• It’s fun!• Different ways of thinking about quantum mechanics sometimes involve

different calculation techniques• Sometimes, these techniques make problems easier

My goals:• Teach you useful techniques• Show you how to use them to solve difficult problems• Explode your brains

11. Interpretation of Quantum MechanicsWhy Discuss Philosophy?

• The state vector |(t) is a linear function of the initial state vector |(t0)• Call the operator that performs this function:

• This operator must preserve probability, so it must be unitary• Some other easy-to-prove identities:

Schrödinger’s Equation for U:• We know that • Therefore

• This, together with the boundary condition U(t0,t0) = 1 defines U(t,t0)

11A. The Time Evolution OperatorDefinition

0,U t t 0 0,t U t t t

†0 0, , 1U t t U t t

0 0, 1U t t 2 1 1 0 2 0, , ,U t t U t t U t t

di t H t t

dt

0 0 0 0, ,i U t t t H t U t t tt

0 0, ,i U t t H t U t t

t

Linearity of Time Evolution operator

• Suppose we have two solutions of Schrödinger’s Equation, |1 and |2• We know Schrödinger’s equation is linear, so that

is also a solution• We therefore

know that

• We see that U is a linear operator• It is also reversible

• Measurement, in contrast, is neither linear nor reversible• Clearly, the time evolution operator only applies when not performing

measurements

1 1 2 2c c

0 0,t U t t t

0 0,U t t t t 1 1 2 2c t c t

1 0 1 0 2 0 2, ,c U t t t c U t t t

0 1 1 0 2 2 0 1 0 1 0 2 0 2, , ,U t t c t c t c U t t t c U t t t

†0 0 0, ,t U t t t U t t t

Finding the Time Evolution Operator

• If H is independent of time, easy to solve these equations:

• Suppose we have a complete set of orthonormal eigenstates of H:• Then insert these states into expression for U:

• If H depends on time, expression gets complicated

0 0, ,i U t t H t U t tt

0 0, 1U t t

0 0, expU t t i H t t

n n nH E

0 0, exp n nn

U t t i H t t 0

0, niE t tn n

n

U t t e

0 0 0

0 0 0

1 2

0

3

, 1t t t

t t t

t t t

t t t

U t t i H t dt i H t dt H t dt

i H t dt H t dt H t dt

Sample ProblemA harmonic oscillator with mass m and angular frequency in 1D is initially

in the state |(t0) at time t0. At a later time t, the energy is measured. What is the probability that it will be measured to have the minimum value ½?

• The probability is just• We need to evolve the initial state to the final state:

• Now we get clever: let H act to the left:

2

0P t

2

0 00 ,P U t t t 2

0 00 exp iH t t t

0 02

00 iE t tP e t 102

22

00 i t tt e 2

00 t

22

20,m xm

P e x t dx

Sample ProblemA spin-1/2 particle is in one of the states |+ or |- at time t0. Find a Hamiltonian that evolves it to the state |+ by time t.

• We need: • Take the inner product of

the first with the second• This is impossible, so NO SOLUTION

• Does this mean one can never create a spin + particle?• Yes, if this particle is the only particle in the universe, and all it has is

spin• If we have some other particle, it is always possible to do something like

†0 0, ,U t t U t t

0 0, and ,U t t U t t

0 1

0 0 0 0, , , and , , ,U t t e e U t t e e

Sample ProblemConsider a superposition detector. This device is initially in the state |S0, but such that when it interacts with a spin |, it will change into state |S-,? when faced with a pure

spin state | + or | – , and state |S+,? when presented with a superposition state, where “?” means that it may represent any quantum state. Show such a device is impossible

• What we want:• We will make no assumptions

about a, b, c, other than that theS+ and S- states are orthogonal

• By linearity:• So impossible

0 0

0 0

1 10 0 2 2

, ,

, ,

, ,

U t t S S a

U t t S S b

U t t S S c

, , 0 , ,S a S c S b S c

1 1 10 0 0 0 02 2 2

, ,U t t S U t t S S

12

, , ,S c S a S b

1 , ,S c S c 12

, , , ,S c S a S c S b 0

• Consider a single spinless particle (in 3D)– This can be generalized

• Given (r,t0), what is (r,t)?• Insert a complete set of states |r0

• Now define the propagator, also called the kernel:• Then:

• We can find the propagator, and use it to get the wave function later in one step

11B. The PropagatorIt’s Reason for Existence

, t t r r

0 0 0 0, ; , ,K t t U t tr r r r

0 0,U t t t r

30 0 0 0 0, ,t d U t t t r r r r r 3

0 0 0 0 0, ,d U t t t r r r r

30 0 0 0 0, , ; , ,t d K t t t r r r r r

Schrödinger’s Equation for the Propagator

• From this equation, easy to see

• Assume we have Hamiltonian• Schrödinger’s Equation

• Since true for all(r0,t0), wemust have

30 0 0 0, ; ,K t t r r r r

30 0 0 0 0, , ; , ,t d K t t t r r r r r

2 2 ,H m V t P r

2

2, , ,2

i t V t tt m

r r r

2

3 2 30 0 0 0 0 0 0 0 0 0, ; , , , , ; , ,

2i d K t t t V t d K t t t

t m

r r r r r r r r r

2

3 3 20 0 0 0 0 0 0 0 0 0, , ; , , , , ; ,

2d t i K t t d t V t K t t

t m

r r r r r r r r r

2

20 0 0 0 0 0, ; , , ; , , , ; ,

2i K t t K t t V t K t t

t m

r r r r r r r

Propagator for Constant H• By definition,

• If H is constant, recall

• It follows that

• Therefore

0

0 0 0, ; , niE t tn n

n

K t t e r r r r

0 0 0 0, ; , ,K t t U t tr r r r

0

0, niE t tn n

n

U t t e

0 *0 0 0, ; , niE t t

n nn

K t t e r r r r

Propagator for No Potential

• Let’s work it out for a free particle in one dimension V(x,t) = 0• The eigenstates are plane waves• The propagator is then

• Each of these ki integrals can be done using:

0 *0 0 0, ; , niE t t

n nn

K t t e r r r r

12

ikxk x e

2 2

2k

kE

m

20 02

0 0, ; ,2

i k t t m ikxikxdkK x t x t e e e

2

0 0exp2 2

dk i kikx ikx t t

m

2 212 22Ax Bx B Ae dx Ae

220

0 00 0

1 2, ; , exp

2 2

i x x mmK x t x t

i t t i t t

2

00 0

0 0

, ; , exp2 2

im x xmK x t x t

i t t t t

Sample ProblemAt t = 0, the wave function of a free particle is given by Find (x,t) at arbitrary time

2 2,0 Axx Nxe

2

00 0

0 0

, ; , exp2 2

im x xmK x t x t

i t t t t

0 0 0 0 0, , ; , ,x t dx K x t x t x t

• In 1D, the final wave functionwill just be

• Now we just substitute in:

• Use the identity:

20

2

2 00 0, exp

2 2Ax im x xm

x t N x e dxi t t

2 210 0 0 02, exp 2

2

mx t N x A im t x imxx t imx t dx

i t

21 2

2 3/2 22Ax Bx B Axe dx BA e

2 23/2

, 2 exp exp2 2 2

imx tm imxx t N imx t A im t

i t A im t t

Sample Problem

2 23/2

, 2 exp exp2 2 2

imx tm imxx t N imx t A im t

i t A im t t

3/2 2 2 2

3/2 exp2 2

Nx m i t m x imx

t A t im tA im t

2

3/2, exp2 11

Nx Axx t

iA t miA t m

At t = 0, the wave function of a free particle is given by Find (x,t) at arbitrary time

2 2,0 Axx Nxe

2 2 2 2 2

3/2 exp21

Nx m x imx A t m x

t A t imiA t m

2

3/2 exp21

Nx imx A

A t imiA t m

• It is hard to find K for large time differences, but easy for small• We can build up large ones out of many small ones• Consider the Hamiltonian in 1D:• We wish to solve:

• For short enough times, we expect V to change relatively little, and K to be non-zero only near x = x0

• Estimate V(x,t) = V(x0,t0)

11C. The Feynman Path Integral FormalismThe Idea Behind it

2

,2

PH V x t

m

2 2

0 0 0 0 0 02, ; , , ; , , , ; ,

2

di K x t x t K x t x t V x t K x t x t

t m dx

2 2

0 0 0 0 0 0 0 02, ; , , ; , , , ; ,

2

di K x t x t K x t x t V x t K x t x t

t m dx

Propagator for Constant H

• Multiply by• Cleverly write this as

• This is same as equation for free propagator,and has the same boundary condition

• It therefore has same solution, at time t1 slightly after t0, at position x1:

• Let t = t1 – t0, then we have

2 2

0 0 0 0 0 0 0 02, ; , , ; , , , ; ,

2

di K x t x t K x t x t V x t K x t x t

t m dx

0 0 0 0 0 0

2 2, ,

0 0 0 02, ; , , ; ,

2i t t V x t i t t V x td

i K x t x t e K x t x t et m dx

0 0 0,i t t V x te

0 0 0 0, ; ,K x t x t x x

1 0 0 0

2

, 1 01 1 0 0

1 0 1 0

, ; , exp2 2

i t t V x t im x xmK x t x t e

i t t t t

2

1 01 1 0 0 0 0, ; , exp ,

2 2

x xm i t mK x t x t V x t

i t t

Wave Function at Time tN

• Since U(t2,t0) = U(t2,t1)U(t1,t0), we can get it at t2 = t0 + 2t

• Iterate it N times to get it at time tN = t0 + Nt

2

1 01 1 0 0 0 0, ; , exp ,

2 2

x xm i t mK x t x t V x t

i t t

2 2 0 0 2 2 1 1 0 0, ; , , ,K x t x t x U t t U t t x

2 2

1 0 2 11 0 0 1 1exp , ,

2 2 2

x x x xm i t m mdx V x t V x t

i t t t

212

10 0 1 1

0

, ; , exp ,2 2

NN

i iN N N i i

i

x xm i t mK x t x t dx dx V x t

i t t

1 2 2 1 1 1 1 0 0, ,dx x U t t x x U t t x 1 2 2 1 1 1 1 0 0, ; , , ; ,dx K x t x t K x t x t

Functional Integrals

• In limit t 0, we are considering allpossible functions xi(t) that start at x0

and end at xN

• Define the functional integral:

• The propagator is now:

0 0

2

1 1lim2

N Nx t N

NNx t

mD x t dx dx

i t

t0 t1 t2 t3 t4 t5 tN

x0 xN

tN-1

…

…

212

10 0 1 1

0

, ; , exp ,2 2

NN

i iN N N i i

i

x xm i t mK x t x t dx dx V x t

i t t

0 0

211

0 00

, ; , exp ,2

N Nx t Ni i

N N i iix t

x xi mK x t x t D x t t V x t

t

The Lagrangian and the Action

• In the limit t 0, the term in round parentheses is a derivative

• The inner sum is the value of a function at various times, added up, and multiplied by the time step– An integral

• That thing in []’sis the Lagrangian

• The integral of theLagrangian is the action

0 0

211

0 00

, ; , exp ,2

N Nx t Ni i

N N i iix t

x xi mK x t x t D x t t V x t

t

0 0

121

0 0 20

, ; , exp ,N Nx t N

N N i i iix t

iK x t x t D x t t mx t V x t

0 0 0

1 210 0 2, ; , exp ,

N N Nx t t

N N x t tK x t x t D x t i mx V x t dt

1, ; , exp , ,F F

I I

x t

F F I I x tK x t x t D x t i L x x t dt

1, ; , expF

I

x

F F I I xK x t x t D x t i S x t

Second Postulate Rewritten:

• The propagator acts on the wave function to make a new wave function• This can be generalized completely to rewrite the second postulate:

• I am being deliberately vague because we won’t ever actually use this version• It is identical with the previous one

1, ; , expF

I

x

F F I I xK x t x t D x t i S x t

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

where S[x(t)] is the classical action associated with the path x(t)

0

10exp ,

t

tt D x t i S x t t

Why This Version of the Postulate?

• The Lagrangian and action are considered more fundamental then the Hamiltonian– Hamiltonian is normally derived from the Lagrangian

• The action is relativistically invariant, the Hamiltonian is not• In quantum field theory, it is far easier to work with the Lagrangian• For some problems in quantum chromodynamics, it is actually the only known

way to do the computation

0

10exp

t

tt D x t i S x t t

Why Not This Version of the Postulate?• To do any problem, you must do infinity integrals – hard even for a computer• I know of no doable problem with this approach

Connection with Classical Physics

• According to this postulate,to go from xI to xF, the particletakes all possible paths – pretty cool

• But which ones contribute the most?• If we consider small, then almost

everywhere, the phase is constantlychanging for even a slight change of path

• Unless small changes in path leave the action stationary– Stationary phase approximation

• This is the same as the classical path!

0S x t

x t

1, ; , expF

I

x

F F I I xK x t x t D x t i S x t

xI

xF

• Quantum mechanics makes predictions about outcomes of measurements• Can be shown: All we need to do is predict

expectation values of operators at arbitrary time• Using the time evolution operator, we relate this to time t0:

• In Schrödinger picture, the state vector changes and the operator is constant• Why not try it the other way?

– Let the state vector be constant and the operator changes• Define the Heisenberg picture:

• Then we have:

11D. The Heisenberg PictureRearranging Where the Work is Done

A t A t

†0 0, ,H SA t U t t A U t t

0 0,t U t t t †0 0 0 0, ,A t U t t AU t t t

0H S t

S S S H H HA t A t A t

Evolution of Operators in Heisenberg• Assume an operator in the Schrödinger picture has no time dependance• In the Heisenberg picture, it evolves according to:• Recall Schrödinger’s equation for U:• Hermitian Conjugate of this expression:

• Take time derivative of AH:

†0 0, ,H SA t U t t A U t t

0 0, ,Si U t t H t U t tt

† †H S S

dA t U A U U A U

dt t t

† †0 0, , Si U t t U t t H t

t

† †S S S S

i iU H t A U U A H t U

† † † †S S S S

i iU H t UU A U U A UU H t U

H H H H

i iH t A t A t H t

,H H H

d iA t H t A t

dt

Second Postulate In Heisenberg

• Note that if A has explicit time dependence, another term must be added• If the Hamiltonian has no explicit time dependence, then H will not evolve, so

• Other postulates must be changed slightly as well• State vector does change, but only during measurement

Postulate 2: All observables A(t) evolve according to

where H(t) is another observable.

,H H H

d iA t H t A t

dt

,d i

A t H t A tdt

A t

t

0H H SH t H t H ,d i

A t H A tdt

Heisenberg vs. Schrödinger

• Schrödinger says the state vector is constant, but the operators change

• To me, this is counterintuitive, since, for example, it is only in measurement that a particle changes

• Since the two have identical predictions, there is no way to know which one is “right”

• I think in Schrödinger but will do calculations in whatever is convenient

Commutation of Operators in Heisenberg• Suppose we have a commutation relation in Schrödinger:• What is the corresponding commutation in Heisenberg?• Recall:• Abbreviate this:• We therefore have

• For example, in 1D, we have• Note that in unequal times, there

is no comparable relationship• At unequal times, many operators

don’t commute with themselves

,S S SA B C

†0 0, ,H SA t U t t A U t t

†SA t U A U

,A t B t A t B t B t A t † † † †

S S S SU A UU B U U B UU A U

† †S S S SU A B U U B A U † ,S SU A B U †

SU C U ,A t B t C t

,X t P t i

,X t P t i

, 0X t X t

Example of Operator Evolution• Consider a free particle in 1D• Let’s find evolution of momentum operator first• And now for position:

• Need to solve these two equations simultaneously• Momentum one is easy:

• Then we solve position one• Note that

• Recall generalized uncertainty principle:• This implies, in this case,

2 2H P m ,

dP iH P

dt

0P t P

10 0X t X t P

m

0 , 0 , 0t

X X t X Pm

0dP

dt

,dX i

H Xdt

2 ,2

iP X

m

, ,2

iP P X P X P

m

2

i iP P

m

1dXP

dt m

i t m

12 ,A B i A B

02

tX X t

m

Sample ProblemHow long can you balance a pencil on its tip before it falls over?

• Not exactly a real problem, but we’ll tackle it anyway• We need expressions for the

kinetic and potential energy• If we treat pencil as a uniform

rod of mass m and length L, then• In a manner similar to how this is handled classically, you define

the momentum corresponding to , and call it P

• Then the Hamiltonian will be

• For small angles

• So

212E I d dt mgh

213I mL

P I d dt

12 cosh L

2 12

1cos

2H P mgL

I

212cos 1

2 21 12 4

1

2H P mgL mgL

I

Sample Problem (2)How long can you balance a pencil on its tip before it falls over?

• The angle variable and itscorresponding momentum satisfy:

• Now work out time derivatives of operators:

2 21 12 4

1

2H P mgL mgL

I

, P i

,d i

A H Adt

,d i

Hdt

21,

2

iP

I , ,

2

iP P P P

I

P

I

,d i

P H Pdt

21

4 ,i

mgL P , ,

4

imgLP P

12 mgL

Sample Problem (3)How long can you balance a pencil on its tip before it falls over?

• Take second derivativeof with respect to t:

• The solution to this is:

• This has boundary values:

• Rewrite second equation in terms of P:• Write (t) in terms of (0) and P(0):

, P i 1d

Pdt I 1

2

dP mgL

dt

2

2

1d dP

dt I dt 2

mgL

I

3

2

g

L

cosh sinht A t B t 3

2

g

L

0 and 0A B

0P I B

10 cosh 0 sinht t P t

I

213I mL

Sample Problem (4)How long can you balance a pencil on its tip before it falls over?

• Look at commutator of at different times:

• Generalized uncertainty relationship:• Therefore:

1, 0 sinh 0 , 0t t P

I

3

2

g

L 1

0 cosh 0 sinht t P tI

213I mL

sinhi

tI

12 ,A B i A B

0 sinh2

t tI

3

3 30 sinh

22

gt t

Lm gL

Sample Problem (5)How long can you balance a pencil on its tip before it falls over?

• Typical pencil:• Substitute in:

• Order of magnitude: if (0) = 1 or (t) = 1, then the pencil has tipped over

• Maximum time for it to balance:• Solve for tmax:

20.0050 kg , 0.170 m , 9.80 m/sm L g

3

3 30 sinh

22

gt t

Lm gL

310 1.18 10 sinh 0.1075 st t

31max1.18 10 sinh 0.1075 s 1t

1max 31

10.1075 s sinh

1.18 10t

0.1075 s 71.9max 7.7 st

The Interaction Picture• Half way between Schrödinger and Heisenberg• Divide the Hamiltonian into two pieces, H0 and H1(t):• Normally, H0 is chosen time-independent and easy to find the eigenstates of• Then operators evolve due to H0 and state vectors due to H1:

Why would we do this?• It is a useful way to do time-dependent perturbation theory

– We will ultimately use this approach, but not use this notation• It is a useful way to think about things

– Very common in particle physics

• Think of the state as “unchanging” until the pion decays

• We will, in this class, nonetheless always work in Schrödinger picture

0 1H H H t

0 ,d i

A H Adt

1

di H t

dt

• Let {|i} be a complete orthonormal basis of a vector space • Let A be any operator in that vector space• Define the trace of A as • In components:• Can be shown: trace is independent of choice of basis:

• Consider trace of a product of operators:

11E. The TraceDefinition

Tr i ii

A A

Tr i ii

A A

Tr iii

A A

i j j ii j

A j i i ji j

A j jj

A Tr A1

Tr i ii

AB AB i j j ii j

A B j i i ji j

B A 1

j jj

BA Tr TrAB BA Tr Tr TrABC BCA CAB

Partial Trace• A trace reduces an operator in vector space to a number• If we have an operator A in a product space of vector spaces, , we can do

a trace over just one of them, say , to get an operator on vector space • Suppose the vector spaces and have basis vectors {|i} and {|j}

respectively• Basis vectors of look like {|i,j} • Define the partial trace as

• This makes Tr(A) an operator on • In components, this is

Tr , ,i i k j k ji j k

A A W

,Tr ik jkijk

A AW

• There is a classical sense of probability that has nothing to do with quantum mechanics:– If I pull a card from a deck of cards, the probability of getting a heart is 25%– We don’t believe it is truly indeterminate, just that we are ignorant

• Quantum mechanics introduces another kind of probability– If a particle has spin + in the x-direction, and we measure the spin in the z-

direction, the probability that it comes out + is 50%• Up to now, we assumed that the quantum state is completely known• What if there are multiple possible quantum states?• Quantum states |i(t) each with probability fi

• The probabilities fi arenon-negative and add to one

• The quantum states will be normalized, but not generally orthogonal

11F. The State Operator / Density MatrixTwo Types of Probability

,i it f

0 , 1i ii

f f

The State Operator

• In principle, this list of possible states/probabilities could be very complicated

• Define the state operator as

Properties of the state operator:• Trace:

• Hermitian (obvious)• Positive semi-definite: for any state vector |:

i i ii

t f t t

,i it f 0 , 1i ii

f f

Tr i j i i jj i

f t t i i j j ij i

f t t i i i

i

f t t ii

f Tr 1

i i ii

f t t 2

i ii

f t 0 0

†

Sample ProblemAn electron is in the spin state + as measured along an axis at an angle randomly chosen in the xy-plane. What is the state operator?

• We are in the normalized positive eigenstate of the operator

• The normalize eigenstate is:• If we knew what

the angle was,the state operator would be

• Since we don’t, and all angles are equally likely, we have to average over all angles:

• Let’s check we got the trace right:

cos sinx yS S S 12 cos sinx y 1

2

0

0

i

i

e

e

11

2 ie

111

2i

i ee

11

2 1

i

i

e

e

2

0 2

d

2

0

1

4 1

i

i

ed

e

2

0

1

4

i

i

ie

ie

12

12

0

0

1 12 2Tr 1

Eigenvectors and Eigenvalues of State Operator

• Like any Hermitian operator, we can find a complete,orthonormal set of eigenstates of with real eigenvalues

• Because it is positive semi-definite, we have

• Trace condition:

written in terms of its eigenvectors:

• Compare to the definition of :Conclusions:• We can pretend is a combination of orthonormal states, even if it isn’t• Any positive semi-definite Hermitian operator with trace 1 can form a valid

state operator

,

i

i i i

i j ij

Tr 1 0 †

0 i i i i i 0i i

1 Tr i ii

ii

1i

i

i ii

i i ii

i i i

i

f

Pure States and Mixed States

• If there is only one non-zero i, then we have a pure state– If it isn’t a pure state, it’s a mixed state

You can prove something is pure in a variety of ways:• You can find the eigenvalues (homework), and show they are 0 and 1• You can find 2 and compare it to :

• For a pure state, i2 = i (because it is zero

or one), for a mixed state it isn’t• One measure of how mixed

the state is is the quantummechanical entropy

• Pure states have S = 0, mixed states have S > 0

i in

0i 1ii

i i ii

2i i i j j j

i j

i j i i j ji j

i j i ij ji j

2i i i

i

2 for pure state

Tr lnBS k lnB i ii

k

2Tr 1 for a pure state

Time Evolution of the State Operator i i i

i

t f t t ,i it f

di t H t

dt

• Whichever state vector it is in, it satisfies Schrödinger’s Equation:

• It follows that:

• Don’t confuse this with the Heisenberg picture:

di t t H

dt

i i ii

d dt f t t

dt dt i i i i i

i

d df t t t t

dt dt

1i i i i i

i

f H t t t t Hi

1

H t t Hi

1,

dt H t

dt i

,H H H

d iA t H t A t

dt

Expectation Values of Operators i i i

i

t f t t ,i it f

i iiA A

• If we knew which state we were in, the expectation value of an operator A is:

• Since we don’t know which state it is, we must weight it by the probabilities:

i ii

A f A i i ii

f A i i j j ii j

f A

i j i i ji j

f A j jj

A Tr A

TrA A

Sample ProblemProve, using thestate operator, that:for operators A that have no time dependence

1,

dA i H A

dt

1

,d

t H tdt i

TrA A

Trd d

A Adt dt

Trd

Adt

1Tr ,i H A

1Tr iH A i HA

1Tr i HA i AH

1Tr ,i H A

1

,i H A

Sample ProblemShow that is constant 2Tr 2Tr Tr

d d d

dt dt dt

1Tr ,H

i

1

Tr H Hi

1Tr H H

i

2Trd

dt

0

Comments on Entropy• We showed in the previous problem that Tr(2) is constant

– Easily could have generalized it to any power– Generalizes to trace of any function of

• The quantum mechanical entropy is not changed by Schrödinger’s Equation• Pure states should always evolve into pure states • It is disputed whether this applies to gravity

– We don’t have a quantum theory of gravity• In principle, if you put something in a black hole, it eventually comes back

out as black body radiation with a lot of entropy

Postulates in Terms of the State Operator

• Note that to get (t), you don’t need |i(t) and fi, you just need (0)• This suggests we could write postulates in terms of the state operator

• The equation above becomes our second postulate

1,

dt H t

dt i

Postulate 1: The state of a quantum mechanical system at time t can be described as a positive semi-definite Hermitian operator (t) in a complex vector space with positive definite inner product

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

where H(t) is an observable.

,d

i t H tdt

Measurement Postulates for the State Operator

• Postulate 3 (measurements correspond to observables) doesn’t need changing• Postulate 4 concerns the probability of getting a result a if you measure A.

,i it f

|i

P a P i P a i 2,i i

i n

f a n , ,i i ii n

f a n a n

, ,i i in i

a n f a n

i i ii

t f t t

, ,n

a n a n

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 operator at time t. Then the probability of getting the result a at time t will be , ,

n

P a a n t a n

Post-Measurement State Operator

• This one is tricky• Need to figure out what the probability that it was in the state i given that the

measurement produced a– Requires a good understanding of conditional probabilities

• Need to figure out the state vector if itwas in the state i after the measurement

• Then find the new state operator

,i it f i i ii

t f t t

&P a i

P a

|P i P a i

P a |if P i a

|

i

P a if

P a

1

, ,|

i in

a n a nP a i

i i ii

f

| 1, , , ,

|i i ii n m

P a if a n a n a m a m

P a P a i

1

, , , ,i i in i m

a n a n f a m a mP a

Post-Measurement State Operator (2)

• This now serves as our final postulate

1

, , , ,n m

a n a n a m a mP a

Postulate 5: If the results of a measurement of the observable A at time t yields the result a, the state operator immediately afterwards will be given by

1, , , ,

n m

t a n a n t a m a mP a

1

, , , ,i i in i m

a n a n f a m a mP a

i i ii

t f t t

Comments on Postulates using State Operator• The postulates in terms of the state operator are equivalent to those in terms of

the state vector

Pros and cons of using the state operator approach:• The irrelevant overall phase in | is cancelled out in • The formalism simultaneously deals with both quantum and ordinary

probability• You have to work with matrices (more complicated) rather than vectors• Requires greater mathematical complexity• Postulates are slightly more complicated

• Whether you believe the postulates should be stated in terms of state operators or not, they are useful anyway

• The state operator can be used even if we don’t write our postulates this way

• It allows us to prove powerful theorems, and simplify what would otherwise be complicated calculations

• Example 1: How do you calculate scattering if the polarization of a spin-1/2 particle is random?

• Example 2: How do you calculate interactions of a particle which is produced in a particular state, but at an unknown time

11G. Working With the State OperatorIt’s Useful

Sample ProblemSuppose a spin- ½ particle is polarized in the |+ state 50% of the time, or the |– state 50% of the time, as measured along an arbitrary axis described by the angles and . What is the state operator ?

ˆ,S n S

• We need to find the normalized eigenstates of the spin operator in an arbitrary direction

• Now we find the eigenstatesof this matrix:

12

ˆ n σ 12 sin cos sin sin cosx y z

12

cos sin

sin cos

i

i

e

e

1 12 2

1 12 2

cos sin,

sin cosi ie e

Sample Problem (2)Suppose a spin- ½ particle is polarized in the |+ state 50% of the time, or the |– state 50% of the time, as measured along an arbitrary axis described by the angles and . What is the state operator ?

• Now find state operator:

• Interestingly, result is independent of angle!

1 12 2

1 12 21 1 1 1 1

2 2 2 2 21 12 2

cos sin1cos sin sin cos

sin cos2i i

i ie ee e

2 1 1 12 2 2

21 1 12 2 2

2 1 1 12 2 2

21 1 12 2 2

cos cos sin1

cos sin sin2

sin cos sin1

cos sin cos2

i

i

i

i

e

e

e

e

1 01

0 12

Calculations for Unpolarized Particles• Suppose the spin of a particle is completely random and uncontrolled• It could be spin up or down on any axis

• As demonstrated, no matter what axis it is on, the state operator is

• Since it doesn’t matter, pick any axis, say the z-axis

• Calculate interaction, say a cross section, for:– Spin up– Spin down

• The unpolarized cross-section is then

• In the Large Hadron collider, for instance, unpolarized protons are collided with unpolarized protons:

1 01

0 12

,

1unpol 2

1unpol 2

Sample ProblemA particle is produced in a quantum state (t0) = 0 at a

completely unknown time t0. At time t, what does the state operator look like?

• If it is in the state |0 at time t0,then at time t it will be in the state

• Write this in terms of eigenstates of the Hamiltonian:

• If we knew the time t0, we would have

• Since we don’t, average over the time t0, from time –½T to +½T

• We will then take the limit T

0 0,t U t t

0

0, niE t tn n

n

U t t e

n n nH E

0

0niE t t

n nn

t e

t t 0 0

0 0n miE t t iE t t

n n m mn m

e e

12 0

12

0 0 0

1n m

T i E E t tn n m mT

n m

e dtT

Sample Problem (2)A particle is produced in a quantum state (t0) = 0 at a

completely unknown time t0. At time t, what does the state operator look like?

• Do the integral:

• In the limit T , first expression numerator is never larger than 2, and denominator goes to – So it vanishes

• If we assume that different states have different energies, expression even simpler

12 0

12

0 0 0

1n m

T i E E t tn n m mT

n m

e dtT

12 0

12

0

2 sin 2if 1

1 if

n m

n mT i E E t n m

n mT

n m

E E TE E

e dt T E ET

E E

12 0

12

0 ,

1lim n m

n m

T i E E tE ETT

e dtT

nm

0 0n nm n m mn m

2

0n n nn

The Spectrum is (Almost) All That Matters• For a particle in a known state emitted at an unknown time

• If it is emitted on one of several states |i at unknown time with probability fi:

• If we measure the energy of the resulting particle, the expression in parentheses is the probability of getting the energy En:

• Hence we know the state operator , i.e., we know everything we can know, if we know the spectrum

• This might be modified if we have degenerate energy states– But only connecting states with the same energy– For example, for photons, polarization information

2

0n n nn

2

i n i n ni n

f

2

n i n ii

P E f

• Consider two spin ½ particles in a state of total spin 0:• These particles can be physically separated by a large distance

– We call these pair of particles an Einstein-Podolsky-Rosen (EPR) pair• We measure one or both spins with Stern-Gerlach devices, at arbitrary anglesWhat happens when we measure the first one’s spin?• The spin of the first particle is indefinite: 50% spin up, 50% spin down• The state afterwards will depend

on the result of the first measurement

11H. SeparabilityEntangled States

12

Source

N

S

N

S

50% :

50% :

Instantaneous Quantum Communication?• According to the postulates (as stated), the system,

including the second particle, changes instantaneouslywhen we preform a measurement– Faster than light

• We call states like this entangled statesCan we somehow use this to communicate faster than light?• Reduce information to bits• Produce a pair of entangled particles, one at sender, one at receiver• Based on each bit, decide to measure or not measure one particle• Measure the other particle and see if the state vector has changed

Source

N

S

N

S

sender receiver

Why It Doesn’t Work in This CaseIf the sender didn’t perform a measurement, then:•State is still in a superposition•When receiver measures, result is uncertain

– 50% spin up, 50% spin downIf the sender did perform a measurement, then:•State is not in a superposition•But we don’t know which state: 50% spin up, 50% spin down•There is no way the receiver can tell the state changedConclusion: The measurement by the sender just converts quantum uncertainty to classical uncertainty

Source

N

S

N

S

sender receiver

12

Faster Than Light Communication• Suppose we have a quantum system distributed in two regions, one for the

sender, one for the receiver• The quantum states for the sender live in vectors space , and for the receiver • The vector space for the whole system is • Let basis of be |vi and of be |wj, so those of will be |vi,wjCan we communicate from sender to receiver?• We could change the Hamiltonian for the sender• We could measure the state for the sender• Would either of these affect a measurement for the receiver?

sender receiverV W

Local Measurements and Hamiltonian

• We will imagine doing a measurement A by the sender or B by the receiver• These measurements will be assumed to be local,

that is, A acts on vector space and B acts only on • The Hamiltonian, similarly, will consist of two pieces,

HV that acts only on , and HW that acts only on • Note that these automatically guarantee that:

sender receiverV WA B

V W

1

1

A A

B B

1 1V WH H H

,A B AB BA 1 1 1 1A B B A A B A B , 0A B

, 1 1 ,1V WH B H H B

VH WH

0 1 ,1WH B 1 1W WH B BH

, 1 ,WH B H B

Can the Hamiltonian Communicate Instantly?

• We can change the state vector byletting it evolve under Schrödinger’s equation

• This causes the expectationvalue of B to change:

• However, the sender can only adjust HV, not HW, so he can’t affect what the receiver measures

• You can’t communicate instantly by modifying the Hamiltonian

sender receiverV WA B

V W

VH WH

, 1 ,WH B H B

,d

i t H tdt

1,

dB i H B

dt

1

1 ,Wi H B

Can Measurement Communicate Instantly? (1)

• The state operator changes when we perform a measurement• Would measuring A cause a change in the expectation value of B?• Since A and B commute, we can simultaneously diagonalize both of them

– Assume that |vi and |wj are eigenstates of A and B respectively• To make the argument simpler, assume no degeneraciesFirst, assume we don’t measure A first

sender receiverV WA B

V W 1

1

A A

B B

, 0A B

, ,

, ,

i j i i j

i j j i j

A v w a v w

B v w b v w

TrB B

,

, ,i j i ji j

v w B v w

,

, ,j i j i ji j

B b v w v w

Can Measurement Communicate Instantly? (2)

This time, let’s assume we do measure A first•The probability of getting the result ai is now•If we knew the result, then thestate vector afterwards is •But we don’t, so wemust take a weightedaverage•Then we can calculate the average measurement result from B:

V W , 0A B , ,

, ,

i j i i j

i j j i j

A v w a v w

B v w b v w

, ,i i j i jj

P a v w v w

,

, ,j i j i ji j

B b v w v w

,

1, , , ,i i j i j i k i k

j ki

v w v w v w v wP a

i ii

P a , ,

, , , ,i j i j i k i ki j k

v w v w v w v w

TrB B , , ,

, , , , , ,l m i j i j i k i k l ml m i j k

v w v w v w v w v w B v w

, , ,

, , , ,il jm i j i k m i k l ml m i j k

v w v w b v w v w , ,

, , , ,i j i k j i k i ji j k

v w v w b v w v w

,

, ,j i j i ji j

B b v w v w

Announcements

1/23

ASSIGNMENTSDay Read HomeworkToday 11I 11.3Monday 11J 11.4, 11.5Wednesday 11K 11.6

Instantaneous Communication?• Bottom line: Neither measurement nor modification of the Hamiltonian

can yield instantaneous communication from the sender to the receiver• In particular, if you are never going to perform measurements in the sender’s

region, then you can ignore all particles and interactions there• Measurements of objects in vector space have expectation values:• Note we are summing on the basis |vi

– This is effectively a partial trace over

• We can, in fact, work with the simplified state operator:• Note: This means when we do an experiment with an

electron, we don’t have to worry about what it has interacted with before

sender receiverV WA B

V W

VH WH

,

, ,j i j i ji j

B b v w v w

Tr W V

Instantaneous Communication?• Have we proven instantaneous (or fast) practical communication is impossible?Yes, assuming:• Particles that carry information are no faster than light• All terms in the Hamiltonian are local (no action at a distance)• All measurements are local• Quantum mechanics is valid

• Note, however, that the hypotheses of quantum mechanics do transmit quantum information instantaneously– Hypotheses 4 and 5 (about measurement)

sender receiverV WA B

VH WH

Sample ProblemA pair of spin-1/2 particles are in the pure spin state (a) What is the state operator, as a matrix?(b) Suppose the second particle is lost forever. What is the effective state operator for the first particle?(c) What is the entropy of the initial and final state operators?

• We have two spin-1/2 particles. Firstthing we need to do is pick a basis:

• The state vector in this basis is then:

• The correspondingstate operator is:

• To find the effective state operator, trace over the second spin:

12

, , , 0

11

12

0

1 12 2

1 12 2

0 0 0 0

0 0

0 0

0 0 0 0

Sample Problem (2)A pair of spin-1/2 particles are in the pure spin state (a) What is the state operator, as a matrix?(b) Suppose the second particle is lost forever. What is the effective state operator for the first particle?(c) What is the entropy of the initial and final state operators?

• To find the effective state operator, trace over the second spin:

• The initial state was a pure state, so it has entropy S = 0• The final state’s entropy can be found from:

12

Tr W V

1 12 2

1 12 2

0 0 0 0Tr Tr

0 0

0 0Tr Tr

0 0 0 0

12

12

0

0

W

lnB i ii

S k W1 12 22 lnBk ln 2BS kW

1 12 2

1 12 2

0 0 0 0

0 0

0 0

0 0 0 0

Quantum Decoherence

• Note that according to Schrödinger’s Equation, entropy is preserved– Pure states evolve into pure states

• However, if particles escape the system, entropy is effectively generated• Hence Schrödinger’s equation can result in effective irreversibility in any

situation where information is lost

• This process is sometimes called quantum decoherence

• Practically, any measurement causes the quantum system to become hopelessly entangled with the environment

• This causes effective entropy, even without the measurement hypotheses

• Consider measuring a particle’s spin in a Stern-Gerlach device• According to quantum, the outcome is probabilistic• Could there be secret additional information

that actually determines the outcome?– Details of the particle– Details of the measuring apparatus– Etc.

• Philosophy of hidden variables – there is some sort of additional information that makes the outcome actually certain– Einstein: “God does not play with dice”– Additional information is called hidden variables

• No assumption that this additional information is accessible• In this picture, quantum uncertainty is just hidden classical uncertainty

11I. Hidden Variables and Bell’s InequalityCan We Get around Probabilities?

N

S

• We will study the results of an EPR experiment withtwo Stern-Gerlach devices at arbitrary angles 1 and 2

Let’s make some reasonable-sounding assumptions about Hidden Variables:• All quantum outcomes are actually certain

– The uncertainties are all classical, representing hidden variables• The results of experiments on one side can’t affect the outcome on the other side

– Can be arranged by doing experiments simultaneously far apart– Einstein Separability

• We will discover that this leads to predictions that contradict quantum mechanics

Einstein Separability and Hidden Variables

Source

N

S

N

S

12

• Each of the SG devices will measure either spin up or spin down• They will either agree or disagree with each other,

so we have some probability that they disagree– According to hidden variables, due to unknown hidden variables

• According to hidden variables, we can talk about what the measurement would have given, even if we didn’t make that measurement– Not true in quantum mechanics!

• According to Einstein separability, performing the measurement on one side can’t effect the outcome on the other side– Not true in Copenhagen interpretation!

Using the Assumptions of Hidden Variables

Source

N

S

N

S

P a b

• Imagine measuring spin 1 in one of the two directions a or c• Imagine measuring spin 2 in one of the two direction b or d• I now make the following logical claim:

• This is logically equivalent to the following claim:

• From which follows the following probability statement:

Bell’s Inequality (Carlson’s Version):

Source

N

S

N

S

a

c

b

d

If a = b and b = c and c = d, then a = d

If a d then a b or b c or c d

P a d P a b P b c P c d

a b

b c

c d

• Hidden variables plus separability implies Bell’s Inequality• What does quantum mechanics predict?

– Homework problem• Assume particular angles• Then we have:

• Quantum mechanics predicts Bell’s Inequality is violated!

Bell’s Inequality vs. Quantum Mechanicsa

c

b

d P a d P a b P b c P c d

9090

45

135 2 12cos a bP a b

2

2

cos 67.5 0.1464 ,

cos 22.5 0.8536

P a b P b c P c d

P a d

0.8536 3 0.1464 0.4393

Quantum mechanics contradicts Bell’s InequalityThree possibilities:1.Hidden variables is incorrect2.Information can move faster than light3.Quantum mechanics makes incorrect predictions

•The third one can be, and has been, tested experimentally– Not exactly with this setup– Uses photons

•The results agree with predictions of quantum mechanics

Testing Hidden Variables

Source

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S

N

S

Can we save hidden variables from extinction?Experiment assumes all photons are measured•Realistically, only a fraction are captured•If this is the explanation, then quantum mechanics would become obviously wrong if we had perfect detectorsPerhaps initial state is modified by setup of the detectors•Solution: Adjust detector at the last moment•Choose it randomly by computerCounter-argument – computers aren’t random•Use fast graduate students and large (astronomical?) scale instead•Do graduate students have free will?

Salvaging Hidden Variables

Source

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S

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S

• Three of our five postulates of quantum mechanics have to do with measurement, and it was mentioned in one other

• These are among our most complicatedand least elegant postulates:

• It is very different from our other postulates– Irreversible– Non-linear– Time asymmetric– Probabilistic

• Worst of all, It was never defined– A theory with undefined expressions is not a theory

• 60% of our postulates concern measurement; 90% of physics concerns only Schrödinger’s equation

11J. MeasurementWhy is it Important?

1, ,

n

t a n a n tP a

• Take a single spin-½ particle in the state• Put it through a Stern-Gerlach device• When does the measurement occur

– When it passes through the magnets?– When it hits the screen?– Later?

• We can test if state vector collapses when you split the beam• Recombine the two beams• Then measure the x-component of the spin• If it is all deflected one way, then state is • With photons, this has definitely been demonstrated

What constitutes a measurement?

N

S

12x

+

N

S

12x

N

S

N

S

• Theorist picture:

• Experimentalist picture:

• Real detectors are built on principles of how particles interact• Particles interact via the Hamiltonian/Schrödinger equation• Maybe these small interactions don’t really cause collapse of the state vector

• Many books (at least those that discuss it), imply or state that it is when effects get macroscopic that measurement occurs– One electron doesn’t count, but one milliamp does

How Does Measuring Actually Work?

12x Measuring

device

50%

50%

• Since measuring devices interact with particles via the Hamiltonian, we should include these measuring devices in our state vectors

• We might think of a simple measuring device M thatmeasures the spin state as having three states:

• The state of the system will be described by the stateof the spin and the state of the measuring device:

• Interactions cause the state of the measuringdevice to change to reflect the measurement:

• Since this is due to some sort of interaction Hamiltonian,there must be a corresponding time evolution operator:

• Recall that the time evolution operator is linear!

Quantum Measurement Devices

0 Not measured yet

Measured, result

Measured, result

M

M

M

, M0, ,M M

0, ,M M

0 0 0 0, , , , , , ,m mU t t M M U t t M M

1 10 0 0 0 02 2

, , , ,m mU t t M U t t M M 12

, ,M M

• When does state vector collapse occur?• Assume first it occurs just after the measurement occurs

• Now assume it occurs just before the measurement occurs

• The final state is the same, whether you think of it occurring before or after the measurement

When Does Measurement Occurs?

10 02

, ,M M

, 50%

, 50%

M

M

12

, ,M M

10 02

, ,M M

, 50%

, 50%

M

M

0,

,

M

M

Path D

etector

• We previously explained that we can demonstrate thatno measurement occurs, because the final state is still

• Measuring Sx then shows us that

• Now add a measuring device• Before measurement, quantum state is • After measurement, state is

• Assume no state vector collapse occurs. What is Sx?

Measurement Without Collapse (1)

N

S+

N

S

N

S

12

12x xS 1

4 x 14 1

2

10 0 02

, , ,M M M

12

, ,M M

x

Path D

etector

• Assume no state vector collapse occurs. What is Sx?

• Interference is lost without collapse of the state vector!• Due to particle becoming entangled with the measuring device

Measurement Without Collapse (2)

N

S+

N

S

N

S

12x xS

12

, ,M M

12 , , , ,xM M M M

x

12 , , , ,M M M M 0

• Consider a system consisting of a source, a measurement device, a computer, and an experimenter

• Initial state, particle is in a spin state, but the measuring device, the computer, and the experimenter don’t know what it is:

• Now, we can assume the collapse occurs at any stage:– Just before the first measurement– Between measurement and computer– Between computer and experimenter– Just after the experimenter gets the data

• What is final state in each case?

How Long Can We Delay Measurement?

Path D

etector

N

S

0 0 0 0, , ,M C E

12

• The steps involved depend on when we think collapse of the state vector

• But the final situation does not• Can we push it later, i.e., can we leave the experimenter in a superposition?

How Long Can We Delay Measurement? (2)

0 0 0 0 0 00

0 0 0 0 0 0

, , , , , , , , , , , ,

, , , , , , , , , , , ,

M C E M C E M C E M C E

M C E M C E M C E M C E

0 0 0 0 00

0 0 00 0

, , , , , , , , , , , ,1, , , , , , , , ,, , ,2

M C E M C E M C E M C E

M C E M C E M C EM C E

0 00

00

, , , , , , , , ,1, , , , , ,, , ,2

M C E M C E M C E

M C E M C EM C E

0

, , , , , ,1, , ,, , ,2

M C E M C E

M C EM C E

• Let’s add a theorist:

• Before the experimentercommunicates to the theorist,

• Then, when the theorist gets the result,

• According to this hypothesis, the experimenter is in a quantum superposition up to the moment the theorist reads the published paper

• Schrödinger’s cat• Wouldn’t the experimenter know she was in a superposition?

How Long Can We Delay Measurement?

Path D

etector

N

S

10 02

, , , , , , , ,M C E T M C E T

0

, , , ,

, , , ,

M C E T

M C E T

• Let’s describe the experimenter as having two internal degrees of freedom– The result of the measurement– Whether they are in a superposition or not

• Each of these two degrees of freedom have at least three states:

A Slightly More Sophisticated View

Path D

etector

N

S

,E D

0 Haven't measured yet

Measured came out

Measured came out

E

E

E

0 Haven't measured/thought about yet

Thought about it, in a definite state

Theought about it, in superposition

D

D

D

• Now, we imagine a process I call “introspection” where the experimenter examines her thoughts and decides if she is in a superposition of thoughts

• In particular, if she is given a pure spin state, introspection will cause

• Because in each case, the experimenter is in a mental pure state• But U is a linear operator, and therefore:

Thinking Quantum Thoughts

Path D

etector

N

S

0 0, , , and , , ,I E I EU t t E D E D U t t E D E D

1 10 02 2

, , , , ,I EU t t E D E D E D E D

• Let’s put the theorist back in:

• Put in a particle in the state• The detector detects, the computer records, the experimenter reads it

• Now the experimenter incorrectly concludes she is not in a superposition

• Only when the theorist asks doesthe state vector collapse

Getting Extreme

Path D

etector

N

S

10 0 0 02

, , , , , , , , , ,M C E D T M C E D T

12

10 02

, , , , , , , , , ,M C E D T M C E D T , , , , ,

, , , , ,

M C E D T

M C E D T

• Solipsism, from dictionary.com:• I think therefore I am, but I’m not so

sure about you

• One can consistently take the attitudethat only I can collapse the state vector

• In this view, the entire universe was in a complicatedsuperposition until November 22, 1961

• It then collapsed. If not for me, you probablywouldn’t exist

• As far as I know, there is nothing wrong with thisas a quantum philosophy

• This seems remarkably egotistical• As far as I know, no one takes this seriously

Quantum SolipsismSolipsism: Noun

1. Philosophy. the theory that only the self exists, or can be proved to exist.

Quantum collapse

device, ca. 2000

• We can delay the state vector collapse as much as possible:• Treat all measuring devices, recording devices, and even experimenters as

complicated quantum objects interacting with the quantum particles• These are governed by a Hamiltonian, which implies that we can only measure

observables• We found no inconsistencies with observations

• What happens if we assume the state vector never collapses?

• Postulate 2 (Schrödinger’s equation) always applies• Postulate 3 (measurement observables) is built in already• Postulate 5 (state vector collapse after measurement) never applies• Postulate 4 will take some discussion …

11K. The Many Worlds HypothesisLet’s Take it One More Step

• We are going to go from five postulates of quantum mechanics to two:

• This one is unchanged from before

• This one used to start, “When you do not perform a measurement, …”

Many Worlds Postulates of Quantum

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

Postulate 2: The state vector evolves according to

where H(t) is an observable.

,i t H t tt

• The fifth postulate talks about the state vector changing due to measurement• This can roughly be explained in terms of decoherence

– Information lost during measurement mimics effective change of state vector (or state operator)

• The FOURTH postulate makes a specific prediction, however• Suppose we take an electron in the spin state• Put it through Stern-Gerlach: • Copenhagen predicts

– 50% takes upper path– 50% takes lower path

• Many Worlds predicts:

• There is no statement or implication of probability, so how can we reproduce this correct prediction (50% each) in Many Worlds?

The Fourth Postulate Revisited

12

+x

Path D

etector

N

S

12

, ,M M

• It seems like, because of superposition, we might never predict anything in many worlds

• Imagine we have spin state• Stern-Gerlach measurement:

• We can make a definite prediction if it is in aneigenstate of what is being measured

• Consider a continuous variable, like the position x• For a general wave function, we can’t predict the result:• The position has uncertainty x

• But if x 0, the result will be definite

• We can predict the result for any variable with zero uncertainty

Can We Ever Predict Things In Many Worlds?

z

Path D

etector

N

S0, ,M M

• What does it mean when you say something has a 50% probability of being spin up?

• It DOES NOT imply anything specific about a single spin• Instead, it implies something about what happens if you repeat it many times

• Perform the experiment N times• Keep track of N+ and N-, the number of times each outcome occurs• Prediction is only in the limit N • We have

• Define the average spin of many particles as

• Then the probability statement is just saying

Probability: What Does it Mean?

lim lim 0.500N N

N N

N N

1

1 N

z zii

S SN

lim 0zNS

lim z zN

S S

• Copenhagen does not make a prediction about the results of one experiment• It is making a prediction about an experiment that is repeated many times• The initial state is not• Instead, it is an N

particle state:• We are not measuring Sz

• We are measuring:

Our goal:• Find the expectation value of this average• Find the uncertainty of this average• Take the limit N

We want to know:• Does the average value match the prediction of Copenhagen?• Does the uncertainty go to zero?

Quantum States for Repeated Measurements

1

1 N

z zii

S SN

12

+x

+ + +x x x

z zS S

2 22z z zS S S

?

0z zS S ?

0zS

• Recall:

• Then we have

• If we just let one Szi act on |, we have

• So we have

Calculating with Many Particles

12zS

12x + + +x x x

1 12 2z xS 1

2 x

+ + + +zi zi x x x xS S 12 + + +x x x x

1

1 N

z zii

S SN

+ +

+

2

x x x

x x x

x x x

N

12x

• First note that:

• Now let’s find the expectation value:

Does the Expectation Value Equal Sz = 0 ? 1

2x + + +x x x

1

1 N

z zii

S SN

+ +

+

2

x x x

x x xz

x x x

SN

12x

z zS S

12x x 0 1x x x x

+ +

+

2

x x x

x x xx x x

x x x

N

1

2

N

x x x xNN

112 0 1N 0z zS S 0

• We now need

Does the Uncertainty Vanish as N ? (1)+ + +x x x

1

1 N

z zii

S SN

+ +

+

2

x x x

x x xz

x x x

SN

2 2z zS S

0x x

1x x x x

2

+ +

+

2

x x x x x x

x x x x x x

x x x x x x

N

2

1 2224

N N

x x x x x x x x x xN N NN

2

zS

2

22

1 04

N N NN

2

2

4zSN

Does the Uncertainty Vanish as N ? (2)

0zS 2

2

4zSN

• The Uncertainty is given by

• Now take N :

• Conclusion: In the limit N , Many Worlds makes the same prediction as Copenhagen

2 22z z zS S S

2

4N

2zS

N

lim 0zNS

• William of Ockham, (c. 1287-1347):

• For those who don’t speak Latin:

• Given a choice between five postulates and two, two is better

• However, it does imply a very complicated view of the universe

Occam’s Razor

“Pluralitas non est ponenda sine necessitate”

“Plurality is not to be posited without necessity”

• What led us here today, according to Copenhagen interpretation:

• What let us here today, accordingto many worlds:

What the Universe Looks Like

Milky Way

No Milky Way

inflation field

No Solar System

Solar System

Earth

No Earth

Eric

No Eric

inflation fieldsin Milky Way

cos No Milky Way

sin sin Solar System

sin cos No Solar System

cos No Milky Way

sin sin sin Earth

sin sin cos No Earth

sin cos No Solar System

cos No Milky Way

sin sin sin cos No Eric sin Eric

sin sin cos No Earth

sin cos No Solar System

cos No Milky Way

Do People Take Many Worlds Seriously?• Many famous physicists take Many Worlds Seriously

Stephen HawkingWizard of

black holes

Murray Gell-Mann

Postulated Quarks

Richard Feynman

Feynman Diagrams

Steven WeinbergElectroweak

Theory

Unknown Name

Expert on 11/11/11

So Many Choices• You have a variety of options, like ordering off a menu• You can mix and match, almost any combination works• My favorite combination, philosophically, marked in yellow• What we use in this class, marked in black

• Using Copenhagen means we don’t have to infinitely repeat experiments

Appetizer:State Vector State Operator

Entrée: Schrödinger Heisenberg Interaction

Salad: Schrödinger’s Eqn. Feynman Path Integral

Dessert: Copenhagen Many Worlds Others