Entangled Photons

Pre­vi­ously in se­ries: De­co­her­ence as Projection

To­day we shall an­a­lyze the phe­nomenon of “en­tan­gled par­ti­cles”. We’re go­ing to make heavy use of po­larized pho­tons here, so you’d bet­ter have read yes­ter­day’s post.

If a par­ti­cle at rest de­cays into two other par­ti­cles, their net mo­men­tum must add up to 0. The two new par­ti­cles may have am­pli­tudes to head off in all di­rec­tions, but in each joint con­figu­ra­tion, the di­rec­tions will be op­po­site.

By a similar method you can pro­duce two en­tan­gled pho­tons which head off in op­po­site di­rec­tions, and are guaran­teed to be po­larized op­po­sitely (at right an­gles to each other), but with a 50% prior prob­a­bil­ity of go­ing through any given po­larized filter.

You might think that this would in­volve am­pli­tudes over a con­tin­u­ous spec­trum of op­po­site con­figu­ra­tions—an am­pli­tude for pho­ton A to be po­larized at 0° and for pho­ton B to be po­larized at 90°, an am­pli­tude for A to be 1° po­larized and for B to be 91° po­larized, etc. But in fact it’s pos­si­ble to de­scribe the quan­tum state “un­known but op­po­site po­lariza­tions” much more com­pactly.

First, note that the two pho­tons are head­ing off in op­po­site di­rec­tions. This jus­tifies call­ing one pho­ton A and one pho­ton B; they aren’t likely to get their iden­tities mixed up.

As with yes­ter­day, the po­lariza­tion state (1 ; 0) is what passes a 90° filter. The po­lariza­tion state (0 ; 1) is what passes a 0° filter. (1 ; 0) is po­larized up-down, (0 ; 1) is po­larized left-right.

If A is in the po­lariza­tion state (1 ; 0), we’ll write that as A=(1 ; 0).

If A=(1 ; 0) and B=(0 ; 1), we’ll write that as

[ A=(1 ; 0) ∧ B=(0 ; 1) ]

The state for “un­known op­po­site po­lariza­tion” can be writ­ten as:

√(1/​2) * ( [ A=(1 ; 0) ∧ B=(0 ; 1) ] - [ A=(0 ; 1) ∧ B=(1; 0) ] )

Note that both terms are be­ing mul­ti­plied by the square root of 12. This en­sures that the squared mod­u­lus of both terms sums to 1. Also, don’t over­look the minus sign in the cen­ter, we’ll need it.

If you mea­sure the A pho­ton’s po­lariza­tion in the up-down/​left-right ba­sis, the re­sult is pretty straight­for­ward. Your mea­sure­ment de­co­heres the en­tan­gle­ment, cre­at­ing one evolu­tion out of the A=(1 ; 0) ∧ B=(0 ; 1) con­figu­ra­tion, and a sec­ond, non­in­ter­act­ing evolu­tion out of the A=(0 ; 1) ∧ B=(1; 0) con­figu­ra­tion.

If you find that the A pho­ton is po­larized up-down, i.e., (1 ; 0), then you know you’re in the A=(1 ; 0) ∧ B=(0 ; 1) blob of am­pli­tude. So you know that if you or any­one else mea­sures B, they’ll re­port to you that they found B in the (0 ; 1) or left-right po­lariza­tion. The ver­sion of you that finds A=(1 ; 0), and the ver­sion of your friend that finds B=(0 ; 1), always turn out to live in the same blob of am­pli­tude.

On the other side of con­figu­ra­tion space, an­other ver­sion of you finds them­selves in the A=(0 ; 1) ∧ B=(1; 0) blob. If a friend mea­sures B, the other you will ex­pect to hear that B was po­larized up-down, just as you ex­pect to meet the ver­sion of your friend that mea­sured B left-right.

But what if you mea­sure the sys­tem in a slanted ba­sis—test a pho­ton with a 30° po­larized filter? Given the speci­fied start­ing state, in the up-down /​ left-right ba­sis, what hap­pens if we mea­sure in the 30° ba­sis in­stead? Will we still find the pho­tons hav­ing op­po­site po­lariza­tions? Can this be demon­strated?

Yes, but the math gets a lit­tle more in­ter­est­ing.

Let’s re­view, from yes­ter­day, the case where a pho­ton pre­vi­ously po­larized in the up-down/​left-right ba­sis en­coun­ters a 30° filter.

Polar3060
A 30-60-90 tri­an­gle has a hy­potenuse of 1, a small side of 12, and a longer side of (√3)/​2, in ac­cor­dance with the Pythagorean The­o­rem.

If a pho­ton passes a 0° filter, com­ing out with po­lariza­tion (0 ; 1), and then en­coun­ters an­other filter at 30°, the vec­tor that would be trans­mit­ted through the 30° filter is

(√3)/​2 * (1/​2 ; (√3)/​2) = (.433 ; .75)

and the po­lariza­tion vec­tor that would be ab­sorbed is

12 * (-(√3)/​2 ; 12) = (-.433 ; .25)

Note that the po­lariza­tion states (1/​2 ; (√3)/​2) and (-(√3)/​2 ; 12) form an or­thonor­mal ba­sis: The in­ner product of each vec­tor with it­self is 1, and the in­ner product of the two vec­tors with each other is 0.

Then we had (√3)/​2 of one ba­sis vec­tor plus 12 of the other, guaran­tee­ing the squared mod­uli would sum to 1. ((√3)/​2)2 + (1/​2)2 = 34 + 14 = 1.

So we can say that in the 30° ba­sis, the in­com­ing (0 ; 1) pho­ton had a (√3)/​2 am­pli­tude to be trans­mit­ted, and a 12 am­pli­tude to be ab­sorbed.

Sym­met­ri­cally, sup­pose a pho­ton had passed a 90° filter, com­ing out with po­lariza­tion (1 ; 0), and then en­coun­tered the same 30° filter. Then the trans­mit­ted vec­tor would be

12 * (1/​2 ; (√3)/​2) = (.25 ; .433)

and the ab­sorbed vec­tor would be

-(√3)/​2 * (-(√3)/​2 ; 12) = (.75 ; -.433)

Now let’s con­sider again with the en­tan­gled pair of photons

√(1/​2) * ( [ A=(1 ; 0) ∧ B=(0 ; 1) ] - [ A=(0 ; 1) ∧ B=(1; 0) ] )

and mea­sure pho­ton A with a 30° filter.

Sup­pose we find that we see pho­ton A ab­sorbed.

Then we know that there was a -(√3)/​2 am­pli­tude for this event to oc­cur if the origi­nal state had A=(1 ; 0), and a 12 am­pli­tude for this event to oc­cur if the origi­nal state had A=(0 ; 1).

So, if we see that pho­ton A is ab­sorbed, we learn that we are in the now-de­co­her­ent blob of am­pli­tude:

( -(√3)/​2 * √(1/​2) * [ A=(-(√3)/​2 ; 12) ∧ B=(0 ; 1) ] )
- ( 12 * √(1/​2) * [ A=(-(√3)/​2 ; 12) ∧ B=(1; 0) ] )

You might be tempted to add the two am­pli­tudes for A be­ing ab­sorbed—the -(√3)/​2 * √(1/​2) and the −1/​2 * √(1/​2)—and get a to­tal am­pli­tude of -.966, which, squared, comes out as .933.

But if this were true, there would be a 93% prior prob­a­bil­ity of A be­ing ab­sorbed by the filter—a huge prior ex­pec­ta­tion to see it ab­sorbed. There should be a 50% prior chance of see­ing A ab­sorbed.

What went wrong is that, even though we haven’t yet mea­sured B, the con­figu­ra­tions with B=(0 ; 1) and B=(1 ; 0) are dis­tinct. B could be light-years away, and un­known to us; the con­figu­ra­tions would still be dis­tinct. So we don’t add the am­pli­tudes for the two terms; we keep them sep­a­rate.

When the am­pli­tudes for the terms are sep­a­rately squared, and the squares added to­gether, we get a prior ab­sorp­tion prob­a­bil­ity of 1/​2—which is ex­actly what we should ex­pect.

Okay, so we’re in the de­co­her­ent blob where A is ab­sorbed by a 30° filter. Now con­sider what hap­pens over at B, within our blob, if a friend mea­sures B with an­other 30° filter.

The new start­ing am­pli­tude dis­tri­bu­tion is:

( -(√3)/​2 * √(1/​2) * [ A=(-(√3)/​2 ; 12) ∧ B=(0 ; 1) ] )
- ( 12 * √(1/​2) * [ A=(-(√3)/​2 ; 12) ∧ B=(1; 0) ] )

In the case where B=(0 ; 1), it has an am­pli­tude of (√3)/​2 to be trans­mit­ted through a 30° filter; be­ing trans­mit­ted through a 30° filter cor­re­sponds to the po­lariza­tion state (1/​2 ; (√3)/​2). Like­wise, a 12 am­pli­tude to be ab­sorbed (po­lariza­tion state (-(√3)/​2 ; 12).)

In the case where B=(1 ; 0) it has an am­pli­tude of 12 to be trans­mit­ted with state (1/​2 ; (√3)/​2). And an am­pli­tude of -(√3)/​2 to oc­cupy the state (-(√3)/​2 ; 12) and be ab­sorbed.

So add up four terms:

( -(√3)/​2 * √(1/​2) ) * [ A=(-(√3)/​2 ; 12) ∧ B=(0 ; 1) ]
breaks down into
( -(√3)/​2 * √(1/​2) ) * (√3)/​2 * [ A=(-(√3)/​2 ; 12) ∧ B=(1/​2 ; (√3)/​2) ] +
( -(√3)/​2 * √(1/​2) ) * 12 * [ A=(-(√3)/​2 ; 12) ∧ B=(-(√3)/​2 ; 12) ]
and
- ( 12 * √(1/​2) ) * [ A=(-(√3)/​2 ; 12) ∧ B=(1; 0) ] )
breaks down into
-( 12 * √(1/​2) ) * 12 * [ A=(-(√3)/​2 ; 12) ∧ B=(1/​2 ; (√3)/​2) ] +
-( 12 * √(1/​2) ) * -(√3)/​2 * [ A=(-(√3)/​2 ; 12) ∧ B=(-(√3)/​2 ; 12) ]

Th­ese four terms oc­cupy only two dis­tinct con­figu­ra­tions.

Ad­ding the am­pli­tudes, the con­figu­ra­tion [ A=(-(√3)/​2 ; 12) ∧ B=(-(√3)/​2 ; 12) ] ends up with zero am­pli­tude, while [ A=(-(√3)/​2 ; 12) ∧ B=(1/​2 ; (√3)/​2) ] ends up with a fi­nal am­pli­tude of √(1/​2).

So, within the blob in which you’ve found your­self, the prob­a­bil­ity of your friend see­ing that a 30° filter blocks both A and B, is 0. The prob­a­bil­ity of see­ing that a 30° filter blocks A and trans­mits B, is 50%.

Sym­met­ri­cally, there’s an­other blob of am­pli­tude where your other self sees A trans­mit­ted and B blocked, with prob­a­bil­ity 50%. And A trans­mit­ted and B trans­mit­ted, with prob­a­bil­ity 0%.

So you and your friend, when you com­pare re­sults in some par­tic­u­lar blob of de­co­hered am­pli­tude, always find that the two pho­tons have op­po­site po­lariza­tion.

And in gen­eral, if you use two equally ori­ented po­lariza­tion filters to mea­sure a pair of pho­tons in the ini­tal state:

√(1/​2) * ( [ A=(1 ; 0) ∧ B=(0 ; 1) ] - [ A=(0 ; 1) ∧ B=(1; 0) ] )

then you are guaran­teed that one filter will trans­mit, and the other filter ab­sorb—re­gard­less of how you set the filters, so long as you use the same set­ting. The pho­tons always have op­po­site po­lariza­tions, even though the prior prob­a­bil­ity of any par­tic­u­lar pho­ton hav­ing a par­tic­u­lar po­lariza­tion is 50%.

What if I mea­sure one pho­ton with a 0° filter, and find that it is trans­mit­ted (= state (0 ; 1)), and then I mea­sure the other pho­ton with a 30° filter?

The prob­a­bil­ity works out to just the same as if I knew the other pho­ton had state (1 ; 0)—in effect, it now does.

Over on my side, I’ve de­co­hered the am­pli­tude over the joint dis­tri­bu­tion, into blobs in which A has been trans­mit­ted, and A ab­sorbed. I am in the de­co­her­ent blob with A trans­mit­ted: A=(0 ; 1). Ergo, the am­pli­tude vec­tor /​ po­lariza­tion state of B, in my blob, be­haves as if it starts out as (1 ; 0). This is just as true whether I mea­sure it with an­other 0° filter, or a 30° filter.

With sym­met­ri­cally en­tan­gled par­ti­cles, each par­ti­cle seems to know the state the other par­ti­cle has been mea­sured in. But “seems” is the op­er­a­tive word here. Ac­tu­ally we’re just deal­ing with de­co­her­ence that hap­pens to take place in a very sym­met­ri­cal way.

To­mor­row (if all goes ac­cord­ing to plan) we’ll look at Bell’s The­o­rem, which rules out the pos­si­bil­ity that each pho­ton already has a fixed, non-quan­tum state that lo­cally de­ter­mines the re­sult of any pos­si­ble po­lariza­tion mea­sure­ment.

Part of The Quan­tum Physics Sequence

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