# Electron spin probabilities

## Homework Statement

After beta- decay electron and antineutrino comes out, electron is moving along z axis and it is moving with velocity v. It's spinor is
$\mid\chi\rangle=A\left(\frac{\sqrt{1+\frac{v}{c}}\sin\frac{\theta}{2}}{\sqrt{1-\frac{v}{c}}\cos\frac{\theta}{2}}\right)$ where A is normalization constant,$\theta$ an angle between z axis and $\overline{\nu}$ movement path (antineutrino's spin is in the same direction as it's movement path)
$\theta=\frac{\pi}{2}$ and $\frac{v}{c}=0,15$
1) Calculate the probability that measured electron's spin will be directed along z axis
2) Calculate the probability that measured electron's spin will be directed along z axis, if the angle of $\overline{\nu}$ trajectory can't be measured and electron's spin is registered by averaging all $\theta$ angles

## Homework Equations

By inserting all values and normalizing the function, spinor is $\mid\chi\rangle=\frac{\sqrt{2}}{2}\left(\begin{array}{c}\sqrt{0,85}\\ \sqrt{1,15}\end{array}\right)$
$\mid\uparrow_z\rangle=\frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\ 0\end{array}\right)$
$S_z= \frac{\hbar}{2}\begin{bmatrix}1 & 0 \\0 & -1 \end{bmatrix}$

## The Attempt at a Solution

1) $\langle\uparrow_z\mid\chi\rangle=c$ then c squared is probability that the spin will be along z axis. Is it good ?
2) $\overline{S_z}=\langle\chi\mid\ S_z\mid\chi\rangle ⇒ \overline{S_z}=\frac{\hbar}{2}$ then $P_+\frac{\hbar}{2}+(1-P_+)(-\frac{\hbar}{2})=\frac{\hbar}{2}\cos\theta$ After that I calculate $P_+$. Is this good or I am missing something? Thanks in advance

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DrClaude
Mentor
$\mid\chi\rangle=A\left(\frac{\sqrt{1+\frac{v}{c}}\sin\frac{\theta}{2}}{\sqrt{1-\frac{v}{c}}\cos\frac{\theta}{2}}\right)$
$\mid\chi\rangle=\frac{\sqrt{2}}{2}\left(\begin{array}{c}\sqrt{0,85}\\ \sqrt{1,15}\end{array}\right)$
I guess that in the first equation, it is not fraction but two elements of a vector. In that case, are you sure that the second equation is correct?

$\mid\uparrow_z\rangle=\frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\ 0\end{array}\right)$
Is that a normalised state?

1) $\langle\uparrow_z\mid\chi\rangle=c$ then c squared is probability that the spin will be along z axis. Is it good ?
It should be $|c|^2$ (you have to account for the fact that $c \in \mathbb{C}$).

2) $\overline{S_z}=\langle\chi\mid\ S_z\mid\chi\rangle ⇒ \overline{S_z}=\frac{\hbar}{2}$ then $P_+\frac{\hbar}{2}+(1-P_+)(-\frac{\hbar}{2})=\frac{\hbar}{2}\cos\theta$ After that I calculate $P_+$. Is this good or I am missing something?
You are asked twice to calculate the probability that the measured electron's spin will be directed along the z axis. Why is your approach different in both cases?

I guess that in the first equation, it is not fraction but two elements of a vector. In that case, are you sure that the second equation is correct?
It shouldn't be a fraction and yes the second one is correct.

Is that a normalised state?
Yes, it is.

It should be $|c|^2$ (you have to account for the fact that $c \in \mathbb{C}$).

You are asked twice to calculate the probability that the measured electron's spin will be directed along the z axis. Why is your approach different in both cases?
Maybe I haven't made it clear enough, but in the first case angle theta is known and in the second one the spin is calculated by averaging all possible theta's

DrClaude
Mentor
Maybe I haven't made it clear enough, but in the first case angle theta is known and in the second one the spin is calculated by averaging all possible theta's