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
22:
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 eiEt
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 ax
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
xa
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 eit
n
t
• The solutions are identical except for an irrelevant phase
iEnt
c
e
n
n
n