Quantum Physics Quantum Physics Help Forum Jun 19th 2019, 04:15 AM #1 Junior Member   Join Date: Feb 2015 Posts: 2 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?   Jun 19th 2019, 02:37 PM   #2

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 Originally Posted by SolidState 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
$\displaystyle cos \left ( \dfrac{ \omega t}{2} \right ) + i ~ sin\left ( \dfrac{ \omega t}{2} \right ) = 0$

$\displaystyle 2~ cos \left ( \dfrac{ \omega t}{2} \right ) = 1$

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.   Jun 20th 2019, 09:25 PM #3 Forum Admin   Join Date: Apr 2008 Location: On the dance floor, baby! Posts: 2,730 Moderation bump. (I know, I know, don't do as I do....) -Dan __________________ Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup. See the forum rules here.  Tags magnet, spin Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post kiwiheretic General Physics 5 Jul 24th 2017 02:36 PM preethipaul Electricity and Magnetism 3 Nov 27th 2013 11:38 AM r.samanta Atomic and Solid State Physics 4 Nov 10th 2009 03:17 AM WanderingMind Periodic and Circular Motion 4 Jun 7th 2009 10:24 PM werehk Electricity and Magnetism 6 Jul 4th 2008 03:33 AM 