# Spinors and Probabilities of Electron-Positron Pair

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"?

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...

• Does the $\otimes$ symbol mean multiplication? Nov 27, 2013 at 21:31
• No, it is a tensorial product (of states). It means that each state has its own independent space, so the total amplitude is just the product of the amplitude of each space. You may also use the Dirac notation with bra and ket : $A = \langle \chi_+(1)|\langle\chi_-(2)| \frac{1}{\sqrt{2}}(|\chi_+(0)\rangle |\chi_-(0)\rangle - |\chi_-(0)\rangle |\chi_+(0)\rangle) = \frac{1}{\sqrt{2}}( \langle \chi_+(1)|\chi_+(0)\rangle\langle\chi_-(2)|\chi_+(0)\rangle - \langle \chi_+(1)|\chi_-(0)\rangle\langle\chi_-(2)|\chi_-(0)\rangle)$ Nov 28, 2013 at 10:22