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Chapter-4 Formalism 4-2 Schrodinger Equation During the early days of in development of QM Schrodinger and Heisenberg led the charge. Schrodinger developed a QM theory (Schrodinger Picture) based on his famous equaton. Heisenberg took a Matrix theory approach. Later the two approached were merged in to one formalism of operators. Paul Dirac pursued the relativistic form of QM.
!!2
2m
d
dx2" (x,t) + V(x)" (x,t) = i! #
#t" (x,t)
Momentum In the Schrodinger Picture In Chapter 3 we saw that momentum is defined as
px=!
i
d
dx or
"p =!
i
"!
"! = (
d
dx,
d
dy,
d
dz)
When p is operated on a momentum wave function
!
p= eikx the momemtum is simply given by
p !p = !
i
d
dx e
ikxx= !k
x= p
x
!
i
"" ei
"k i"x= !
i (
d
dx,
d
dy,
d
dz) e
i(kxx+k
yy+k
zz)= ! (k
x,k
y,k
z) = !
"k ="p
4-3 Normalization of the Wave Function Due to the probability interpretation of the probability density ψ∗ψ we must insist on otainign a unit probability over all space.
all space
! " * (x,t) " (x,t) dx = #(x,t)all space
! dx = 1
The continuity equation in physics insures the continuity of flux lines (E&M) or in quantum theory that probability density is continuous.
!!t
"(x,t) +!#i
!j (x,t) = 0
"(x,t) $ % (x,t)* % (x,t)
!j (x,t) =
"
2im% (x,t)* &
!!x
% (x,t) ' &% (x,t)
!!x
% (x,t)*(
)*+
,-
!
!t"(x,t)
!
!x
!j (x,t) ~ "i
!j (x,t) divergence of probability current
Finite surface !
!
!t"(x,t) = 0
!#i
!j (x,t) = 0 at x = $
Time independence of the Normalization
!
!t"(x,t) = 0
!!t " # * (x,t)# (x,t)dx = "
!!t# * (x,t) # (x,t) + # * (x,t)
!!t# (x,t)
$%&
'() dx
!!t# (x,t) =
1
i!
*!2
2m
d
dx2# (x,t) + V(x)# (x,t)
$
%&'
() from Schrodinger Eq.
!!t# * (x,t) =
*1
i!
*!2
2m
d 2
dx2# * (x,t) + V(x)# * (x,t)
$
%&'
() from Schrodinger Eq.
!!t" # * (x,t)# (x,t)dx( )
+(x,t )
" #$$$ %$$$= "
# (x,t) *1
i!
$%&
'()
*!2
2m
d
dx2# * (x,t) + V(x)# * (x,t)
$
%&'
()
# * (x,t) + 1
i!
$%&
'()
*!2
2m
d
dx2# (x,t) + V(x)# (x,t)
$
%&'
()
$
%
&&&&&
'
(
)))))
dx
!!t" +(x,t)dx = "
+!
2im# (x,t)
d 2
dx2# * (x,t) *
1
i!# (x,t)V * (x)# * (x,t)
cancel if V =V *
" #$$$$ %$$$$
*!2im
# * (x,t) d 2
dx2# (x,t) +
1
i!# * (x,t)V(x)# * (x,t)
cancel if V =V *
" #$$$$ %$$$$
$
%
&&&&&&
'
(
))))))
dx
!!t" +(x,t)dx =
!
2im" +# (x,t)
d
dx2# * (x,t)
*# * (x,t) d
dx2# (x,t)
$
%
&&&&
'
(
))))
dx
!!t" +(x,t)dx = *
!
2im"
*d
dx# (x,t)
d
dx# * (x,t)
$%&
'()+
d
dx# (x,t)
d
dx# * (x,t)
cancel
" #$$$ %$$$
+d
dx# * (x,t)
d
dx# (x,t)
$%&
'()*
d
dx# * (x,t)
d
dx# (x,t)
cancel
" #$$$$ %$$$$
$
%
&&&&&&
'
(
))))))
dx
!!t" +(x,t)dx =
*!2im
0
,
"d
dx# (x,t)
d
dx# * (x,t)
$%&
'() * # * (x,t)
d
dx# (x,t)
$%&
'()
$
%&'
()dx = 0
!!t" +(x,t)dx = *
!
2im# * (x,t)
d
dx# (x,t)
$%&
'()* # * (x,t) # (x,t)
d
dx
$%&
'()
$
%&'
()
wave functio and derivaive must vanish at x=0 and inf inity
" #$$$$$$$$$ %$$$$$$$$$
|,0= 0
!!t " +(x,t)dx =
!!t" +(x,t) dx -
!!t
+(x,t) = 0 QED
In general !!t
+(x,t) +.i j(x,t) = 0
The probability density should be constant unless there is a divergence of the probability current from the region of interest. Probability density must be conserved over all space. 4-4 Expectation Values and Variances Expectation values are weighted averages over the quantum probability density ρ = ψ(x,t)* ψ(x,t)
< x > = ! *
V
" x ! dV
< p > = ! *
V
"!
i
d
dx
#$%
&'(! dV
< f (x,p) > = ! *
V
" f (x,!
i
d
dx)! dV
We define the variance ΔA of quantity A as:
!A2 != "A
2=< ((A# < A >)
2>!= $ *
a
b
% ((A# < A >)2$ dx
!x2 !&!"x
2=< ((x# < x >)
2>!= $ *
a
b
% ((x# < x >)2$ dx
4-5 Ehrenfest Theorem The Ehrenfest Theorem connects the time evolution of quantum variables to expectation values. That is we can write the classical equations in terms of expectation values.
p = mdx
dt ! < p > = m
d < x >
dt
F =dp
dt= "
dV
dx !
d < p >
dt = "
dV
dx
4-6 Operators Observables in quantum theory are represented by Operators. The operator A acts on a function f(x) and changes it to g(x). Operators may be differential, rotations, translations, matrix, etc.
A !f (x) = g(x)
Operators A and B do not in general commute AB ! BA . The commutator between two operators A and B is written as
[ A,B ] = AB - BA
In Quantum Mechanics only the expectation value of a commutator has physical meaning
< [A,B] > = ! " * [A,B] " dx
The momentum operator p =!
i
d
dxdoes not commute with the position operator for example:
< [x,p] > = < xp > ! < px > =!
i" * x#
d
dx " dx
$%&
'()!!
i" *#
d
dx (x" ) dx
$%&
'()
< [x,p] > =!
i" * x#
d
dx " dx
$%&
'()!!
i" * (x#
d
dx" +" ) dx
$%&
'()= !!
i" *"dx# = i! QED
Hermitian Operators A = At = A*T
< [x,p] > = i!
Expectation values of an Hermitian operator are real and linked to measurement of quantum variables. Physical variables must be represented by Hermitian operators.
Show < A > = < A >t U sing (AB)t
= (BA)t
! " * A"dx=
?!
! " * A"dx( )t
= ! (" * A"AB
"#$ %$)t dx = ! (A" )t (" *)t dx = ! " * At" dx QED
The position operator x is hermitian and
< x > = ! " * x" dx
< x >t= ! " * x"( )
t
dx = ! x"( )t
" *( )t
dx = ! " tx
t " dx = ! " * x " dx QED
We can show the momentum operator is Hermitian using integration by parts.
< p > =!"
+"
# $ * !
i
d
dx$
%&'
()*
dx = !"
+"
#!
i
d
dx($ *$ )dx
=0$ =0 at±"
" #$$$ %$$$
!!"
+"
# $!
i
d
dx$ *%
&'()*
dx =
!"
+"
#!
i
d
dx$
%&'
()*
*
$dx
<p>t" #$$$ %$$$
Any operator A can be redefined as Hermitian AH as
A
H= 1/ 2 (A + A
t )
Hamiltonian Operator H =
p2
2m+V(x) = ih
!
!t
The Hamiltonian operator is hermitian because it is made of hermitian operators x and p. Thus the energy given by the expectation value E = <H> is a real physical observable. Any operator O(x,p,H) would be hermitian also. Bracket Notation
ψ(x,t) = |! >!=a
b
"#$
%&'
ψ∗(x,t) = <! |!= a*
b*( )
the expectation value of an operator can be denoted by
<O > = <! |O |! > = " ! *O !dx
Hermitian operators satisfy
<ψ |O|ψ> = <ψ |O|ψ>t.
4-8 Uncertainty Principle between Non-commuting Operators If two quantum operators don’t commute,
[A,B] ! 0 they can not be simultaneously measured.
The measurement of A disturbs B and vice versa. One can show that if two operators that don’t commute,
[A,B] ! 0 , then an uncertainty exists eg.
!
A !
B"
1
2i|< [A,B] >|
This important result indicates that we can not simultaneously measure the value of non-commuting operators without some uncertainty in the measurement.
From < [x,p] >= i! we have
!x!p "
!
2
A sharp packet in space would consist of many wavelengths thus broad in mometum!
The operators H= H = i!
!
!t and t do not commute " < [H,t ] >= i!
!E!t "!
2
A sharp packet in time would consist of many frequencies thus broad in energy! 4-9 Eigenstates and Eigenvalues The Hamiltonian operator H represents the quantum system, ! (x,t) the wave function represents an arbitrary state of the system. After seperation of variables we obrtain two eigenvalue equations. We can then solve for the can solve for the eigenmodes !
n.
H(x,x ') ! (x) "(t) = i! ##t
! (x) "(t)
1
! (x) "(t)"(t)H(x,x ') ! (x) =
1
! (x) "(t)! (x)i! #
#t "(t)
1
! (x) H(x,x ') ! (x) =
1
"(t)! (x)i! #
#t "(t) = E separation cons tan t (time and space)
Eigenvalue Eqs.
i! ##t
"(t) = E "(t) $ "(t) = Ae% i
E
!t
H !n(x) = E
n !
n(x) $ Time Independent Schrodinger Eq.
Each operator A has a preferred set of stationary states (constant in time) that can best describe its expectation value. These states and values are called eigenstates |ψ(x)> and eigenvalues λ, satisfying the eigenvalue equation
A |! (x) > = " |! (x) >
The action of
A |! (x) > is to rescale
|! (x) > by an amount λ called eigenvalue.
Also:
An |! > = A "A "A.. |! > = # " # " #.. |! >= #n |! >
F(A) |! > = a0+ a
1A + a
2A
2+ ......( ) |! > = F(#) |! >
Example, let H =1 1
1 1
!
"#$
%& and solve for |'
n> =
a
b
!
"#$
%&
1 1
1 1
!
"#$
%&a
b
!
"#$
%& = E
a
b
!
"#$
%&
1( E 1
1 1( E
!
"#$
%&a
b
!
"#$
%& = 0
det1( E 1
1 1( E
!
"#$
%&= 0 or trivial solution
(1( E)(1( E) (1= 0 ) E 2 ( 2E = 0 * (E ( 2)E = 0 two roots
1) Let E = 0 in equation
1 1
1 1
!
"#$
%&a
b
!
"#$
%& = 0 * a + b = 0 * ) E = 0 and |'
1> =
1
2
1
(1
!
"#$
%&
2) Let E = 2
(1 1
1 (1
!
"#$
%&a
b
!
"#$
%& = 0 * a ( b = 0 * E = 2 and |'
2> =
1
2
1
1
!
"#$
%&
Zero Dispersion The exectation value of an operator A has zero dispersion when measured in its eigenstate.
Pr oof : Let A |!n>= " |!
n>
#A2= < A
2> $ < A >
2
= <!n
| A2 |!
n> $ <!
n| A |!
n>( )
2
= <!n
| "2 |!n> $"2
<!n
|!n>( )
=1
! "## $##
2
= 0 QED
Orthogonality Eigenfuncttions of an hermitian operator corresponding to two different eigenvalues are necessarily orthogonal to each other,
<!
1|!
2>= 0 .
Pr oof :
Let A |!1> = "
1|!
1> and A |!
2>= "
2|!
2> where "
1# "
2 and real
<!2
| A |!1> =<!
2| "
1|!
1>
right
! "# $# = <!
2| "
2
*
left
! "# $#|!
1> $ "
1% "
2
*( ) <! 2| !
1>= "
1% "
2( ) <! 2| !
1>
but "1# "
2 for hermitian operators $ <!
2|!
1> = 0 orthogonal
Degenerate Eigenstates EIgenstates with the same eigenvalue are call degenersate eigenstates. For example a group of states has the same energy. Any group of degenerate states can be arranged in to a set of orthogonal states by Gram=Schmidt Orthogonalization.
Gram - Schmidt Procedure
Suppose we start with a set of degenerate eigenvectors. | e1>, | e
2>, | e
3>,... | e
n>
with eigenvector !.
1) normalize the first ket | e1
'> =
| e1>
|| e1>|
2) Find the overlap of the 2nd vector on the first and subtract it off and normalize.
| e2
'> = | e
2> " < e
1
' | e2>| e
1
'>
Then < e1
' | e2
'> = 0,
3) | e3
'> = | e
3> " < e
1
' | e3>| e
1
'> " < e
2
' | e3>| e
2
'>
Then < e2
' | e1
'> = < e
3
' | e1
'> =< e
3
' | e2
'>= 0
________________________________________________________
Example :
Let | e1> =
1
2
#
$%&
'(, | e
2> =
1
1
#
$%&
'( ,find a set of orthonormal vectors.
1) | e1' > =
1
5
1
2
#
$%&
'(
2) < e1' | e2 >=
1
51 2( )
1
1
#
$%&
'( =
3
5
| e2
'> =
1
1
#
$%&
'("
3
5
1
5
1
2
#
$%&
'(=
1
1
#
$%&
'("
3
5
1
2
#
$%&
'(=
1
5
2
"1
#
$%&
'( )
normalize
!1
5
2
"1
#
$%&
'(
Superposition Principle- Expansion in Eigenfunctions The eigenfunctions of a system forms an orthonormal basis set by which any arbitrary vector can be expanded.
|! > = cn
|"n>
n
#
The probability coefficients are given by c
n= < !
n |" > w probabilities P
n= c
n
2
Pr oof :
Let H |!n> = E
n|!
n> , where |!
n> forms an orthonormal basis set in S.
Assume state |" > in S is expanded in a superposition of basis states |!n>:
|! > = cn
|"n>
n
# $ Can we find the coefficients cn ?
< "m
|! > = < "m
| cn
|"n>
n
#%&'()*= c
n< "
m |"
n>
n
# = cn +
mn=
n
# cm
cm= < "
m |! > QED (quad et demonstatum)
e1
e2
e2’
|e1’. e2|
4-10 Measurement Theorems The quantum measurement of an operator A is made by taking its expectation value < A >!=!< ! | A |! > . This action is interpreted as the average of a series of measurements on an ensemble of identical states. Zero Dispersion Theorem The operator dispersion <ΔA> is zero if |! > resides in eigenstate |!
n> .
The quantum measurement of an operator A residing in one of its eigenstates has zero dispersion.
Pr oof : Let A |!n> = a |!
n>
< A > = <!n
| a |!n> = a <!
n|!
n> = a
< A2> = <!
n| a2 |!
n> = a2
"A2= < A2
> # < A >2 = 0 no dispersion QED
Spectral Theorem
The measurement of a quantum operator in arbitrary state !(x,t) can be written as an expansion
in its probability coefficients cn and eigenvalues "
n as < A > = | c
nn
# |2 "n .
Pr oof :
Let |!(x,t) > = cn
n
# |$n> where A |$
n> = "
n|$
n>
< A > = < !(x,t) | A |!(x,t) > = cm
*m
# <$m
|%&'
()*
A cn
n
# |$n>
%&'
()* = c
m* c
nm,n
# <$m
| A |$n>
= cm
* cn
m,n
# <$m
| "n
|$n> = c
m* c
nm,n
# "n+
mn = | c
nn
# |2 "n QED
< A > == | cn
n
! |2 "n
Compatibility of Commuting Operators
Operators that commute can have simultaneous eigenfunctions.
Pr oof :
Let AB = BA and A |! > = "a |! >
A B |! > = B A |! > = B"a
|!n>= "
a B |!
n>
A B |!n>( ) = "
a B |!
n>( ) apparently B |! > is also an eigenfunction of A!
B |!n> # |!
n> or B |!
n> = "
b|!
n>
Thus |!n> is a simul tan ious eigen function of A and B but with different eigenvalues "
a and "
b
4-11 Continuous eigenvalues When solving the Schrodinger Equation we can find that the eigenstates and eigenvalues are continuous in nature.
Consider the eigenvalue equation for the momentum operator of a free particle :
p |! > = p |! >
!
i
d!
dx= p! "
d!
!=
i
!pdx " !(x) = !(0)e
i!
px = !(0)eikx
Their are no constra int s on momentum #$ % p % +$
For a free particle the eigenstate of momenum |!
p> = | eikx
> is continuous.
For a free particle the eigenstate of momenum
|!
p> = | eikx
> is continuous. Momentum Space Wave Functions
In momentum space
p = p x = !
i
d
dp
!(p,t) =1
2" ! #$
+$
% & (x,t) e# i
p
!x#
E
!t
'()
*+,dx Momentum space wave function
4-12 Stationary States Stationary states are steady state solutions to the Schrodinger Equation. These solutions of eigenvalue equations yielding discrete eigenvalues and eigenstate are often encountered in bound state problems. We demonstrated that if V=V(x) and not a function of time then through separtion of variables We obtain 2 eigenvalue equations in space and time
Eigenvalue Eqs.
i! !!t
"(t) = E "(t) # "(t) = Ae$ i
E
!t
H %n(x) = E
n %
n(x) # Time Independent Schrodinger Eq.
Here the most general solution to the problem is an expansion of ψ(x,t) in a series of discrete eigenstates with associates probability amplitudes cn .
!(x,t) = cn"
nn
# e$ i
En!
t superposition principle
Pr obn= c
n
2
probability forand ! to bein eigenstate "n
a measure of < E > to yield En
Example 1:
You have solved H |! >= E |! > for the energy eigenstates |!1,2,3
>, and corresponding
energies E1,2,3
of a system. The system is prepared in the initial state |! >:
|! > = A 2 |!1> +
1
2 |!
2> "i |!
3>
#
$%&
'( where E
1= 2eV, E
2= 3eV, E
2= 5eV
(a) Normalize |! > ?
A2 2 )2 +1
2)
1
2+ ("i)(i)
#
$%&
'(= 1 A =
1
5.5
|! > = A 2
5.5c1
!
|!1> +
1
2
1
5.5
c2
" #$ %$
|!2> +
"i
5.5c3
!
|!3>
#
$
%%%
&
'
(((
(b) What average energy do you measure for the system?
< E > = 2
5.5
2
2eV +1
2
1
5.5
2
3eV +"i
5.5
2
5eV = 1
5.54 )2eV +
1
2)3eV +1)5eV
#$%
&'(= 47.5 eV
(c) What is the probability for being found in the 3rd energy state ?
P3= c
3
2
= "i
5.5
2
= 1
5.5
Example 2 :
Let H =0 !! 0
"
#$%
&' and |( >=
a
b
"
#$%
&'
H |( >= E |( >
0 !! 0
"
#$%
&'a
b
"
#$%
&'= E
a
b
"
#$%
&'
)E !! )E
"
#$%
&'a
b
"
#$%
&'= 0
det)E !! )E
"
#$%
&'= 0 for nontrivial solution
E2 ) ! 2
= 0 E = +!,)!
1) E1 = +! *
)! !! )!
"
#$%
&'a
b
"
#$%
&'= 0 a = b
|(1>=
1
2
1
1
"
#$%
&'
1) E2 = +! *
)! !! )!
"
#$%
&'a
b
"
#$%
&'= 0 a = )b
|(2>=
1
2
1
)1
"
#$%
&'
( (x,t) = c1|(
1> e
) iE
1
!t
+ c2
|(2> e
) iE
2
!t
where | c1|2 + | c
2|2= 1
Problem #9 ! Time developement of an Operator
i!""t
|# >= H |# > and A not an exp licit function of time.
d
dt< A > =
d
dt<# | A |# > = <
d#dt
| A |# > + <# |"A
"t|# >
"A
"t=0
" #$$ %$$
+ <# | A |d#dt
>
H |# >= ! d
dt|# > $ |
d#dt
> =1
i!H |# > and <
d#dt
| = !1
i!H <# |
d
dt< A > = !
1
i!<# | H
%&'
()*
A |# > + <# | A1
i!H |# >
%&'
()*=
i
!<# | HA ! AH |# >
The commuator with the hamiltonian gives the time developemet of the operator !
d
dt< A > = [H,A]
Note if [H,A] = 0 then d
dt< A > $ < A > + cons tan t of motion
Problem #8 Consider A |!1,2
> = "1,2
|!1,2
> and B |#1,2
> = $1,2
|#1,2
>
|!1> =
1
132 |#
1> + 3 |#
2>( ) |#1
> =1
132 |!
1> + 3 |!
2>( )
|!2> =
1
133 |#
1> % 2 |#
2>( ) |#2
> =1
133 |!
1> % 2 |!
2>( )
1) At time zero < A > = "1 & |! > = 1 |!
1> + 0 |!
2> System in state |! > = |!
1>
2) Measure < B > = <! | B |! > = <!1
| B |!1> =
1
13
'
()*
+,
2
2 < #1| + 3 < #
2|( )B 2 |#
1> + 3 |#
2>( )
< B > = 1
13 (4 < #
1| B |#
1> +6< #
1| B |#
2>
=$2<#
1|#
2>=0
! "## $##+ 6 < #
2| B |#
1>
$1<#
1|#
2>=0
! "## $##+ 9 < #
2| B |#
2>) =
4
13$
1+
9
13$
2
|# > = 2
13|#
1> +
3
13|#
2> After measurement < B > system is in this state!
|# > = |!1>
3) Remeasure < A >
If < B > = $1 with p =
4
13 then |# > = |#
1> =
1
132 |!
1> + 3 |!
2>( )
< A > = < #1| A |#
1> =
1
132 <!
1| + 3 <!
2|( ) A 2 |!
1> + 3 |!
2>( ) =
1
13(4"
1+ 9"
2)
P"1
= p 4
13
'()
*+,=
4
13
'()
*+,
4
13
'()
*+,=
16
169
If < B > = $2 with p =
9
13 then |# > = |#
2> =
1
133 |!
1> % 2 |!
2>( )
< A > = < #2
| A |#2> =
1
133 <!
1| % 2 <!
2|( ) A 3 |!
1> % 2 |!
2>( ) =
1
13(9"
1+ 4"
2)
P" 2= p
9
13
'()
*+,=
9
13
'()
*+,
9
13
'()
*+,=
81
169
Ptot
= P"1
or P"2
=16
169+
81
169 =
97
169
Summary!of!Formalism
1)!!Write!!Schrodinger !equation!H ! = i!""t
! !!where!H = T +V
2)!!If !V # V (t)!then!! !(x,t) =! !(x) !e$ iE
!t
!by!separation!of !var iables.!!!
3)!!Solve!!H ! = E ! !!to! find !orthonormal!set !of !eigenvectors! ! = % 1 , % 2 , % 3 , % 4 ....!!!
!!!!!!!!!!!!!!!!!!!!!!!!!and !corresponding!eigenvalues!!!!E1,!!!!E2 ,!!!E3,!!!E4 ,.....
4)!By!the!expansion!theorem!!!&!! !(x,t) =!
n
' % 1 e$ iE
!t
5)!Find !the!set !commuting!and !non $ commuting!operators!in!the!eigenspace!H!!! !H ,!!A 's
Commutindingeg.![!H ,A]=0
"#$ %$ !!......... B 'sNotCommutingeg.![!H ,B]#0
& !
6)!The!set !of !Commuting!operators!in!H ,A..!!share!eigenvectors! % i ,!!i = 1,2,3!...!
7)!Commuting!operators!A 's!!have!a!diagonal !representation!!!!Aij !=! % i A % j
( j % j
"#%=
(1 0 0
0 (2 0
0 0 ( 3
)
*
++
,
-
.
.
8)!Non $ commuting!operators!B 's!!are!not !digonal !in!system!H
Dij !=! % i D % j
#/ j % j
"#%=
D11 D12 D13
D23 D22 D23
D31 D32 D33
)
*
++
,
-
.
.
9)!The!A 's!!commute!with!themselves!because!they!are!diagonal.!!!But !!generally!![A,B] # 0.!
10)!!Because!of !the!Time!Developement !Theorem!![H ,A] = i!d A
dt!!!!H !and !A!are!stationary!operators
whose!eigenvalues!are! fixed.!!!Operators!B!are!non $ stationary!and !!changing!in!time.
11)!If ! !(0) =! cn % n
n
' !!known!then!by!the!Spectral !Theorem!!! H =! cn2En
n
' !,!! A =! cn2(n
n
'
12)!Successive!measurements!of !a!stationary!operator !have!zero!dispersion.!!!!0 A = A2 $ A
2= 0
13)!Simul taneous!measurements!of !stationary!operators!yield !no!uncertainty,!but !simul taneous!
measurements!of !non $ stationary!operators!are!uncertain.!!
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!0 A !0 B 1!!1
2i|< [A,B] >|0 !!!!!!!.
#10 - Let [H,A] ! 0 and
|"1> =
1
2(| u1> + | u2 >) where A |"
1> = #
1|"
1>
|"2> =
1
2(| u1> $ | u2 >) where A |"
2> = #
2|"
1>
Also H | u1>= E1| u1> | u1> =
1
2|"
1> + |"
2>( )
H | u2 >= E2
| u2 > | u2 > = 1
2|"
1> $ |"
2>( )
Let |% > = |"1> at t = 0
|% (t) > =1
2 ( | u1> e
$ iE1
!t
+ | u2 > e$ i
E 2
!t
)
< A > = <! (t) | A |! (t) > = 1
2 ( < u1| e
+ iE1
!t
+ < u2 | e+ i
E 2
!t
) "A "( | u1> e# i
E1
!t
+ | u2 > e# i
E 2
!t
)
< A > = 1
2< u1| A | u1> + < u2 | A | u1> e
+ iE 2#E1
!t
+ < u1| A | u2 > e# i
E 2#E1
!t
+ < u2 | A | u2 >$
%&'
()
< u1| A | u1> =1
2< *
1| + < *
2|( )A |*
1> + |*
2>( )
=1
2< *
1| A |*
1> +
1
2< *
2| A |*
1>
+1<*
1|*
1>=0
" #$$ %$$+
1
2< *
1| A |*
2>
+2<*
1|*
1>=0
" #$$ %$$+
1
2< *
2| A |*
2> =
1
2 +
2+ +
1( )
< u2 | A | u2 > =1
2< *
1| # < *
2|( )A |*
1> # |*
2>( ) =
1
2 +
2+ +
1( )
< u2 | A | u1> =1
2< *
1| # < *
2|( )A |*
1> + |*
2>( ) =
1
2 +
2# +
1( )
< u1| A | u2 > =1
2< *
1| + < *
2|( )A |*
1> # |*
2>( ) =
1
2 +
2# +
1( )
< A > = 1
2
1
2 +
2+ +
1( ) +1
2 +
2# +
1( )e+ i
E 2#E1
!t
+1
2 +
2# +
1( )e# i
E 2#E1
!t
+1
2 +
2+ +
1( )$
%&'
()
< A > = 1
2 +
2+ +
1( ) +1
4 +
2# +
1( ) e+ i
E 2#E1
!t
+ e# i
E 2#E1
!t$
%&'
() =
1
2 +
2+ +
1( ) +1
2 +
2# +
1( ) cos(E2 # E1
!t)
< A > =1
2 +
2+ +
1( ) +1
2 +
2# +
1( ) cos(E2 # E1
!t)
Since [H,A] , 0 - d
dt< A > , 0 not acons tan t of motion
Similarity!Transformations
Let !!H =1 1
1 1
!"#
$%&!!Find !the!eigenvetors!and !eigenvalues!of !H !and !the
similarity!transformation!which!diagonalizes!H ??
1 1
1 1
!"#
$%&a
b
!"#
$%&= E
a
b
!"#
$%&!!!'!!
1( E 1
1 1( E!"#
$%&a
b
!"#
$%&!= 0!!!'!!E
2 ( 2E +1(1 = 0!!'!!E(E ( 2) = 0
E = 0!!!!'!!1 1
1 1
!"#
$%&a
b
!"#
$%&= 0!!'!!a + b = 0!!!!!!!!!!|)
1>!=
1
2
1
(1!"#
$%&
E = 2!!!!'!!(1 1
1 (1!"#
$%&a
b
!"#
$%&= 0!!'!!a ( b = 0!!!!!|)
2>!=
1
2
1
1
!"#$%&
Similarity!Transformation
H |) n >= E |) n >!!'!!!SHS(1
HD
!S |) n >
new !basisvevtor
"#$ %$= ES |) n >
new !basisvector
"#$ %$
