# Contribution to angular momentum $L_z$ - due to rotation of probability fluid?

I'm doing a course on QM and this concept is entirely new to me:

"The eigenvalue $m\hbar$ of $L_z$ can be understood as the result from the rotational motion of probability fluid around the z-axis. "

Then I thought, suppose we have a volume element $dV$ at some point $r$ away, and I'm trying to find the contribution to $L_z$.

I think this is related to $J_{\phi} = \frac{m\hbar}{\mu r sin\theta}|\psi|^2$. Is the contribution

$$J_{\phi} \space dA = J_{\phi} \space r^2 sin\theta \space d\theta \space d\phi= \frac{m\hbar}{\mu} r \space d\theta \space d\phi$$.

Am I right? Also, how do I obtain $m\hbar$?

• Which reference is this from? – user7757 Jun 16 '14 at 17:16
• i59.tinypic.com/10d9jd5.png I got this question from a very old exam paper in quantum mechanics. – user44840 Jun 16 '14 at 17:20
• Your $J_{\phi}=\frac{\hbar}{\mu}|\psi|^2\frac{1}{r \sin{\theta}}\frac{\partial \alpha}{\partial\phi}$ And $L_z=-i \hbar \frac{\partial}{\partial \phi}(\sqrt{\rho} e^{i \alpha})$ Use value of J, to find value of the partial derivative in $L_z$ – user7757 Jun 16 '14 at 17:35
• I don't understand. – user44840 Jun 16 '14 at 17:50
• It's pretty simple, $J_{\phi}$ and $L_z$ are the same things upto a constant, they are just the operator $\frac{\partial}{\phi}$ acting on the azimuthal part of the wavefunction, you just need to evaluate the ratio of the two which comes out to be $m \hbar$ – user7757 Jun 16 '14 at 17:54

As stated in your reference, the probability current is given by $$\vec J[\varphi(\vec r)] = \frac{\hbar}{2\mu i}[\varphi^*\nabla\varphi - \varphi\nabla\varphi^*],$$ where $\mu$ is the mass of the particle. In the above, it is defined through its action on any arbitrary dummy wave function $\varphi$. Specifically for the case of a wave function which separates the radial from the angular part, $\varphi_{nlm}(\vec r)=R_l(r)Y_{lm}(\theta,\phi)$, which is the solutions to central potential problems, e.g. the hydrogen atom, one can evaluate $\vec J$ and finds $$\vec J[\varphi_{nlm}(\vec r)] = \frac{\hbar m}{\mu r\sin\theta}|\varphi_{nlm}(\vec r)|^2\vec e_\phi,$$ a current which only has an azimuthal component. This can be seen as $Y_{lm}(\theta,\phi)=P_{lm}(\cos\theta)e^{im\phi}$: The $i$ in the exponent has a different sign for $\varphi$ and $\varphi^*$, so the $\phi$ derivative is the only one which doesn't cancel.

Now, the exercise you were given aims at a semiclassical derivation of the result. The $L_z$ component of orbital angular momentum can be written as $L_z=r_\perp p_\phi$ where $r_\perp=r\sin\theta$ is the distance from the $z$-axis and $p_\phi$ is the azimutal momentum. This hold if the motion is such that the electron cloud is rotating around the $z$-axis. Momentum is mass times velocity. In general, a current can be written as $\vec J=\rho\vec v$ where $\rho$ is the density. Here, $\rho=|\varphi|^2$, the probability density. Thus, we may write the velocity as $\vec v=\vec J/|\varphi|^2$, so $p_\phi=\mu v_\phi=\mu J_\phi/|\varphi|^2$. Inserting this, we get $$L_z = r\sin\theta\mu J_\phi/|\varphi|^2\qquad\Rightarrow\qquad J_\phi = \frac{L_z}{\mu r\sin\theta}|\varphi|^2.$$

However, note that this only holds if we assume that the motion if around the $z$-axis. Above, we saw that this indeed was the case for the atomic (central potential) wave function $\varphi_{nlm}$. Comparing the results, we see that $L_z$ is replaced by $\hbar m$, so that must be the eigenvalue of the $L_z$ operator in the state $\varphi_{nlm}$.

Please note that all of this is quite hand-wavy. It is no doubt what the exercise wants you to do but please think of it as a motivation for why it might be a good idea to define the probability current the way it is defined, or why the angular momentum $z$-component is quantized in integer multiples of $\hbar$. Also, thinking of the electron cloud as a rotating fluid is not very helpful, IMO. It's yet another semiclassical model which captures some aspect of quantum mechanics which the particle-orbiting-a-nucleus-(just-like-a-solar-system!)-model can't (absence of trajectories), but it is bound to fail in describing other aspects.

Finally, let me address some of the confusion you expressed in the comments of the original post: I think claiming that $J_\phi$ and $L_z$ are the same operator is a bit overdoing it. In fact, a would call the probability current an operator b/c in quantum mechanics, "operator" usually means "linear operator" and the probability current is not linear.

• Ok, I understand your semi-classical argument, it's clear and easy to grasp. Is the contribution to the current $r\sin\theta\mu J_\phi/|\varphi|^2$? And $m\hbar$ is just the eigenvalue of the contribution? – user44840 Jun 18 '14 at 17:33
• $\hbar m$ is the eigenvalue of the $L_z$ operator corresponding to the eigenstate $\varphi_{nlm}$. You could also have gotten that by applying $L_z=-i\hbar(r_x\partial_y-r_y\partial_y)$ directly to the wave function. Here, the intuitive/classical idea of what the current should be leads to the same result. – Jonas Greitemann Jun 18 '14 at 18:03
• Yeah I get that, but I'm interested to know how to answer the question – user44840 Jun 18 '14 at 18:23
• I'm sorry, which question exactly are you referring to? – Jonas Greitemann Jun 18 '14 at 18:38
• Specifically the 7 mark bit on "What is the contribution to the current at the infinitesimal element $dV$" and "How do you obtain $m\hbar$?" – user44840 Jun 18 '14 at 18:52