4. The Postulates of Quantum Mechanics

Recall our position and momentum operators P and R•They have corresponding eigenstate r and 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

•This will generalize to any observable we might want to measure•When we aren’t doing measurements, we expect Schrödinger’s Equation to work

4A. Revisiting Representations

, R r r r P k k k

22 r r

r

2 2

k k

k

Vectors Space and Schrödinger’s Equation4B. The Postulates

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

• Positive definite just means• At the moment, we don’t know what this space is

0 unless 0

Postulate 2: When you do not perform a measurement, the state vector evolves according to

where H(t) is an observable.

,i t H t tt

• Recall that observable implies Hermitian• H(t) is called the Hamiltonian

The Results of MeasurementPostulate 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 orthonmormal 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

,n

P a a n t

• This is like saying the probability it is at r is proportional to |(r)|2

The State Vector After You MeasurePostulate 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

• The measurement is assumed to take zero timeTHESE ARE THE FIVE POSTULATES

• We haven’t specified the Hamiltonian yet

• The goal is not to show how to derive the Hamiltonian from classical physics, but to find a Hamiltonian that matches our world

1, ,

n

t a n a n tP a

21,

2H V t

m P R

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

• The postulates as stated assume discrete eigenstates• It is possible for one or more of these

labels to be continuous

• If the residual labels are continuous, just replace sums by integrals

• 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)

Continuous Eigenvalues, , aa nna n a n

, ,

2

,P a d a t

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

1

2

1 2 ,P d d t

Postulate 5b: If the results of a measurement of the observ-able A at time t yields the result in the range 1 < < 2, the state vector immediately afterwards will be given by

2

11 2

1, ,t d d t

P

Consistency of Postulates 1 and 24C. Consistency of the Postulates

• Postulate 1 said the state vector is always normalized; postulate 2 describes how it changes– Is the normalization preserved?

• Take Hermitian conjugate• Mulitply on left and right to make these:

• Subtract:

i t H tt

†i t t Ht

†andi t t t H t i t t t H tt t

†i t t i t t t H t t H tt t

†i t t t H H tt

0t tt

Consistency of Postulates 1 and 5• Postulate 1 said the state vector is

always normalized; postulate 5 des-cribes how it changes when measured– Is the normalization preserved?

1, ,

n

t a n a n tP a

2

1, ,

n

t t a n a n tP a

1, , , ,

m n

t a m a m a n a n tP a

1, , , ,

m n

t a m a m a n a n tP a

1, ,mn

m n

t a m a n tP a

21

,n

a n tP a

1P a

P a

Probabilities Sum to 1?• Probabilities must be positive and sum to 1

a

P a 2

,a n

a n t , ,a n

t a n a n t 1t t

Postulate 4 Independent of Basis Choice?

• Yes

Postulate 5 Independent of Basis Choice?

• Yes

Sample Problem

2 2

2 0 0 0

0 1 1 0

0 1 1 0

0 0 0 2

S

1 12 2

1 2 3 41 12 2

0 01 0

0 0, , ,

0 0

0 10 0

v v v v

A system is initially in the state

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?

21

2

23

24

2

0

2

2

1 11 22 2

t

12

12

0

0

t

2

,n

P a a n t • We need eigenvalues and eigenvectors

2 2 22

1 3 42P v t v t v t 2 2 21 1

220 3

4

2

20P v t 21 12 4 2 3 1

4 42 , 0P P

Sample Problem (2)

1 12 2

1 2 3 41 12 2

0 01 0

0 0, , ,

0 0

0 10 0

v v v v

(b) For each outcome in part (a), what would be the final state vector?

21

2

23

24

2

0

2

2

12

12

0

0

t

1, ,

n

t a n a n tP a

• If 0:

2 34

14

2 ,

0

P

P

2 2

1

1 4t v v t 1

2 22 v

2t v

• If 22:

1 1 3 3 4 4

1

3 4t v v t v v t v v t

2 11 33 3t v v

2 1 11 3 423 2

0v v v

Comments on State Afterwards:

• 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

21

2

23

24

2

0

2

2

20 : t v 2 2 11 33 32 : t v v

How Measurement Changes Things4D.Measurement and Reduction of State Vector

Whenever you are in an eigenstate of A, and you measure A, the results are certain, and the measurement doesn’t change the state vector•Eigenstates with different eigenvalues are orthogonal•The probability of getting result a is then

•The state vector afterwards will be:

•Corollary: If you measure something twice, you get the same result twice, and the state vector doesn’t change the second time

A a

2,

n

P a a n

, 0 ifa n a a

2

,

,a n

a n

,

, ,a n

a n a n

1

1

, ,n

a n a nP a

,

, ,a n

a n a n

Sample ProblemA single spin ½ particle is described by a two-dimensional vector space. Define the operators

The system starts in the state

If we successively measure Sz, Sz, Sx, Sz, what are the possible outcomes and probabilities, and the final state?

1 0 0 1,

0 1 1 02 2z xS S

1

0

100%

12measure +

, z

, z 100%

12measure +

, z

• We need eigenvalues and eigenvectors for both operators

• Eigenvalues for both are ½1 0

, , ,0 1

z z

1 11 1, , ,

1 12 2x x

• It starts in an eigenstate of Sz

• So you get + ½ , and eigenvector doesn’t change

Sample Problem (2)If we successively measure Sz, Sz, Sx, Sz, what are the possible outcomes and probabilities, and the final state?

50%

12measure

, x

, z

12measure + : , z

1 0, , ,

0 1z z

1 11 1

, , ,1 12 2

x x

• When you measure Sx next, we find that the probabilities are:• Now when you measure Sz, the probabilities are:

212, ,P x z

50%

12measure +

, x

212, ,P z x

50%

50%

50%

50%

12measure : , z

12measure : , z

12measure : , z

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

Expectation values 4E. Expectation Values and Uncertainties

• If you measure A and get possible eigenvalues {a}, the expectation value is

• There is a simpler formula:

• Recall:

• The old way of calculating P:

• The new way of calculating P:

a

A A aP a 2

,a n

A a a n , ,a n

a a n a n , ,a n

A a n a n , ,A a n a a n

A A A

23/2 3 32 ,id e d k rk r r P k k k

3 *i d P r r r

Uncertainties• In general, the uncertainty in a measurement is the root-mean-

squared difference between the measured value and the average value

• There is a slightly easier way to calculate this, usually:

22

a

A a A P a 2A A

2 2 22A A AA A 2 22 1A A A A 2 22A AA A

22 2A A A

Generalized Uncertainty Principle• We previously claimed and we will now prove it• Let A and B be any two observables• Consider the following mess:• The norm of any vector is positive:

12xx p

A A i B B

0 0A A i B B A A i B B

2 22 2 , 0A A B B i A A B B • The expectation values are numbers, they commute with everything

• Now substitute

• This is two true statements, the stronger one says:

2 22 2 ,A B i A B

,B A

2 22 ,A B A B i A B 1

2 ,A B i A B

12 ,A B i A B

Example of Uncertainty Principle

• Let’s apply this to a position and momentum operator

• This provides no useful information unless i = j.

12 ,i j i jR P i R P

12 ,A B i A B

12 ijii 1

2 1ij 12 ij

12

12

12

x

y

z

X P

Y P

Z P

Sample ProblemGet three uncertainty relations involving the angular momentum operators L

,

,

,

x y z

y z x

z x y

L L i L

L L i L

L L i L

12 ,x y x yL L i L L

12 zii L 1

2 zL

1 1 12 2 2

1 1 12 2 2

,

,

y z y z x x

z x z x y y

L L i L L ii L L

L L i L L ii L L

12

12

12

x y z

y z x

z x y

L L L

L L L

L L L

General Expression 4F. Evolution of Expectation Values

• How does A change with time due to Schrödinger’s Equation?• Take Hermitian conj. of Schrödinger.• The expectation value will change

d dA A

dt dt

di H

dt

di H

dt

d A dA A

dt t dt

i A iHA AH

t

,d i A

A H Adt t

• In particular, if the Hamiltonian doesn’t depend explicitly on time: 0d

Hdt

Ehrenfest’s Theorem: PositionSuppose the Hamiltonian is given by

How do P and R change with time?•Let’s do X first:

•Now generalize:

,d i

X H Xdt

,d i A

A H Adt t

21

2H V

m P R

2 , ,2

i iX V X

m P R

2 2 2 ,

2 x y z

iP P P X

m

, ,2 x x x x

iP P X P X P

m

2 x x

i iP P

m

1d

dt mR P

1x

dX P

dt m

Ehrenfest’s Theorem: Momentum• Let’s do Px now:

• Generalize

Interpretation: R evolution: v = p/m P evolution: F = dp/dt

,d i A

A H Adt t

,x x

d iP H P

dt

, x

iV P R

, ,xV P i Vx

R r R r i V V

x x

r r r r

Vi

x

r

r x

d iiP V

dt X

R

, xV P i VX

R R

dV

dt P R

1d

dt mR P

Sample ProblemThe 1D Harmonic

Oscillator has Hamiltonian

Calculate X and P as functions of time

2 2 21 1

2 2H P m X

m

dV

dt P R

1d

dt mR P

1dX P

dt m

2 21

2

d dP m X

dt dX 2m X

• We need 1D versions of these:

• Combine these:

• Solve it:

• A and B determined by initial conditions:

2

2

1d dX P

dt dt m

2mX

m

22

2

dX X

dt

cos sinX A t B t

cos sinB A

P t tm m

0

0

t

t

A X

mB P

Finding Solutions 4G. Time Independent Schrödinger Equation

• Before we found solution when H was independent of time• We did it by guessing solutions with separation of variables:

• 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

• It remains only to solve the time-independent Schrödinger Equation

t t

di t H t

dt

d ti t H

dt

d ti

t dt

H E

di Edt lni Et iEtt e

H E iEtt e

Solving Schrödinger in General

Given |(0), solve Schrödinger’s equation to get |(t)•Find a complete set of orthonormal eigenstates of H:

– Easier said than done– This will be much of our work this year

•These states are orthonormal•Most general solution to time-independent Schrödinger equation is

•The coefficients cn can then be found using orthonormality:

n n nH E

niE tn n

n

t c e

i t H tt

n m nm

0n n m mm

c m nmm

c nc

0n nc

Sample ProblemAn infinite 1D square well with allowed region 0 < x < a has initial wave function

in the allowed region. What is (x,t)?

3

3, 0 2x t a a x

a

n n nH E

• First find eigenstates and eigenvalues• Next, find the overlap constants cn

2 2 2

2

2sin ,

2n n

nx nx E

a a ma

0n nc

x a

a

> assume(n::integer,a>0);sqrt(6)/a^2* integrate(sin(n*Pi*x/a)*(a-abs(2*x-a)),x=0..a);

12

2 2

4 6 sin n

n

• sin(½n) = 1, 0, -1, 0, 1, 0, -1, 0, …

30

2 3sin 2

a nxa a x

a a a

Sample Problem (2)An infinite 1D square well with allowed region 0 < x < a has initial wave function

in the allowed region. What is (x,t)?

3

3, 0 2x t a a x

a

• Now, put it all together

x a

a 12

2 2

4 6 sinn

nc

n

niE tn n

n

t c e

12

2 21

4 6 sin 2, sinniE t

n

n nxx t e

n a a

IRe ma a

2 21

22 22

odd

8 3 1, 1 sin exp

2

n

n

nx i n tx t

n a maa

Irrelevance of Absolute EnergyIn 1D, adding a constant to the energy makes no difference

•Are these two Hamiltonians equivalent in quantum mechanics?

•These two Hamiltonians have the same eigenstates

•The solutions of Schrödinger’s equation are closely related

•The solutions are identical except for an irrelevant phase

H H

n n n n n nH E H E

niE tn n

n

t c e ni E tn n

n

t c e niE titn n

n

e c e

itt e t