How Does Electron Spin Orientation Relate to Its Velocity and Antineutrino Path?

[Tegalad]

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]

$\mid\chi\rangle=A\left(\frac{\sqrt{1+\frac{v}{c}}\sin\frac{\theta}{2}}{\sqrt{1-\frac{v}{c}}\cos\frac{\theta}{2}}\right)$

$\mid\chi\rangle=\frac{\sqrt{2}}{2}\left(\begin{array}{c}\sqrt{0,85}\\ \sqrt{1,15}\end{array}\right)$

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?

 

[Tegalad]

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
Thank you for your reply :)

 

[DrClaude]

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

I had understood that. So, can you answer my questions?

 

[Tegalad]

I had understood that. So, can you answer my questions?

Now I see that I made a mistake, because I used a spinor with a given theta, so I would get the same result.

 

[DrClaude]

Now I see that I made a mistake, because I used a spinor with a given theta, so I would get the same result.

I don't understand what you mean, so I can't tell if your result will be correct.