# Finding Spin Expectation Values At Any Time t > 0

• Leechie
In summary: So, the method used is basically to find the projection of the spinor in the direction ##n## onto the eigenspinors in that direction and then use the expectation value equation to find the expectation value in that direction?Thanks. I think I'm finally starting to get my head round this now. So, the method used is basically to find the projection of the spinor in the direction ##n## onto the eigenspinors in that direction and then use the expectation value equation to find the expectation value in that direction?Yes, that is essentially the method used. You find the projection of the spinor onto the eigenspinors in a specific direction, and then use the expectation value equation to find the expectation value in
Leechie

## Homework Statement

Write down a spinor that represents the spin state of the particle at any time t > 0. Use the expression to find the expectation values of ##S_x## and ##S_y##

## Homework Equations

The particle is a spin-##\frac 1 2## particle, the gyromagnetic ratio is ##\gamma_s \lt 0##, and the magnetic field points in the ##z## direction.

The initial spin state is: ##| A \rangle_{initial}=\frac 1 5 \begin{bmatrix}3\\4\end{bmatrix}##

## The Attempt at a Solution

This is where I've got so far:
$$| A \rangle_{initial}=\frac 1 5 \begin{bmatrix}3\\4\end{bmatrix}=a_u|\uparrow_n \rangle+a_d|\downarrow_n \rangle$$
Finding ##|\uparrow_n \rangle## and ##|\downarrow_n \rangle## using ##|\uparrow_n \rangle=\begin{bmatrix}\cos(\theta / 2)\\e^{i\phi}\sin(\theta / 2)\end{bmatrix}## and ##|\downarrow_n \rangle=\begin{bmatrix}-e^{-i\phi}\sin(\theta / 2)\\\cos(\theta / 2)\end{bmatrix}##. The magnetic field points in the ##z## direction so ##\theta = 0## and ##\phi = 0##:
$$|\uparrow_z \rangle=\begin{bmatrix}\cos(0)\\e^0\sin(0)\end{bmatrix}=\begin{bmatrix}1\\0\end{bmatrix} \\ |\downarrow_z \rangle=\begin{bmatrix}-e^0\sin(0)\\\cos(0)\end{bmatrix}=\begin{bmatrix}0\\1\end{bmatrix}$$
Finding the coeffecients ##a_u## and ##a_d##:
$$a_u=\langle \uparrow_z | A \rangle=\frac 1 5 \begin{bmatrix}1&0\end{bmatrix} \begin{bmatrix}3\\4\end{bmatrix}=\frac 3 5 \\ a_d=\langle \downarrow_z | A \rangle=\frac 1 5 \begin{bmatrix}0&1\end{bmatrix} \begin{bmatrix}3\\4\end{bmatrix}=\frac 4 5$$
So:
$$| A \rangle_{initial}=\frac 1 5 \begin{bmatrix}3\\4\end{bmatrix}=\frac 3 5 \begin{bmatrix}1\\0\end{bmatrix} + \frac 4 5 \begin{bmatrix}0\\1\end{bmatrix}$$
Using the the equation for spin at any time ##| A \rangle=a_u e^{-iE_ut/\hbar}|\uparrow_n\rangle + a_d e^{-iE_dt/\hbar}|\downarrow_n\rangle## and since ##\gamma_s \lt 0## the energy eigenvalues are ##E_u=+\frac {\hbar \omega} 2## and ##E_d=-\frac {\hbar \omega} 2## I get:
$$| A \rangle=\frac 3 5 e^{-i\omega t/2}\begin{bmatrix}1\\0\end{bmatrix} + \frac 4 5 e^{+i\omega t/2}\begin{bmatrix}0\\1\end{bmatrix}$$
And so the spinor I get to is:
$$| A \rangle=\frac 1 5\begin{bmatrix}3 e^{-i\omega t/2}\\4 e^{+i\omega t/2}\end{bmatrix}$$
For the expectation value of ##S_x## I get:
\begin{align} \langle S_x \rangle & =\langle A | \hat {\mathrm S}_x | A \rangle \nonumber \\ & =\frac 1 5 \begin{bmatrix}3 e^{+i\omega t/2} & 4 e^{-i\omega t/2}\end{bmatrix} \frac \hbar 2 \begin{bmatrix}0&1\\1&0\end{bmatrix} \frac 1 5 \begin{bmatrix}3 e^{-i\omega t/2} \\ 4 e^{+i\omega t/2}\end{bmatrix} \nonumber \\ & = \frac 1 5 \begin{bmatrix}3 e^{+i\omega t/2} & 4 e^{-i\omega t/2}\end{bmatrix} \frac \hbar {10} \begin{bmatrix}4 e^{+i\omega t/2} \\ 3 e^{-i\omega t/2} \end{bmatrix} \nonumber \\ & = \frac \hbar {50} \left( 12e^{i\omega t} + 12e^{i\omega t} \right) \nonumber \\ & = \frac {12\hbar} {25} e^{iwt} \nonumber \end{align}
Could someone tell me if I'm along the right lines with this. I've been working on this for so long now I'm starting to lose sight of how this should workout.
Thanks

Leechie said:
\begin{align} \langle S_x \rangle & =\langle A | \hat {\mathrm S}_x | A \rangle \nonumber \\ & =\frac 1 5 \begin{bmatrix}3 e^{+i\omega t/2} & 4 e^{-i\omega t/2}\end{bmatrix} \frac \hbar 2 \begin{bmatrix}0&1\\1&0\end{bmatrix} \frac 1 5 \begin{bmatrix}3 e^{-i\omega t/2} \\ 4 e^{+i\omega t/2}\end{bmatrix} \nonumber \\ & = \frac 1 5 \begin{bmatrix}3 e^{+i\omega t/2} & 4 e^{-i\omega t/2}\end{bmatrix} \frac \hbar {10} \begin{bmatrix}4 e^{+i\omega t/2} \\ 3 e^{-i\omega t/2} \end{bmatrix} \nonumber \\ & = \frac \hbar {50} \left( 12e^{i\omega t} + 12e^{i\omega t} \right) \nonumber \\ & = \frac {12\hbar} {25} e^{iwt} \nonumber \end{align}
You made a mistake going from line 4 to line 5 line 3 to 4 in there, so the final result is not correct. Otherwise, it looks fine.

Last edited:
Leechie
Leechie said:
Finding ##|\uparrow_n \rangle## and ##|\downarrow_n \rangle## using ##|\uparrow_n \rangle=\begin{bmatrix}\cos(\theta / 2)\\e^{i\phi}\sin(\theta / 2)\end{bmatrix}## and ##|\downarrow_n \rangle=\begin{bmatrix}-e^{-i\phi}\sin(\theta / 2)\\\cos(\theta / 2)\end{bmatrix}##. The magnetic field points in the ##z## direction so ##\theta = 0## and ##\phi = 0##:
$$|\uparrow_z \rangle=\begin{bmatrix}\cos(0)\\e^0\sin(0)\end{bmatrix}=\begin{bmatrix}1\\0\end{bmatrix} \\ |\downarrow_z \rangle=\begin{bmatrix}-e^0\sin(0)\\\cos(0)\end{bmatrix}=\begin{bmatrix}0\\1\end{bmatrix}$$

Just an observation. The formulas you are using are valid for the "z-basis" where ##\begin{bmatrix}1\\0\end{bmatrix}## is the z-spin-up and ##\begin{bmatrix}0\\1\end{bmatrix}## is the z-spin-down. So, ##a_u = \frac35## and ##a_d = \frac45## immediately.

In other words, by definition:$$| A \rangle_{initial}=\frac 1 5 \begin{bmatrix}3\\4\end{bmatrix}=\frac35 |\uparrow_z \rangle + \frac45 |\downarrow_z \rangle$$

Leechie
I had a feeling there was something wrong somewhere, I'll take another look. Thanks for your help.

PeroK said:
Just an observation. The formulas you are using are valid for the "z-basis" where ##\begin{bmatrix}1\\0\end{bmatrix}## is the z-spin-up and ##\begin{bmatrix}0\\1\end{bmatrix}## is the z-spin-down. So, ##a_u = \frac35## and ##a_d = \frac45## immediately.

In other words, by definition:$$| A \rangle_{initial}=\frac 1 5 \begin{bmatrix}3\\4\end{bmatrix}=\frac35 |\uparrow_z \rangle + \frac45 |\downarrow_z \rangle$$

Thanks PeroK. When I was working through that I realized it just led back to the initial spin state because of the z-basis. Is the method I used a general way of calculating a spinor in any direction ##n##?

Leechie said:
Thanks PeroK. When I was working through that I realized it just led back to the initial spin state because of the z-basis. Is the method I used a general way of calculating a spinor in any direction ##n##?

Yes, the formulas you used would give you (expressed in the z-basis) the eigenspinors in the direction ##n##.

Leechie
PeroK said:
Yes, the formulas you used would give you (expressed in the z-basis) the eigenspinors in the direction ##n##.

Thanks. I think I'm finally starting to get my head round this now.

PeroK

## 1. What is the spin expectation value?

The spin expectation value is a measure of the average spin of a particle in a specific direction. It is calculated by taking the dot product of the spin operator and the spin state vector, and is represented by the symbol ⟨S⟩.

## 2. How is the spin expectation value calculated?

The spin expectation value is calculated by taking the dot product of the spin operator and the spin state vector. This is represented by the formula ⟨S⟩ = ⟨Ψ|S|Ψ⟩, where Ψ is the spin state vector and S is the spin operator.

## 3. Can the spin expectation value be measured at any time t > 0?

Yes, the spin expectation value can be measured at any time t > 0. This is because the spin operator is a time-independent operator, meaning it does not change with time. Therefore, the spin expectation value remains constant at any given time t > 0.

## 4. Why is it important to find the spin expectation value at any time t > 0?

Finding the spin expectation value at any time t > 0 allows us to understand the behavior of the spin of a particle over time. It can provide insights into the dynamics of a system and can be used to make predictions about the future behavior of the particle.

## 5. How does the spin expectation value change over time?

The spin expectation value remains constant over time, as long as the particle is in an isolated system. This is because the spin operator is a time-independent operator and does not change with time.

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