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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?

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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...

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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 $l=0$.

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  • $\begingroup$ If sir, as you wrote, $L_z$ is well-defined, how does it manipulate the average value of the other components to $0$? I am not getting this fact:( $\endgroup$ – user36790 Jun 16 '15 at 19:50
  • $\begingroup$ The three components of angular momentum are related via a commutation relation, $$[L_x,L_z]=i\hbar L_z$$ (and its cyclic permutations), which has consequences as profound as the more usual one between position and momentum, $[x,p]=i\hbar$. From there to $⟨L_x⟩=0$ is a simple calculation, but if you don't see the connection then I'd recommend further reading on commutation relations and their implications. It is hard to overstress how important this relation is, and how completely away it is from classical mechanics. $\endgroup$ – Emilio Pisanty Jun 16 '15 at 20:33
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Honestly, it's because the ideas of "parallel" and "antiparallel" and "perpendicular" don't quite apply to quantum angular momentum.

The angular momentum of a quantum particle is not really a vector in the sense you're probably used to thinking about it. What I mean is, it's not really something that points in a direction and has a magnitude. It's more complicated than that. But it turns out that a convenient way to describe the angular momentum of a quantum state is with three operators, which happen to transform under physical rotations just like components of a vector would. So we call these operators $\hat{L}_x$, $\hat{L}_y$, and $\hat{L}_z$, to remind ourselves that in many ways they act like components of a vector.

Now, one of the properties of a classical vector (which should be fairly obvious) is that, if it points along a certain axis, then its component along that axis is equal to its magnitude. But in the case of quantum angular momentum, we can't meaningfully talk about whether it "points along a certain axis" because quantum angular momentum doesn't really have a direction, in the classical sense.

However, we can talk about its component along a certain axis, namely $L_z$, and its magnitude, $\sqrt{L^2}$. Both those are measurable quantities, and in fact simultaneously measurable. So you can measure both of them at once, and if you find that the component of quantum angular momentum along a certain axis is equal to the magnitude of the quantum angular momentum, you might say, as a suggestive analogy, that the quantum angular momentum "points" along that axis.

What your book is showing you is that this can never happen for $l \neq 0$. You will never find the eigenvalue of $L_z$ to be equal to the eigenvalue of $\sqrt{L^2}$, because of the $+1$ in $\sqrt{l(l + 1)}\hbar$. And therefore, if you take "component along an axis equals the magnitude" as your definition of "(anti-)parallel to that axis", that means you will never find the quantum angular momentum to be (anti-)parallel to the axis you measure it on.

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  • $\begingroup$ I know sir, it's a pseudo-vector! I am not understanding how $L_z = m_l \hbar$. Beiser hasn't shown any deduction or so. Can you tell, sir? $\endgroup$ – user36790 Jun 16 '15 at 18:48
  • $\begingroup$ @user36790 Note that $\mathbf L$ is a pseudo-vector in the technical sense, but what David is saying is rather different. $\endgroup$ – Emilio Pisanty Jun 16 '15 at 19:32
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    $\begingroup$ @user36790 as Emilio mentioned, the point of my answer is not to say that $\mathbf{L}$ is a pseudovector. I'm saying that it is a "quantum quantity" which is neither a classical vector nor a classical pseudovector, nor any other kind of object you're used to thinking about from classical mechanics. $\endgroup$ – David Z Jun 17 '15 at 4:42
  • $\begingroup$ The most correct way of expressing it, I think, is that since the angular momentum operator is the generator of rotations, it is an operator-valued two-form. In three dimensions, two-forms and one-forms are isomorphic, which is why two-forms are most times called pseudovectors. $\endgroup$ – Robin Ekman Jun 17 '15 at 13:33
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I think the statement is nonsensical. In quantum mechanics the vector $\mathbf L$ is not a world-vector. Its components are not numbers, they are operators. However, the expectation value of $\mathbf L$ in a given state, $\langle \mathbf L \rangle$ is a world-vector and it can be aligned with any given axis. (The magnetic field is a red herring.)

It doesn't make sense to talk about the direction of $\mathbf L$ since its components aren't numbers. What the statement really means to say probably is that $\langle \mathbf L\rangle ^2 \neq \langle \mathbf L^2 \rangle$.

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