# 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

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