Angular momentum of an atom within a rigid body in motion

[james fairclear]

TL;DR Summary

Considering an atom within a rigid body, does the angular momentum of an electron within the atom vary when the body is put in motion?

Considering an atom within a rigid body, does the angular momentum of an electron within the atom vary when the body is put in motion?

My intuition is that, whether considered in a classical sense or quantum sense, the speed of a given electron in its motion within an atom will be constant and invariant (like c) regardless of whether the rigid body is stationary on Earth or moving away from Earth at 0.99c.

 

[PeroK]

Summary:: Considering an atom within a rigid body, does the angular momentum of an electron within the atom vary when the body is put in motion?

Considering an atom within a rigid body, does the angular momentum of an electron within the atom vary when the body is put in motion?

My intuition is that, whether considered in a classical sense or quantum sense, the speed of a given electron in its motion within an atom will be constant and invariant (like c) regardless of whether the rigid body is stationary on Earth or moving away from Earth at 0.99c.

The state of the electron in an atom is usually described relative to the nucleus. If you consider a system where the nucleus is not at rest, then the state of the electron must include an element relating to the state of motion of the nucleus.

 

[PeterDonis]

the speed of a given electron in its motion within an atom

Is meaningless since the electron in an atom is not in a state that has a definite speed. The electron has orbital angular momentum but you cannot think of it in a classical sense as being due to a well-defined orbital speed about the nucleus.

 

[james fairclear]

Thank you for your response.

Does the quantity of the angular momentum of an electron within the atom vary when the body is put in motion?

 

[PeterDonis]

Does the quantity of the angular momentum of an electron within the atom vary when the body is put in motion?

It depends on what you mean by "the quantity of the angular momentum of an electron within the atom".

In the usual formulation of non-relativistic QM, the operator referred to by "angular momentum" (or more precisely "orbital angular momentum") tells you the electron's orbital angular momentum relative to the nucleus, so it doesn't depend on the state of motion of the nucleus.

I suppose one could construct a different operator that would include the motion of the nucleus and refer the orbital angular momentum to some chosen point in space relative to which the nucleus is moving, but I've never seen that done.

 

[PeterDonis]

If you consider a system where the nucleus is not at rest, then the state of the electron must include an element relating to the state of motion of the nucleus.

Can you give a reference where this is actually done? I've never seen this done with respect to individual electrons in an atom, only with respect to entire atoms.

 

[vanhees71]

In non-relativistic physics it's easy to take into account the motion of the nucleus. For the hydrogen atom you can even solve the energy eigenvalue problem analytically. You just work in center-of-math and relative coordinates as in classical mechanics. This can be found in most introductory QM-textbook.

 

[PeterDonis]

You just work in center-of-mass and relative coordinates as in classical mechanics.

Yes, and this is equivalent to evaluating the orbital angular momentum of an electron in an atom relative to the nucleus of the atom, regardless of the state of motion of the nucleus.

 

[vanhees71]

Nearly! The cm-motion is of coarse that of a free particle. The relative motion is equivalent to the motion of a particle of the reduced mass in the Coulomb interaction potential. Rewriting the solution in terms of the original p- and e-coordinates, you'll find that they are in entangled states. Tomorrow I'll look for a nice AJP paper about this.

 

[PeroK]

Can you give a reference where this is actually done? I've never seen this done with respect to individual electrons in an atom, only with respect to entire atoms.

I was referring to what @vanhees71 posted above. That the electron wave function must depend on the motion of the atom.

That said, I may have misunderstood what the OP meant by "invariant".

 

[PeterDonis]

the electron wave function must depend on the motion of the atom.

I'm not sure that's what @vanhees71 was saying. Hopefully he'll be able to post a reference to a paper that will have the actual math.

 

[vanhees71]

Let's take the hydrogen atom in its most simple form, i.e., non-relativistic, no spin. Its Hamiltonian is that of a proton and an electron interacting via the Coulomb potential (in the following I leave out the hats for operators; all observables are understood to be represented by self-adjoint operators, and I use Heaviside-Lorentz units for the Coulomb potential):
$$H=\frac{1}{2m_{\text{p}}} \vec{p}_1^2 + \frac{1}{2m_{\text{e}}} \vec{p}_2^2 -\frac{e^2}{4 \pi |\vec{x}_1-\vec{x}_2|}.$$
We are looking for the energy eigenvalues and eigenfunctions. For that it's always good to use all the symmetries to find a complete set of independent compatible observables, i.e., the symmetries of the problem.

Here we can borrow from the classical analogue of the problem. We have a closed system of two particles interacting via a central interaction potential. Thus the full Galileo group is a symmetry. From this it's clear that it is convenient to use the center-of-mass coordinates $\vec{R}$ and relative coordinates $\vec{r}$
$$\vec{R}=\frac{1}{M} (m_{\text{p}} \vec{x}_1 + m_{\text{e}} \vec{x}_2), \quad \vec{r}=\vec{r}_2-\vec{r}_1$$
with $M=m_1+m_2$.

Now we need the canonical momenta to these new position vectors. That's determined by finding $\vec{P}$ and $\vec{p}$ such that
$$[R_i,P_j]=\mathrm{i} \hbar \delta_{ij}, \quad [r_i,p_j]=\mathrm{i} \hbar \delta_{ij}, \quad [r_i,P_j]=[R_i,p_j]=0.$$
It's easy to see that from these commutation relations one has uniquely
$$\vec{P}=\vec{p}_1+\vec{p}_2, \quad \vec{p}=\frac{1}{M}(m_1 \vec{p}_2-m_2 \vec{p}_1).$$
In the new variables the Hamiltonian reads
$$H=\frac{1}{2M} \vec{P}^2+ \frac{1}{2 \mu} \vec{p}^2 -\frac{e^2}{4 \pi |\vec{r}|}.$$
Here $\mu=m_1 m_2/M$ is the reduced mass. Obviously a complete compatible set of observables is
$$\vec{P}, \quad H_{\text{rel}}, \quad \vec{L}_{\text{rel}}^2, \quad L_{\text{rel}3}$$
with
$$H_{\text{rel}}=\frac{1}{2 \mu} \vec{p}^2 -\frac{e^2}{4 \pi |\vec{r}|}, \quad \vec{L}_{\text{rel}}=\vec{r} \times \vec{p}.$$
It's clear that the diagonalization of $H_{\text{rel}}$ works as with the approximation, where the proton is just approximated as "infinitely heavy", i.e., sitting at rest in the origin and just providing an external Coulomb potential for the electron. Here $H_{\text{rel}}$ describes the motion of a quasi-particle of mass $\mu$ in such an external Coulomb potential.

The rest is nicely discussed in the following AJP article by Tommasini et al

https://arxiv.org/abs/quant-ph/9709052

It nicely discusses the energy eigenstates as entangled states between electron and proton, while it's a product state of the free center-mass motion and the relative motion.

 

[james fairclear]

The state of the electron in an atom is usually described relative to the nucleus. If you consider a system where the nucleus is not at rest, then the state of the electron must include an element relating to the state of motion of the nucleus.

On the basis that the atom (in my understanding) is bound within a rigid body in uniform translatory motion why would you consider the motion of the electron separately to the motion of the nucleus ?

Would you expect the value for angular momentum of the electron to change?

 

[PeroK]

On the basis that the atom (in my understanding) is bound within a rigid body in uniform translatory motion why would you consider the motion of the electron separately to the motion of the nucleus ?

Would you expect the value for angular momentum of the electron to change?

A rigid body in uniform translatory motion is a fundamentally classical concept. Whereas, electrons bound to an atom requires a QM model.

Ultimately, yes, as basic QM is non-relativistic. If you consider an atom moving at relativistic speed then you would need to update your atomic model. For a free electron, for example, you must replace the non-relativistic Schrodinger equation with the Dirac equation.

 

[vanhees71]

On the basis that the atom (in my understanding) is bound within a rigid body in uniform translatory motion why would you consider the motion of the electron separately to the motion of the nucleus ?

Would you expect the value for angular momentum of the electron to change?

If you want to understand the atom as a bound state of an atomic nucleus and electrons, the model of a rigid body is not sufficient but you have to describe the nucleus and the electrons as a dynamical system of interacting charged particles. The atom as a whole is described by its center of mass, which indeed moves uniformly (as long as you can neglect the interactions of the atom with any other particles around it). You can then choose the rest frame of the center of mass and consider the much more complicated dynamical quantum mechanical problem. Here the energy eigenvalue problem is important, leading to a description of the atom in terms of discrete-energy levels (bound states), describing the static configurations of the electrons around the nucleus. One observable conclusion is the prediction of the characteristic spectrum of the light emitted by the atom.

 

[james fairclear]

Let's take the hydrogen atom in its most simple form, i.e., non-relativistic, no spin. Its Hamiltonian is that of a proton and an electron interacting via the Coulomb potential (in the following I leave out the hats for operators; all observables are understood to be represented by self-adjoint operators, and I use Heaviside-Lorentz units for the Coulomb potential):
$$H=\frac{1}{2m_{\text{p}}} \vec{p}_1^2 + \frac{1}{2m_{\text{e}}} \vec{p}_2^2 -\frac{e^2}{4 \pi |\vec{x}_1-\vec{x}_2|}.$$
We are looking for the energy eigenvalues and eigenfunctions. For that it's always good to use all the symmetries to find a complete set of independent compatible observables, i.e., the symmetries of the problem.

Here we can borrow from the classical analogue of the problem. We have a closed system of two particles interacting via a central interaction potential. Thus the full Galileo group is a symmetry. From this it's clear that it is convenient to use the center-of-mass coordinates $\vec{R}$ and relative coordinates $\vec{r}$
$$\vec{R}=\frac{1}{M} (m_{\text{p}} \vec{x}_1 + m_{\text{e}} \vec{x}_2), \quad \vec{r}=\vec{r}_2-\vec{r}_1$$
with $M=m_1+m_2$.

Now we need the canonical momenta to these new position vectors. That's determined by finding $\vec{P}$ and $\vec{p}$ such that
$$[R_i,P_j]=\mathrm{i} \hbar \delta_{ij}, \quad [r_i,p_j]=\mathrm{i} \hbar \delta_{ij}, \quad [r_i,P_j]=[R_i,p_j]=0.$$
It's easy to see that from these commutation relations one has uniquely
$$\vec{P}=\vec{p}_1+\vec{p}_2, \quad \vec{p}=\frac{1}{M}(m_1 \vec{p}_2-m_2 \vec{p}_1).$$
In the new variables the Hamiltonian reads
$$H=\frac{1}{2M} \vec{P}^2+ \frac{1}{2 \mu} \vec{p}^2 -\frac{e^2}{4 \pi |\vec{r}|}.$$
Here $\mu=m_1 m_2/M$ is the reduced mass. Obviously a complete compatible set of observables is
$$\vec{P}, \quad H_{\text{rel}}, \quad \vec{L}_{\text{rel}}^2, \quad L_{\text{rel}3}$$
with
$$H_{\text{rel}}=\frac{1}{2 \mu} \vec{p}^2 -\frac{e^2}{4 \pi |\vec{r}|}, \quad \vec{L}_{\text{rel}}=\vec{r} \times \vec{p}.$$
It's clear that the diagonalization of $H_{\text{rel}}$ works as with the approximation, where the proton is just approximated as "infinitely heavy", i.e., sitting at rest in the origin and just providing an external Coulomb potential for the electron. Here $H_{\text{rel}}$ describes the motion of a quasi-particle of mass $\mu$ in such an external Coulomb potential.

The rest is nicely discussed in the following AJP article by Tommasini et al

https://arxiv.org/abs/quant-ph/9709052

It nicely discusses the energy eigenstates as entangled states between electron and proton, while it's a product state of the free center-mass motion and the relative motion.

The article by Tommasini seems to only deal with a stationary system. I am interested in whether the angular momentum of an electron within an individual atom changes when that atom is in motion.

 

[james fairclear]

A rigid body in uniform translatory motion is a fundamentally classical concept. Whereas, electrons bound to an atom requires a QM model.

Ultimately, yes, as basic QM is non-relativistic. If you consider an atom moving at relativistic speed then you would need to update your atomic model. For a free electron, for example, you must replace the non-relativistic Schrodinger equation with the Dirac equation.

What would be the QM equivalent descriptive term for "A rigid body in uniform translatory motion"?

So in consideration of a rigid body moving at relativistic speed would you expect the angular momentum of an electron within an atom of that body to vary?

 

[vanhees71]

An example for a system which are described approximately as a rigid body in quantum mechanics are the rotational states molecules, but this holds of course only as long as the involved excitation energies are low enough not to excite vibrational modes. For a nice treatment of the quantized spinning top, see Landau and Lifshitz vol. 3.

 

[PeroK]

What would be the QM equivalent descriptive term for "A rigid body in uniform translatory motion"?

Rigid body is problematic (see below). The nearest to uniform translatory motion for a particle would be a minimum uncertainty Gaussian wave-packet, probably.

So in consideration of a rigid body moving at relativistic speed would you expect the angular momentum of an electron within an atom of that body to vary?

A rigid body is not compatible with relativity either. There's a thread about this at the moment.

https://www.physicsforums.com/threads/the-paradox-of-relativity-length-contraction.1010138/

What you have now is a mixture of concepts from three different physical theories: Classical Mechanics, Special Relativity and (non-relativistic) QM. That makes your question fairly meaningless.

 

[james fairclear]

If you want to understand the atom as a bound state of an atomic nucleus and electrons, the model of a rigid body is not sufficient but you have to describe the nucleus and the electrons as a dynamical system of interacting charged particles. The atom as a whole is described by its center of mass, which indeed moves uniformly (as long as you can neglect the interactions of the atom with any other particles around it). You can then choose the rest frame of the center of mass and consider the much more complicated dynamical quantum mechanical problem. Here the energy eigenvalue problem is important, leading to a description of the atom in terms of discrete-energy levels (bound states), describing the static configurations of the electrons around the nucleus. One observable conclusion is the prediction of the characteristic spectrum of the light emitted by the atom.

A rigid body in uniform translatory motion is a fundamentally classical concept. Whereas, electrons bound to an atom requires a QM model.

Ultimately, yes, as basic QM is non-relativistic. If you consider an atom moving at relativistic speed then you would need to update your atomic model. For a free electron, for example, you must replace the non-relativistic Schrodinger equation with the Dirac equation.

So considering a rigid body in motion at 0.9c can we envisage in a classical sense the electrons of a constituent atom moving at a speed close to c but taking the usual (stationary body) time period per orbit of the nucleus due to the extra distance they are traveling through space?

 

[PeterDonis]

can we envisage in a classical sense the electrons of a constituent atom

No. Electrons in atoms are not classical. They do not have classical trajectories.

 

[james fairclear]

No. Electrons in atoms are not classical. They do not have classical trajectories.

If we consider a rigid body firstly at rest. Within a constituent atom of the body we can consider the probability of an electron cloud and its effect being in a given location within the vicinity of the nucleus.

If we then consider the body in motion at 0.9c. Since a constituent atom of the body will also have a component of motion of 0.9c in the direction of motion of the body and will therefore have further to travel through space than when the body was at rest can we expect a lesser probability of an electron cloud and its effect being in a given location within the vicinity of the nucleus or will it be the same?

 

[james fairclear]

A rigid body in uniform translatory motion is a fundamentally classical concept. Whereas, electrons bound to an atom requires a QM model.

Ultimately, yes, as basic QM is non-relativistic. If you consider an atom moving at relativistic speed then you would need to update your atomic model. For a free electron, for example, you must replace the non-relativistic Schrodinger equation with the Dirac equation.

Comparing a body in motion with the same body at rest can we expect a lesser probability of an electron cloud and its effect being in a given location within the vicinity of the nucleus or will it be the same?

 

[PeterDonis]

If we then consider the body in motion at 0.9c.

Motion is relative. The state you called "body at rest" is in motion at 0.9c in some frame. There is no way to detect "motion" in this sense by any change in physical parameters. So it will have no effect on an atom's electron cloud. QM still has to obey the principle of relativity.

 

[james fairclear]

To be more specific. Consider 2 identical stationary synchronised clocks A and B at a given location on earth. Clock B is set in motion relative to clock A in an Eastwardly direction at 0.9c. There are 2 physical parameters that are now different in clock B compared to Clock A:

1. Its mass has increased
2. Its tick rate has decreased

Considering any constituent atom of clock B can we expect a lesser probability of an associated electron cloud and its effect being in a given location within the vicinity of the nucleus or will it be the same?

 

[weirdoguy]

1. Its mass has increased

No it did not. Relativistic mass is not used in modern physics.

 

[PeroK]

There are 2 physical parameters that are now different in clock B compared to Clock A:

1. Its mass has increased
2. Its tick rate has decreased

Both these assertions are false. There is no physical difference associated with relative velocity. Clocks A and B remain physically identical. This is essentially the first postulate of Special Relativity: that there is no such thing as a state of absolute motion.

 

[james fairclear]

No it did not. Relativistic mass is not used in modern physics.

My understanding is that particles with mass cannot be accelerated to c due to their mass increasing exponentially. Is there now a different explanation?

 

[james fairclear]

Both these assertions are false. There is no physical difference associated with relative velocity. Clocks A and B remain physically identical. This is essentially the first postulate of Special Relativity: that there is no such thing as a state of absolute motion.

But surely if the mass doesn't increase then it will be possible to accelerate the clock to a speed > c?

In the Hafele Keating experiment the clock moving eastwards indicated an earlier time than the stationary ground clock, so surely it must have been ticking at a slower rate whilst in motion?

 

[PeroK]

My understanding is that particles with mass cannot be accelerated to c due to their mass increasing exponentially. Is there now a different explanation?

It's the kinetic energy that increases exponentially. Kinetic energy is frame-dependent (as it always has been), and an increase in kinetic energy does not represent a "physical" change.

The idea of ascribing increasing KE to increasing mass is not very useful. Especially as it gives the impression that the particle is physically changing, which it is not doing.

 

[PeroK]

In the Hafele Keating experiment the clock moving eastwards indicated an earlier time than the stationary ground clock, so surely it must have been ticking at a slower rate whilst in motion?

Not necessarily.

There are several factors in the H-K experiment. The ground clock is only stationary relative to the surface of the Earth. There is no sense in which the ground clock is absolutely stationary. In fact, the westbound clock recorded more time than the ground clock.

Ultimately, an attempt to explain the differences in times due to the concept of absolute states of motion does not work. And, contradicts the principle of relativity.

Instead, all clocks tick physically at the same rate, but they take different paths through spacetime. The clock that takes the shortest path through spacetime records the least time (Eastbound); the ground clock takes a longer path; and the westbound clock takes the longest path.

 

[PeroK]

PS One problem with the idea that "this clock is really moving and is physically ticking slowly" is that we can always change reference frames. Let's take an example.

For simplicity let's assume that the Earth is not spinning on its axis. Clock A remains at rest on the surface of the Earth and clock B moves round the world. Clock B will show less time than clock A when they are back together.

Now, let's take a frame of reference where the Sun is at rest and the Earth is orbitting the Sun. Clock A is moving on this orbit. And, if clock B moves in the opposite direction to the orbit, then initially clock B slows down! In this frame of reference, it's clock A that is initially "ticking slower than clock B". This contradicts the idea that clock B is really, physically ticking slower than clock A. Note that when clock B gets to the far side of the Earth it must move faster than clock A (in the Sun's frame) and when it gets back to clock A the net result is that it shows less time.

This shows that explanations based on the notion of absolute motion cannot be correct. Instead, the two clocks took different paths through spacetime, with clock B taking the shorter path overall. The lengths of the spacetime paths are invariant (the same in every reference frame) and that is a much better explanation. In other words, it's not that clock A or B ticked slower, but that less time passed for one than for the other.

 

[weirdoguy]

My understanding is that particles with mass cannot be accelerated to c due to their mass increasing exponentially.

I think that this insights article is a good summary of why relativistc mass is not used in most of the scientific contexts: https://www.physicsforums.com/insights/what-is-relativistic-mass-and-why-it-is-not-used-much/

 

[PeterDonis]

surely if the mass doesn't increase then it will be possible to accelerate the clock to a speed > c?

Your reasoning is backwards. Particles with nonzero rest mass cannot move at $c$ because of the geometry of spacetime. Particles with nonzero rest mass move on timelike worldlines, and particles that move at $c$, such as photons, move on null worldlines. Those are two fundamentally different kinds of things and you can't change one into the other.

The "mass increase" you mention (to the extent it makes sense at all--note that there are limitations to the concept of "relativistic mass", as explained in the Insights article that has already been linked to) is a side effect of the above, not a cause of it.

 

[james fairclear]

Not necessarily.

There are several factors in the H-K experiment. The ground clock is only stationary relative to the surface of the Earth. There is no sense in which the ground clock is absolutely stationary. In fact, the westbound clock recorded more time than the ground clock.

Ultimately, an attempt to explain the differences in times due to the concept of absolute states of motion does not work. And, contradicts the principle of relativity.

Instead, all clocks tick physically at the same rate, but they take different paths through spacetime. The clock that takes the shortest path through spacetime records the least time (Eastbound); the ground clock takes a longer path; and the westbound clock takes the longest path.

Regardless of the different paths taken how is it possible for 3 previously synchronised clocks to indicate different times without their respective tick rates having changed at some point? To my knowledge there is no evidence to support your claim that "all clocks tick physically at the same rate"?