This is quoted from Arthur Beiser's Concepts of Modern Physics:

Why is only one quantity of $\mathbf{L}$ quantized? The answer is related to the fact that $\mathbf{L}$ can never point in any specific direction but instead is somewhere on a cone in space such that $L_z$ is $m_l\hbar$. Were this not so, the uncertainty principle would be violated. If $\mathbf{L}$ were fixed in space, so that $L_x,L_y$ as well as $L_z$ had definite values, the electron would be confined to a definite plane, say in the $xy$ plane all the time. This can occur only if the electron's momentum component $p_z$ in the $z$ direction is infinitely uncertain, which of course is impossible if it is to be a part of hydrogen atom.

However, since in reality only one component of $L_z$ of $\mathbf{L}$ together with its magnitude $L$ have definite values & $|L| > |L_z|$, the electron is limited to have a single plane. Thus there is a built-in uncertainty in the electron's $z$ coordinate. The direction of $\mathbf{L}$ is not fixed, & so on the average, $L_x$ & $L_y$ are are $0$, although $l_z$ always has the specified value $m_l\hbar$.

  1. How does the $z$ component of momentum becomes infinitely uncertain if the electron lies in $xy$ plane? How does it lead to the impossibility of its presence in the hydrogen atom?

  2. Can anyone explain me the second paragraph of the above quote? What is it saying?

  3. Why is the average of $L_x, L_y$ is zero but $L_z$ is always $m_l\hbar$?

  • $\begingroup$ This book's title is lying to you, this is not "Concepts of modern physics", the excerpts in your lastest questions are one of the worst manglings of modern quantum mechanics I've read lately. $\endgroup$
    – ACuriousMind
    Commented Jun 16, 2015 at 19:52
  • 2
    $\begingroup$ Beiser's introductory works are moderately adequate for a first brush with quantum theory, but in general I would urge you to move away from it as soon as you can digest more advanced textbooks. (Try Griffiths, Landau & Lifshitz, Cohen-Tannoudji, Feynman, or others.) Beiser is full of passages which bravely attempt to fit QM back in the classical mould, giving the misleading impression that this can be done at all. In the particular case of the angular momentum 'cone', it is a semi-useful analogy but it's not worth the associated misconceptions. $\endgroup$ Commented Jun 16, 2015 at 20:41
  • $\begingroup$ The choice to measure $L_z$ is one of those "without loss of generality" things. That could be any axises, but having chosen it you lose quantization on the other axis. Until you chose again. $\endgroup$ Commented Jun 16, 2015 at 20:42
  • $\begingroup$ @dmckee There is no loss of quantization. A measurement of L_x will still return quantized results, I believe you mean that the measurement is no longer deterministic. $\endgroup$
    – Mebert
    Commented Jun 17, 2015 at 11:36
  • $\begingroup$ @Mebert Not quite. I meant that having measured in the $z$ direction if you write the state in terms of the eigenstates of $L_x$ you have a compound expression, and further that if there is an energy difference (say caused by a magnetic field $B\hat{x}$) that the composition will oscillate. The former is in essence what you said, but that last bit goes deeper than just noticing that you can no longer predict what you'll get from a subsequent measurement in $x$. $\endgroup$ Commented Jun 17, 2015 at 14:31

3 Answers 3

  1. If the electron is confined to the $x-y$-plane, it's $z$-position is fixed, i.e. certain, and hence the $z$-momentum infinitely uncertain by the uncertainty relation.

  2. That paragraph is trying to say that, if the magnitude of $\vec L$ is larger than $L_z$, then $\vec L$ is not fixed, and if angular momentum is not fixed, i.e. conserved, then the motion does not lie in a plane (perpendicular to it). This is a classical argument that is completely beside the point of quantum mechanics. The paragraph is completely non-sensical from a quantum mechanics view-point since it seems to treat $\vec L$ as some vector in space when only one of its three components even take definite values at the same time. (That there are other sources taking this approach does not make it less non-sensical, before anyone asks)

  3. For an eigenstate of the operator $L_z$, one may always ask what the expectation values of $L_x$ and $L_y$ on it are. It follows from the uncertainty principle that the expectation values of $L_x$ and $L_z$ on an eigenstate of $L_z$ must vanish, since we have the three relations $$ \sigma_x \sigma_y \ge \frac{\hbar}{2} \lvert \langle L_z \rangle \rvert $$ $$ \sigma_z \sigma_x \ge \frac{\hbar}{2}\lvert \langle L_y \rangle \rvert $$ $$ \sigma_y \sigma_z \ge \frac{\hbar}{2}\lvert \langle L_x \rangle \rvert $$ for $\sigma_i$ the standard deviation of $L_i$. Now, in an eigenstate of $L_z$, $\sigma_z = 0$, so the absolute values of the averages of $L_x$ and $L_y$ are bounded by $0$ (given that $\sigma_i$ is always finite), and hence must be zero themselves.

  • $\begingroup$ But if the expectation values are zero then can't we say that the components on the other two axis are just zero? $\endgroup$ Commented Oct 31, 2020 at 13:51

Any one of operators Lx, Ly, or Lz can be called quantized.

There exists a set of states which are eigenstates of Lz; the matrix Lz is diagonal but Ly and Lx are not. There exists a set of states which are eigenstates of Ly; the matrix Ly is diagonal but Lx and Lz are not. There exists a set of states which are eigenstates of Lx; the matrix Lx is diagonal but Lz and Ly are not.

However, there is no state (except the singlet $|l=0, lz=0>$) which is an eigenstate of more than one of these operators, because these operators do not commute with each other. Stated another way, you can't simultaneously diagonalize two matrices unless they commute with each other.

Quick proof: Suppose there was a state $|l; lx, ly> where (l>0)$ that was simultaneously an eigenstate of Lx and Ly. Then $$ [Lx,Ly] |l;lx,ly>=(LxLy-LyLx)|l;lx,ly>=(lxly-lylx)|l;lx,ly>=0|l;lx,ly>=0 $$ But [Lx,Ly]=ihLz (not 0) so the assumption of the existence of a simultaneous eigenstate of Lx and Ly was false.

  1. If the particle is known to be in the $xy$-plane, then $\Delta z = 0$, and so (by the uncertainty principle) $\Delta p_z = \infty$. Roughly speaking, this means that there's a good chance that the particle's momentum would be greater than "escape momentum", and thus that it could escape from the hydrogen atom. (I'm kind of dubious about this argument myself, but I think that's what's being said.)

  2. I think you have a typo here; my guess it is should read

    ... the electron is not limited to have a single plane.

    (Bolding mine.) What's being said here is that if all three components of $\bf{L}$ were definite, the particle would be confined to a plane; but since they're not, the particle is not confined to a single plane.

  3. What's being said here is that it is possible for a particle to be in a definite state of $L_z$; and if you were to prepare such a state and measure $L_z$, you would always get $m_l \hbar$. However, such a state is not a state of definite $L_x$ or $L_y$; and so you'll get a range of possible values if you try to measure these quantities. If you prepare this same $L_z$ state many times, and then measure its $L_x$ value, the distribution of $L_x$ values would average out to zero in the long run.

    Of course, you could equally well define a state of definite $L_x$, in which case $L_y$ and $L_z$ would not have definite values, and in which the expectation values of $L_y$ and $L_z$ would both be zero. Really, the best way to put it is that for any direction in space, there exist states for which (a) the total magnitude of the angular momentum has a definite value, and (b) the component of the angular momentum in the given direction has a definite value. But if we look at these sets of states for two different directions in space, they will not be the same. In linear-algebra-speak, they each form a basis for the space of states, but a basis vector from one set will generally not be in the other set.

(I also agree with the comment above that this passage is not a marvel of clarity.)

  • $\begingroup$ You are correct on the typo re: 2. $\endgroup$ Commented Jun 16, 2015 at 20:46

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