Lecture Notes PH 411/511 ECE 598 A. La Rosa Portland State University

INTRODUCTION TO QUANTUM MECHANICS _____________________________________________________________________________________

CHAPTER-11 QUANTUM ENTANGLEMENT The annihilation of the positronium

Except for the cartoons presented in the section “Preliminary Comparison” at the very

beginning, and the section “my own view” given at the very end, this chapter is taken completely from the The Feynman Lectures, Vol III, Chapter 18. The annihilation of the positronium process with the consequent generation of photons is described by Feynman in great detail, accounting for the conservation of energy, linear momentum, angular momentum and parity. Although the word entanglement is not mentioned explicitly, the “Einstein-

Podolsky-Rosen paradox” is mentioned in the description. Feynman is on the argument side that

states that there is no paradox, and that indeed, measurement on one side affect the result of

measurement made at another far away location.

I. PRELIMINARY COMPARISON

1. The figure shows a box enclosing buckets (apparently many) containing paint of different colors, plus two particles (assumed to be initially colorless). Based on the (hypothetical) results described below, it appears that there are many containers of different colors inside the box; only two colors are shown in the figure). But, actually, it is unknown what exactly is going on inside the box.

Window Window

Box

Figure 1. Two particles inside a box

2. What is known is that, after shaking the box, the two particles leave the box in opposite directions (linear momentum conservation). The particles are detected, respectively, by two observers A and B.

A

B

Figure 2. After a “shaking process” the particles leave in opposite.

After shaking the box we observe either

A

B

Red Red

(23)

Or

A B

Blue Blue

Figure 3. Two possible outcomes from the “shaking process” experiment.

The “shaking the box” experiment is repeated many times. The case on the left is observed in 50% of the cases.

Do the particles acquire their color before leaving the box?

Do the balls have no color after leaving the box? Do they acquire a color only right after they are detected by A (or by B)?

3. Observer B places a green color filter before detecting the particle. Afterwards, he/she observes a green color particle.

A B

Green

filter

Figure 4.

What color particle would A detect?

Will A ‘s measurement be influenced by what B is doing?

An old fashion quantum practitioner (I used to be one of them) would say that,

- B’s observations should NOT affect A’s measurements;

- Whether B makes measurements or not,

if A’s detector is set to measure red or blue particles,

then 50% of the times A will detect a blue color particle and 50% a red particle.

However, it turns out that, what is observed is what is shown in Fig. 5 below.

A B

Purple Green

filter

Figure 5.

- That is to say, B’s measurements affect the measurements of A.

4. If B does not make any measurement (nor he/she places any filter),

Then, indeed, when the A’s detector is set to measure red or blue particles,

50% of the times A will detect a blue color particle and 50% a red particle.

A

B

Figure 6.

II. The annihilation of the positronium Ref: The Feynman Lectures, Vol III, page 18-5 http://feynmanlectures.caltech.edu/III_toc.html

Positronium, “Atom” made up of an electron and a positron.

It is a bound state of an e+ and an e−, like a hydrogen atom, except that a positron replaces the proton.

This object has—like the hydrogen atom—many states.

Like the hydrogen, the ground state is split into a “hyperfine structure” by the interaction of the magnetic moments.

The electron and positron have each spin ½ , and they can be either parallel or antiparallel to any given axis.

States are indicated by: (electron’s spin, positron’s spin) (In the ground state there is no other angular momentum due to orbital motion.)

The states of compound systems (i.e. systems composed of more than one particle) are subjected also to the conditions of symmetry conditions: symmetric or antisymmetric.

So there are four states: Three are the sub-states of a spin-one system, all with the same energy;

(+ ½ , + ½) m = 1

2

1 [ (+ ½ , - ½) + (- ½ , + ½) ] m = 0 (1)

(- ½ , - ½) m = -1

and one is a state of spin zero with a different energy.

2

1 [ (+ ½ , - ½) - (- ½ , + ½) ] m = 0 (2)

However, the positronium does not last forever.

The positron is the antiparticle of the electron; they can annihilate each other.

The two particles disappear completely—converting their rest energy into radiation,

which appears as -rays (photons).

In the disintegration, two particles with a finite rest mass go into two or more objects which have zero rest mass.

Case: Disintegration of the spin-zero state of the positronium. It disintegrates into two γ-rays with a lifetime of about 10−10 seconds. The initial and final states are illustrated in Fig. 7 below.

Initial state: we have a positron and an electron close together and with spins antiparallel, making the positronium system.

2

1 [ (+ ½ , - ½) - (- ½ , + ½) ] spin zero state

Final state: After the disintegration there are two photons going out with equal magnitude but opposite momenta,

(because the total momentum after the disintegration must be zero; we are taking the case of the positronium being at rest).

Angular distribution of the outgoing photons

Since the initial state (a) has spin zero, it has no special axis; therefore that state is symmetric under all rotations.

The final state (b) (constituted by photons) must then also be symmetric under all rotations.

That means that all angles for the disintegration are equally likely (3)

The amplitude is the same for a photon to go in any direction.

Of course, once we find one of the photons in some direction the other must be opposite.

Figure 7. Annihilation of positronium and emission of two-photons. We are interested in the polarization state of the outgoing photons.

Polarization of the photons

The only remaining question is about the polarization of the (4) outgoing photons.

In Fig. 7(b), let's call the directions of motion of the two photons the plus and minus Z-axes. See Fig. 8 below.

Photon polarization states: We can use any representations we want for the polarization states of the photons.

We will choose for our description right and left circular polarization. In the classical theory, right-hand circular polarization has equal components in x and y which are 90∘ out of phase.

Figure 8. Classical picture of the electric field vector.

In the quantum theory, a right-hand circularly (RHC) polarized photon has equal amplitudes to be |x⟩ polarized or |y⟩ polarized, and the amplitudes are 90∘ out of phase. Similarly for left-hand circularly (LHC) polarized photon.

|R ⟩ = 2

1 [ |x⟩ + i |y⟩ ] RHC photon a state

(5)

|L ⟩ = 2

1 [ |x⟩ - i |y⟩ ] LHC photon a state

Case-1: Emitted photons in the RHC states

If the photon going upward is RHC, then angular momentum will be conserved if the downward going photon is also RHC.

Each photon will carry +1 unit of angular momentum with respect to its momentum direction, which means plus and minus one unit about the z-axis.

The total angular momentum will be zero. The angular momentum after the disintegration will be the same as before. See Fig. 9 below.

Fig. 9 Positronium annihilation along the z-axis. The final

state is indicated as |R1R2⟩.

Case-2: Emitted photons in the LHC states There is also the possibility that the two photons go in the LHC state.

Figure 10. Another possibility for positronium annihilation

along the z-axis. The final state is indicated as |L1 L2⟩.

Relationship between the two decay modes mentioned above What is the relation between the amplitudes for these two possible decay modes? The answer comes from using the conservation of parity.

The parity of a state |⟩, relative to a given operator action, is related to whether,

F~ |⟩ = |⟩ even parity or (6) F~ |⟩ = - |⟩ odd parity

Before the decay: Theoretical physicists have shown in a way that is not easy to explain that the spin-zero ground state of positronium must be odd. We will just assume that it is odd, and since we will get agreement with experiment, we can take that as sufficient proof.

After the decay: Let's see then what happens if we make an inversion of the process in Fig. 9.

In the QM jargon, we say “let’s apply the operator P~

to the state”.

Here P~

stands for the inversion operator.1

When we apply P~

to the state described in Fig. 3 , the two photons reverse directions and polarizations. The inverted picture looks just like Fig. 10.

Let,

|R1R2⟩ stand for the final state of Fig. 9 (7) in which both photons are RHC,

and

|L1L2⟩ stand for the final state of Fig. 10 (8) in which both photons are LHC.

We notice that an inversion of the photon state in Fig. 9 results inan arrangement equal ti the one in Fig. 10; and vice versa. That is,

P~

|R1R2⟩ = |L1L2⟩

(9) P

~|L1L2⟩ = |R1R2⟩

So neither the state |R1R2⟩ nor the state |L1L2⟩ conserve the parity.

So, how to build a state |⟩ such that P~

|⟩ = - |⟩ ?

Answer,

|F ⟩ = | R1 R2⟩ − | L1 L2⟩ (10)

1 An inversion operation means that we should imagine what the system would look like if we were to move each

part of the system to an equivalent point on the opposite side of the origin. When we change x,y,z into −x,−y,−z, all polar vectors (like displacements and velocities) get reversed, but not the axial vectors (like angular momentum—or any vector which is derived from a cross product of two polar vectors). Axial vectors have the same components after an inversion.

for an inversion changes the R's into L's and gives the state

P~

|F⟩ = |L1 L2⟩ −| R1 R2⟩ = − |F ⟩ (11)

So the final state |F⟩ has negative parity, which is the same as the initial spin-zero state of the positronium. Accordingly,

| F ⟩

Before After

Figure 11. Decay of the spin-zero positronium state.

| F ⟩

-

Before After

| R1 R2⟩ | L1 L2⟩

Figure 12. Decay of the spin-zero positronium state. The state |F ⟩ = | R1 R2⟩ −| L1

L2⟩ is the only final state that conserves both angular momentum and parity.2

2 Notice | R1 R2⟩ −| L1 L2⟩ ≠ [ |R1 ⟩ −|L1 ⟩ ] [ |R2 ⟩ −|L2 ⟩ ] = - 4 |y1 ⟩ |y2 ⟩

What does the final state |F ⟩ = | R1 R2⟩ −| L1 L2⟩ mean physically? Even though the state |F ⟩ is the output state, it does not prevent us from asking for the amplitude probability to obtain a state different state.

For example, what is the amplitude probability to obtain the two RHC photons state: | R1 R2⟩ ?

< R1 R2| F ⟩ = ⟨ R1 R2| [ | R1R2⟩ − | L1 L2⟩ ]

= ⟨ R1 R2| R1 R2⟩ − ⟨ R1 R2| | L1 L2 ⟩

= ⟨ R1| R1 ⟩ ⟨ R2| R2⟩ − ⟨ R1| L1⟩ ⟨ R2| L2⟩

1 . 1 − 0 . 0

⟨ R1 R2| F ⟩ = 1 (12)

Similarly, what is the amplitude probability to obtain the two LHC photons state: | L1 L2⟩ ?

⟨ L1 L2| F ⟩ = -1 (13)

The results (9) and (10) mean the following:

If we observe the two photons in two detectors which can be set to count separately the RHC or LHC photons, we will always see two RHC photons together, or two LHC photons together.

That is, if you stand on one side of the positronium and someone else stands on the opposite side, you can measure the polarization and tell the other guy what polarization he will get.

You have a 50-50 chance of catching a RHC photon or a LHC photon; whichever one you get, you can predict that the other will get the same.

Case: What happens if we observe the photon in counters that accept only linearly polarized light?

Let’s assume that it is somewhat easy to measure the polarization of -rays. Suppose that

i) you have a counter that only accepts light with x-polarization,

and

ii) that there is a guy on the other side that also looks for linear polarized light with, say, y-polarization.

What is the chance to pick up the two photons from an annihilation?

What we are asking is the amplitude that |F⟩ will be in the state |x1 y2⟩.

Before After

| x1 ⟩

| y2 ⟩

x

y

z

Fig. 13 Experimental set up to observe the output state |x1 y2⟩.

⟨ x1 y2|F ⟩ = ?

⟨ x1 y2|F ⟩ = ⟨ x1 y2| ( | R1 R2⟩ −| L1 L2⟩ )

= ⟨ x1 y2| R1 R2 ⟩ − ⟨ x1 y2| L1 L2 ⟩ (14)

Now although we are working with two-particle amplitudes for the two photons, we can handle them just as we did the single particle amplitudes, since each particle acts independently of the other. That means,

⟨ x1 y2|R1 R2⟩ = ⟨x1|R1⟩ ⟨y2|R2⟩ (15)

Using |R ⟩ = 2

1 [ |x ⟩ + i |y ⟩ ]

|L ⟩ = 2

1 [ |x ⟩ - i |y ⟩ ]

we obtain ⟨ x1 y2| R1 R2⟩ = + i/2

Similarly, we find that ⟨ x1 y2|L1 L2⟩ = − i / 2.

Subtracting these two amplitudes according to (14), we get,

⟨x1 y2|F ⟩ = + i . (16)

So there is a unit probability that if you get a photon in your x-polarized detector, the other guy will get a photon in his y-polarized detector (see Fig. 13).

Case: Now suppose that the other guy sets his counter for x-polarization the same as yours. That is,

i) you have a counter that only accepts light with x-polarization,

and

ii) that there is a guy on the other side that also looks for linear polarized light with x-polarization.

What is the chance to pick up the two photons from an annihilation?

What we are asking is the amplitude that |F⟩ will be in the state |x1 y2⟩.

Answer: He would never get a count when you got one. If you work it through, you will find that ⟨x1 x2| F ⟩ = 0. (17)

Case: It will, naturally, also work out that if you set your counter for y-polarization, the guy in the other side will get coincident counts only if he is set for x-polarization. Using polarized beam splitters

Now all this leads to an interesting situation. Suppose you were to set up something like a piece of calcite which separates the photons into x -polarized and y -polarized beams

You put a counter in each beam. Let's call one the x -counter and the other the y -counter.

The guy on the other side does the same thing.

The results above indicate that,

You can always tell him which beam his photon is going to go into.

Whenever you and he get simultaneous counts, you can see which of your detectors caught the photon and then tell him which of his counters had a photon.

Let's say that in a certain disintegration you find that a photon went into your x -counter; you can tell him that he must have had a count in his y -counter.

| F ⟩

Before After

| x1 ⟩

| y2 ⟩

| y1 ⟩

| x2⟩

x

y

z

Figure 14. Photon detection with polarized beam splitters at each side.

Now many people who learn quantum mechanics in the usual (old-fashioned) way find this disturbing. They would like to think that,

Once the photons are emitted it goes along as a wave with a definite character.

Since “any given photon” has some “amplitude” to be x-polarized or to be y-polarized,

there should be some chance of picking it up in either the x- or y-counter and that this chance shouldn't depend on what some other person finds out about a completely different photon.

“Someone else making a measurement shouldn't be able to change the probability that I will find something.”

Our quantum mechanics says, however, that,

by making a measurement on photon number one, you can predict precisely what the polarization of photon number two is going to be when it is detected.

This point was never accepted by Einstein, and he worried about it a great deal—it became known as the “Einstein-Podolsky-Rosen paradox.” But when the situation is described as we have done it here, there doesn't seem to be any paradox at all; it comes

out quite naturally that what is measured in one place is correlated with what is measured somewhere else.

The argument that the result is paradoxical runs something like this:

(1) If you have a counter which tells you whether your photon is RHC or LHC, you can predict exactly what kind of a photon (RHC or LHC) he will find.

(2) The photons he receives must, therefore, each be purely RHC or purely LHC, some of one kind and some of the other.

(3) Surely you cannot alter the physical nature of his photons by changing the kind of observation you make on your photons. No matter what measurements you make on yours, his must still be either RHC or LHC.

(4) Now suppose he changes his apparatus to split his photons into two linearly polarized beams with a piece of calcite so that all of his photons go either into an x-polarized beam or into a y-polarized beam. There is absolutely no way, according to quantum mechanics, to tell into which beam any particular RHC photon will go. There is a 50% probability it will go into the x-beam and a 50% probability it will go into the y-beam. And the same goes for a LHC photon.

(5) Since each photon is RHC or LHC—according to (2) and (3)—each one must have a 50-50 chance of going into the x-beam or the y-beam and there is no way to predict which way it will go.

(6) Yet the theory predicts that if you see your photon go through an x-polarizer you can predict with certainty that his photon will go into his y-polarized beam. This is in contradiction to (5) so there is a paradox.

Nature apparently doesn't see the “paradox,” however, because experiment shows that the prediction in (6) is, in fact, true.

In the argument above, Steps (1), (2), (4), and (6) are all correct, but Step (3), and its consequence (5), are wrong; they are not a true description of nature.

Argument (3) says that by your measurement (seeing a RHC or a LHC photon) you cannot determine which of two alternative events occurs for him (seeing a RHC or a LHC photon), and that even if you do not make your measurement you can still say that his event will occur either by one alternative or the other. But this is not the way Nature works.

Her way requires a description in terms of interfering amplitudes, one amplitude for each alternative.

A measurement of which alternative actually occurs destroys the interference,

but if a measurement is not made you cannot still say that “one alternative or the other is still occurring.”

If you could determine for each one of your photons whether it was RHC and LHC, and also

whether it was x-polarized (all for the same photon) there would indeed be a paradox. But you

cannot do that—it is an example of the uncertainty principle.

Do you still think there is a “paradox”? Make sure that it is, in fact, a paradox about the behavior of Nature, by setting up an imaginary experiment for which the theory of quantum mechanics would predict inconsistent results via two different arguments. Otherwise the “paradox” is only a conflict between reality and your feeling of what reality “ought to be.”

Do you think that it is not a “paradox,” but that it is still very peculiar? On that we can all agree. It is what makes physics fascinating.

My view

| x1 ⟩ x

y

z

| F ⟩

(2) (1)

Figure 15. The output state of the twin photons after the positronium decay

Although the state | F ⟩ is the only one that fulfills the conservation of parity, quantum mechanics allows us to calculate the amplitude probability to obtain another state upon making the system (the two outgoing photons) to interact with some (polarizers) apparatus. For example, we can calculate the amplitude probability ⟨x1 y2|F ⟩ of detecting photon (1) in state | x1 ⟩ and photon (2) in state | y2 ⟩.

| x1 ⟩

| y2 ⟩

x

y

z

| R ⟩

(2)

| x ⟩

| L ⟩ )

(2)

(2)

(1)

(1)

(1)

| y ⟩

| R ⟩

| L ⟩ )

Figure 16. Pictorial representation of the different quantum states

that can be obtained after a measurement. Only three quantum states are indicated (with different color lines) in the figure.

The states shown in Fig. 16 are different possible states after a measurement is performed on the two photons. Notice each possible state is composed of one photon going to one side and another going to the other side. We say “the photons are entagled”. A different color is used in the figureto represent each different entangled state. One state (depicted in green) shows RHC polarization (on the left side) and RHC polarization (on the right side). Another state shows y-polarization (on the photon going left) and x-polarization (on the photon going right). Etc.

If this experiment (the positroniun decay ) is repeated N times , with the

detectors set to detect | R ⟩ polarization on (1) and | R2 ⟩ polarization on (18) (2) then a fraction |⟨ R1 R2| F ⟩ |2 of N will give such a expected result.

(Correspondingly, imilar for any other specific state) It is tempting to affirm that, Fig. 16 shows the paths available for a given twin of photons, and that in a particular single experiment all they have to do is to choose one available (19) path. But that cannot be true.

We know that the output state is | F ⟩ is the only one satisfying all the

conservation laws after the decay. However, Fig. 16 shows output (20) states that do not satisfy parity; therefore they are not outcome results allowed. Those paths in Fig. 16 are states after a measurement is made.

Accordingly, we should present Fig. 16 but with only one outcome state, as in Fig. 17.

| x1 ⟩

| y2 ⟩

x

y

z

| L ⟩ )

(2)

(1)

| L ⟩ )

| x1 ⟩

| y2 ⟩

x

y

z

| x ⟩ (1)

| y ⟩ (2)

| x1 ⟩

| y2 ⟩

x

y

z

| R ⟩

(2)

(1)

| R ⟩

Figure 17. Possible states after making a corresponding measurement. Statement (20) is very strong. It prevent us from affirming, for example, the following: Once a decay occur (a single event) a twin of photons can go either path red, or blue or green, each with its own weight of amplitude probabilities. That latter can not be true, since each of the paths shown violate the parity conservation. Instead So the measurement on one side forces the state | F ⟩ to collapse on a particular state. The collapsing implies that the other side collapses too (21) in the corresponding twin state. In a single observation (i.e. just one positronium decay) , if one of the observers detects the photon in the red state, then the observer on the other side will detect the photon on the red state . This is because the the quantum state is composed of the twin photon (even though they are far apart in space). If observer on the right side is set to observe only blue states, the the observer on the left will detect only the corresponding blue state.