How does the expectation value of the spin operator evolve over time?

[Garlic]

Dear PF,

As an exercise I am to find out how the expectation value of the spin operator evolves over time.

There was a hint, stating that it is enough to show that
$$
e^{i \frac{\phi ( \hat{n} \cdot \sigma )}{2}} \sigma_i e^{- i \frac{\phi ( \hat{n} \cdot \sigma )}{2}} = [R_{ \hat{n} }]_{ij} \sigma_j
$$
Where $R_{ \hat{n} }$ is a 3x3 matrix that describes a rotation with the angle Φ around the n axis.

In the last exercise we showed that putting a vector between these exponantial terms, we get a rotated vector.

In the given equation the free parameter is the angle, where the spin expectation value would evolve over time.

-So I don't quite get how showing that equation holds would tell us how the spin expectation value evolves over time.

Also: In the exercise it was written S-vector, I was surprised because usually it is asked to find the expectation value of S_z.
The expectation value of S_z would be ±hbar/2.
And when we search for the expectation value of S-vector, I think of a "vector inside a unit sphere" pointing in a specific direction, and this spin vector would somehow rotate over time.

-Am I thinking in the right way?

-In Griffiths I could not find much information about the time evolution of spin over time. Do you know a source where I can find more explanations about this subject?

Thank you very much for your time,

Garli

 

[vanhees71]

I'm not sure that your description is accurate enough to answer it thoroughly. You have at least to tell us with respect to which state you are supposed to evaluate the expectation value.

In the second part you claim that the expectation value of $S_z$ is constant $\pm \hbar/2$. This indicates that you choose $\hat{n}=\hat{e}_z$. Then $\hat{S}_z$ of course doesn't change under the unitary transformation and, if you then take the expectation values wrt. the eigenvectors of $\hat{S}_z$ you get the corresponding eigenvalue as the expectation value of $\hat{S}_z$ of course.

To give a more specific answer, we'd need the precise formulation of the question, and it also looks as if this is more appropriate for the home-work section of Physics Forums!

 

[Hans de Vries]

As an exercise I am to find out how the expectation value of the spin operator evolves over time.
There was a hint, stating that it is enough to show that
$$
e^{i \frac{\phi ( \hat{n} \cdot \sigma )}{2}} \sigma_i e^{- i \frac{\phi ( \hat{n} \cdot \sigma )}{2}} = [R_{ \hat{n} }]_{ij} \sigma_j$$
Where $R_{ \hat{n} }$ is a 3x3 matrix that describes a rotation with the angle $\phi$ around the $n$ axis. ...So I don't quite get how showing that equation holds would tell us how the spin expectation value evolves over time.

If the spin-vector $\vec{s}$ is given by,

$\vec{s}~=~\left[\begin{array}{c}\xi^*\sigma_x\,\xi \\ \xi^*\sigma_y\,\xi \\ \xi^*\sigma_z\,\xi\end{array}\right]$

then the rotated spin-vector $\vec{s}'$ is given by

$\vec{s}' ~=~ R_{ \hat{n} }\left[\begin{array}{c}\xi^*\sigma_x\,\xi \\ \xi^*\sigma_y\,\xi \\ \xi^*\sigma_z\,\xi\end{array}\right]$

Alternatively you can define a new set of matrices $\sigma_x',\sigma_y',\sigma_z'$ using the right-hand side of your equation

$\sigma_i' ~=~ [R_{ \hat{n} }]_{ij}\, \sigma_j$.

In this case the rotated spin-vector $\vec{s}'$ is given by:

$\vec{s}' ~=~ \left[\begin{array}{c}\xi^*\sigma_x'\,\xi \\ \xi^*\sigma_y'\,\xi \\ \xi^*\sigma_z'\,\xi\end{array}\right]$

Yet another alternative is that you rotate the spinor itself using the left-hand side terms of your equation:

$\xi'~=~ e^{- i \frac{\phi ( \hat{n} \cdot \sigma )}{2}}\,\xi~~~~$ and $~~~~~\xi^*{'}~=~ \xi^* e^{i \frac{\phi ( \hat{n} \cdot \sigma )}{2}}$

Now the rotated spin-vector $\vec{s}'$ is given by:

$\vec{s}' ~=~ \left[\begin{array}{c}\xi^*{'}\,\sigma_x\,\xi' \\ \xi^*{'}\,\sigma_y\,\xi' \\ \xi^*{'}\,\sigma_z\,\xi'\end{array}\right]$

This explains your formula.

Finally it is good to know why a spin-vector component $s_x$ is given by

$s_x=\xi^*\sigma_x\,\xi$.

This is because $\sigma_x$ is the product of two operators:

$\sigma_x ~=~ (i\sigma_o)\,(-i\sigma_x)$

1) The operator $-i\sigma_x$ rotates the spinor by 180$^o$ around the $x$-axis.
2) The operator $i\sigma_o$ rotates the spinor by -180$^o$ around its own axis.

Now consider a normalized spinor $\xi$ which is directed parallel along the $x$-axis. The net rotation in this case is zero and thus:

$\xi^*\sigma_x\,\xi~=~\xi^*\xi ~=~ 1$

If the normalized spinor $\xi$ is directed anti-parallel along the $x$-axis then the total rotation is 360$^ o$ degrees which amounts to a sign change and thus:

$\xi^*\sigma_x\,\xi~=~-\xi^*\xi ~=~ -1$

This explains why the Pauli matrices can be used to project-out the elements of the spin-vector $\vec{s}$

 

[Garlic]

Thank you for your quick replies :)

I have found a link where it says that the associated time evolution unitary operator $U(t,0)$ is similar to the operator $R_n(α)$ which rotates a vector by an angle α around the axis defined by the vector $\bar{n}$.
I diddn't understand how a time independent rotation by $R_n$ could explain a time evolution of a spin state. Now I see that the rotation caused by $U(t,0)$ has a dependency on the time so the angle α is time dependent in this case α=α(t).
Source: https://ocw.mit.edu/courses/physics...all-2013/lecture-notes/MIT8_05F13_Chap_07.pdf

To give a more specific answer, we'd need the precise formulation of the question

My apologies. Because the text was in a different language it was hard for me to formulate it well.
The spin state was not given, only the S-vector was defined as $S= \frac{ hbar }{2} σ_i$ where the $σ_i$ are the pauli matrices.

If the spin-vector →ss→\vec{s} is given by

Interesting. In our undergrad course we defined the spin vector without the spinors. Thanks to you today I have learned about them :) I find the concept very interesting.