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Old Jun 19th 2019, 04:15 AM   #1
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Spin and Magnet

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?
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Old Jun 19th 2019, 02:37 PM   #2
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Originally Posted by SolidState View Post
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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Last edited by topsquark; Jun 21st 2019 at 03:56 AM.
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Old Jun 20th 2019, 09:25 PM   #3
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Moderation bump. (I know, I know, don't do as I do....)

-Dan
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