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 $|s\rangle$


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.

  • $\begingroup$ What happens if we have two $\uparrow$ spins with $\alpha$ being real and $\beta$ being imaginary? $\endgroup$ – That1guy Apr 22 '15 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. $\endgroup$ – AV23 Apr 22 '15 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? $\endgroup$ – That1guy Apr 22 '15 at 7:14
  • $\begingroup$ Or would only having $\uparrow$ mean that the result is a vector? $\endgroup$ – That1guy Apr 22 '15 at 7:24
  • $\begingroup$ You can take $\alpha = x + iy$ and $\beta = 0$. $\endgroup$ – AV23 Apr 22 '15 at 7:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.