1
$\begingroup$

Given a spin state: $|s\rangle$ = some linear combination of $|\uparrow\rangle + |\downarrow\rangle$ possibly with an imaginary component. How do you get from the definition of a magnetic momentum operator $\hat{\mu}_e = g\mu_B\hat{\sigma}$ to the expectation value of the electron spin magnetic moment?

$g$ is the gyrmoagnetic factor and is approximately 2.0023.

$\mu_B =\frac{e\hbar}{2m_o}$ is the Bohr magneton.

$\hat{\sigma}$ is the Pauli spin matrix.

I feel like this is the operation

$\langle s| \hat{\mu}_e |s\rangle$

If it is, I need an example walk-through with some arbitrary complex $|s\rangle$

Improve this question
$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

Let $$|s\rangle = \alpha|\uparrow\rangle + \beta|\downarrow\rangle$$ We assume that $s$ is normalized i.e. $\langle s | s\rangle = 1 \implies |\alpha|^2+|\beta|^2 = 1$. Then the expectation value of $\hat{\mu}_e$ is: $$\langle\hat{\mu}_e\rangle = \langle s|\hat{\mu}_e|s\rangle$$ $$\implies \langle\hat{\mu}_e\rangle = |\alpha|^2\langle\uparrow| \hat{\mu}_e |\uparrow\rangle + |\beta|^2\langle\downarrow| \hat{\mu}_e |\downarrow\rangle + \alpha^{\ast}\beta\langle\uparrow| \hat{\mu}_e |\downarrow\rangle + \alpha\beta^{\ast}\langle\downarrow| \hat{\mu}_e |\uparrow\rangle$$ Now, $\hat{\mu}_e = g\mu_B\hat{\sigma}$. We use this together with $\langle \uparrow|\hat{\sigma}|\uparrow\rangle = 1$, $\langle \downarrow|\hat{\sigma}|\downarrow\rangle = -1$, and $\langle \uparrow|\hat{\sigma}|\downarrow\rangle = \langle \downarrow|\hat{\sigma}|\uparrow\rangle = 0$, to get: $$\langle\hat{\mu}_e\rangle = g\mu_B(|\alpha|^2 - |\beta|^2)$$ Note that this expression for the expectation value is consistent with interpreting $|\alpha|^2$ and $|\beta|^2$ as probabilities of finding the spin to be $\uparrow$ and $\downarrow$ respectively, as required.

Improve this answer
$\endgroup$
6
  • $\begingroup$ What happens if we have two $\uparrow$ spins with $\alpha$ being real and $\beta$ being imaginary? $\endgroup$ Commented Apr 22, 2015 at 7:08
  • $\begingroup$ The result is applicable for general complex coefficients. Two $\uparrow$ spin terms with different coefficients would just add up to one $\uparrow$ spin term with the (complex) sum of the coefficients. But it is only $\uparrow$ and $\downarrow$ together that describe all possible spin states. $\endgroup$ Commented Apr 22, 2015 at 7:13
  • $\begingroup$ Ah, I could just consider it $\alpha= x+ yi $ and $\beta= 0$ and it would apply to a specific observation. Yes? $\endgroup$ Commented Apr 22, 2015 at 7:14
  • $\begingroup$ Or would only having $\uparrow$ mean that the result is a vector? $\endgroup$ Commented Apr 22, 2015 at 7:24
  • $\begingroup$ You can take $\alpha = x + iy$ and $\beta = 0$. $\endgroup$ Commented Apr 22, 2015 at 7:29

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.