Is it an instance of rotational invariance of a classical hamiltonian being broken in quantum mechanics?

Question

Consider a spin-$\frac{1}{2}$ particle (say, an electron) subjected to a uniform external magnetic field $\textbf{B}$, and the part of its Hamiltonian associated with its spin degree of freedom: $$H=-\frac{e}{m_e}\textbf{S}\cdot\textbf{B}\tag{1}$$ where $e, m_e$ are respectively the charge and the mass of the electron, and $\textbf{S}$ is the spin operator.

Classically, $H=-\boldsymbol{\mu}\cdot\textbf{B}$, being a scalar product of two vectors, is a scalar. Therefore, $H$ is invariant under rotations because a scalar is invariant under rotations.

Quantum mechanically, for $H$ to be invariant under rotations, one must have $[H,S_i]=0$ for $i=x,y$ and $z$. In the present case, $[H,S_i]\neq 0$, $\forall$ $i=x,y,z$.

Question Does it mean that a system which is rotationally invariant in classical mechanics, can become rotationally non-invariant in quantum mechanics? Since it looks very unlikely, I would like to know what is the mistake in my argument.

Answer 1

Therefore, $H$ is invariant under rotations because a scalar is invariant under rotations.

No, it's not. It's rotationally invariant if and only if you rotate the magnetic field together with the system. If the field is externally imposed and you don't rotate it with the system, then rotating the system will change the dynamics.

The same is true in quantum mechanics. This is wrong:

Quantum mechanically, for $H$ to be invariant under rotations, one must have $[H,S_i]=0$ for $i=x,y$ and $z$. In the present case, $[H,S_i]\neq 0$, $\forall$ $i=x,y,z$.

Instead, for rotational invariance you need $H$ to commute with the generator of that rotation, which needs to be the total angular momentum $\mathbf J=\mathbf L+\mathbf S$ for the entire system, including the magnetic field; once you do that, you find that if $\mathbf B$ rotates with $\mathbf J$ via the standard vector rules, then $$ [\mathbf J, \mathbf S\cdot \mathbf B]=0. $$

(It's maybe worth doing this in detail. Thus, make $\mathbf B$ transform according to the standard $[L_i,B_j]=i\hbar \epsilon_{ijk}B_k$, which is what you get if $\mathbf B$ is a dynamical variable that's part of your system and subject to the rotations generated by $\mathbf L$, and then calculate \begin{align} [\mathbf J, \mathbf S\cdot \mathbf B] & = \hat{\mathbf {e}}_i \left[ J_i, S_j B_j\right ] \\& = \hat{\mathbf {e}}_i \left( \left[ J_i, S_j\right] B_j + S_j\left[ J_i, B_j\right] \right) \\& = \hat{\mathbf {e}}_i \left( \left[ S_i, S_j\right] B_j + S_j\left[ L_i, B_j\right] \right) \\& = \hat{\mathbf {e}}_i \left( i\hbar \epsilon_{ijk}S_k B_j + S_j\: i\hbar \epsilon_{ijk}B_k \right) \\& = \hat{\mathbf {e}}_i i\hbar \:\epsilon_{ijk}\left(S_k B_j + S_jB_k \right) \\ & = 0, \end{align} assuming Einstein summations, where the last term vanishes because you're contracting an asymmetric tensor, $\epsilon_{ijk}$, with the symmetric combination $S_k B_j + S_jB_k$. If you do anything other than this, then the cancellation will fail and you'll be left with a nonzero commutator.)

And, in the opposite direction, the operator-commutator shenanigans are in no way unique to QM - you just need to swap commutators for Poisson brackets and you get exactly the same structures within classical mechanics. Once you define a (classical) intrinsic angular momentum $\mathbf S$ that generates rotations for $\boldsymbol \mu$, then all the structures are preserved exactly.

Answer 2

Emilio's answer is great, but you might be interested in this (maybe less rigorous) point of view. For the record, this is inspired in this nice paper (particularly section 2.1.3) covering some finer points of Noether's theorem. You should go and read it because it probably explains the issue better than I can.

One way to put it is that the physics should be independent of the coordinate system always. This should come as no surprise: it would be very strange indeed if different coordinate systems saw different physical events occurring. This means that the Hamiltonian must be a scalar; in your example, it should be the dot product of two vectors. Think of what it would mean if the Hamiltonian were not a scalar: it would have to be something like $H = -\mu_z B$ in every coordinate system; that is, everyone sees a magnetic field in their $z$ direction. Clearly this doesn't correspond to the real world.

However, the mathematics (by which I mostly mean the formulas and equations) is not necessarily independent of coordinate system. If it is, we have a symmetry or invariance. To tell whether it is, we need to know what variables are dynamical, and which are external and held fixed. If the magnetic field is an external parameter, it might have some components: say, in some particular frame, $\mathbf{B} = (2, 5, -3)$ so that $H = -(2\mu_x + 5\mu_y -3\mu_z)$. In another coordinate system, this formula will look different; maybe we have the $z'$ axis aligned with the field, so that $H = -\sqrt{38}\mu_z'$.

If, however, $\mathbf{B}$ is a dynamical variable, we must write the Hamiltonian as $H = -(B_x \mu_x + B_y \mu_y + B_z \mu_z)$, since we don't have actual values for the $B_i$. Now comes the clincher: if we go to another coordinate system, $H$ looks exactly the same, i.e., the equations will look exactly the same. Only now we have rotational invariance and conservation of angular momentum