Spinors and Probabilities of Electron-Positron Pair

Question

Question:

An electron and positron are moving in opposite directions, and are in the spin singlet state. Two Stern-Gerlach machines are orientated in some arbitrary direction; one along unit vector $\hat{s}_1$ (which measures the electron's position) and one along unit vector $\hat{s}_2$ (which measure's the positron's position).

1) Find the electron and positron eigenspinors in terms of spherical coordinates $\theta_1$ and $\phi_1$ (corresponding $\hat{s}_1$) and $\theta_2$ and $\phi_2$ (corresponding $\hat{s}_2$).

2) Calculate the probabilities of obtaining every possible spin outcome (both particles up; electron up, positron down, etc), and simplify by using directional cosines to replace the angles.

Attempt: Both the electron and positron are spin one-half particles. Let us define

$$\hat{s}_1=\hat{i}\sin\theta_1\cos\phi_1+\hat{j}\sin\theta_1\sin\phi_1+\hat{k}\cos\theta_1$$ $$\hat{s}_2=\hat{i}\sin\theta_2\cos\phi_2+\hat{j}\sin\theta_2\sin\phi_2+\hat{k}\cos\theta_2$$

If we look at the electron, we can then constract the spin matrix $S_1$, which represents the spin angular momentum along the $\hat{s}_1$ position:

$$S_1=S\cdot \hat{s}_1=S_x\sin\theta_1\cos\phi_1+S_y\sin\theta_1\sin\phi_1+S_z\cos\theta_1 $$ $$\rightarrow S_1=\frac{\hbar}{2}\begin{pmatrix} \cos\theta_1 & e^{-i\phi_1}\sin\theta_1\\ e^{i\phi_1}\sin\theta_1 & -\cos\theta_1 \end{pmatrix}$$

Here, I have utilized the Pauli matrices $S_x$, $S_y$, and $S_z$. From here, one finds the eigenvalues to be

$$\lambda=\pm \frac{\hbar}{2}$$

Plugging in, we obtain the normalized eigenspinors $\chi_+^1$ and $\chi_-^1$, corresponding to the spin up and spin down directions, respectively:

$$\chi_+^1=\begin{pmatrix} \cos(\theta_1/2) \\ e^{i\phi}\sin(\theta_1/2) \end{pmatrix}$$ $$\chi_-^1=\begin{pmatrix} e^{i\phi_2}\sin(\theta_1/2) \\ -\cos(\theta_1/2) \end{pmatrix}$$

We can now find the generic spinor $\chi^1$:

$$\chi^1=\left (\frac{a+b}{\sqrt{2}}\right)\chi_+^{(1)}+\left (\frac{a-b}{\sqrt{2}}\right)\chi_-^{(1)}$$

However, would I basically obtain the same expressions for the positron, only with different angles: i.e.,

$$\chi_+^2=\begin{pmatrix} \cos(\theta_2/2) \\ e^{i\phi}\sin(\theta_2/2) \end{pmatrix}$$ $$\chi_-^2=\begin{pmatrix} e^{i\phi_2}\sin(\theta_2/2) \\ -\cos(\theta_2/2) \end{pmatrix}$$

I know that both the electron and positron are spin $1/2$, but does that mean their eigenspinors would look nearly identical?

As well, I have a question concerning part 2. I know that, if we are measuring, say, $S_y$, then the probability of obtaining an up spin up would be $[(\chi_+^{(y)})^\dagger \chi]^2$. However, here, we have a two particle system, so does that mean the probability of obtaining, say, an electron spin up and a positron spin down would be $[(\chi_+^{(1)})^\dagger \chi^1]^2\cdot [(\chi_-^{(2)})^\dagger \chi^2]^2$, i.e., you multiply the separate probabilities? Also, what does it mean to express the probability in "directional cosines"?

Thank you in advance.

Answer 1

Spin eigenvectors are the same for the electron and the positron.

The transition amplitude between the singlet state $s$, and for instance, a state up for electron and down for positron may be written (up to a complex unit phase) :

$A = ((up_1)^\dagger \otimes (down_2)^\dagger) s \\=((\chi_+(\theta_1, \phi_1))^\dagger \otimes \chi_-(\theta_2, \phi_2))^\dagger) \frac{1}{\sqrt{2}}(\chi_+(0, 0) \otimes \chi_-(0, 0) - \chi_-(0, 0) \otimes \chi_+(0, 0))$

$= \frac{1}{\sqrt{2}} (\chi_+(\theta_1, \phi_1)^\dagger\chi_+(0, 0)\quad \chi_-(\theta_2, \phi_2)^\dagger\chi_-(0, 0) - \chi_+(\theta_1, \phi_1)^\dagger\chi_-(0, 0)\quad \chi_-(\theta_2, \phi_2)^\dagger\chi_+(0, 0)) $

The probability is $P = |A|^2$

Directional cosinus are just cosinus of angles of spin with some axes (for instance $\cos \theta, \cos \phi$), etc...