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In quantum mechanics, Ehrenfest's theorem states that $\langle p_x\rangle = m \frac{d}{dt}\langle x\rangle$. My question is, does there exist a similar relationship between $\langle L_z\rangle$, the expectation value of the z-component of the orbital angular momentum operator, and the time derivative of $\langle\theta\rangle$, the expectation value of the position operator $\hat{\theta}$ in spherical coordinates?

If not, is there any way to relate $\langle L_z\rangle$ to the time derivatives of expectation values of one or more operators? For instance if you knew what $\langle x\rangle$, $\langle y\rangle$, $\langle z\rangle$, $\langle p_x\rangle$, $\langle p_y\rangle$, and $\langle p_z\rangle$ are as a function of time, would that give you enough information to determine $\langle L_z\rangle$?

And are the answers to these questions affected at all by whether the particle has spin or not? By the way, this question was inspired by the comment section of this answer.

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  • See my answer to your preceding question. – dahemar Nov 06 '16 at 19:12
  • I would also add that you are interested in a fundamental relation, that is, something unrelated to a particular approximation or system, and therefore necessarily stemming from the full Hamiltonian of a system (for example, the full Hamiltonian of quantum electrodynamics for electron+photon systems), or from using some (to me unknown) trick. – glS Nov 06 '16 at 19:37
  • @glS Well, I'm not talking about quantum field theory here, just nonrelativistic quantum mechanics. – Keshav Srinivasan Nov 06 '16 at 19:42
  • @DavidHerreroMarti Well, your answer was about spin, and it was about a specific system. My question here is about orbital angular momentum and arbitrary systems. – Keshav Srinivasan Nov 06 '16 at 19:48
  • @KeshavSrinivasan Ok for the arbitrary systems part, but note that my answer was also valid for angular momentum. – dahemar Nov 06 '16 at 19:57
  • Ehrenfest's theorem very generally states a result for all observables, cf. the second boxed formula in the Wikipedia article. Without assuming a particular form of the Hamiltonian, you won't be able to say anything more. – ACuriousMind Nov 06 '16 at 22:04
  • @ACuriousMind Well, but the relation between the expectation value of the momentum operator and the time derivative of the expectation value of the position operator holds for all Hamiltonians. So it's not inconceivable that similar thing could happen if you apply Ehrenfest's theorem in this case. – Keshav Srinivasan Nov 06 '16 at 23:37
  • No, it does not hold for "all" Hamiltonian, it holds for Hamiltonians of the form $\frac{1}{2}p^2 + V(x)$. – ACuriousMind Nov 07 '16 at 00:10
  • @ACuriousMind Yeah, point taken. In any case, for all Hamiltonians of that same form (except with $V(\vec{r})$ instead of $V(x)$) is there a relationship which holds between $\langle L_z \rangle$ and the time derivative of $\langle \hat{\theta} \rangle$? – Keshav Srinivasan Nov 07 '16 at 00:49
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    @KeshavSrinivasan A bit of personal research effort would go a long way here: The first paper that turns up on googling "Ehrenfest theorem angular momentum" is http://www.physics.drexel.edu/~bob/Manuscripts/Ehrenfest.pdf. See Sec. 3.6, pg.5, it deals precisely with angular momentum and spin. – udrv Nov 07 '16 at 02:56
  • @udrv That paper discusses relating the time derivative of $\langle L_z \rangle$ to the expectation value of torque. My question is about relating $\langle L_z \rangle$ to the time derivative of $\langle \theta \rangle$. – Keshav Srinivasan Nov 07 '16 at 03:18
  • Well then, at the bottom of the same page is listed this AJP paper on "Ehrenfest theorem for angular displacement", http://scitation.aip.org/content/aapt/journal/ajp/51/10/10.1119/1.13383. Enjoy! – udrv Nov 07 '16 at 05:49
  • @udrv Thanks, that paper answers my question. If you want to post the result given at the end of that paper as an answer, I'm happy to accept it. – Keshav Srinivasan Nov 07 '16 at 09:05
  • Welcome, glad it worked out, but I only did a search this time :) If you'd like, please feel free to write your own answer, the topic is pretty interesting. – udrv Nov 07 '16 at 16:27
  • @udrv OK, I wrote an answer. – Keshav Srinivasan Nov 12 '16 at 16:18
  • Great. One suggestion: From those eom-s follow the corresponding eom-s for $e^{\pm i {\hat \theta}}$, and each of those involve just $e^{\pm i {\hat \theta}}$ respectively. Otoh, the derivative of the exponential can be related to the derivative of the exponent, see QMechanic's 2nd answer to this question: http://physics.stackexchange.com/questions/15742/derivative-of-the-product-of-operators-and-derivative-of-exponential. Perhaps some joint manipulation of those can beget something even better? – udrv Nov 12 '16 at 18:39

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Thanks to @udrv, I found the answer in this journal paper. Let's work in cylindrical coordinates $(r,\theta,z)$. Let $\cos\hat{\theta}$ and $\sin\hat{\theta}$ be defined by Taylor series, and let $\hat{L}_z = m\hat{r}^2 \hat{\omega}_z$. Then we can write the result in two forms:

$$\frac{d}{dt}\langle \cos\hat{\theta} \rangle = \langle -\frac{1}{2}(\hat{\omega}_z \sin\hat{\theta}+ \sin\hat{\theta}\hat{\omega}_z))\rangle$$

$$\frac{d}{dt}\langle \sin\hat{\theta} \rangle = \langle\frac{1}{2}(\hat{\omega}_z \cos\hat{\theta}+ \cos\hat{\theta}\hat{\omega}_z))\rangle$$

The reason why why we can't simply use the operator $\hat{\theta}$ is that $\hat{L}_z$ is only a Hermitian operator if its domain is restricted to periodic functions, and $\hat{\theta}$ maps periodic functions to non-periodic functions. So if we want to keep things within the domain of $L_z$ we need to work with an operator $f(\hat{\theta})$ where $f$ is a periodic function. And the simplest periodic functions which make $f(\hat{\theta})$ a Hermitian operator are sine and cosine. (It needs to be Hermitian if we want our Ehrenfest result to be between observable quantities.)

EDIT: The paper also provides a more general result for arbitrary periodic functions $f$ with period $2\pi$:

$$\frac{d}{dt}\langle f(\hat{\theta}) \rangle = \langle \frac{1}{2}(\hat{\omega}_z f'(\hat{\theta})+ f'(\hat{\theta})\hat{\omega}_z)\rangle$$

where again $f(\hat{\theta})$ and $f'(\hat{\theta})$ are defined via Taylor series.

Note that while this formula is true for all such functions $f$, in order it to be a result between observable quantities $f(\hat{\theta})$ needs to be a Hermitian operator. I posted a question here to find out what functions $f$ make $f(\hat{\theta})$ Hermitian.

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  • They are indeed the simplest such functions, but that means relatively little. If you can extend this to an arbitrary $\cos(n\theta)$, then you have a complete set and you're in business. Otherwise, this is nice but incomplete. – Emilio Pisanty Nov 12 '16 at 18:11
  • @EmilioPisanty If you're trying to get an equation for general functions $f$, the paper provides such a result. I just edited my question to include it. – Keshav Srinivasan Nov 13 '16 at 00:38