##r-##independent angular momentum in quantum mechanics

[hokhani]

TL;DR

Why in quantum mechanics the angular momentum is independent of the $r$?

Angular momentum as generator of rotation in defined by $L=r\times p$. However, none of the angular momentum wave functions depends on the $r$. They only depend on the angles.

 

[anuttarasammyak]

Inserting
$$\mathbf{p}=-i\hbar \nabla$$
in formula of L, you will prove that L has no r dependence.

 

[haushofer]

TL;DR: Why in quantum mechanics the angular momentum is independent of the $r$?

Angular momentum as generator of rotation in defined by $L=r\times p$. However, none of the angular momentum wave functions depends on the $r$. They only depend on the angles.

Intuitively: if r and v point in the same or opposite direction, L is zero. So only movement in the direction perpendicular to r contributes to L. Those are the angles.

By the way, this is also what makes the l=0 states of e.g. hydrogen clasically difficult to understand: it involves movement of a charge along a line through the nucleus. QMically, it means a spherically symmetric wavefunction. Bye bye classical orbits :P

 

[hokhani]

Intuitively: if r and v point in the same or opposite direction, L is zero. So only movement in the direction perpendicular to r contributes to L. Those are the angles.

By the way, this is also what makes the l=0 states of e.g. hydrogen clasically difficult to understand: it involves movement of a charge along a line through the nucleus. QMically, it means a spherically symmetric wavefunction. Bye bye classical orbits :P

On the contrary, I think the particular case $l=0$ is more understandable from classical view. It describes the particle at rest. However, the quantum particle is not localized at a particular point. We can find it everywhere with the same probability.

 

[PeterDonis]

I think the particular case $l=0$ is more understandable from classical view. It describes the particle at rest.

No, the particle can't be at rest because it has nonzero kinetic energy. There is no classical state that corresponds to the $l = 0$ quantum state.

The quantum particle is not localized at a particular point. We can find it everywhere with the same probability.

This is not correct; the $l = 0$ wave function does not have an equal amplitude everywhere.

 

[hokhani]

Inserting
$$\mathbf{p}=-i\hbar \nabla$$
in formula of L, you will prove that L has no r dependence.

So, for a constant $L_z=m\hbar$ we expect that getting far from the $z-$axis the linear momentum decreases.

 

[anuttarasammyak]

We are tempted to say such and such momentum at such and such place, but QM uncertainty relation or commutation relation
$$[x,p_x]=i\hbar$$
make it impossible.

 

[hokhani]

We are tempted to say such and such momentum at such and such place, but QM uncertainty relation or commutation relation
$$[x,p_x]=i\hbar$$
make it impossible.

Here, it is not $[x,p_x]$ but we have something for example like $[x,p_y]=0$.

 

[PeterDonis]

Here, it is not $[x,p_x]$ but we have something for example like $[x,p_y]=0$.

No, we don't. The fact that $L = r \times p$ does not mean that $r$ and $p$ are orthogonal.

 

[hokhani]

No, we don't. The fact that $L = r \times p$ does not mean that $r$ and $p$ are orthogonal.

For rotation around the z axis $p$ and $r$ are orthogonal, or at least non orthogonal components don't contribute.

 

[PeterDonis]

For rotation around the z axis

Who said we were just talking about rotation around the $z$ axis?

non orthogonal components don't contribute.

Non orthogonal components don't contribute to $L$, but that doesn't mean they don't exist. Go read the statement of mine that you quoted again, carefully. And then keep reading it again and again until it sinks in.

 

[hokhani]

Who said we were just talking about rotation around the $z$ axis?

Thanks again for your help! To build on what I mentioned in post #6, the case I had in mind was specifically a rotation around the $z$ axis.

 

[PeterDonis]

the case I had in mind was specifically a rotation around the $z$ axis.

First, I'm not sure I see the point of such a restriction, because it rules out practically every scenario where there's any significant physics in the angular momentum operators. For example, you can't even analyze a hydrogen atom this way.

Second, even if we leave that aside, the responses you have been given in this thread still apply to this restricted case.

 

[anuttarasammyak]

So, for a constant Lz=mℏ we expect that getting far from the z−axis the linear momentum decreases.

Pursuiting the calculation I showed
$$\hat{L}_z=-i \hbar \frac{\partial}{\partial \phi}$$
I am not sure whether it is consistent with your picture.

 

[hokhani]

Pursuiting the calculation I showed
$$\hat{L}_z=-i \hbar \frac{\partial}{\partial \phi}$$
I am not sure whether it is consistent with your picture.

I meant $\hat{L}_z e^{im\phi}=m\hbar e^{im\phi}$, in classical view, represents the rotation around the z-axis with angular momentum $m\hbar$. Since the particle is not localised, again by classical view, we expect the rotation occurs in all circles around the z-axis. Then, at greater $r$ we have less magnitude of $p$. I think writing momentum operator in polar coordinates as $\hat{p}=-i\hbar(\hat{p}_r \hat{r}+\frac{\partial}{r\partial \phi}\hat{\phi})$ also approves this: $\hat{p} e^{im\phi}=\frac{\hbar m}{r} e^{im\phi}\hat{\phi}$.

 

[PeterDonis]

in classical view

We are not talking about classical physics here. We are talking about quantum mechanics. Whatever behavior you might expect classically here is irrelevant to analyzing what QM says.

Since the particle is not localised, again by classical view

This is nonsense. You're mixing classical and quantum concepts.

 

[jedishrfu]

The poster's original post has been answered. It’s a good time to close this thread. Thank you all for participating in this forum.