A deep discrepancy between the classical and quantum models of ##L##

[hokhani]

TL;DR

Classical model confines the particle while quantum mechanics doesn't.

Take as an example the quantum state $\gamma_{3,3}$ of angular momentum where $\vec{L}$ makes the angle $\frac{\pi}{6}$ with $z-$ axis. This angular momentum state in classical model, describes a circular motion in a plane perpendicular to $\vec{L}$. So, in all the possible motions, the particle cannot have polar angle less than $\frac{\pi}{3}$. However, in quantum model this motion is described by $\gamma_{3,3}\propto \sin^3{\theta}$, i.e., the particle can be present in every polar angle except for zero. It's weird that we have the angular momentum operator as $\hat{L}=\hat{r} \times \hat{p}$ which is like classical model (except for replacing the operators with quantities) but the two models seem not compatible.

 

[Dale]

What’s the issue? Isn’t this the same with all observables? For a quantum system you get a probability distribution with different possible values and classically you get only the expectation value.

 

[hokhani]

What’s the issue? Isn’t this the same with all observables? For a quantum system you get a probability distribution with different possible values and classically you get only the expectation value.

I would like to know what causes in the wave picture that particle to penetrate in the classically forbidden regions?

 

[Demiurge]

I would like to know what causes in the wave picture that particle to penetrate in the classically forbidden regions?

A 'particle' is a misreprentation that causes widespread confusion to this day.

It's closer to a pixel on the screen. Much closer than a marble ball is. Or ever will be. The essence, behavior and findamental nature of the so called particles is very close to that of pixels on a screen. What happens when they are not 'lit up'? Where are they? Math can provide probalities

 

[Nugatory]

I would like to know what causes in the wave picture that particle to penetrate in the classically forbidden regions?

Nothing. When QM allows something that is classically forbidden, the explanation is that it isn't really forbidden and the classical picture is inaccurate.

 

[PeterDonis]

Take as an example the quantum state $\gamma_{3,3}$ of angular momentum where $\vec{L}$ makes the angle $\frac{\pi}{6}$ with $z-$ axis.

There is no such thing. Quantum mechanics is not classical mechanics. The fact that a state has definite $L^2$ and $L_z$, which is what I take your idiosyncratic notation of $\gamma_{3,3}$ to refer to, does not mean it has a definite angular momentum vector $\vec{L}$ in QM.

It's weird that we have the angular momentum operator as $\hat{L}=\hat{r} \times \hat{p}$ which is like classical model (except for replacing the operators with quantities)

In other words, like the classical model except for being completely different. I don't think you fully grasp just how much of a difference having operators (QM) instead of "quantities" (classical) makes.

the two models seem not compatible.

Of course not. As above, QM is not classical mechanics. It was developed in the first place in order to explain phenomena that classical mechanics cannot explain. You should not expect it to be "compatible" with classical mechanics.

 

[Dale]

I would like to know what causes in the wave picture that particle to penetrate in the classically forbidden regions?

Is there a classically forbidden region for angular momentum?

 

[PeterDonis]

Is there a classically forbidden region for angular momentum?

I think the OP is confused about models and is using the term "classically forbidden" not to refer to any actually valid physics model, but to the mishmash of QM and classical that he has in his head.

In a correct classical model, any angular momentum is of course allowed. But if we specify an angular momentum vector $\vec{L}$, then that is the angular momentum, which is part of the state of the object. Assuming that the other properties of the object are known (in this case its mass and moment of inertia are sufficient), there is just one motion that is consistent with that $\vec{L}$. There is not a "region" of motions that is "classically allowed" and another that is "classically forbidden".

In a correct quantum model, as noted, $\vec{L}$ is an operator, not a state. The state the OP has specified is one with particular angular momentum quantum numbers, which I think are intended to be $l = m = 3$. But this state is not a "motion" of the particle. It's a quantum state. So talking about "classically forbidden" motions makes no sense in this context.

 

[Dale]

But if we specify an angular momentum vector L→, then that is the angular momentum, which is part of the state of the object.

Right. Just because the state has some value doesn’t mean that other values or states are forbidden. It’s like saying this specific human is 180 cm tall therefore 182 cm is classically forbidden.

It is hard to justify calling that a “deep discrepancy”

 

[hokhani]

A far as I know, the root of difference between QM and classical mechanics is in the wave nature of quantum particles. So, to deal with quantities, we have to use operators which act on all the positions that a particle may be present. In this view, when a particle has specific $L^2$ and $L_z$ (for example the state $\gamma_{3,3}$), it is expected that the wave of particle to rotate on all the possible circles with constant $L^2$ and $L_z$. For the circular motions which satisfy this condition, the particle may be present everywhere except at some intervals of polar angles which I idiosyncratically termed as classically forbidden regions. Off course, for larger values of $m$ and particularly for $|l,l\rangle$ the wave function is proportionate with greater powers of $\sin \theta$ which is consistent with this analysis. But these wave functions, don't forbid entirely the presence of particle except at $\theta=0$. Also, for $|l,0\rangle$ there is no classical limitations for the presence of particle everywhere, even at $\theta=0$, which is completely compatible with quantum mechanical wave function $\gamma_{l 0}$ where is proportional to powers of $\cos \theta$. Finally, it seems this attitude isn't correct.

 

[Dale]

As I said above: For a quantum system you get a probability distribution with different possible values and classically you get only the expectation value.

I think that is the root of your concern.

 

[hokhani]

As I said above: For a quantum system you get a probability distribution with different possible values ...

You know that $L^2 \gamma_{l,m}(\theta, \phi)=l(l+1)\hbar^2 \gamma_{l,m}(\theta, \phi)$ and $L_z \gamma_{l,m}(\theta, \phi)=m \hbar\gamma_{l,m}(\theta, \phi)$ which means that at each $(\theta, \phi)$ the wave $\gamma_{l,m}(\theta, \phi)$ is so that we have $L^2=l(l+1) \hbar^2$ and $L_z=m\hbar$. Therefore, although we have different possible values for wave function at various points, the values of $L^2$ and $L_z$ are respectively $l(l+1) \hbar^2$ and $m\hbar$ everywhere. If we classically look for such points where these quantities are constant there, these points are not located at polar angles less than $\frac{\pi}{3}$ (in the case that $l=3$ and $m=3$).

 

[PeterDonis]

A far as I know, the root of difference between QM and classical mechanics is in the wave nature of quantum particles.

No, that's not correct. The root of the difference between QM and classical mechanics is that QM only predicts probabilities. But there is nothing that restricts those probabilities to be the probabilities of "particles" being detected at some particular place and time.

to deal with quantities, we have to use operators

Yes.

which act on all the positions that a particle may be present.

No. I don't know where you are getting this from. Operators act on quantum states, not positions.

when a particle has specific $L^2$ and $L_z$ (for example the state $\gamma_{3,3}$), it is expected that the wave of particle to rotate on all the possible circles with constant $L^2$ and $L_z$.

No, this is not correct. You are reasoning from wrong premises, so it's no wonder you are getting nonsensical conclusions.

I think you need to re-learn basic QM from a textbook. Ballentine, Chapter 7, has a good treatment of how angular momentum is modeled in QM.

 

[PeterDonis]

Therefore, although we have different possible values for wave function at various points, the values of $L^2$ and $L_z$ are respectively $l(l+1) \hbar^2$ and $m\hbar$ everywhere. If we classically look for such points where these quantities are constant there, these points are not located at polar angles less than $\frac{\pi}{3}$ (in the case that $l=3$ and $m=3$).

This is all word salad that has nothing to do with how QM is actually done.

 

[hokhani]

let's look at the problem quite quantum mechanically. The wave function $\gamma_{3,3}$ describes a free particle and so we expect the probability of the presence of particle to be identical everywhere while this wave function depends on the polar angle $\theta$!

 

[renormalize]

The wave function $\gamma_{3,3}$ describes a free particle and so we expect the probability of the presence of particle to be identical everywhere while this wave function depends on the polar angle $\theta$!

Can you please explicitly display both your free-particle wave function $\gamma_{3,3}$ and the probability denisity $\gamma_{3,3}^{*}\gamma_{3,3}$ so we may see how they depend on $\theta\,$?

 

[Dale]

The wave function γ3,3 describes a free particle

What? That doesn’t seem right.

so we expect the probability of the presence of particle to be identical everywhere

That is definitely not right.

 

[PeterDonis]

The wave function $\gamma_{3,3}$ describes a free particle

Not the one you wrote down, no. That describes a particle in a bound state.

The angular momentum operators of course can be applied to free particles, but that doesn't mean that any eigenstate of those operators that you write down is a free particle state.

 

[hokhani]

Not the one you wrote down, no. That describes a particle in a bound state.

The angular momentum operators of course can be applied to free particles, but that doesn't mean that any eigenstate of those operators that you write down is a free particle state.

For the free particle the three operators Hamiltonian, $L^2$ and $L_z$ commute togethers. So, it seems that $\gamma_{3,3}$ is also an eigenfunction of the free particle. Could you please explain more?

 

[renormalize]

For the free particle the three operators Hamiltonian, $L^2$ and $L_z$ commute togethers. So, it seems that $\gamma_{3,3}$ is also an eigenfunction of the free particle. Could you please explain more?

It is not true that $\psi$ is a free-particle wavefunction just because $\hat{H},\hat{L}^2,\hat{L}_z$ commute when acting on $\psi$. An obvious counterexample is the hydrogen atom, wherein the electron is bound to a proton:

(https://people.chem.ucsb.edu/metiu/horia/OldFiles/QM2015/Ch11QM.pdf)

 

[Matterwave]

For the free particle the three operators Hamiltonian, $L^2$ and $L_z$ commute togethers. So, it seems that $\gamma_{3,3}$ is also an eigenfunction of the free particle. Could you please explain more?

It might be helpful if you actually wrote out what $\gamma_{3,3}$ is.

 

[bhobba]

I would like to know what causes in the wave picture that particle to penetrate in the classically forbidden regions?

You need to go to Quantum Field Theory for an intuitive understanding of that. Everything is an excitation of quantum fields that permeate the universe. The excited fields of one system can occupy the same space as those of a different excitation of the same or different fields.

Thanks
Bill

 

[PeterDonis]

I would like to know what causes in the wave picture that particle to penetrate in the classically forbidden regions?

You need to go to Quantum Field Theory for an intuitive understanding of that.

I'm not sure that's actually necessary. The key point is, as has already been pointed out, viewing the quantum system as a "particle", as though it were a tiny billiard ball, is simply wrong. Even in non-relativistic QM, the system is described by a wave function. Even if we view the wave function as giving us probability amplitudes for "particle detections" at particular points in space, that still does not mean you can think of the system, even heuristically, as a particle with a definite trajectory in between measurements that just happens to be unknown. We already know that kind of local hidden-variable model can't be right, because QM can violate the Bell inequalities.

 

[bhobba]

A far as I know, the root of difference between QM and classical mechanics is in the wave nature of quantum particles.

Actually, as far as we can tell today, it is there is no separation between particles and fields. Everything is an excitation of quantum fields that permeate the universe, and they follow unintuitive, mathematically sophisticated rules.

Ordinary non-relativistic QM, from a mathematical modelling perspective, is asking what would happen if, instead of objects having definite values or properties, they were modelled by diagonal matrices, with their possible values on the diagonal and a powerful theorem called Gleason's Theorem. That is not how it is presented in introductory texts, but once you understand linear algebra (eigenvalues, eigenvectors, etc.) and have enough background for a more advanced treatment like Ballentine (mentioned previously), the details become clearer.

Thanks
Bill

 

[hokhani]

It is not true that $\psi$ is a free-particle wavefunction just because $\hat{H},\hat{L}^2,\hat{L}_z$ commute when acting on $\psi$. An obvious counterexample is the hydrogen atom, wherein the electron is bound to a proton:
View attachment 374355
(https://people.chem.ucsb.edu/metiu/horia/OldFiles/QM2015/Ch11QM.pdf)

Right, I didn't notice the degeneracy. Thanks.

 

[hokhani]

Not the one you wrote down, no. That describes a particle in a bound state.

The angular momentum operators of course can be applied to free particles, but that doesn't mean that any eigenstate of those operators that you write down is a free particle state.

Thanks, I think since $[H,L^2]=0$ we have $H |lm\rangle = \sum_m {\alpha_m |l,m\rangle}$ and since $[H,L_z]=0$ we have $H |lm\rangle = \sum_l {\beta_l |l,m\rangle}$. Isn't it?

 

[PeterDonis]

Thanks, I think since $[H,L^2]=0$ we have $H |lm\rangle = \sum_m {\alpha_m |l,m\rangle}$ and since $[H,L_z]=0$ we have $H |lm\rangle = \sum_l {\beta_l |l,m\rangle}$. Isn't it?

I have no idea until you tell me what states you mean by $\ket{l, m}$. Or, for that matter, what $H$ is, since $H$ is different for a free particle vs. a particle in a potential.

Since $H$ commutes with $L^2$ and $L_z$, and $L^2$ and $L_z$ commute with each other, we can say that there is some complete set of states that are eigenstates of all three operators. But exactly which states those are depends on the specific scenario. If you already know you have a free particle, i.e., that $H = p^2 / 2m$, then you have one set of states that are common eigenstates of $H$, $L^2$, and $L_z$. But if $H$ is something else, you will have a different set of states that are common eigenstates of those three operators.

 

[hokhani]

I have no idea until you tell me what states you mean by $\ket{l, m}$. Or, for that matter, what $H$ is, since $H$ is different for a free particle vs. a particle in a potential.

$|l,m\rangle$ are the common eigenstates of $L^2$ and $L_z$. Also, $H$ is the free particle Hamiltonian (or any Hamiltonian with spherical symmetry that certainly commutes with both $L^2$ and $L_z$).

 

[PeterDonis]

$|l,m\rangle$ are the common eigenstates of $L^2$ and $L_z$.

But without knowing $H$, we don't know which states these are.

Also, $H$ is the free particle Hamiltonian (or any Hamiltonian with spherical symmetry that certainly commutes with both $L^2$ and $L_z$).

These are physically different cases. Which one do you want to talk about? If we don't know which specific Hamiltonian you mean, we don't know what states we're talking about.

 

[hokhani]

But without knowing $H$, we don't know which states these are.


These are physically different cases. Which one do you want to talk about? If we don't know which specific Hamiltonian you mean, we don't know what states we're talking about.

Since $[H,L^2]=0$ we have $L^2H|l,m\rangle=HL^2|l,m\rangle=l(l+1)H|l,m\rangle$. So, $H|l,m\rangle$ is an eigenstate of $L^2$ with eigenvalue $l(l+1)$ and so it is linear combination of $|l,m\rangle$ as $H|l,m\rangle=\sum_m \alpha_m |l,m\rangle$. Similarly, since $[H,L_z]=0$ we would have $H|l,m\rangle=\sum_l \beta_l |l,m\rangle$.