Saturday, 6 June 2009

Quantum Entanglement, entanglement with 2 electron singlet

(and now for something completely different... Quantum mechanics and quantum entanglement)

Quantum Entanglement:
Entanglement with 2 element quantum singlet.

Having recently emailed Leonard Susskind (Stanford CA) with a question regarding entanglement of a 3rd element with a 2 element system in the quantum singlet state (an entangled pair).

Leonard Susskind is the Felix Bloch professor of theoretical physics at Stanford University in the field of string theory and quantum field theory http://en.wikipedia.org/wiki/Leonard_Susskind

Susskind is widely regarded as one of the fathers of String Theory.

So naturally, I was somewhat surprised when I actually got a detailed response, and the answer is very interesting and informative too! Like most things in Quantum mechanics, the result is not something that could be guest or anticipated, it's not something that could be expected using classical intuition or experience.

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Let’s begin with the initial state consisting of two electrons in a singlet state |ud-du> .( I should include a factor of one-over-square-root-of-2 to normalize the state but it’s too hard to type in Word so I will leave it to you.) Now add another electron in the up state— |u). Also, in order to keep track of photons, lets indicate that there are no photons present with the notation |0}. The initial state is,

[ |ud-du> |u) ] |0}

Now let’s rearrange the symbols so as to group electrons 2 and 3 together. This is just a notational device. The initial state is

[ |u> |du) - |d>|uu) ] |0}

Now use the trivial identity

|du) = ½ |du-ud) + ½ |du+ud)

to rewrite it

[ |u> |du-ud) ] |0} + ½ [ |u> |du+ud) ] |0}

-[ |d>|uu) ] |0}

Now think of electron 2 and 3 as close one another, while electron 1 is off in the distance. In the first term 2-3 are in a singlet state. Nothing happens to it. In the other two terms the 2-3 system is orthogonal to the singlet. The 2-3 system will interact to emit a photon, and decay to the singlet in those cases, although the photons will be in different states in the two terms. Denoting the presence of the photon by |γ} and |γ’}, the final state will be

[ |u> |du-ud) ] |0} + ½ [ |u> |du-ud) ] |γ}

-[ |d>|du-ud)] |γ’} (note, there is missing sqrt2)

So what do we have in the end? A superposition of states in which the 2-3 system is in the singlet and

1. electron 1 is up with no photon

1. electron 1 is up with a photon in state |γ}

1. electron 1 is down with a photon in state |γ’}.

If there is no photon or a photon in state |γ} then we know that electron 1 is up. If there is a photon in state |γ’} then e-1 must be down.

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Having such a brilliantly and concise response was fantastic!

I've gone on to calculate the trivial equations to determine the probability in each state, as follows:

Given the probabilities of the 3 possibilities must add up to 1, taking the coefficients as lambda, the probability is given by lambda.lambda* which for real numbers is just lambda^2, so:

[ |u> |du-ud) ] |0} + ½ [ |u> |du-ud) ] |γ}
-[ |d>|du-ud)] |γ’} (note, there is missing sqrt2)

a. electron 1 is up with no photon

`|      1      | 2| ------------|      = 1/4 = p1 = 0.25|  2 sqrt(2)  |`

b. electron 1 is up with a photon in state |γ}

p2 = 0.25

c. electron 1 is down with a photon in state |γ’}.
`|      1      | 2| ------------|      = 1/2 = p3 = 0.5|  sqrt(2)    |`