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I'm currently stuck with a question involving what I think is sort of the Larmor precession. I have an electron in a uniform, magnetic field pointing in the $x$-axis and I'm supposed to calculate the eigenstate of $S_z$ at time $t$.

I'm trying to follow an example in Griffiths book but I'm running into a problem with $S_z$ and the magnetic field. The hamiltonian is defined as, for this problem, $H=-k\,B\cdot S$ where $B\cdot S$ denotes a dot product. This dot product is $0$ since they are perpendicular to each other, how should I confront this problem?

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  • $\begingroup$ You need to realise that $S$ is an operator, not just a vector. $\endgroup$
    – user154997
    Commented Oct 6, 2017 at 15:42
  • $\begingroup$ Related: physics.stackexchange.com/q/294228/154997 $\endgroup$
    – user154997
    Commented Oct 6, 2017 at 15:43
  • $\begingroup$ Also related: physics.stackexchange.com/q/300327/154997 $\endgroup$
    – user154997
    Commented Oct 6, 2017 at 15:44
  • $\begingroup$ One more: physics.stackexchange.com/q/233873/154997 $\endgroup$
    – user154997
    Commented Oct 6, 2017 at 15:45
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    $\begingroup$ Under the given hypotheses, your Hamiltonian is $H=-kBS_x$. Now in the basis $(|+1>,|-1>)$ of the eigenvectors of $S_z$, the operator $S_x$ has a well-known matrix. Once you have written it down, you can diagonalise it to find the eigenvalues and eigenvectors of $H$. Then you can evolve any initial state. $\endgroup$
    – user154997
    Commented Oct 6, 2017 at 16:07

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This is incorrect:

This dot product is $0$ since they are perpendicular to each other.

While the expected value of the spin might lie along the $z$ axis, the spin is still a vector-valued operator, which means that when contracted with the regular vector $\mathbf B$ you're left with the linear combination \begin{align} \hat{\mathbf S}\cdot\mathbf B = \hat{S}_xB_x+\hat{S}_yB_y+\hat{S}_zB_z, \end{align} and that's an operator since the $\hat{S}_i$ are operators and the $B_i$ are numbers.

This then gives you your hamiltonian, and you just work the time evolution like you would with any arbitrary hamiltonian.

in your specific case $\mathbf B$ is along $x$, which means that $\hat H = - kB\hat S_x$. If you want to work in the $S_z$ eigenbasis, that hamiltonian has the matrix representation $$ \hat H = - kB\hat S_x =-kB\frac12 \begin{pmatrix} 0&1\\1&0\end{pmatrix}. $$ For further details, refer to your textbook.

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  • $\begingroup$ thank you for the explanation, right now I'm basically trying to understand why I can use the matrix Sx when I have spinn "in z direction". What part of the theory haven't I understood? $\endgroup$
    – A.Maine
    Commented Oct 7, 2017 at 9:47
  • $\begingroup$ Well, that depends on what you mean by "the Sx matrix". If you mean its diagonal form, then no, that's inappropriate. However, you can represent any spin component, along an arbitrary direction, as a non-diagonal matrix in the Sz basis. (or you can work in the Sx eigenbasis, in which case the hamiltonian is diagonal but your initial state is a superposition of basis states.) If any of that sounds confusing, you probably need to go back to the textbook. $\endgroup$ Commented Oct 7, 2017 at 9:54
  • $\begingroup$ I'm working with Grifiths book currently, trying to read up what everything means, I'm a bit confused, not going to lie, is there a way you could give me an example if I wanted to represent a spin component, for example in x-direction, with the Sz basis? $\endgroup$
    – A.Maine
    Commented Oct 7, 2017 at 10:07

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