# Angular momentum in the 3D Schrodinger eqn with a central force

1. Apr 13, 2014

### hideelo

I guess this question can apply in all the generality of the 3D Schrodinger eqn. with a central force, the case I'm thinking of however is the the hydrogen atom.

When solving the equation, we derive the quantization of the angular momentum, which has me thinking that before we begin quantizing it, shouldnt we first ask if it has any angular momentum?

My impression (and it might be wrong) is that if we have an accelerating charge, it must radiate photons and lose energy etc. the reason why this does not happen in the atom is that the electron waves are stationary. Meaning, that the electron does not revolve around the nucleus, rather there are stationary probability waves centered around the nucleus. If that is in fact the case it shouldnt have any velocity since it's not moving

The problem with that is that if it doesnt have velocity then it shouldnt have momentum either (unless it can somehow have momentum without velocity). But if it doesnt then have momentum the wavelength from the de broglie eqn should be infinite.

On the other hand if it does have velocity (and corresponding momentum, wavelength, and angular momentum) shouldnt it also have centripetal acceleration? If that is in fact the case, it seems to me that it is not a stationary wave, rather its an accelerating particle, and should be radiating, losing energy...

How do I make sense of this?

2. Apr 13, 2014

### ZapperZ

Staff Emeritus
The problem here is that you are equating "angular momentum" in the QM case with the classical angular momentum.

First of all, note what happens in the QM case when the angular momentum is actually zero. It is in the s-orbital. Look at the symmetry or geometry of that orbital. Now, move up to l=1, 2, 3, etc. and examine the geometry of the orbital. Now, see if you can deduce why these are considered to have a non-zero angular momentum in the QM case. It has everything to do with the "symmetry" of the orbital, not about things "orbiting" around the nucleus.

The same way one shouldn't confuse the term "spin" in QM to imply something spinning, these orbital angular momentum should also not be confused with the classical angular momentum.

Zz.

3. Apr 13, 2014

### hideelo

So, are you saying that they can have QM angular momentum without ordinary (mv) momentum, because the prob density itself is moving?

4. Apr 13, 2014

### ZapperZ

Staff Emeritus
No, I'm saying don't pay attention to the label! "Spin" angular momentum does not mean it is spinning. "Orbital" angular momentum doesn't mean something is orbiting!

Zz.

5. Apr 13, 2014

### hideelo

so can we attribute momentum to something that is not moving?

6. Apr 13, 2014

### WannabeNewton

Hi hideelo! In my opinion this is an excellent question. Thankfully the explanation is actually quite simple. I assume your question is modulo spontaneous emission because electrons in orbitals can radiate via this QED mechanism. The charge density and current are given by $\rho = \psi^* \psi$ and $\vec{j} \propto \psi^* \vec{\nabla} \psi - \psi \vec{\nabla} \psi^*$ respectively, for a wave-function $\psi$. Electrons that are in stationary states of $H$ for the hydrogen atom thus have time-independent $\rho$ and $\vec{j}$ and as such do not radiate.

The electron being in a stationary state does not mean that it is not moving, I have no idea where you got that idea. The whole concept of the electron orbiting the nucleus or moving around the nucleus relies on the notion of a classical well-defined trajectory for the electron, which is entirely non-existent in the basic formalism of QM; if there was such a trajectory then Maxwell's equations would imply a radiation field attached to the electron but such a trajectory is nonsensical in QM as already stated. You have to abandon the (incorrect) connection you have made between stationary states of the electron and the electron's movement. This relates back to Zz's comment about your misconceptions regarding the angular momentum operator $J$ in QM.

7. Apr 13, 2014

### hideelo

Hi WannabeNewton! I'm afraid you give me and my question more credit than either deserve. I am a sophmore physics major and taking my first course in "modern physics". I wish my question was as sophisticated as the one you suspected I was asking, but it is in fact a much simpler one ;-)

I "know" (from an assertion in another textbook) that accelerating charges radiate, and I am now wondering if this is in fact generally true.

I am trying to figure out how the electrons in the hydrogen atom behave. My impression was, that in order to avoid having them radiate and lose energy we devised these stationary states for which $\Psi$ had no time dependence. It is possible that I was mistaken in equating this with not moving. But if it is moving, and not falling inward, then it must be moving tangentially, in which case it is accelerating which brings us back to the question if accelerating charges must really always radiate.

8. Apr 13, 2014

### The_Duck

Your idea of the motivation here is a little off. The way QM avoids the electron collapsing into the nucleus is that there is a ground state--a state of minimum energy, such that no state of lower energy exists. Radiation from this state is forbidden by energy conservation.

The "stationary" in "stationary state" does not mean the electron is motionless. It means that the state itself does not change in time. But the electron has nonzero expected kinetic energy in any stationary state (or indeed in any state at all).

"Accelerating charges radiate" is a classical idea that you shouldn't try to apply in QM unless you are in some sort of semiclassical limit. Rather, in QM an electron has some probability to emit a photon at any time, as long as it can satisfy all conservation laws (energy, momentum, etc.) when it does so.

9. Apr 13, 2014

### WannabeNewton

It is true that a charge in circular motion in classical electrodynamics will continuously emit radiation. But this depends on the classical notion of a particle traveling in a well-defined (and in this case circular) trajectory.

Just a bit of semantics here but we do not "devise" the stationary states with any intention of eliminating the classical radiation due to charges in circular orbits. The stationary states fall right out of the time-independent Schrodinger equation for the hydrogen atom and are simply states of the electron with well-defined energy; note that there is no stationary state of the electron with vanishing energy-there is a non-vanishing ground state energy.

Yes this misconception is what is making you confused. An electron being described by a stationary wavefunction does not mean that the electron is not moving. As I have noted above, the concepts of "moving" and "not moving" are intangible in QM for general states.

See above. You are (understandably) getting hung up on the classical notions of "moving tangentially" and "accelerating" based on the existence of well-defined spatial trajectories of particles. These do not exist in QM for general states. What we can talk about are averages of observables and the time derivatives of these observable averages. For an electron in a hydrogen atom stationary state, it is easy to see that $\frac{d}{dt}\langle P \rangle = 0$ where $P$ is the momentum operator and $\langle P \rangle$ is its average in the stationary state of the electron. Does this mean that the electron is not moving? No because as has already been explained, in QM the statement "not moving" does not have the meaning it does in classical mechanics, unless we are in a momentum eigenstate (which we are not) and go to the frame in which the associated momentum eigenvalue vanishes.