• Support PF! Buy your school textbooks, materials and every day products Here!

Correlation coefficient of two-particle half-spin entangled state

  • Thread starter Skiggles
  • Start date
  • #1
2
0
Hi, I can't get my head around this question.

Homework Statement


The Bell state:
[tex]|\psi\rangle = \frac{1}{\sqrt{2}}\left(|\uparrow_1\rangle|\downarrow_2\rangle +|\downarrow_1\rangle |\uparrow_2\rangle \right)[/tex]

Find the correlation coefficient of the measurement of the spins in the directions [tex]z,\phi[/tex].

Homework Equations


[tex] C(\phi) = \langle S_{z1} S_{\phi 2} \rangle [/tex]

S takes the values [tex]\pm 1[/tex]

The Attempt at a Solution


I know that I need to calculate the probability of the states being parallel and subtract the probability that the states are anti-parallel. But how do you calculate those?

[tex]Pr(\uparrow_{z1}\uparrow_{\phi 2}) = \left|\langle\uparrow_{z1}\uparrow_{\phi 2} \left |\frac{1}{\sqrt{2}}\left(|\uparrow_1\rangle|\downarrow_2\rangle +|\downarrow_1\rangle |\uparrow_2\rangle \right)\right| \uparrow_{z1}\uparrow_{\phi 2} \rangle \right|^2[/tex]

Where do I go from here?

Thanks
 
Last edited:

Answers and Replies

  • #2
2
0
I think I might be on to something with this, but am not quite there yet:

[tex] \hat{S}_z =\frac{\hbar}{2} \begin{pmatrix}1 & 0 \\ 0 & -1 \\ \end{pmatrix}[/tex]

Rotate by angles [tex]\theta, \phi[/tex] to get

[tex] \hat{S}_\phi = \frac{\hbar}{2} \begin{pmatrix} \cos \theta & \sin \theta e^{-i\phi} \\
\sin \theta e^{i\phi} & -\cos \theta \end{pmatrix}[/tex]

Can I use this somehow to get the correlation coefficient?

Unfortunately the notation is getting a bit mixed up because in this post [tex]\phi = 0[/tex] and in the post above [tex]\phi[/tex] is used in place of [tex]\theta[/tex].
 

Related Threads on Correlation coefficient of two-particle half-spin entangled state

  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
1
Views
1K
Replies
1
Views
2K
Replies
0
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
5
Views
3K
Replies
10
Views
1K
Replies
1
Views
2K
Replies
3
Views
497
Replies
1
Views
1K
Top