Entanglement and two particle system

[J.Asher]

Hello, I am planning to talk about the entanglement for the presentation but now I am a bit confused of between entanglement and many-particle system. Actually, I have only studied two-particle system when dealing the total angular momentum. let me assume that there are a photon and an electron then the total angular momentum might be 3/2. So the possible states describing the system would be |3/2, 3/2>, |3/2,1/2>, |3/2, -1/2>, |3/2, -3/2>, |1/2, 1/2>, |1/2, -1/2>. Here, for example if I write down |3/2, 1/2> in terms of the production states of particle 1 and 2, I can say c_1|1,1>|1/2,-1/2> + c_2|1,0>|1/2,1/2> (c_1 and c_2 are normalized constants). If I write down again nicely :D
|3/2,1/2> = c_1|1,1>|1/2,-1/2> + c_2|1,0>|1/2,1/2>.
But it look totally the same as the wave function of the entangled system.
So can I say that If I measure 1h/2pi by operating z component momentum upon photon, then it is obviously the electron lies on the state |1/2, -1/2> ... ?
Or is it impossible to operate the momentum operator of particle 1 or 2 upon this |3/2,1/2>.
(Actually, I am also confused of meaning of |3/2,1/2>, what exactly does is says...?)

Thx for reading.

 

[torquil]

If the Hamiltonian describing your system has rotational invariance, then every state of a particular energy (energy eigenstate), will also have a particular value for those two angular-momentum numbers, which in your case is 3/2 and 1/2.

For a given z-axis in space, |3/2,1/2> means a state with total angular momentum 3/2 and with angular momentum 1/2 about the z-axis (in units of hbar). I.e. the projection of the angular momentum vector along the z-axis is 1/2.

In your case, |3/2,1/2> denotes a two-particle state. It is then possible to write that state as a linear combination of products of single-particle states, as you have done in your post.

When a two-particle state cannot be written as one product of one-particle states, it is said to be entangled. So if both your coefficients c1 and c2 are non-zero, the two-particle state is entangled.

For a two-particle state, it is always (in theory) possible to measure properties of one of them. When this is done you will obtain a quantum number for that particle, and at the same time the full two-particle state will collapse in the usual way, and you then know without measuring some thing about the other particle.

In your example, that would mean that if you measure the z-component of spin on one particle to be 1, then the z-component of angular momentum for the other particle must be -1/2 as you say.

That was my understanding, but don't take it as gospel...

These lecture notes are pretty good:
http://www.uio.no/studier/emner/matnat/fys/FYS4110/h10/undervisningsmateriale/LectureNotes2010.pdf"

 

[J.Asher]

When a two-particle state cannot be written as one product of one-particle states, it is said to be entangled. So if both your coefficients c1 and c2 are non-zero, the two-particle state is entangled.
What does it mean? I think if c1 and c2 are non-zero, it does means that the two-particle state can be described in a product of two single-particle states...?