Spin and Magnet

Feb 2015
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0
Suppose in a magnetic field which is in the z-direction we send a magnet with its N-pole directed in the x-direction. We know the magnet would rotate and become in line with the magnetic field. However, if we send a pure spin state \(\displaystyle |S_x,+>\) it can depart in the two opposite directions +z and -z, with the same probability without changing its state. If we simulate \(\displaystyle |S_x,+>\) with a magnet which its N-pole directed in the x-direction then we expect the spin also to rotate, in other words, change its state to \(\displaystyle |S_z,+>\). Could anyone please help me where I am wrong?
 

topsquark

Forum Staff
Apr 2008
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On the dance floor, baby!
Suppose in a magnetic field which is in the z-direction we send a magnet with its N-pole directed in the x-direction. We know the magnet would rotate and become in line with the magnetic field. However, if we send a pure spin state \(\displaystyle |S_x,+>\) it can depart in the two opposite directions +z and -z, with the same probability without changing its state. If we simulate \(\displaystyle |S_x,+>\) with a magnet which its N-pole directed in the x-direction then we expect the spin also to rotate, in other words, change its state to \(\displaystyle |S_z,+>\). Could anyone please help me where I am wrong?
I want to make an apology: This thread has been in the Moderation queue for a few days and when I first replied I didn't realize that. I came back today to make a couple of small changes and found that out.

The Hamiltion for an electron in a constant magnetic field is \(\displaystyle H = - \dfrac{ge}{2 mc} \vec{B} \cdot \vec{S}\).

Since the magnetic field is constant and along the z axis we have \(\displaystyle B = B_z \hat{k}\). (I'm going to go ahead and use "B" instead of
" \(\displaystyle B_z\) " for convenience.) So we have
\(\displaystyle H = -\dfrac{ge}{2 mc} B S_z = \dfrac{geB}{2mc} \left ( \begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix} \right )\)
(I'm using the \(\displaystyle | \pm >\) z basis.)

Now, the time evolution operator is \(\displaystyle U(t, 0) = e^{-iHt/ \hbar}\) so applying this to the electron spin state we get
\(\displaystyle | \psi (t) > = e^{-iHt/ \hbar} | S_x ; + > \)

so we need to know \(\displaystyle H |S_x ; + >\).

\(\displaystyle H | S_x ; + > = \dfrac{geB}{2mc} \left ( \begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix} \right ) ~ \dfrac{ \hbar }{2} \dfrac{1}{ \sqrt{2} } \left ( |+> + ~ |-> \right ) \) (where, again, the \(\displaystyle | \pm >\) are taken to be the z direction kets.)

\(\displaystyle = \dfrac{geB \hbar}{4 \sqrt{2} ~ mc} \left ( \begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix} \right ) ~ \left ( |+> + ~ |-> \right ) = \dfrac{geB \hbar}{4 \sqrt{2} ~ mc} \left ( \begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix} \right ) \left ( \left ( \begin{matrix} 1 \\ 1 \end{matrix} \right ) + \left ( \begin{matrix} 1 \\ -1 \end{matrix} \right ) \right )\)

\(\displaystyle = \dfrac{geB \hbar}{4 \sqrt{2} ~ mc} \left ( \left ( \begin{matrix} 1 \\ 0 \end{matrix} \right ) + \left ( \begin{matrix} 0 \\ 1 \end{matrix} \right ) \right )\)

Thus
\(\displaystyle H | S_x ; + > = \dfrac{geB \hbar}{4 \sqrt{2} ~ mc} \left ( \left ( \begin{matrix} 1 \\ 0 \end{matrix} \right ) + \left ( \begin{matrix} 0 \\ 1 \end{matrix} \right ) \right ) = \dfrac{geB\hbar}{4 \sqrt{2} ~ mc} \left ( |+> + ~ |-> \right ) \)

So, putting all of this together:
\(\displaystyle | \psi (t) > = e^{-it (geB/(4 \sqrt{2} ~ mc))} |+> + ~ e^{it (geB/(4 \sqrt{2} ~ mc))} |->\)

Let's make it pretty: Define \(\displaystyle \omega = \dfrac{geB}{2 \sqrt{2}}\). Then we have
\(\displaystyle | \psi (t) > = e^{-i \omega t / 2} |+> + e^{i \omega t / 2} |->\). Normalizing: \(\displaystyle | \psi (t) > = \dfrac{1}{\sqrt{2}} \left ( e^{-i \omega t / 2} |+> + e^{i \omega t / 2} |-> \right )\)

Let's use this form for the moment. Setting t = 0 we get \(\displaystyle | \psi (t) > = \dfrac{1}{\sqrt{2}} \left ( |+> + ~ |-> \right )\) , which is (in the direction of) \(\displaystyle |S_x; +>\) as required by the problem statement.

To see the time evolution it is simpler to remove the exponentials. From Euler's theorem we have that
\(\displaystyle e^{i \omega t/2} = cos \left ( \dfrac{ \omega t}{2} \right ) + i ~ sin\left ( \dfrac{ \omega t}{2} \right )\)
\(\displaystyle e^{- i \omega t/2} = cos \left ( \dfrac{ \omega t}{2} \right ) - i ~ sin\left ( \dfrac{ \omega t}{2} \right )\)

so for \(\displaystyle | \psi (t) >\) we get
\(\displaystyle | \psi (t) > = \dfrac{1}{\sqrt{2}} \left ( cos \left ( \dfrac{ \omega t}{2} \right ) - i ~ sin\left ( \dfrac{ \omega t}{2} \right ) \right ) |+> + \dfrac{1}{\sqrt{2}} \left ( cos \left ( \dfrac{ \omega t}{2} \right ) + i ~ sin\left ( \dfrac{ \omega t}{2} \right )\right ) |->\)

So at what time do we have \(\displaystyle | \psi (t) > = |+>\)? This will happen when
\(\displaystyle cos \left ( \dfrac{ \omega t}{2} \right ) - i ~ sin\left ( \dfrac{ \omega t}{2} \right ) = 1\)

and
\(\displaystyle cos \left ( \dfrac{ \omega t}{2} \right ) + i ~ sin\left ( \dfrac{ \omega t}{2} \right ) = 0\)

Adding these two gives
\(\displaystyle 2~ cos \left ( \dfrac{ \omega t}{2} \right ) = 1\)

solving this gives \(\displaystyle t = \dfrac{ 2 \pi }{3 \omega}\).

-Dan
 
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topsquark

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Apr 2008
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On the dance floor, baby!
Moderation bump. (I know, I know, don't do as I do....)

-Dan