quantum mechanics - Why can't the angular momentum vector be parallel or anti-parallel to the applied magnetic field?

This is the excerpt from my book, Arthur Beiser's Concepts of Modern Physics:

An atom with a certain value of $\displaystyle{m_l}$ will assume the corresponding orientation of its angular momentum $\mathbf{L}$ relative to an external magnetic field if it finds itself in such a field. However, we note that $\mathbf{L}$ can never be aligned exactly parallel or anti-parallel to $\mathbf{B}$ because $L_z$ is always smaller than $\sqrt{l(l + 1)\hbar}$ of the total angular momentum.

Why can't $\mathbf{L}$ be parallel to the applied magnetic field? I am not getting the reason showed by the author. Can anyone help me conceive the reason stated above?

Answer

As the other answers have said, the true reason for this result is that the angular momentum of a quantum system cannot really be thought of as a classical vector with a magnitude and a direction, and well-defined values on all three components simultaneously. What Beiser is trying to do is to go against this direction as far as possible, and see how reasonable a semiclassical model will come out of that effort. The result is something that works surprisingly well, but it has a bunch of weird corners like the one you just found.

Within this semiclassical model, the reason that the angular momentum cannot point completely along a given axis (say $z$) is that the other two components will always have a nonzero uncertainty, which is forced upon them by the commutation relations. This uncertainty, in turn, makes the $x$ and $y$ components have nonzero expected magnitude, and this makes the semiclassical angular momentum vector point slightly off-axis.

To get a bit more technical, consider a state with nonzero $l$ and maximal $m=l$ along the $z$ axis. The square of the angular momentum is $$L_x^2+L_y^2+L_z^2=\mathbf L^2,$$ and there is a number of things you can say about it:

• For one, since $L_z$ is well defined, then $L_z^2=l^2\hbar^2$ is also well defined.


• Because $L_z$ is well defined, the expectation values of the other two components is known to be zero: $⟨L_x⟩=-i⟨[L_y,L_z]⟩/\hbar=-i⟨L_yL_z-L_zL_y⟩/\hbar=-i m⟨L_y-L_y⟩/\hbar=0$.


• This means, in turn, that the uncertainty in $L_x$ and $L_y$ is exactly the expected value of their squares: $\Delta L_x^2=⟨(L_x-⟨L_x⟩)^2⟩=⟨L_x^2⟩$.

This uncertainty, on the other hand, must be nonzero, and you can in fact give a good bound for it: $$ \Delta L_x \Delta L_y \geq \frac14 \left|⟨[L_x,L_y]⟩\right|^2 = \frac\hbar 4 \left|⟨L_z ⟩\right|^2 =\frac14\hbar^2 l. $$ This is further simplified by the fact that $\Delta L_x$ must equal $\Delta L_y $ by symmetry reasons (choose your favourite argument), so the two results above mean that the expected squares of the $x$ and $y$ components are bounded below by a nonzero amount: $$⟨L_x^2⟩=⟨L_y^2⟩\geq\frac14\hbar^2 l.$$

If you insist on trying to use a semiclassical model, then this means that you cannot ignore the magnitude of $L_x$ and $L_y$, even if their expected value is zero. The usual resolution is to say "well, they're somewhere in this circle" and to pretend that the problem doesn't exist. This yields the pretty 'cone' pictures that feature so prominently in Beiser...

^Source

but in the end you're still ignoring the fact that you cannot draw the angular momentum, because it can never* have all three of its components be well-defined.

^{* Unless $l=0$.}