant). Let’s call this constant E. So now if we solve the differential equation on the LHS of the equation:
Edt
di
1
Rearrange:
iE
dt
d
1
Separate variables: dtiEd
2
Doing the indefinite integral gives: /)( iEtAet , where A is the integration constant (we will set A=1 for simplicity, and just absorb it into the position dependence) Now the right hand side is
EVdx
d
m
2
22 1
2
Which becomes: EV
dx
d
m
2
22
2
[2.5]
This is the time-independent Schrödinger equation (TISE) and its solution will depend on the potential that is present. Properties of Solutions (x) to the TISE and of )()(),( txtx
1) The solutions are stationary states. i.e. 2
),( tx is independent of time. The
wavefunction varies with time, but the probability of being at a given location is independent of time.
Why? 2//2
))(*(*),( iEtiEt eetx
In fact, this time independence is not only true for the probability, but for ANY
expectation value dxOOO
ˆ*ˆˆ of any operator O for these
wavefunctions. In particular, this means that <x> is constant, so <p> = 0nothing happens! 2) Since we can define an operator that measures the total energy of a state, these solutions have a well-defined total energy that is constant. We call the energy operator the ‘Hamiltonian’, just like in classical mechanics.
Classically, )(2
),(2
xVm
ppxH [2-10]
Quantum mechanically, we have defined the momentum operator dx
d
ip
ˆ , so
)(2
ˆ2
22
xVdx
d
mH
[2-11]
So now if we rewrite the TISE, we see that EH ˆ [2-12]
3
Thus
EdxEdxHH *ˆ*ˆ [2-13]
So our special constant E is just the expectation value of the Hamiltonian operator, i.e. the total energy.
2222 *ˆ*ˆ*ˆˆ*ˆ*ˆ EdxEdxHEdxEHdxHHdxHH
So, 0ˆˆ 22222 EEHHH every measurement of energy gives the same
answer. 3. We can form a general solution to the time dependent Schrodinger equation by forming a linear combination of solutions to the TISE. i.e. if the TISE has solutions …. With energies E1, E2, E3 …., then
/33
/22
/11
321 )(),(,)(),(,)(),( tiEtiEtiE extxextxextx
So a general solution to the TDSE is
1
/)(),(n
tiEnn
nexctx [2-15]
We can then solve for the specific cn using boundary conditions. Now we can solve any potential V(x) using the following rules: 1. Solve TISE to get n(x) and En 2. Use boundary conditions for the potential to solve for the initial (t=0) state
nnn xcx )()0,(
3. General solutions to the TISE are
n nnn
tiEnn txcexctx n ),()(),( /
Note that this general solution is NOT stationary, thus probability and expectation values do vary with time. Example: )()()0,( 2211 xcxcx is a linear combination of two stationary states.
Solve for 2
),(),,( txtx and describe the motion, assuming that cn and n are real.
/
22/
1121 )()(),( tiEtiE excexctx
4
/)(cos2
)()()()(
)()()()(),(
1221212
222
21
21
/)(/)(2121
22
22
21
21
/22
/11
/*2
*2
/*1
*1
/22
/11
*/22
/11
2
1212
2121
2121
tEEcccc
eecccc
excexcexcexc
excexcexcexctx
tEEitEEi
tiEtiEtiEtiE
tiEtiEtiEtiE
This is a probability that fluctuates with a frequency of /)( 12 tEE about a value
of 2
222
21
21 cc , i.e. not a stationary state
We can simplify our procedure via four theorems (Problems 2.1 and 2.2) a) If the solutions are normalizable, then E must be real. b) We can always find a real solution to the TISE c) If V(x) is even (i.e. V(-x) = V(x)) then (x) are either all odd or all even. d) For any normalizable solution to the TISE, E must be greater than the minimum value of V(x). 2.2 The Infinite Square Well.
elsewhere
axxV
)0(0)(
Obviously the particle moves freely between x=0 and a and turns around at either end. Since V(x) is infinite outside the box, then (x)=0 outside the box, and the probability of the particle being outside the box is also zero. Solution 1) Solve TISE for n and En: inside the well, V(x)=0, so
E
dx
d
mV
dx
d
mTISE
2
22
2
22
22
[2-20]
Or 22
2 2
mE
dx
d [2-21]
Define a wavevector 2
2
mEk . Note k>0, since E>Vmin=0
So our equation becomes: 22
2
kdx
d which is just the classical harmonic oscillator,
which has a solution kxBkxA cossin [2-22] 2) Use boundary conditions to solve for A and B. For this potential, we know (x)=0 outside the box, therefore our two conditions are (0)=0 and (a)=0.
5
a
nkknkakaAa
BBA
n
0sin)(
000cos0sin)0(
This means that the wavevector, k, is quantized, n is positive or negative integers, or zero.
So this means the wavefunctions are ...2,1,0,sin)(
n
a
xnAxn
Since sin(-x)=sin(x) we can ignore the negative n solutions, and just use n = 0, 1, 2, …,
so ...2,1,0,sin)(
n
a
xnAxn
But this means that 00sin)(0 Ax : we can discard this solution.
...3,2,1,sin)(
n
a
xnAxn
[2-24]
Since 2
22222
2 22
2
ma
n
m
kE
mE
a
nk n
nn
[2-27]
Simply due to the boundary conditions, then we have induced quantization of the energy. **note that these solutions for En obey the theorem that En>Vmin.
Lastly we need to solve for A by normalization.
dxn2
1
Since the wavefunction is zero for x<0 and x>a, we can reset the integration limits.
a
AaA
nn
aaA
a
xn
a
nx
Adxa
xnAdx
a
o
aa
n
2
2)0sin(2sin
421
2sin
4
1
2sin1
22
0
222
0
2
So finally, ...3,2,1,sin2
)(
n
a
xn
axn
[2-28]
6
Properties of Solutions to Infinite Square Well: 1. The wavefunctions are alternately odd and even: n=odd, wavefunction even and n=even,
wavefunction odd, 2. number of nodes(not counting walls)= n-1, antinodes =n 3. n are orthogonal, something which is often true in this course
nmdxnm ,0* [2-29]
Proof:
a
nm dxa
xn
a
xm
adx
0
* sinsin2
Use trig. Identity: )cos(2
1)cos(
2
1sinsin bababa
So
0)(
sin)(
)(sin
)(
1
)(cos
)(cos
1
0
00
*
a
aa
nm
a
xnm
nm
a
a
xnm
nm
a
a
dxa
xnmdx
a
xnm
adx
By L’Hopital’s rule: if m=n, the denominator in the first term 0 and the integral diverges
So in fact mnnm dx * QED
Kronecker delta
nm
nm
0
1mn
4. We can express any function f(x) as a linear combination of n(x), i.e. they form a closed
set. This completeness holds for almost all V(x), not just the infinite square well.
So
11
sin2
)()(n
nn
nn a
xnc
axcxf
, which is just Fourier analysis. [2-32]
This means that we can use Fourier’s trick to solve for Cn. Just multiply by m* and integrate
11
*
1
** )(n
mmnnn
nmnn
nnmm ccdxcdxcdxxf
So dxxfc nn )(* [2-34]
What is cn? Approximately the “amount of n in ”
7
2
nc is the probability of measuring the system to be in a state with energy En.
As in chapter 1, the sum of the probabilities must be one for a normalized wavefunction,
1
21
nnc [2-38]
Proof: If the wavefunction is normalized at t=0, and cn is time-independent,
m n nnmnnm
m nnmnm
nnn
mmm cccdxccdxccx
2***
*2
)0,(1
______________ The expectation value for energy is
1
2
nnn EcH [2-39]
Proof: TISE says that nnn EH
m n nnnmnnnm
m n nnnmnnm
m nnmnm
nnn
mmm
EcEcccdxEcc
dxHccdxcHcH
2*2**
**
*
2.3 The Harmonic Oscillator Classically, we know the equation for the harmonic oscillator comes down to Hooke’s Law
2
2
dt
xdmmakxF , and you will have solved it several times in your studies to find
solutions, tBtAtx cossin)( , with m
k and potential 2
2
1kxV .
This quadratic potential is very useful, since we can approximate almost any potential that was a local minimum as a harmonic oscillator by using a Taylor expansion about the position of the minimum, x0
200000 ))((''
2
1))((')()( xxxVxxxVxVxV
At the minimum, V’( 0x ) = 0, subtract the constant V( 0x ) and ignoring higher order terms,
200 ))((''
2
1)( xxxVxV for virtually any oscillatory motion
8
Quantum mechanically: Use 222
2
1
2
1xmkxV in the TISE and solve for
Exm
dx
d
m 22
2
22
2
1
2
[2-44]
This is not a simple differential equation to solve. There are two ways to do this: an elegant algebraic “ladder technique” and a brute force power series method. Both give the same answer, but the “ladder technique” is much simpler, and will come in handy in later chapters. Another benefit of the ladder technique is that it almost completely allows us to eliminate the need for doing integration! So let’s look at the Algebraic technique first. 2.3.1 Algebraic Solution of SHO
From chapter 1, dx
dip
, so,
xd
dp
2
222
The TISE then becomes
Exmm
p
22
2
2
1
2, or Exmp
m 22
2
1
Thus our Hamiltonian becomes 22
2
1xmp
mH . The next step is to try to factor it.
If they were numbers, we could just write:
))((22 vivviuvu , since uv = vu, i.e. u and v commute Since p and x are operators px may not be equal to xp, i.e. p and x may not commute. Define a commutator: [A, B] =AB – BA = the commutator of A and B. If [A,B] = 0, then we say that A and B commute, if 0],[ BA then they do not commute.
So what is [x,p]? Let them act on a test function, f(x), and remember dx
dip
fifidx
dfxi
dx
dfxixf
dx
di
dx
dfxipxfxpfxfpx )(,
So ipx , [2-51] Canonical commutation relation
9
So going back to the Hamiltonian, 22
2
1xmp
mH
mxmppximxmppximxpimxmpxmipxmip 222222 )(],[)()(
So, 22
1
2
1 xmipxmip
mmxmipxmip
mH
Let’s make a shorthand notation for the two terms in ( ) Define two operators as
)(2
1xmip
ma
Raising, or “STEP UP” operator
)(2
1xmip
ma
Lower, or “STEP DOWN” operator
Then
2
1aaH (Note, a+ and a- do not commute, so be careful with the order!
1],[ aa , so we can also write
2
1aaH
In general
2
1 aaH . We can choose which one we want to use for convenience.
So now the TISE that we need to solve is Eaa
2
1 [2-57]
Let’s make a hypothesis. If EH , i.e. is a solution to the TISE with energy E, Then aEaH )( , i.e. a is a solution to the TISE with energy E’= )( E And aEaH )( , i.e. a is a solution to the TISE with energy E’= )( E This means that if we know any one wavefunction and its energy E, we can generate all others using the raising and lowering operators. Can we prove this hypothesis? Yes, using our commutators.
10
Proof:
2
1
2
1
2
1aaaaaaaaaaaH
1],[ aa aaaa 1 ,
So
))((
)2
1(
2
1)1(
aEaEaaH
aaaaaaaaH
EH
QED
AND
2
1
2
1
2
1aaaaaaaaaaaH
1],[ aa 1 aaaa ,
So
))((
)2
1(
2
1)1(
aEaEaaH
aaaaaaaaH
EH
QED
So how do we get our starter wavefunction? Remember we require that E > Vmin, so at some point we can no longer lower the energy any more with the lowering operator. There must be some minimum energy ground state 0 with E=E0. If we act on this ground state with the lowering operator we must get zero. i.e. 00 a : this gives us a differential equation for y0. If we can solve it, we can then use
the raising operator to get all other wavefunctions (eigenstates) of the SHO. (See fig 2.5) Let’s do the solution.
11
0)(2
1)((
2
1)(
2
10 0
0000
xm
dx
d
mxm
dx
dii
mxmip
ma
Rearranging, 0
00
0
xm
dx
dxm
dx
d
And again, xdxmd
0
0 : integrate both sides
2/
0
2
0
2
2ln xmAeC
xm
Now we just need to normalize it:
0
/2/22
0
22
21 dxeAdxeAdx xmxm
Standard integral a
dxe xa
20
22
, where /ma
So normalization requires 4/1
22
/221
m
Am
Am
A
And our final solution is
2/
4/1
0
2xmem
[2-59]
Now we can find the ground state energy, using 000 EH
0002
1 Eaa
0000 2
1 Eaa
, but 00 a , so
22 0000
EE [2-60]
Now all other wavefunctions and energies can be found.
2
1;0 nEaA n
nnn , where An is a normalization factor.
How do we get the normalization factor An? Well 11 nnnnn caa and 11 nnnnn daa
12
Look at our operators, a±. a+=a-* and a-=a+
*.
For any two operators [2-64a,b]
gdxfadxgaf ** )()( and
gdxfadxgaf ** )()(
Normalization:
22
1
2
1*
1* )()()()( nnnnnnnnn ddxddxdddxaa
Use [2-64] with f, g = n, then
ndnd
ndxndxndxaadxaa
nn
nnnn
nH
nnn
n
n
2
2**
2
1
* )()()()(
Similarly, reversing operator order, 1 ncn
So 11 nn na and 1 nn na
We can now solve for all possible wavefunctions:
321
)(
3;
21
)(
2;
10
32
20
21
20
1
aaaaa
Or in general, 0!
1 nn a
n [2-67]
These states are orthonormal: i.e. mnnm dx
*
Proof:
dxmdxaadxaadxaa
dxndxaa
nmnmnmnm
nmnm
****
**
)()()()(
)(
This is true only if n=m or if 0*
dxnm .
This means that we can also use Fourier’s trick for a general solution/linear combination of wavefunctions.
13
2.3.2 Analytical Method: Now let’s solve the SHO the hard way
TISE Exm
dx
d
m 22
2
22
2
1
2
Introduce new variables to simplify the equation:
xm
(units of distance) and E2
(dimensionless)
Then xdm
d
Multiply the TISE by /2 :
Ex
m
dx
d
m
2
2
22
2
And now the constants can be removed from the TISE, as we can write it as
22
2
d
d or
)( 22
2
d
d [2-72]
If we solve this differential equation for K, we will get values for allowed energies, E. _____________________________
What would we get for large values of ? 22 K , so 2
2
2
d
d
This has a simple solution: 2/2/ 22 BeAe Remember every wavefunction must be normalizable, therefore B=0, otherwise the wavefunction will blow up as goes to infinity.
Our solution for large is then 2/2 Ae . ______________________________ So then let’s assume that for our general solution for any value of x, that we can assume a
solution of the form 2/2
)()( eh where )(h is some polynomial. We need to substitute this form into the TISE and then check if it is a valid solution. First we need to find what the second derivative is:
2/2/ 22
)(
ehed
dh
d
d
14
)(12
)()(
22
22/
2
2
2/22/2/2/2/2
2
2
2
2
22222
hd
dh
d
hde
d
d
ehed
dhehe
d
dhe
d
hd
d
d
[2-78]
)( 22
2
d
d
2/222
22/ 22
)()(12
ehKh
d
dh
d
hde
And therefore 0)(122
2
hKd
dh
d
hd
Now if we assume that h() is a polynomial,
0
)(j
jjah [2-79]
Find derivatives to sub into differential equation
02
24322
2
23210
0
1
)2)(1(...)3(4)2(32
...3210
j
jj
j
jj
ajjaaad
hd
aaaajad
dh
Substitute into equation.
0)1(2)2)(1(
0)1(2)2)(1(
02
02
j
jjjj
j
jj
jj
jj
aKjaajj
aKjaajj
In order for this to be true, the value in [ ] must be zero for each power of j. So that means that jj aKjajj )12()2)(1( 2
Or: jj ajj
Kja
)2)(1(
122 [2-81]
So if we know a0 we can derive all even aj:
,....2
1
)4)(3(
5
)4)(3(
5;
2
102402 a
KKa
Kaa
Ka
And if we know a1 we can derive all odd aj:
15
,....)3(2
3
)5)(4(
7
)5)(4(
7;
)3(2
313513 a
KKa
Kaa
Ka
So we have two sets of polynomials, heven and hodd which are completely independent, and
oddeven hhh )(
One problem is that this wavefunction blows up as goes to infinity, so we need to impose some conditions. 1) Parity H(x)=H(-x): this means we have either all odd or all even but not both, i.e. either a0
or a1 must be zero. 2) The remaining series must truncate at some j, i.e. aj+2=0 at some j: let’s call this special
j=n.
12012 nKKn and therefore
2
1nEn , the same as we got the
easy way!
Our new polynomial is )()2)(1(
)(2 2 nja
jj
jna jj
[2-84]
So if n=0, only a0 term, a1=0 (only evens, a2=0 2/00
2 ea
n=1, a0=0=aeven, 11 )( ah , a3=0 2/11
2 ea
n=2, a1=aodd=0; 2202 )( aah 2/2
202
2
)( eaa
If we substitute in xm
and you get the same as found for the operator method.
Hermite polynomials: 0
)(
a
hH
(even) or
1
)(
a
hH
(odd)
So H0=1, H1=2
After normalizing, 2/4/1
2
)(!2
1)(
eH
n
mx nnn
[2-85]
16
2.4 The Free Particle V(x)=0 for all x
Classically, this would imply no force constant velocity
dx
dVF
What happens quantum mechanically?
TISE: 22
2
2
22
2k
dx
dE
dx
d
m
where
mEk
2 is real since E>Vmin=0
This just looks like our solution inside the infinite square well, but we have no boundaries this time. Use: ikxikx BeAe
tiiEt exextx )()(),( / where m
k
m
kE
22
222
So then
t
m
kxikt
m
kxik
BAeAetx 22),(
[2-93]
This is just the equation for a wave with speed m
kv
2
The A term is traveling to the right (+x) while the B term is traveling to the left (-x)
17
A given wave is travelling one way or the other, so either A or B is zero
Lets redefine k:
mEk
2 , + if the wave moves to the right and – if it moves to the
left.
Then
t
m
kkxi
Aetx2
22
),(
Properties of the wave: m
kvkp
k q 2;;
2
Classically, the speed is qvm
k
m
pv 2
for a particle
There is a paradox in the disagreement between the classical particle speed and the quantum mechanical wave speed! Another problem is that if we try to normalize the wave,
dxAdx2
* : not normalizable. This means that these solutions are not
possible: the free particle can not exist in any one of these stationary states. Solution: create a wave packet made up of these states to describe the motion of the free particle.
A given wavepacket has a range of k, E and v since it is made up of a number of waves.
dkektx mtkkxi 2/2
)(2
1),(
[2-100]
18
This is like the linear combinations we made before, except here there is an integral
rather than a sum, and dkkcn )(2
1
What if we are given (x,0) and are asked to find (x,t)
At t=0, dkekx ikx
)(
2
1)0,(
This is just a problem in Fourier analysis (math methods):use Planchard’s theorem
f(x) of ansformFourier trf(x) of ansformFourier tr Inverse
)(2
1)()(
2
1)(
dkexfkFdxekFxf ikxikx
[2-102]
So for a free particle, if we know(x,0) we can write
dkektx mtkkxi 2/2
)(2
1),(
where dxexk ikx
)0,(
2
1)(
[2-103]
Now the particle wave function is a wavepacket, not a wave, so it is the group velocity, not the velocity of an individual wave that matters e.g. in water, vg=v/2, ripple moves at half the speed of an individual wave. So what is vg? Assuming that vg is well-defined means that (k) must be sharply peaked at some value k0.
...2
)('')(')()(
200
0000
kk
kkkk ,
where 0’ is the first derivative wrt k, evaluated at k0, etc If (k) is sharp, then we can ignore all terms above first order, and center our wavepacket at k=k0 (i.e. redefine s=k-k0)
So dsesktx tsxski )'()(0
000)(2
1),(
At t=0, dseskx xski )(0
0)(2
1)0,(
And at t≠0,
)0,'(
)')(0
)'(
0
00000 )(2
1),(
tx
txski
phase
tki dsesketx
Remember, m
k
2
2
19
And so m
k
dk
dv
kkg
0
0
(classical v) and
m
k
kvphase 2
0
(quantum v)
So let’s compare our solutions so far: Infinite Square Well Harmonic Oscillator Free Particle normalizable normalizable not normalizable n discrete n discrete k continuous
n
nnc n
nnc dkk k )(
Why is the free particle so different to the others? For the infinite square well, and SHO, E<V at some pointthere are classical turning points, but for a free particle V=0 so E >V for all points Scattering states: if E > V(x) on one or both sides of the potential, the system is in a scattering state, otherwise it is in a bound state (can’t escape) See figure 2.12 in text. In real life, most potentials go to zero at infinite separation (e.g. Coulomb, gravity), so if the energy is positive, you have a scattering state, and if the energy is negative, you have a bound state. The SHO and infinite square well are bound states while the free particle is in a scattering state. 2.5 Delta function potentials: (either well or barrier)
Math review: Dirac delta function 0)(x
0)(x
0)(
x and 1)(
dxx
a)(x
a)(xax
0)(
)()()( afdxaxxf
i.e. in “picks” out the value of the function at x = a.
Delta function well )()( xxV , >0.
TISE: Ex
dx
d
m )(
2 2
22
20
Bound states E<0, scattering states E>0 Bound States (E<0): for x<0, V(x) = 0
222
2
2
22 2
2
mE
dx
dE
dx
d
m ( is real since E<0) [2-116]
This has a simple solution:
xx BeAe for x<0 [2-118] As x goes to negative infinity, the first term blows up, so we must have A=0. So xBAe (x<0) [2-119] For x>0, we work it out the same way. xx GeFe This time we must set G=0 to prevent the wavefunction from blowing up as x goes to plus infinity. So xFe for x > 0. [2-120] What happens at x=0? Well the two wavefunctions must match at x=0 (boundary conditions).
Normally we also require dx
dto match at the boundary as well, but in this case we can’t,
since V at x=0. So: FBFeBAe )0()0(
Then )0(
)0(
x
x
BAe
BAex
x
[2-122]
We now need to solve for B by normalizing the wavefunction.
0
2022
0
220
222
221
xxxx ee
BdxeBdxeBdx
/22
21
22 mEB
BB
21
So to find B we need to know what the allowable values of energy, E, are. To do this, we need to look at the discontinuity in wave function at x=0.
This means that we will have to look at a very small region as within a distance ± of x=0 Let’s look at the TISE, and integrate it from – to + and then take the limit as goes to zero.
dxxEdxxxdx
dx
d
mdxd
)()()(2
)0(/
2
22
to- from curveunder (area 0
00
2
)(lim)0(lim2
dxxE
dx
d
dx
d
m
Evaluate first derivatives:
0 as 2/32/3
eedx
d
dx
d x
0 as 2/32/3
eedx
d
dx
d x
22
So now our equation gives 0)2(2
)0(2/32
em
m
m 0
2
There is only one value of , so this means there is only one value of E only one bound state solution.
2
222
22
m
mE
[2-129] 2/)(
xmx em
ex
Scattering (continuum) states (E > 0) Again, as before, except where x=0, the TISE is
222
2
2
22 2
2k
mE
dx
dE
dx
d
m
(k is real since E>0)
Now the solutions are:
ikxikx BeAe for x<0 ikxikx GeFe for x>0 (G from right, F reflected by well) These terms are oscillatory, so we can’t make any assumptions about what happens at infinity. We need to make some assumptions: initial wave only comes in from left to right, no wave comes in from x=+∞ G=0 Then we have the terms meaning: A incident from left, B reflected by well toward left and F transmitted through the well to the right. We have the same problems as before, is continuous at the well, but its derivative is not. is continuous therefore at x=0, we get A + B = F
ikBikAdx
d
x
0
ikFdx
d
x
0
23
If we do as we did for the bound states, and integrate the TISE as we go through zero, we get
BA
and
BAFikdx
d
dx
d
m
dx
d
dx
d
)0(
)(lim
above, from and,
)0(2
lim
0
20
So )()(2
2BAFikBA
m
We can now rearrange to get an expression for the transmitted intensity, F, in terms of the incident and reflected intensities.
22
21
21
imB
imAF
Define k
m2
Then )21()21( iBiAF To express the transmitted amplitude in terms of only the incident amplitude, use F=A+B, so substitute in B=F-A
iA
F
1
and also the reflected intensity is
i
AiB
1.
To get the reflection and transmission probabilities (R and T),
Reflection Probability:
2
22
2
2
2
21
1
1
m
EA
BR
[2-141]
24
Transmission Probability:
E
mA
FT
2
222
2
21
1
1
1
[2-141]
Properties: 1) R+T=1, 100% probability of being either reflected or transmitted. 2) Higher energy means higher probability of transmission, lower probability of reflection. 3) These wavefunctions can’t be normalized, so we must make a wavepacket. This means that energy, R and T are AVERAGE values, not exact Delta function barrier )(xV Solutions are found the same way, but there are no bound states. R and T for the scattering states are exactly the same as for the Delta function well. Classically, we would expect zero transmission through a barrier, but in quantum mechanics we can have a finite tunneling probability. 2.6 Finite Square Well:
ax
axaVoxV
(
0
)(
There are both bound states and scattering states Bound states for 00 EV and scattering states for E >0.
Bound states: expect exponential decay outside the well, oscillating inside Region 1 (x<a): V=0 , E<0
22
2
2
22
2
dx
dE
dx
d
m
where
mE2 is real
Solution: xx BeAe ; A0 otherwise (x) as x - So x
I Bex )( Region III (x>a),: V=0, E<0
25
Solution: xx GeFe ; G 0 otherwise (x) as x So x
III Fex )( In region II: -a<x<a, -V0 <E <0 (V = -V0)
22
2
02
22
2l
dx
dEV
dx
d
m
where
)(2 0VEml
is real
Oscillatory solution: )cos()sin( lxDlxC Apply boundary condictions at x = ±a, wavefunction and its derivative are continuous Also, since V(x) = V(-x) we can choose either even or odd solutions, i.e. (x) = ±(-x)
Even:
0
0
)(
)cos()(
x
ax
ax
x
lxD
Fe
x
x
[2-151]
At x=a: continuity of )cos(laDFe a continuity of d/dx )sin(lalDFe a
Dividing these two gives : )tan(lal where
mE2 and
)(2 0VEml
So this is a formula for all allowed values of energy E that can be solved numerically or graphically:
Let z=la and 00 2mVa
z
: then a
zmVl
20
2022 2
So 20
2 zza
Then we can rewrite our condition as zzzz tan20
2
or 1tan2
0
z
zz (transcendental equation) [2-156]
Solve Graphically: (drawn on board in class)
26
27
Limiting solutions: (solve exactly) Wide, deep well, V0, a are very large, z0 is very large, solutions are zn~n/2 (n=odd)
2
222
0 )2(2 am
nVEn
: like an infinite square well with a2a,
En+V0 is the distance from the bottom of the well Shallow, narrow well: V0, a are small large, z0 gets smaller and smaller until for z0</2 there is only one bound state: there is always at least ONE bound state in the well. Scattering states: E>0, all 3 regions are oscillatory
reflected
ikx
incident
ikxI BeAe where
mEk
2
)cos()sin( lxDlxCII where
)(2 0VEml
dtransmitte
ikxIII Fe (transmitted wave only, no wave incident from right to left.
This gives four boundary condition equations, two at x=a and two at x = -a. x = a (1) laDlaCBeAe ikaika cossin since sin(x) = sin(x) (2) laDlaClikBeikAe ikaika sincos x = a (3) ikaFelaDlaC )cos()sin(
(4) ikaikFelaDlaCl sincos Now solve for B, C, D, F in terms of A
(1)+(3) )cos(2
)(
la
eFBAeD
ikaika
(5)
(2)+(4)
)cos(2
)(
lal
eBFAeikC
ikaika
(6)
Substitute (5) and (6) into (1) and (2) and solve for A in both:
)tan(1
)tan(1)( 2
lal
ik
elal
ikBF
A
ika
and lalik
lalikBFA
tan
tan)(
28
Set these two equations for A equal and solve
Fklkl
laiB 22
2
)2sin( [2-167]
)2sin(2
)()2cos(
22
2
lakl
lkila
AeF
ika
[2-168]
Transmission probability: 2
2
A
FT
Transmission coefficient
)(2
2sin
)(41
10
2
0
20 VEm
a
VEE
V
T
We get perfect transmission when T = 1, so when sin( ) = 0
So nVEma
)(22
0
or 2
222
0 )2(2 am
nVEn
(again infinite well with a2a)
Generalizations of wells, barriers: (discuss in class)
C H A P T E R
3
C H A P T E R
4
C H A P T E R
5
1
Chapter 5: Identical Particles How do we combine wavefunctions for systems in which there are more than one particle? e.g. Two identical particles at locations 21 and rr
at time t, have the wave
function ),,( 21 trr
which satisfies the time dependent Schrödinger equation
),,(22
H where 2122
2
221
1
2
trrVmm
Ht
i
To form 1
one differentiates with respect to the coordinates of particle 1,
e.g. kz
jy
ix
ˆˆˆ111
1
These wavefunctions are also normalized, i.e. 123
132 rdrd
And the probability of finding particle 1 in volume 13rd
and of finding particle 2 in
volume 23rd
at time t is 221 ,, trr
If we further simplify this, so that the potential becomes time-independent, then we can
separate variables as in chapter 2, and /2121 ),(),,( iEterrtrr
This is a solution to the time independent Schrödinger equation,
ErrVmm
),(22 21
22
2
221
1
2
e.g. Hydrogen atom consists of a heavy proton and a light electron, and the Coulomb potential depends only on the separation BETWEEN them, i.e. rVV , where
21 rrr
. In this special case, the Schrödinger equation can be rewritten in terms of
this separation, r
, and the position of the center of mass, 21
2211
mm
rmrmR
.
Obviously, we need to rewrite 2122
21 and ,, rr
in terms of R and
r
2
Solve for 1r
rrrrrr
1221
rm
Rrmm
mRr
rmrmmRmmmm
rmrmrm
mm
rmrmR
121
21
21212121
21211
21
2211
Where the reduced mass, 21
21
mm
mm
Similarily, solve for 2r
2121 rrrrrr
rm
Rrmm
mRr
rmrmmRmmmm
rmrmrm
mm
rmrmR
221
12
12212121
22211
21
2211
Define: kZjYiXR ˆˆˆ
where 21
2211
mm
xmxmX
etc.
Define: kzjyixr ˆˆˆ
where x = x1x2 etc. Then to get the gradient functions,
rR
x
m
xXmxx
x
Xx
X
x
21
21111
And
rR
x
m
xXmxx
x
Xx
X
x
12
12222
3
rRrR
rRrrRR
mm
mmm
2
222
2
21
22211
21
2
Similarily: rRrR mm
2
222
2
22 2
So now if we substitute these into the TISE, we get:
ErVmmmmmm
ErVmmmmmm
ErVmm
rR
rRrRrRrR
)(11
2
11
2
)(22
22
)(22
21
22
21
2
21
22
1
222
12
2
2
222
21
2
22
2
221
1
2
And further, 111
21
12
21
mm
mm
mm
So we can simplify this down,
ErV
mm rR )(22
22
2
21
2
or
ErVmm rR
)(
2)(22
22
21
2
The part in the box depends only on the center of mass position, while the rest depends only on the particle separation, so the Hamiltonian can be separated into two independent TISE’s.
RR Emm
2
21
2
)(2
and
rr ErV )(2
22
, where E = ER + Er
4
In Chapter 4 we ignored the motion of the center of mass (i.e. the first equation) when we solved for the hydrogen atom wavefunctions using the second equation with =me. How much error did this introduce?
Error is approximately 0.054%or 1044.51 4
p
eee
m
mmmE
In quantum 2 we will learn how to treat these small “perturbative” effects. Bosons and Fermions We have two particles, a and b, with wave functions )(ra
and )(rb (ignoring spin).
The two particle wavefunction can be written as the product of these, i.e.
)()(),( 2121 rrrr ba
If these particles are identical, we can not tell them apart, so we don’t know which particle is at which position. In this case, there are two possible new wavefunctions:
)()()()(),( 212121 rrrrArr abba
For bosons (particles with integer spins) we use e.g. photons, mesons For fermions (particles with half-integer spins) we use e.g. electrons, protons. So for fermions, two particles can not exist with the same wavefunction, i.e. in the same state. Why? If both particles are in the state )(ra
, then we would have
0)()()()(),( 212121 rrrrArr aaaa
This statement is known as the Pauli exclusion principle, and applies to all identical fermions. Lets define an “exchange operator”, P, which exchanges/switches the particles.
),(),( 1221 rrfrrPf
Obviously P2 = 1 and the eigenvalues of P are ±1 for functions which are symmetric or antisymmetric under exchange. For identical particles, m1=m2 and 0,),(),( 1221 HPrrVrrV
5
This means we can have simultaneous eigenfunctions of the parity and Hamiltonian operators.
),(),( 1221 rrrr
+ for bosons, the wave function is symmetric under exchange. for fermions, the wave function is antisymmetric under exchange. Exampe: Two non-interacting particles in one dimensional infinite square well. From Chapter 2, the one-particle states are given by:
2
222
2
222
2 ;
2 ;sin
2)(
maKKn
ma
nE
a
xn
ax nn
The two-particle state depends on the type of particles.
1) Distinguishable particles: then KnKnExxxx nnnnnn22
212121 212121
);()(),(
The ground state has n1=n2=1, and KEa
x
a
x
ax 2 ;sinsin
2)( 11
2111
The first excited state has 2 degenerate wavefunctions:
KEa
x
a
x
ax 5 ;
2sinsin
2)( 12
2112
KEa
x
a
x
ax 5 ;sin
2sin
2)( 21
2121
2) Indistinguishable bosons:
KEa
x
a
x
a
A
a
x
a
x
aa
x
a
x
aAx 2 ;sinsin
4sinsin
2sinsin
2)( 11
21122111
Find A by normalizing:
2
11sinsin
16
0 0
2212122
2
Aa
x
a
xdxdx
a
Aa a
In this case, we get the same as for distinguishable particles,
;sinsin2
)( 2111
a
x
a
x
ax
6
First excited state: non degenerate, only one possible combination E=5K
2
12sinsin
22sinsin
2)( 1221
21
Aa
x
a
x
aa
x
a
x
aAx
a
x
a
x
a
x
a
x
ax 1221
212
sinsin2
sinsin2
)(
3) If indistinguishable fermions, we can’t have n1=n2 therefore the ground state is non-degenerate, E=5K
a
x
a
x
a
x
a
x
axx 1221
21122
sinsin2
sinsin2
)()(
Exchange Forces Assure that the one dimensional wavefunctions , )(xa , )(xb , are orthonormal, 1
dimension wavefunctions. Then the two particle wavefunction is Case 1: distinguishable particles )()(),( 2121 xxxx ba
Case 2: Indistiguishable bosons: )()()()(2
1),( 212121 xxxxxx abba
Case 3: Indistiguishable fermions: )()()()(2
1),( 212121 xxxxxx abba
What is the expectation value of the square of the separation between the particles for each of these three cases?
We want to find 2122
21
221
2 2 xxxxxxx
Case 1: a
ba xdxxdxxxx 2
1
22
212
121
21 )()(
(i.e. it is the same as the
expectation value of x2 in state a.
Similarly b
ab xdxxdxxxx 2
1
12
122
222
22 )()(
And: ba
xxxx 21
7
Therefore, babahabledistinguis
xxxxx 2222
Case 2:
ba
lorthonorma
baab
lorthonorma
abba
abba
xx
dxxxdxxxxdxxxdxxxx
dxxdxxxdxxdxxxx
22
,0
222*
111*2
1
,0
222*
111*2
1
1
22
212
121
1
22
212
121
21
2
1
])()()()()()()()(
)()()()([2
1
Similarly:
abxxx 222
2 2
1
And 2
21 abbaxxxxx , where dxxxxx baab )()(*
Therefore, 2222 22 abbabaxxxxxx
Case 3: 2222 22 abbabaxxxxxx
In summary:
222 2 abhabledistinguis
xxx
This means that identical bosons are closer together than distinguishable particles, while identical fermions are farther apart than distinguishable particles (due to Pauli exclusion principle). If the particles are far enough apart that their wavefunctions do not overlap, then
0abx , and we do not need to consider the Pauli exclusion principle.
This function, abx is known as the overlap integral. It makes the system act as though
there is an attractive (bosons) or repulsive (fermions) force between the particles. This is called the exchange force, but it is purely geometric, and not really a force.
8
Hydrogen Covalent Bonding: Symmetry Considerations Consider the hydrogen molecule, H2: it consists of two hydrogen atoms each with one electron (fermion). These two fermions tend to push away from each other due to the Pauli exclusion principle, leaving bare protons in the center of the molecule. This would suggest that the molecule should break apart. In reality, though, we get covalent bonding. Why? We have not considered the electron spin in any of this! The complete wavefunction for the molecule is: )()( sr
. The Pauli exclusion principle implies that this ENTIRE wavefunction must be antisymmetric. The two electron spins combine in either a singlet or triplet state.
Singlet: 2
10,0 :
This spin state is antisymmetric, therefore the radial (r) is symmetric, and the molecule bonds together.
Triplet:
1,1
2
10,1
1,1
:
These spin states are symmetric, therefore the radial (r) is antisymmetric, and the molecule pulls apart as described above. Therefore in H2 the two electrons always combine in a singlet state, S=0 General Atoms A neutral atom with atomic number Z has a nucleus with charge of +Ze surrounded by Z electrons (mass m, charge –e). The Hamiltonian for this atom is:
counting doublefor scompensate 1/2 theelectronsbetween repulsion Coulomb
2
01
2
0
22
4
1
2
1
4
1
2
Z
kj kj
Z
j jj
rr
e
r
Ze
mH
Electrons are fermions, so the total wavefunction must be antisymmetric under the exchange of any two electrons. No two electrons can occupy the same state.
EH where).....,,().....,,( 321321 ZZ ssssrrrr
This problem can only be solved exactly for the case of hydrogen, with Z=1.
9
Helium: Z=2
21
2
02
2
0
22
2
1
2
0
21
2
4
12
4
1
2
2
4
1
2 rr
e
r
e
mr
e
mH
If we ignore the last term, the wave function separates:
)()(),( 2'''121 rrrr mlnnlm
)6.13(1
42 2
2
2
2
20
2
2eV
n
ZE
n
Z
n
ZemE Bohrn
is the energy for each
Z
a
Zema 0
2
20
2
)(
4
is the “Bohr radius” of each
This means the total energy is E = 4 (En + En’).
Helium Ground State: arrea
rrrr /)(2321001100210
218
)()(),(
(symmetric) with
a ground state energy of E0 = 8 (-13.6 eV) = -109eV. Since the radial state is symmetric, the spin state must be antisymmetric (singlet, spins opposite). In reality, E0 = -78.955 eV, where the difference arises due to the electron repulsion term that we had ignored. Helium Excited States: one electron in ground state, one in excited state nlm100
Two possibilities: 1) symmetric spatial, antisymmetric spin-singlet- (parahelium) higher energy since electrons are closer together.
)()()()(),( 211002100121 rrrrArr nlmnlm
2) Antisymmetric spatial, symmetric spin-triplet (orthohelium) lower energy )()()()(),( 211002100121 rrrrArr nlmnlm
10
Parahelium | Orthohelium Note equivalently named states are higher for Para than for Ortho
11
eThe Periodic Tabl
Due to the Pauli Exclusion Principle, in atoms with Z electrons, only 2 electrons can go into a state with a given set of quantum numbers n,l,ml: one each of spin up and spin down (i.e. different ms). The complete set of quantum numbers (n,l,ml,ms) must be different for each electron.
For a given shell (n), there are n2 spatial orbitals, so counting spin, there can be 2n2 electrons per shell. (n=1 holds 2, n=2 holds 8, n=3 holds 18 e-, etc)
He: Z=2, n=1 fills the first row, move on to second row. Li: Z=3, 2 in n=1 shell, 1 in n=2 shell. N=2 can have l=0 or l=1 which have same energy if we ignore the repulsive term, but with the repulsive term, l=0 (lowest) l is favored. In general, the innermost electrons see the full charge of the Ze nucleus, and tend to screen the outer electrons from the nuclear charge. Electrons with higher l see less attractive force, and thus have higher energy.
Li: 2 x [] and 1 x [] Be: Z=4 has 2 x [] and 2 x [] B: Z=5 has 2 x [], 2 x [],and 1 x [x]
12
This continues up to Ne (Z=10) at which point the n=2 shell is filled, and we move on to the next row in the periodic table. Na (Z=11) and Mg (Z=12) fill the n=3, l=0 subshell Al Ar, next six fill the n=3, l=1 subshell
4th row in table. Screening effect is now so strong that K (Z=19) and Ca (Z=20) occupy the n=4, l=0 subshell. Sc Zn, fill n=3, l=2 Ga Kr fill n=4, l=1 Then move to n=5, l=0 etc, but filling order becomes complicated.
Nomenclature for Electron Occupancy l=0 s (sharp), l=1 p (principal), l=2 d(diffuse), l=3 f(fundamental), l=4 g, l=5 h, l=6 i, l=7 k (i.e. skip j), l=8 l etc.
Carbon (Z=6) ground state is
12
2
02
2
01
2 )2()2()1(
ln
ln
ln
pss
Nomeclature for total anular momentum (use capitals)
JS L12 : carbon ground state is 0
3P (S=1, L=1, J=0)
Hund’s Rules for J, L, S (Assume L-S or Russell-Saunders coupling) (best for lighter elements) 1) The lowest energy state has the highest S consistent with Pauli exclusion principle 2) For a given S, the lowest energy state has the highest L consistent with overall antisymmetry. 3) If (n,l) subshell is less than half full, then J = |LS|, if it is more than half full J = L + S See http://en.wikipedia.org/wiki/Electron_configurations_of_the_elements_(data_page) http://www.mpcfaculty.net/mark_bishop/complete_electron_configuration_help.htm
13
14
Solids In solids, the loosely bound valence electrons become free to travel throughout the material. They are typically modeled in one of two ways. 1) Free Electron (Sommerfeld Electron Gas) Theory: no potential energy except at the boundaries of the solid 2) Bloch Theory: electrons feel a periodic attractive potential from the positive atomic cores. 1) Electron Gas: Consider a rectangular prism of material with edge lengths Lx, Ly,Lz. In this case the potential V(x,y,z)= 0 for 0<x<Lx;0<y<Ly; 0<z<Lz and ∞ elsewhere. The wavefunction can be separated (x,y,z)=X(x)Y(y)Z(z), and the TISE can be broken into three parts:
EEEEZEdz
Zd
mYE
dy
yd
mXE
dx
Xd
m zyXzyx ;2
;2
;2 2
22
2
22
2
22
Clearly the solutions to all three parts have the same form.
15
zzzzzz
yyyyyy
xxxxxx
mEkzkBzkAZ
mEkykBykAY
mEkxkBxkAX
2);cos()sin(
2);cos()sin(
2);cos()sin(
Applying the boundary conditions means the wavefunctions 0 at x=0 and x=Li At x=0, this means Bx=By=Bz=0 At x=Li this means that ,....3,2,1;;; nnLknLknLk zzzyyyxxx
After normalizing, this becomes
z
x
y
x
x
x
zyxnnn L
zn
L
yn
L
xn
LLLzyx
sinsinsin
8
2222222222
2
22)(
2 z
z
y
y
x
xzyxnnn L
n
L
n
L
n
mm
kkkk
mE
zyx
This situation is best pictured in k-space. Imagine a lattice of points spaced at distances /Lx along the kx axis,/Ly along the ky axis, and/Lz along the kz axis. An allowed stationary state of the gas exists at each intersection of this grid. Each allowed state
occupies a volume of k-space equal to VLLL zyx
33 , where V is the volume of the
solid. If the solid contains N atoms (on the order of Avogadro’s number) each with q (1,2 or 3) free electrons, then we can calculate the radius, kF, of the octant of the sphere they occupy in k-space. (Since ki >0, only the positive octant can be occupied). Two
electrons can fit into each allowed state, each of which occupies volume V
3, so
3/123/123
3 3/323
4
8
1
VNqkV
Nqk FF
This VNq / is known as the free electron density.
16
The surface of the sphere outlines the level below which the k-space is full for the ground state, and is known as the Fermi surface. The energy corresponding to this surface is called the Fermi energy,
mm
kE F
F 2
)3(
2
3/22222
To find the total energy of all the electrons, imagine, rather than adding individually, assuming a continuum and integrating. The total number of states within a octant of a spherical shell of radius k, width dk is
dkkV
V
dkk2
23
2
2
)4(8
1
stateper volume
shell of volume
And the energy for that shell is 2
422
2
22
22 m
dkVkdkk
V
m
kdE
So for an electron gas,
17
3/23/522
25
2
2
0
42
2
310522
VNqm
k
m
Vdkk
m
VE F
k
TOT
F
How can we explain why the solid does not collapse due to the inward attraction between atoms? Consider what happens if the solid expands by an amount dV
V
dVEdE
VNqmdV
dE
TOTTOT
TOT
3
2
3103
2 3/53/522
2
What happens to an ideal gas when it expands? dW =PdV dW is energy, so in comparison, the above equation gives a equivalent quantum mechanical outward pressure, completely due to the fermions via the Pauli exclusion principal. If the electrons were bosons, they could all exist within a single state, and this outward pressure would not exist.
2
52
3/53/522
2
103
2
3
2
3103
2
m
k
V
EP
VNqmdV
dE
FTOT
TOT
This picture of the free electron gas implies that all solids are conductors. However, we know that there are also insulators and semiconductors. So in order to account for the complexity of materials, we need to make our model a bit more difficult—introduce the periodicity of the lattice potential. Band Structure We will confine ourselves to one dimension for simplicity. V(x+a)=V(x), periodic potential.
Bloch’s Theorem: if V(x+a)=V(x) then constant);()( Kxeax iKa
Obviously, 22
)()( xax
18
In order to eliminate problems due to reaching the end of the sample, assume the sample to be a ring, after a very large number, N, repeats. I.e. string with the end connected to the beginning.
)2,1,0(2
1)()( nNa
nKexNax iNKa
This symmetry means that we only need to solve the Schrodinger equation in one unit cell of the lattice ax 0 in order to find the solution for all space. For our potential, we will use a “Dirac Comb” one dimensional lattice of delta functions with lattice spacing a.
1
0
)()(N
j
jaxxV (the Nth spike at x=-a, then the potential wraps around)
In the region 0 < x<a, V=0, so the TISE is an easy solution that we have seen before.
22
22
2kE
dx
d
m
with
mEk
2 as usual.
The solution for this is (in the region 0<x<a)
)cos()sin( kxBkxA
Now Use Bloch’s theorem to extend the solution. Look at the cell to the LEFT of the origin.
))(cos[)](sin[)( axkBaxkAex iKa ( -a < x <0)
Now put in boundary conditions: At x=0, must be continuous:
)sin(
)cos(]cos[]sin[
ka
kaeBAkaBkaAeB
iKaiKa
As x=0, d/dx is discontinuous, and analyzing as for delta function barriers in Chapter 2,
Bm
kaBkaAkekA iKa2
2)sin()cos(
19
Substitute in )sin(
)cos(
ka
kaeBA
iKa , and we get:
kak
mkaKa sin)cos()cos(
2
Let kaz and 2
am , then
)sin()cos()cos( zz
zKa
: clearly, the right-hand-side of the equation must lie
between -1 and +1, since cos(Ka) has values only in this range. Any z that give values outside this range are forbidden. The resultant forbidden z (i.e. forbidden k) correspond to forbidden energies, or “gaps” in the allowed energies. The allowed energies form “bands” with almost continuous energies, since Ka=2n/N, and N~1023. This entire erivation is assuming one electron. Of course we actually have Nq electrons. q=1 the first band is half full q=2 the first band is filled q=3, fill 1st band and halffill 2nd band, etc. In real materials, bands are more complex than this, but the same general idea applies. An insulator has a full band, a conductor has a partially full band, while a semiconductor has a few holes in a filled band or a few electrons in an empty band usually due to impurity doping. In the free electron model, all energies are allowed, there are no gaps, and every material is a conductor. Quantum Statistical Mechanics All of our work so far has been at temperatures of absolute zero, so all electrons have been in their ground states. As the temperature increases, the electrons begin to occupy excited states. In order to find the probability of a given state being occupied, we need to used quantum statistical mechanics. Assume: in thermal equilibrium any state with the same total energy is equally probable. Temperature is a measure of the total energy of the system.
20
Start with a simple example: 3 non-interacting particles (A, B and C) of mass m in a one-dimensional infinite square well.
)(2
2222
22
CBACBA nnnma
EEEE
where ni=1,2,3,…
Choose a given energy, how many states are there?
e.g. 363)(2
363 2222
22
CBA nnnma
E
There are 13 possible ways this can happen: a) nA=nB=nC=11 which we can write in shorthand as (11,11,11) (1 permutation) b) two=13, one=5: (13,13,5) (5,13,13) (13, 5, 13) (3 permutations) c) two = 1, one = 19 (1,1,19) (19,1,1) (1,19,1) (3 permutations) d) one=5, one = 7, one = 17: (5,7,17) (5, 17, 7), (7,5,17) (7,17,5) (17,5,7) (17,7,5) (6 permutations) If the particles are distinguishable, then we can write N=nA+nB+nC = total occupancy of a given state. 3 in least probable state. 2 in 1 in 2 in 1 in 1 in 1 in 1 in most probable state So if we pick a particle at random from the three, what is the probability of that particle having a given energy? E1: must be in the third configuration, 3 states /13 total states, 2 particles in each state
have E1 13
2
13
3
3
21 P
E2 – E4: zero probability: none of the particles are ever in these states.
21
E5: either 1/3 chance in configuration 3 (3 states/13) or 1/3 chance in configuration 4 (6
states/13) 13
3
13
6
3
1
13
3
3
15 P
E6 not allowed
E7: only 1/3 state in configuration 4 (6/13) 13
2
13
6
3
17 P
E8 –E10 not allowed
E11: configuration 1 all three particles, only 1 of 13 states 13
3
13
1
3
311 P
E12 not allowed
E13 2 out of three in second configuration (3/13) 13
2
13
3
3
213 P
E14 – E16 not allowed
E17: only 1/3 state in configuration 4 (6/13) 13
2
13
6
3
117 P
E18 not allowed
E19: 1/3 chance in configuration 3 (3 states/13) 13
3
13
3
3
119 P
Note that the total probability sums to one, as required. If the particles are indistinguishable fermions, then we can’t have more than one particle in a given state (n), assuming a symmetric spin state. Thus the system must be in the 4th configurations, P5=P7=P17=1/3 If the particles are indistinguishable bosons, then there is one allowed state for each of the 4 configurations. (4 possibles)
6
1
3
2
4
11 P ;
6
1
3
1
4
1
3
1
4
15 P ;
12
1
3
1
4
17 P ;
4
1
3
3
4
111 P ;
6
1
3
2
4
113 P ;
12
1
3
1
4
117 P ;
12
1
3
1
4
119 P
Again P=1 counting of states depends on particle type At equilibrium, particles are distributed in the most probable configuration.
22
General Case: Arbitrary potential, V Energies, E1, E2 etc with degeneracies d1,d2 ... Put N particles into this potential, how do they distribute themselves? Question: how many possible ways (i.e. distinct states) with a configuration (N1, N2, N3, ...), i.e. with N1 particles with energy E1, N2 particles with energy E2, etc. The answer depends on the particle type. Dintinguishable: binomial coefficient (N chose N1) tells you how many ways to choose N1 particles from N candidates.
!)!(
!
111 NNN
N
N
N
How many ways to arrange these N1 particles into the degenerate number of states d1?
!)!(
!
11
11
NNN
dN N
Similarly for each of the other states, except each time we have fewer particles to choose from.
For N2: !)!(
)!(
221
212
NNNN
dNN N
So the total number of distinct states
121
3321
321
221
21
11
121
!!...),(
!)!(
)!(
!)!(
)!(
!)!(
!...),(
321
n n
Ni
NNN
N
dNNNQ
NNNNN
dNNN
NNNN
dNN
NNN
dNNNQ
n
Identical Fermions: only one particle per state, only one N particle state in which a certain combination of one-particle states can exist.
The number of ways to choose Nn occupied states within bin n=!)!(
!
nnn
n
n
n
dNd
d
N
d
23
121 )!(!
!,....),(
n nnn
n
NdN
dNNQ
Identical bosons: hardest, no limit on how many particles in a given slot, skip math
1
21 )!1(!
)!1(,....),(
n nn
nn
dN
dNNNQ
The most probable configuration:in thermal equilibrium every state with a given total energy E and N particles is equally likely. The most probable (N1,N2, …) is the one that can be made the largest number of different ways, i.e. Q (N1,N2…) is a maximum.
This is of course subject to the constraints that:
1nn NN and
1nnn ENE
So how do we maximize Q while maintaining these constraints. Remember the technique of Lagrange Multipliers. For a given function F(x1,x2…) subject to constrains f1(x1,x2,…)=0, f2(x1,x2,…)=0, etc Make a new functions 2211321321 ...),,,...,,( ffFxxxG
To maximize this function, set all derivatives with respect to xi and i to zero and solve. For our function, Q, since we have products, lets use ln(Q) rather than Q in our function G, since the maximum of ln(Q) and the maximum of Q occur at the same point, and taking the logarithm with turn the products into sums, simplifying the math.
11
)ln(n
nnn
n NEENNQG . Here and are the Lagrange
multipliers.
Set 0
nN
G
Distinguishable particles:
121 !
!...),(n n
Ni
N
dNNNQ
n
111
!lnln!lnn
nnn
nn
nnn NEENNNdNNG
If we assume Nn>>1, we can use Stirling’s formula for large N {ln(N!)=NlnN – N}
24
Then we can simplify G:
ENNENNNNdNNNNGn
nnnnnnnn
1
lnlnln
)(01)ln(lnln nEnnn
n
nnnnn
nedNE
N
NNdNd
N
G
Identical Fermions: 1
nE
nn
e
dN
Identical Bosons: 1
1
nEn
ne
dN
What is the physical significance of and ? Need to know Nn,dn,En to calculate them exactly. Let’s do an example Ideal gas in three dimensional infinite square well (e.g. electron gas)
2222222222
2
22)(
2 z
z
y
y
x
xzyxn L
n
L
n
L
n
mm
kkkk
mE
As for free electron gas, ith either 1 or 2s+1 state per 3/V unit of k-space,
Integrate energies over octant of k space: dkkV
V
dkkdk 2
23
2
2
)4(8
1
For distinguishable particles:
1st constraint: 2/322/3
20
22
2
22
mV
Ne
mVedkke
VN kE
2nd constraint:
2
3
2
22
3
2
2/322/3
20
22
VE
mV
Nesubstitite
me
VdkkeE
VE kE
k
25
Note: it doesn’t matter whether the occupancy is 1 or 2s+1, as it would cancel out. This equation is correct for all spins.
Classically: Kinetic energy of gas molecule: Tk
TkN
E
BB
1
2
3
Define the chemical potential TkT B )( . Then: n()= number of particles in a particular state with a given energy:
TkBen /)()( Maxwell-Boltzmann classical distribution for distinguishable particles
1
1)(
/)(
TkBen Fermi-Dirac Distribution for identical fermions
1
1)(
/)(
TkBen Bose-Einstein distribution for identical bosons.
As the temperature approaches absolute zero, the Fermi-Dirac distribution,
0)();(
1)();(0/)(
/)(
ne
neTk
Tk
B
B
The chemical potential at absolute zero is the Fermi energy: all states below are filled, all avoe are empty.
Ideal gas, distinguishable particles, TNkE B2
3
26
TmkV
NTkTkT
BBB
22ln
2
3ln)()(
For fermions and bosons, it is much more complicated. Blackbody Spectrum Photons are bosons (S=1) and m=0 To deal with massless particles we need 4 new assumptions due to relativity: 1. E=h2. ck //2 3. Can’t have S=0, only S = ±1 4. the total number of photons increases as the temperature increasesno more 1st constraint, = 0
Most probable occupation number is 1
1/
TkBe
N
3
2
22
2 2)2(
2 c
dVdkk
Vdk
spin
Planck’s Blackbody Spectrum (energy density): )1(
)(/32
3
TkBecV
Nd
