Transcript quant4

4. The Postulates of Quantum Mechanics
4A. Revisiting Representations
Recall our position and momentum operators R and P
• They have corresponding eigenstate r and k
R r r r , P k  k k
If we measure the position of a particle:
• The probability of finding it at r is proportional to
• If it is found at r, then afterwards it is in state
If we measure the momentum of a particle:
• The probability of momentum k is proportional to
• After the measurement, it is in state
 r   r 
2
2
  r
 k   k 
2
2
  k
• This will generalize to any observable we might want to measure
• When we aren’t doing measurements, we expect Schrödinger’s Equation to
work
4B. The Postulates
Vector Space and Schrödinger’s Equation
Postulate 1: The state vector of a quantum mechanical system
at time t can be described as a normalized ket |(t) in a
complex vector space with positive definite inner product
   0 unless
• Positive definite just means
• At the moment, we don’t know what this space is
 0
Postulate 2: When you do not perform a measurement, the
state vector evolves according to

i
 t   H t   t  ,
t
where H(t) is an observable.
• Recall that observable implies Hermitian
• H(t) is called the Hamiltonian
The Results of Measurement
Postulate 3: For any quantity that one might measure, there is
a corresponding observable A, and the results of the
measurement can only be one of the eigenvalues a of A
• All measurements correspond to Hermitian operators
• The eigenstates of those operators can be used as a basis
Postulate 4: Let {|a,n} be a complete orthonormal basis of
the observable A, with A|a,n = a|a,n, and let |(t) be the
state vector at time t. Then the probability of getting the
result a at time t will be
2
P  a    a, n   t 
n
• This is like saying the probability it is at r is proportional to |(r)|2
The State Vector After You Measure
Postulate 5: If the results of a measurement of the observable
A at time t yields the result a, the state vector immediately
afterwards will be given by
1

 t  
a, n a , n   t 

P a n
• The measurement is assumed to take zero time
THESE ARE THE FIVE POSTULATES
• We haven’t specified the Hamiltonian yet
1 2
H
P  V  R, t 
2m
• The goal is not to show how to derive the Hamiltonian from classical
physics, but to find a Hamiltonian that matches our world
Comments on the Postulates
• I have presented the Schrödinger picture with state vector postulates with
the Copenhagen interpretation
• Other authors might list or number them differently
• There are other, equally valid ways of stating equivalent postulates
– Heisenberg picture
– Interaction picture
– State operator vs. state vector
• Even so, we almost always agree on how to calculate things
• There are also deep philosophical differences in some of the postulates
– More on this in chapter 11
Continuous Eigenvalues
• The postulates as stated assume discrete eigenstates
a, n a, n   aa nn
• It is possible for one or more of these
 ,  ,           
labels to be continuous
• If the residual labels are continuous, just replace sums by integrals
P  a    d a,   t 
2
• If the eigenvalue label is continuous, the probability of getting
exactly  will be zero
• We need to give probabilities that  lies in some range, 1 <  < 2.
• We can formally ignore this problem
– After all, all actual measurements are to finite precision
– As such, actual measurements are effectively discrete
(binning)
Modified Postulates for Continuous States
Postulate 4b: Let {|,} be a complete orthonmormal basis
of the observable A, with A|, = |,, and let |(t) be the
state vector at time t. Then the probability of getting the
result  between 1 and 2 at time t will be
2
P 1     2    d  d  ,   t 
2
1
Postulate 5b: If the results of a measurement of the observable A at time t yields the result  in the range 1 <  < 2,
the state vector immediately afterwards will be given by
2
1

 t  
d  d  ,  ,   t 

1
P 1     2 
4C. Consistency of the Postulates
Consistency of Postulates 1 and 2
• Postulate 1 said the state vector is always normalized; postulate 2
describes how it changes

i
 t   H  t 
– Is the normalization preserved?
t
• Take Hermitian conjugate

†

i

t


t
H




• Mulitply on left and right to make these:
t



i  t 
  t     t  H   t  and  i 
 t    t    t  H †  t 
t
 t

• Subtract:



i  t 
 t   i 
 t    t    t  H  t    t  H †  t 
t
 t

i

 t   t    t   H  H †   t 
t

 t   t   0
t
Consistency of Postulates 1 and 5
• Postulate 1 said the state vector is
always normalized; postulate 5 describes how it changes when measured
– Is the normalization preserved?
2
1


 t   t  
a, n a, n   t 

P a n
 t



1
P a
 a, n
n
1 



  t  a, m a, m   a, n a, n   t  


P a  m
 n

1

  t  a, m a, m a , n a , n   t 

P a m n
1
1

  t  a, m  mn a, n   t  
a, n   t 


P a m n
P a n
a, n   t 
2
P a

1
P a
Probabilities Sum to 1?
• Probabilities must be positive and sum to 1
 P  a   
a
a
n
a, n   t 
2
    t  a, n a, n   t     t    t   1
a
n
Postulate 4 Independent of Basis Choice?
• Yes
Postulate 5 Independent of Basis Choice?
• Yes
Sample Problem
A system is initially in the state
 t  
1
2
 
  
1
2

 2
1

When S2 is measured,
(a) What are the possible outcomes and
corresponding probabilities,
(b) For each outcome in part (a), what
would be the final state vector?
P  a    a, n   t 
2
n
• We need eigenvalues and eigenvectors
P 2
2
  v1   t 
P  0   v2   t 
2
 v3   t 
2


1 2
2

1
4
2
2 0 0 0
 12 


 1 
0
1
1
0
2
2

 2
S

 t    
0 1 1 0


0
0
0 0 0 2
 
0
0
1
0
 1 
 1 
 
 
0
0
2 
2 





v1 
, v2   1  , v3   1  , v4 
0
0
2
2
 
 
 
 




0
 
1
0
0
1  2 2
 v4   t 
P 2
2

2
 
1
2

3
4
2


1 2
2
 02  34
, P 0 
1
4
2  0
3  2
2
4  2
2
Sample Problem (2)
(b) For each outcome in part (a), what
would be the final state vector?
 t   
1
P a
 t
• If 0:
 t   
• If


 a, n
a, n   t 
n
 v2
1
v2 v2   t   2 v2
14
 t   
22:
 t   

2
3
v1 
1
3
1
2
 t 
 12 
 1 
 
 2
0
0
 
P 2
v
1
1
2
 v3
1
2
 v4 0

1
4
3
4
,
2
3  2
2
4  2
2
0
0
1
0
 1 
 1 
 
 
0
0
2 
2 





v1 
, v2   1  , v3   1  , v4 
0
0
2
2
 
 
 
 




0
1
0
0
1
v1 v1   t   v3 v3   t   v4 v4   t 
34
2
3

P  0 
v3

2
1  2
2  0

Comments on State Afterwards:
2
2
:
 t



2
3
v1 
1
3
v3
0:
 t


 v2
• The final state is automatically normalized
• The final state is always in an eigenstate of the observable, with
the measured eigenvalue
• If you measure it again, you will get the same value and the state
will not change
• When there is only one eigenstate with a given eigenvalue, it must
be that eigenvector exactly
– Up to an irrelevant phase factor
1  2
2  0
2
3  2
2
4  2
2
4D.Measurement and Reduction of State Vector
How Measurement Changes Things
Whenever you are in an eigenstate of A, A   a  and you measure A,
the results are certain, and the measurement doesn’t change the state vector
• Eigenstates with different eigenvalues are orthogonal a, n   0 if a  a
• The probability of getting result a is then
2
2
P  a    a, n    a, n     a, n a, n      1
n
a, n
a, n
• The state vector afterwards will be:
1

 
a, n a, n    a, n a, n   

a, n
P a n
• Corollary: If you measure something twice, you get the same result
twice, and the state vector doesn’t change the second time
Sample Problem
• We need eigenvalues and
eigenvectors for both operators
• Eigenvalues for both are  ½
1
, z    ,
0
1  1
, x 
 ,
2  1
 0
, z   
1
1 1
, x 
 
2  1
• It starts in an eigenstate of Sz
• So you get + ½ , and
eigenvector doesn’t change
measure + 12
  , z
100%
  , z
A single spin ½ particle is described by a
two-dimensional vector space. Define the
operators
1 0 
0 1
Sz  
 , Sx  

2  0 1
2 1 0
1
  
0
If we successively measure Sz, Sz, Sx, Sz,
what are the possible outcomes and
probabilities, and the final state?
The system starts in the state
100%
measure + 12
  , z
Sample Problem (2)
1
 0
, z    , , z   
0
1
1  1
1 1
, x 
  , , x 
 
1
2 
2  1
If we successively measure Sz, Sz, Sx, Sz,
what are the possible outcomes and
probabilities, and the final state?
• When you measure Sx next, we find that the probabilities are: P     , x , z
• Now when you measure Sz, the probabilities are: P     , z , x 2  1
2
measure +
  , z
1
2
  , x
measure 
  , x
1
2
50%
measure + 12 :
  , z
50%
measure  12 :
  , z
50%
measure  12 :
  , z
50%
measure  12 :
   , z
2
 12
Commuting vs. Non-Commuting Observables
• The first two measurements of Sz changed nothing
– It was still in an eigenstate of Sz
• But when we measured Sx, it changed the state
• Subsequent measurement of Sz then gave a different result
The order in which you perform measurements matters
• This happens when operators don’t commute, AB  BA
• If AB = BA then the order you measure doesn’t matter
• Order matters when order matters
Complete Sets of Commuting Observables (CSCO’s)
• You can measure all of them in any order
• The measurements identify | uniquely up to an irrelevant phase
4E. Expectation Values and Uncertainties
Expectation values
• If you measure A and get possible eigenvalues {a}, the expectation value is
a   aP  a 
a
• There is a simpler formula:
a   a  a, n    a  a, n a, n     A a, n a, n 
2
a
n
a
• Recall: A a, n  a a, n
n
a
n
a A   A
• The old way of calculating expectation value of p:
  k     2 
3/2
3
d re
 r  , p 
 ik r
 d k  k 
3
• The new way of calculating expectation value of p:
p  i  d 3r *  r    r 
2
k
Uncertainties
• In general, the uncertainty in a measurement is the root-meansquared difference between the measured value and the average
2
2
2
value
 a     a  a  P  a    A  a 
a
• There is a slightly easier way to calculate this, usually:
 a 
2
 A  2 Aa  a
2
2
 A  2a A  a 1  A  2 A  A
 a 
2
2
2
 A  A
2
2
2
2
2
Generalized Uncertainty Principle
•
•
•
•
We previously claimed  x  px   12
and we will now prove it
Let A and B be any two observables
    A  a   i  B  b  
Consider the following mess:
The norm of any vector is positive:
   A  a   i  B  b    A  a   i  B  b    0
  0
   A  a       B  b    i   A  a  ,  B  b    0
2
2
2
2
• The expectation values are numbers, they commute with everything
 2  a    2  b   i   A, B  
2
2
• Now substitute   b ,   a
2  a   b     a  b  i  A, B
2
2
 a  b   12 i  A, B 
• This is two true statements, the stronger one says:
a  b  12 i  A, B 
Example of Uncertainty Principle
 a  b   12 i  A, B 
• Let’s apply this to a position and momentum operator
 ri   p j   12
i  Ri , Pj   12 ii  ij 
1
2
 ij 1  12  ij
• This provides no useful information unless i = j.
 x  px   12
 y   p y   12
 z  pz   12
Sample Problem
Get three uncertainty relations involving the angular momentum operators L
 Lx , Ly   i Lz
 Ly , Lz   i Lx
 Lz , Lx   i Ly
lx  l y  12 i  Lx , Ly   12 ii Lz  12 Lz
l y  lz  12 i  Ly , Lz   12 ii Lx  12 Lx
lx  l y 
1
2
Lz
l y  lz 
1
2
Lx
lz  lx  12 i  Lz , Lx   12 ii Ly  12
lz  lx 
1
2
Ly
Ly
4F. Evolution of Expectation Values
General Expression
d
 H 
• How does A change with time due to Schrödinger’s Equation? i
dt
d
• Take Hermitian conj. of Schrödinger.
i
   H
• The expectation value will change
dt
d
d
d
A
d

A 
 A   
 A   
   A 
dt
dt
t
dt
 dt


i
A
i
 HA   
   AH 
t
d
i
A 
dt
 H , A 
A
t
• In particular, if the Hamiltonian doesn’t depend explicitly on time:
d
H 0
dt
Ehrenfest’s Theorem: Position
Suppose the Hamiltonian is given by
d
i
A
A   H , A 
1 2
dt
t
H
P V R
2m
How do P and R change with time?
• Let’s do X first:
d
i
i
i
i
 Px2  Py2  Pz2 , X 
 P 2 , X   V  R  , X  
X  H , X  
dt
2m
2m
i

Px  Px , X    Px , X  Px
2m
• Now generalize:
d
1
R 
P
dt
m
i  i

2m

Px  Px
d
1
X 
Px
dt
m
Ehrenfest’s Theorem: Momentum
• Let’s do Px now:
d
i
Px 
dt
 H , Px 

i
V  R  , Px 
d
i
A 
dt
A
 H , A 
t








i
V
r

r

V
r

r


 
    
V  R  , Px    r   i V  R  ,    r 
  

x
x


x 


V  r 
d
ii

V R
Px 
V R
i
  r   V  R  , Px   i
X
dt
X
x
• Generalize
d
P  V  R 
dt
Interpretation:
• R evolution: v = p/m
• P evolution: F = dp/dt
d
1
R 
P
dt
m
Sample Problem
The 1D Harmonic
Oscillator has Hamiltonian
1 2 1
H
P  m 2 X 2
2m
2
Calculate X and P as
functions of time
d
P  V  R 
dt
d
1
R 
P
dt
m
• We need 1D versions of these:
d
d 1
P  
m 2 X 2  m 2 X
dt
dX 2
d
1
X 
P
dt
m
2
m
d
d
1


• Combine these:
X  
P  
2
m
dt
dt  m

• Solve it:
X  A cos t   B sin t 
B
A
P 
cos t  
sin t 
m
m
• A and B determined by initial conditions:
2
d2
2
X



X
2
dt
X
A X
B
m

t 0
P
t 0
4G. Time Independent Schrödinger Equation
Finding Solutions
d
i
 t   H  t 
dt
• Before we found solution when H was independent of time
• We did it by guessing solutions with separation of variables:
 t    t  
 i d  t  
 d  t  
 
H

   E
i
    t   H  
dt 

   t  dt 
• Left side is proportional to |, right side is independent of time
• Both sides must be a constant times |
• Time Equation is not hard to solve
iEt
d

t

e


i ln   Et
i
 Edt

• It remains only to solve the timeindependent Schrödinger Equation
H  E
  t   eiEt 
Solving Schrödinger in General

 t   H  t 
t
Given |(0), solve Schrödinger’s equation to get |(t)
• Find a complete set of orthonormal eigenstates of H: H  n  En  n
– Easier said than done
– This will be much of our work this year
 n  m   nm
• These states are orthonormal
• Most general solution to time-independent Schrödinger equation is
i
  t    cn e iEnt  n
n
• The coefficients cn can then be found using orthonormality:
 n  0   n
c
m
m
 m   cm nm  cn
m
cn   n   0 
Sample Problem
An infinite 1D square well with allowed
region 0 < x < a has initial wave function
 a
H  n  En  n
  x, t  0   3 a3  a  a  2 x 
in the allowed region. What is (x,t)?
• First find eigenstates and eigenvalues
• Next, find the overlap constants cn
cn   n   0 
xa
2 2 2
2

nx

n


 n  x 
sin 
 , En 
a
2ma 2
 a 
a ax
 12 a x
  nx 
  nx  
2
  nx  3
dx  1
sin 
dx 

sin 
a  a  2 x   2 6  0 2 sin 



2

3
a
0
2
a
a
 a 
 a  
 a
 a  a
1a
a
2

1
xa
1
  x

  nx 
  nx  
  nx 
  nx   
 2 6 
cos 

sin

cos

sin






 

2 2
2 2
 na
 a   n
 a   0   na
 a   n
 a   12 a 



2 6
• sin(½n) = 1, 0, -1, 0, 1, 0, -1, 0, …
 2 2 sin  12  n 
 n
a
Sample Problem (2)
An infinite 1D square well with allowed
region 0 < x < a has initial wave function

  x, t  0   3 a 3 a  a  2 xa 3
 a
cn 

in the allowed region. What is (x,t)?
• Now, put it all together
n

  x, t   
n 1
8 3
  x, t   2
 a
4 6 sin  12 n 
 n
2
2
e
iEn t
2
  nx 
sin 

a
a


2 2
n 1

1

nx
i

n t


2
1 sin 

 exp  
2 
2 
 a 
n odd n
 2ma 
 2 n2
xa
a Re   
  t    cn e iEnt  n
4 6 sin  12 n 
a Im   
Irrelevance of Absolute Energy
In 1D, adding a constant to the energy makes no difference
H  H  
• Are these two Hamiltonians equivalent in quantum mechanics?
• These two Hamiltonians have the same eigenstates
H  n  En  n
 H   n   En     n
• The solutions of Schrödinger’s equation are closely related
  t    cn e iEnt  n
n
  t   e it 
   t    cn e
i  En  t
 n  eit
n
 t 
• The solutions are identical except for an irrelevant phase
iEnt
c
e
n
n
n