5
$\begingroup$

For an electron, are the Hilbert spaces for the spin angular momentum and the orbital angular momentum the same or are they different? If they're different, how do we justify the operator $L\cdot S$ in spin-orbit coupling?

Also for coupling of 2 spins, what is the Hilbert space being considered? Is it $\mathbb{C}^2\otimes\mathbb{C}^2$ as I think it should be or otherwise? If its what I think it should be, what is the meaning of the operator $S_1\cdot S_2$?

$\endgroup$
4
$\begingroup$

For both the situations you're considering, the vector spaces are different, and the joint state space is the tensor product of the individual vector spaces.

To describe operators on this space, we simply use the tensor product of operators: if $\hat{L}_z:\mathcal H_\mathrm{orb} \to \mathcal H_\mathrm{orb}$ and $\hat{S}_z:\mathcal H_\mathrm{spin} \to \mathcal H_\mathrm{spin}$, then their tensor product $$ \hat{L}_z\otimes \hat{S}_z:\mathcal H_\mathrm{orb}\otimes \mathcal H_\mathrm{spin} \to \mathcal H_\mathrm{orb} \otimes \mathcal H_\mathrm{spin} $$ is uniquely defined by its action on product states $$ (\hat{L}_z\otimes \hat{S}_z)|\psi⟩\otimes |\phi⟩ = (\hat{L}_z|\psi⟩)\otimes (\hat{S}_z|\phi⟩) $$ and by linearity.

On top of that structure, we often consider vector operations on the vector characters of those operators, including in particular their dot product $$ \hat{\mathbf{L}} \stackrel{\otimes}{\cdot} \hat{\mathbf{S}} = \sum_{j=1}^3 \hat{L}_j\otimes \hat{S}_j. $$ This is a legitimate dot product in that one can show that it does not depend on the basis with respect to which the components are taken, because each component transforms as a vector and so the usual proof techniques still apply.

Now, in practice, we normally drop the explicit tensor-product marks $\otimes$ unless we really need the clarity, because the structure is typically clear from the context (so that a product like $\hat{L}_z\hat{S}_z$ is generally unambiguous) and the explicit marks add notational bulk and therefore make everything harder to read. Thus, what you'll typically see is notation of the form $$ \hat{\mathbf{L}} \cdot \hat{\mathbf{S}} = \sum_{j=1}^3 \hat{L}_j \hat{S}_j. $$ in which the tensor products between operators that act on different sectors of the state space are implicit.

$\endgroup$
  • $\begingroup$ Yes, you're right and that's exactly what I thought. However, when I read about exchange interaction, it had the following steps: $H=-\sum\limits_{i,j}J_{ij}S_i.S_j+g\mu_B\sum\limits_iS_i.B=g\mu_B\sum\limits_iS_i.(B+B')$, where $B'=-\frac{2}{g\mu_B}\sum\limits_j(J_{ij}S_j)$. As far as I know, a magnetic field isn't considered an operator in this case and even if it is, it isn't in the same space as the spin $B'$. Hence, if I take your definition to be correct(which intuitively seems so), then this derivation must be wrong which leads to a wrong formulation of ferromagnetism $\endgroup$ – Souradeep Aug 13 '18 at 16:28
  • $\begingroup$ @Souradeep the operator status of the magnetic field depends on context; in the formalism you're quoting, you're working in an effective formulation in which the QED aspects you refer to have been eliminated, so that the identification of $B'$ as an operator dependent on the $S_j$ is perfectly legitimate. Nothing in your previous comment looks wrong. (Unless what bothers you is the addition of $B$ with $B'$? There $B$ is just a $c$-number scalar, which trivially resolves to that scalar times any identity matrix you need to include. There's nothing wrong with $B+B'$.) $\endgroup$ – Emilio Pisanty Aug 13 '18 at 17:23
  • $\begingroup$ Oh, I get it. So when one writes S.B it implies $\sum\limits_iS_i\otimes\mathbb{1}B_i$. Thanks a lot for the clarification. Truly appreciate it. $\endgroup$ – Souradeep Aug 14 '18 at 19:21
  • $\begingroup$ Precisely - but note that it is in fact much more extensive. If you've got $n$ spins and you want a fully-rigorous tensor-product notation for $S_i$, what you really should be writing is $1\otimes \cdots \otimes 1\otimes S_i \otimes 1\otimes \cdots \otimes 1$, with $i-1$ factors on the left and $n-i-2$ factors on the right, because your Hilbert space is the $n$-fold tensor product of the single-spin state space and if you want to be completely rigourous you need to specify that the action on the other factos is the identity. This is what I meant when I said that this "adds notational bulk". $\endgroup$ – Emilio Pisanty Aug 14 '18 at 19:40
  • $\begingroup$ Yeah, that's exactly the formalism I used for the above derivation after the above issue was clarified. Thanks a lot for the understanding. $\endgroup$ – Souradeep Aug 16 '18 at 13:48

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.