Total angular momentum of Positronium

[Academic]

Homework Statement

Positronium is formed by stopping anti-electrons in matter. It is found that the bound system is formed in two distinct states, both of which have orbital angular momentum $ L = 0 $. Consider the possible spin configuration of this system, to determine the expected total angular momentum for these two positronium states.

Homework Equations

Not sure. Selection rules?

The Attempt at a Solution

Here is what I wrote down, lol:

\mid +,\uparrow \rangle \otimes \mid -,\downarrow \rangle

\mid +, \downarrow \rangle \otimes \mid -,\uparrow \rangleIm not sure if these are right or relevant though. It seems to me the answer is zero, does that sound right? Is there a way I can show this?

 

[diazona]

Remember the basics: there are two kinds of angular momentum, orbital and spin. They combine (not just simple addition, but group addition) to produce the total angular momentum. However, you've been told that there is no orbital angular momentum, so you don't have to worry about that. This problem is purely about spin.

So the first step is to write down the possible spin states, which you've already started to do. However, the way you've been writing them down, there are two more possible spin states that you're missing. What are they?

Once you've done that, you'll have four basis spin states. The actual spin state of the system will be a linear combination of those four. One such linear combination is different from the other three. Can you find it? Think about what you know about addition of angular momenta, especially regarding other systems you've studied which have two spin-1/2 particles.

 

[Academic]

I can't figure out what the two other states you mention are.

Is it these?

<br /> \mid -,\uparrow \rangle \otimes \mid +,\downarrow \rangle <br />

\mid -, \downarrow \rangle \otimes \mid +,\uparrow \rangle

If so I don't really see how this is different than the two states I gave. I don't understand the outer product very well.


You then say that the actual state is a linear combination of the basis states, which makes sense to me. But then you say that one linear combination is different than the other three... Is there four different linear combinations? That confuses me.

 

[diazona]

I can't figure out what the two other states you mention are.

Is it these?

<br /> \mid -,\uparrow \rangle \otimes \mid +,\downarrow \rangle <br />

\mid -, \downarrow \rangle \otimes \mid +,\uparrow \rangle

No, those are just the same two states you already had.

Think classically: if you have two distinct particles, and each of them can be in one of two distinct boxes, there are four possible states. (2\times 2 = 4) You know how to enumerate all the possible states in that case, right? Just do the same thing here. The positron and electron are your particles, and spin-up and spin-down are the "boxes".

 

[Academic]

Oh right. The total angular momentum is zero, but the spin angular momentum could be non-zero. That means I could have states,
\mid -, \uparrow \rangle \otimes \mid +, \uparrow \rangle
\mid -, \downarrow \rangle \otimes \mid +, \downarrow \rangleSo my state is a linear combination of these four states.
( \mid +, \uparrow \rangle \otimes \mid -,\downarrow \rangle) + ( \mid +, \downarrow \rangle \otimes \mid -, \uparrow \rangle) + (\mid +,\uparrow \rangle \otimes \mid -,\uparrow \rangle) + (\mid +,\downarrow \rangle \otimes \mid -,\downarrow \rangle)

Then to get the expectation value of total angular momentum I take the inner product of that state with a spin operator. ??

 

[diazona]

Right, your state will be a linear combination of those four.
a_0\vert \uparrow_+ \uparrow_-\rangle + a_1\vert\uparrow_+ \downarrow_-\rangle + a_2\vert\downarrow_+\uparrow_-\rangle + a_3\vert\downarrow_+\downarrow_-\rangle
(I trust you understand my abbreviated notation) For any given state (any given values of the coefficients a_i), you can calculate the expectation value of the total angular momentum as
\sqrt{\langle a\vert J^2\vert a\rangle}
where \vec{J} is the total angular momentum operator. It can be written as \vec{J} = \vec{L} + \vec{S}, the sum of the orbital and spin angular momentum operators.