QM Eigenstates and the Notion of Motion

[strangerep]

I'd be interested to hear more arguments against my statement that nothing moves in an energy eigenstate, but that's of course off-topic here.

I'll risk a quick off-topic answer here, since I think it's straightforward QM, not vague "interpretation" stuff.

In QM (e.g., Ballentine p81), for a free particle, $H = \frac12 \, M\, V\cdot V + E_0$. So in the ground state $|E_0\rangle$ we have $H|E_0\rangle = E_0 |E_0\rangle,$ hence $V^2 |E_0\rangle = 0$.

I'm reasonably sure that a zero eigenvalue for the $V^2$ operator corresponds to "not moving".

But in a higher energy eigenstate, we get a non-zero eigenvalue for $V^2$, which suggests "moving".

(I leave to the Mentors' discretion whether to convert this into a new thread in the QM forum.)

 

[Nugatory]

I'll risk a quick off-topic answer here, since I think it's straightforward QM, not vague "interpretation”….

That I agree with, but it seems from #4 of the offending thread that the OP there was thinking about a bound electron in the nucleus’s Coulomb potential. Is there a similarly quick and convincing way of extracting a value of $v^2$?

 

[dextercioby]

What is the working theoretical definition of "motion" for a quantum system? We saw tracks in the cloud chamber, so from a practical standpoint "motion" is a variation in time of space coordinates assigned to the lab the cloud chamber resides in.

 

[A. Neumaier]

I'd be interested to hear more arguments against my statement that nothing moves in an energy eigenstate

Nothing moves in a normalizable energy eigenstate.

But plane waves are counterexamples to your general statements, giving unnormalizable energy eigenstates at arbitrary kinetic energies.

In general, a system has a partly continuous spectrum iff it has moving parts.

What is the working theoretical definition of "motion" for a quantum system? We saw tracks in the cloud chamber, so from a practical standpoint "motion" is a variation in time of space coordinates assigned to the lab the cloud chamber resides in.

A subsystem is in motion iff the kinetic energy operator of its center of mass has a nonzero expectation value.

In QM (e.g., Ballentine p81), for a free particle, $H = \frac12 \, M\, V\cdot V + E_0$. So in the ground state $|E_0\rangle$ we have $H|E_0\rangle = E_0 |E_0\rangle,$ hence $V^2 |E_0\rangle = 0$.

But in a higher energy eigenstate, we get a non-zero eigenvalue for $V^2$, which suggests "moving".

For a free particle, the only bound state is the ground state. The spectrum consists of the interval $[E_0,\infty]$. The solutions of the time independent Schrödinger equation for $E>E_0$ are plane waves with frequency $\omega=(E-E_0)/\hbar$, corresponding to uniform motion with kinetic energy $E-E_0$.

 

[vanhees71]

Nothing moves in a normalizable energy eigenstate.

If the energy eigenstate is not normalizable, then you can't prepare the system in this state, of course. Then you must build "wave packages" of these scattering states to approximate it, and thus it's not an energy eigenstate anymore, and there indeed something may move.

Here we discuss bound states like the hydrogen atom. In a hydrogen atom in its ground state nothing moves.

 

[Morbert]

I'd be interested to hear more arguments against my statement that nothing moves in an energy eigenstate, but that's of course off-topic here.

...

If the energy eigenstate is not normalizable, then you can't prepare the system in this state, of course.

From an instrumentalist perspective, in the same way we associate a state with a preparation, we should associate the posed question with an experiment. E.g. "Does the system, when prepared in an energy eigenstate, move?" could be interpreted as "What is the probability that sequential determinations of the position of the centre of mass of the system all return the same result?"

 

[vanhees71]

Let $A$ be an arbitrary observable, represented by the self-adjoint operator, $\hat{A}$. Then the operator that represents the time derivative $\dot{A}$ is (independently of the chosen picture of time evolution!)
$$\mathring{\hat{A}}=\frac{1}{\mathrm{\hbar}}[\hat{A},\hat{H}].$$
Thus the expectation value of the time derivative is
$$\frac{\mathrm{d}}{\mathrm{d} t} \langle A \rangle=\langle E|\mathring{\hat{A}}|E \rangle=\frac{1}{\mathrm{\hbar}} E (\langle E|hat{A}|E \rangle-\langle E |\hat{A}|E \rangle=0,$$
where I have used that
$$\hat{H} |E \rangle=E |E \rangle, \quad \langle E|\hat{H}=E \langle E|,$$
which follows from the self-adjointedness of $\hat{H}$.

 

[gentzen]

Let $A$ be an arbitrary observable, represented by the self-adjoint operator, $\hat{A}$. Then the operator that represents the time derivative $\dot{A}$ is...

Of course the time derivative of the expectation value of an arbitrary observable is zero for a stationary state. Sure, the explicit calculation verifying this is still nice. But verifying something obvious does not settle the current disagreement. It even gives the impression that you don't understand what the disagreement is about.

 

[vanhees71]

Obviously, I don't understand, where there can be a disagreement about the fact that energy eigenstates are stationary states. It follows even simpler from the Schrödinger equation. If $|\Psi(0) \rangle=|\Psi_0 \rangle=|E \rangle$ ,where $|E \rangle$ is an energy eigenvector, then it follows
$$|\Psi(t) \rangle=\exp(-\mathrm{i} \hat{H} t) |\Psi(0) \rangle=\exp(-\mathrm{i} \hat{H} t) |E \rangle=\exp(-\mathrm{i} E t) |E \rangle.$$
The state thus is
$$\hat{\rho}(t)=|\Psi(t) \rangle \langle \Psi(t)|=|E \rangle \langle E|=\text{const}.$$
So in which sense do you think that "something is moving" even if the system is in an energy eigenstate?

 

[Morbert]

Let $A$ be an arbitrary observable, represented by the self-adjoint operator, $\hat{A}$. Then the operator that represents the time derivative $\dot{A}$ is (independently of the chosen picture of time evolution!)
$$\mathring{\hat{A}}=\frac{1}{\mathrm{\hbar}}[\hat{A},\hat{H}].$$
Thus the expectation value of the time derivative is
$$\frac{\mathrm{d}}{\mathrm{d} t} \langle A \rangle=\langle E|\mathring{\hat{A}}|E \rangle=\frac{1}{\mathrm{\hbar}} E (\langle E|hat{A}|E \rangle-\langle E |\hat{A}|E \rangle=0,$$
where I have used that
$$\hat{H} |E \rangle=E |E \rangle, \quad \langle E|\hat{H}=E \langle E|,$$
which follows from the self-adjointedness of $\hat{H}$.

I don't strongly disagree with associating motion with a changing expectation value so long as the association is made explicit. But I don't think motion is exclusively associated with the expectation value. E.g. This paper https://arxiv.org/abs/gr-qc/9210010 shows how we might model the noise in sequences of measurements on the same system (as opposed to repeated measurements, each on different members of an ensemble) with equations of motion and a Langevin force. We can see that this sense of motion would emerge even if the system is always prepared in the ground state.

 

[Fra]

I missed the background to the thread, the link to the quotation i the first post seems broken so perhaps i am missing something..

Obviously, I don't understand, where there can be a disagreement about the fact that energy eigenstates are stationary states.

As others ask how is motion defined?
Is it certain observables or our information that "moves"?

Motion to me means change of some parameter with timr with respect to some space. Position in 3d space seems like standard mening if not qualified.

The stationarity of E-eigenstates imo refers to our information. So a stationary state often captures a stationary information state that reflects via the conjugate variable a periodic motion.

But i supposed its something more delicate that disscussed?

For examole ib the ground state of hydrogen, it seems there are radial oscillations, described by the stationary s orbital with angular symmetry? Do you label this "motion" or not? It seema like a matter of definition.

/Fredrik

 

[gentzen]

I missed the background to the thread, the link to the quotation i the first post seems broken so perhaps i am missing something..

The original discussion was probably the one stopped shortly after

Then, how can an energy eigenstate describe a moving particle? It's a stationary state!

But $\langle p^2 \rangle \neq 0$. I guess it comes down to what "moving" means

I guess vanhees71 started this thread (because the other thread got closed), then tried to delete it again, but already got the answer from strangerep. (Or maybe a moderator deleted vanhees71's post, still the answer and this thread survived.)

 

[vanhees71]

If we know that the system is in an exact energy eigenstate then this implies the information that it's a stationary state and thus that nothing is changing with time. In pop-sci language I'd say it means "nothing moves".

This thread was somehow initiated from an off-topic side discussion in the advisor launch. I've also seen that they forgot to copy one of my postings from there. Its content is what I've posted now in #9.

 

[DrClaude]

The original discussion was probably the one stopped shortly after

I guess vanhees71 started this thread (because the other thread got closed), then tried to delete it again, but already got the answer from strangerep. (Or maybe a moderator deleted vanhees71's post, still the answer and this thread survived.)

This thread is an offshoot of https://www.physicsforums.com/threads/electrons-at-absolute-zero-do-they-still-move.1050269/

Although I don't get why this should be of importance.

 

[Fra]

If we know that the system is in an exact energy eigenstate then this implies the information that it's a stationary state and thus that nothing is changing with time. In pop-sci language I'd say it means "nothing moves".

I see, with that definition what you say makes sense to me.

(I thought the discussion was about some duality between different information states, and such as states vs changes of the same state, or conjugate spaces and which one is more primary, but it seems not.)

/Fredrik

 

[A. Neumaier]

Thus the expectation value of the time derivative is
$$\frac{\mathrm{d}}{\mathrm{d} t} \langle A \rangle=\langle E|\mathring{\hat{A}}|E \rangle=\frac{1}{\mathrm{\hbar}} E (\langle E|\hat{A}|E \rangle-\langle E |\hat{A}|E \rangle=0,$$

But in the statistical interpretation that you always advocate, this only says that in an energy eigenstate, nothing happens on the average. In each single realization, something might happen.

 

[gentzen]

If we know that the system is in an exact energy eigenstate then this implies the information that it's a stationary state and thus that nothing is changing with time.

Such statements may be fine as long as it is a purely academic discussion with no risk to ever lead to mistakes (or "costly" disputes) in practical applications. Practice tends to be messy, and induces the desire to make things simpler than they actually are. One such discussion I still have in mind was about "The ‘stable’ velocity for the electron for an orbit around the nucleus equals, $v=\sqrt{\frac{eQ}{4\pi \epsilon_0 mr}}$ (3.103)" on page 83 of Thomas Verduin's "Quantum Noise Effects in e-Beam Lithography and Metrology" PhD thesis. The discussion was about the moment transfer between the primary electron and a secondary electron "kicked out" of an inner shell at a certain ionization energy. Because that electron in the inner-shell is in a stationary shell, "somebody" was convinced that assigning a velocity to it (before the collision) and using that "classical" information for computing the moment transfer was "completely wrong". So even after I fixed some artifacts of that specific "formulation" above, it felt still "too classical" to him. I am an instrumentalist in those matters, the formulas have to work well in their specific scenarios, and interpolate somewhat reasonably between scenarios. Focusing on one specific aspect over all others simply risks to generate totally wrong results for some scenarios. (And believe me, the formulas can get messy, no matter how you compute it.)

Moment without movement. Just like in Bohmian mechanics. Since nearly all physicists agree that this is a shortcoming of Bohmian mechanics, my guess would be that vanhees71 is wrong in this specific case. I just can't believe that Bohmian mechanics should be right in this respect. It will often be the ground state of an harmonic oscillator, and of course oscillate is what it will do.

Perhaps vanhees71 is right, but then I would like to see how he argues that Bohmian mechanics is correct in this "moment without movement" aspect. I don't fully understand it yet, I plan to read a two paper series by Peter Holland on that problem at some point, but I am fully aware that vanhees71 has something completely different in mind.

Currently, I am actually working on something even more complicated, namely crystal effects in SEM. And because there the question which classical computation are appropriate for which part arose again, I worked out a closely related riddle arising in Bohmian mechanics. If you want to understand my position, look at Demystifier's initial reaction when I tried to explain my resolution. (As a mathematician, I believe in "conservation of difficulty" and that questions can ultimately be resolved, but I certainly don't believe that the solution will always be the most simple one that first comes to your mind, before you tried to work-out the tricky parts in detail.)

Can you be more specific about your limit? I mean, can you express it with math, or with pictures, rather than with words?

Take a "simple" window function, for example a (suitably shifted) Hann function ("raised cosine window"): $w_L(x):=\sin^2(\pi x/L) \chi_{(-L,0)}(x)$. Take $\phi_L(x,0):=w_L(x)\exp(ikx)$ to be the initial wavefunction, for the 1D scenario with a potential barrier starting at $x=0$. This initial wavefunction is now used both for the distribution of particles, and for solving the time dependent Schrödinger equation. For a given $L$, we get (a distribution of) trajectories, and those trajectories may cross each other, because of the time dependence. And now we investigate the limit for $L \to \infty$, especially the limit of the phase distribution of $\phi_L(x,t)$, where we are allowed to suppress a global phase factor like $\exp(-i\omega t)$.

 

[vanhees71]

Such statements may be fine as long as it is a purely academic discussion with no risk to ever lead to mistakes (or "costly" disputes) in practical applications. Practice tends to be messy, and induces the desire to make things simpler than they actually are. One such discussion I still have in mind was about "The ‘stable’ velocity for the electron for an orbit around the nucleus equals, $v=\sqrt{\frac{eQ}{4\pi \epsilon_0 mr}}$ (3.103)" on page 83 of Thomas Verduin's "Quantum Noise Effects in e-Beam Lithography and Metrology" PhD thesis. The discussion was about the moment transfer between the primary electron and a secondary electron "kicked out" of an inner shell at a certain ionization energy. Because that electron in the inner-shell is in a stationary shell, "somebody" was convinced that assigning a velocity to it (before the collision) and using that "classical" information for computing the moment transfer was "completely wrong". So even after I fixed some artifacts of that specific "formulation" above, it felt still "too classical" to him. I am an instrumentalist in those matters, the formulas have to work well in their specific scenarios, and interpolate somewhat reasonably between scenarios. Focusing on one specific aspect over all others simply risks to generate totally wrong results for some scenarios. (And believe me, the formulas can get messy, no matter how you compute it.)

Well, I'm convinced that one can understand this in correct quantum-mechanical terms. In the classical theory there is no "stable velocity for the electron for an orbit around the nucleus". That problem was famously solved with the formulation of modern quantum mechanics, getting rid of all inconsistencies of the pseudo-quantum theory a la Bohr.

Perhaps vanhees71 is right, but then I would like to see how he argues that Bohmian mechanics is correct in this "moment without movement" aspect. I don't fully understand it yet, I plan to read a two paper series by Peter Holland on that problem at some point, but I am fully aware that vanhees71 has something completely different in mind.

I don't think that Bohmian mechanics helps in any way here. I don't know, how Bohmian mechanics describes a bound electron though. Do you have a reference, where this is discussed?

Currently, I am actually working on something even more complicated, namely crystal effects in SEM. And because there the question which classical computation are appropriate for which part arose again, I worked out a closely related riddle arising in Bohmian mechanics. If you want to understand my position, look at Demystifier's initial reaction when I tried to explain my resolution. (As a mathematician, I believe in "conservation of difficulty" and that questions can ultimately be resolved, but I certainly don't believe that the solution will always be the most simple one that first comes to your mind, before you tried to work-out the tricky parts in detail.)

You can, of course, make things more complicated than necessary by using Bohmian mechanics, which adds unobservable elements ("trajectories") to quantum theory, which are not useful for anything to get the most efficient way of calculating a given physical problem.

 

[Demystifier]

The question is whether something moves in the energy eigenstate. This question by itself is vague, so it doesn't have a unique answer. There are, however, several inequivalent ways how this question can be made precise. Among them, there are at least 2 different precise versions in which the answer is - yes.

1. Prepare the system in an energy eigenstate and then perform a single measurement of the velocity defined by the operator $\hat{v}$. Can the measurement outcome $v$ be different from zero? The answer is - yes it can.

2. Use weak measurement to measure the trajectory of the particle, without affecting its wave function. The trajectory obtained this way corresponds to a moving trajectory, which in fact looks exactly like the Bohmian trajectory.

 

[vanhees71]

The question is whether something moves in the energy eigenstate. This question by itself is vague, so it doesn't have a unique answer. There are, however, several inequivalent ways how this question can be made precise. Among them, there are at least 2 different precise versions in which the answer is - yes.

1. Prepare the system in an energy eigenstate and then perform a single measurement of the velocity defined by the operator $\hat{v}$. Can the measurement outcome $v$ be different from zero? The answer is - yes it can.

I think the precise definition that "something is moving" is that its state changes with time, and in this sense if the system is prepared in an energy eigenstate there's "nothing moving", because then the state $\hat{\rho}(t)=|E \rangle \langle E|$ is time-independent (argued within the Schrödinger picture of time evolution for simplicity).

Ad 1. Of course, you can get $v \neq 0$, but this doesn't imply that anything "is moving": The velocity is simply indetermined before the measurement, and measuring it gives with a probability given by the prepared state some value, which may be different from 0.

2. Use weak measurement to measure the trajectory of the particle, without affecting its wave function. The trajectory obtained this way corresponds to a moving trajectory, which in fact looks exactly like the Bohmian trajectory.

Do you mean the "trajectory" of a charged particle as seen in a cloud chamber? Here indeed the state of the particle changes, and it's not in an energy eigenstate. It looses energy by ionizing the vapour molecules in the cloud chamber! What this has to do with Bohmian trajectories is not clear to me.

 

[Demystifier]

I think the precise definition that "something is moving" is that its state changes with time

This definition is precise, but not directly measurable.

 

[Demystifier]

Do you mean the "trajectory" of a charged particle as seen in a cloud chamber?

No, the weak measurement is something completely different.

 

[Fra]

I think the precise definition that "something is moving" is that its state changes with time, and in this sense if the system is prepared in an energy eigenstate there's "nothing moving",

I'd say it's the information of future as defined by the agents expectation that doesn't change with time, this is fine. But even if it did change with time - in an unitary way - the total information sort of doesnt change anyway if you include "knowledge of the hamiltonian" as the agents information.

But I think that stationary state of the information itself sometimes encode via conjudate variables that the future itself is expected to be uncertain and thus in a way "changing" or "moving" but not necessarily in a way that implies a defined trajectory. wether its chaotic beyond tracking or just uncertain makes no difference ot the agent i think, the result is the same - unpredictable "change".

/Fredrik

 

[gentzen]

Well, I'm convinced that one can understand this in correct quantum-mechanical terms. In the classical theory there is no "stable velocity for the electron for an orbit around the nucleus". That problem was famously solved with the formulation of modern quantum mechanics, getting rid of all inconsistencies of the pseudo-quantum theory a la Bohr.

(Edit: In the end, it is the application of some Virial theorem which hides behind that paragraph which includes the sentence with that "stable velocity for the electron for an orbit around the nucleus".)
The problem for me is this focus of "correct" on consistent application of quantum mechanics. If this insistence in the end leads to simulations with systematic deviations from what one can measure in experiments, for parts which could have been understood (and "simulated") well enough in terms of a mixture of classical and quantum terms (not in the sense of Bohr's old pseudo-quantum theory, but still in terms of Bohr newer mixing of classical images with quantum theory), then I get the impression that a better mastery of interpretational issues (in an instrumentalistic sense) by physicists would be a good thing.

This is similar as if you would design an optical system, and somebody would always protest against the use of geometrical optics to design most parts of the system, always insists that he is convinced that a proper understanding in wave-optical terms is possible, and in the end fails to optimize the relevant properties of the optical system.

I don't think that Bohmian mechanics helps in any way here. I don't know, how Bohmian mechanics describes a bound electron though. Do you have a reference, where this is discussed?

Where is the problem? Bound states are normalizable, they give a "canonical" solution to the time dependent Schrödinger equation, so you just can apply "standard Bohmian mechanics". But the trajectories in that solution are all constant, i.e. nothing is moving at all.

You can, of course, make things more complicated than necessary by using Bohmian mechanics, which adds unobservable elements ("trajectories") to quantum theory, which are not useful for anything to get the most efficient way of calculating a given physical problem.

What you get is an "exact model" where you can investigate your confusing problem, and try to work-out some of the tricky parts in detail. And by "conservation of difficulty," those details often also help you better understand what might be "some possible" way of calculating a given physical problem. A drawback of such an "exact model" is that it often prefers "some specific way" of calculating over other ways, and the most efficient way of calculating might have been one of the other ways.

 

[strangerep]

Let $A$ be an arbitrary observable, represented by the self-adjoint operator, $\hat{A}$. Then the operator that represents the time derivative $\dot{A}$ is (independently of the chosen picture of time evolution!)
$$\mathring{\hat{A}}=\frac{1}{\mathrm{\hbar}}[\hat{A},\hat{H}].$$
Thus the expectation value of the time derivative is
$$\frac{\mathrm{d}}{\mathrm{d} t} \langle A \rangle=\langle E|\mathring{\hat{A}}|E \rangle=\frac{1}{\mathrm{\hbar}} E (\langle E| \hat{A}|E \rangle-\langle E |\hat{A}|E \rangle=0,$$
where I have used that
$$\hat{H} |E \rangle=E |E \rangle, \quad \langle E|\hat{H}=E \langle E|,$$
which follows from the self-adjointedness of $\hat{H}$.

I had a similar thought about this as already expressed by @A. Neumaier. Just because the expectation value vanishes doesn't mean that (e.g.,) the variance also vanishes, since it involves
$$\mathring{\hat{A}}^2 ~\propto~ [\hat{A},\hat{H}] \, [\hat{A},\hat{H}]
~=~ \hat{A} \hat{H} \hat{A}\hat{H} - \hat{A} \hat{H} \hat{H}\hat{A}
~-~\hat{H} \hat{A} \hat{A}\hat{H} +\hat{H} \hat{A} \hat{H}\hat{A} $$and you don't get zero in general when you put this inside $\langle E| \dots |E \rangle$ .

This is related to what happens if one queries the magnitude of velocity squared. Its expectation value does not vanish, even though the expectations of the individual velocity components do.

 

[WernerQH]

I think the precise definition that "something is moving" is that its state changes with time, and in this sense if the system is prepared in an energy eigenstate there's "nothing moving", because then the state $\hat{\rho}(t)=|E \rangle \langle E|$ is time-independent (argued within the Schrödinger picture of time evolution for simplicity).

The word "stationary" can be seriously misleading in the description of statistical ensembles. True, you have $\langle x \rangle = 0$ for a harmonic oscillator in an energy eigenstate (with the phase maximally uncertain). But looking at the formula $$
\langle x(t) x(0) \rangle = \textstyle
{\hbar \over m \omega} \langle n + \frac 1 2 \rangle \cos (\omega t) \ ,
$$ can you really teach your students that an oscillator doesn't move when it is in the "stationary" state $n = 3$ ?

 

[vanhees71]

This is a autocorrelation function, describing fluctuations. In which sense do you think it describes that the oscillator is "moving"?

 

[WernerQH]

In which sense do you think it describes that the oscillator is "moving"?

Strange question. Doesn't $x$ usually denote the position of a harmonic oscillator?

 

[vanhees71]

Yes, and its expectation value is time-independent if the oscillator is in an energy eigenstate. What you considered is a autocorrelation function, describing fluctuations around this stationary expectation value.

It's as with a gas at a finite temperature in its rest frame in global thermal equilibrium. Nothing moves, and on average the gas molecules are at rest, but of course they are still flucuating around. The average velocity (the expectation value of a time derivative) is 0, but its autocorrelation function or simply $\langle v^2 \rangle \neq 0$.

 

[WernerQH]

What you considered is a autocorrelation function, describing fluctuations around this stationary expectation value.

I think your view is too myopic. The "fluctuations" are not random, but strictly periodic, and in perfect agreement with what you'd expect classically. (Assuming of course that the oscillator remains unperturbed.)

 

[vanhees71]

What I expect from classical mechanics is rather described by a coherent or squeezed state than an energy eigenstate!

 

[Fra]

WernerQHs point was what I thoughtr the thread was about. What is a periodic "movement" in Q space, is stationary in the conjugate P space. That way, one can argue that what is stationary or not, depends on hte agents way of encoding things. Of course the fourier transform is good because it makes encoding periodic phenomena simple or even stationary.

/Fredrik

 

[gentzen]

This is a autocorrelation function, describing fluctuations. In which sense do you think it describes that the oscillator is "moving"?

Strange question. Doesn't $x$ usually denote the position of a harmonic oscillator?

Yes, and its expectation value is time-independent if the oscillator is in an energy eigenstate. What you considered is a autocorrelation function, describing fluctuations around this stationary expectation value.

Wow, so the disagreement really comes down to how you (vanhees71) interpret the word "moving". For me, the most important part of "moving" is that properties (like position) in a stationary state are not necessarily constant over time. But you basically seem to have no objections to this, you just want to call that "fluctuations" instead of "moving".

It seems like for you, "moving" implies a more systematic way of being not-constant, like for example changing position continuously over time with a certain velocity. I don't believe in that one either. WernerQH's example is nice, even so for me feels just as "unphysical" (only to the other extreme) as the constant non-moving trajectories of stationary states in Bohmian mechanics.

It's as with a gas at a finite temperature in its rest frame in global thermal equilibrium. Nothing moves, and on average the gas molecules are at rest, but of course they are still flucuating around. The average velocity (the expectation value of a time derivative) is 0, but its autocorrelation function or simply $\langle v^2 \rangle \neq 0$.

And here you gave a nice detailed description making it clear that you don't insist on everything being always constant over time, but that it is mostly the absence of systematic movement which is important to you.

 

[vanhees71]

For me moving means that the state of a system is time-dependent. In QT energy eigenstates are time-independent and thus the system is "not moving". It's the solution for the problem of instable atoms in the classical picture, where accelerating electrons crash into the nuclei within a very short time due to radiative energy loss. In QT that's not the case: Neglecting the quantization of the electromagnetic field you get electronic energy eigenstates which are the stationary states, and the atoms are thus stable in these states forever. The expectation values of the time-derivative of any observable is 0, but of course the observables don't take determined values and thus fluctuate around their time-independent expectation values, and thus the expectation values of squares of time-derivatives of observables are not 0, although "nothing is moving".

 

[Fra]

For me moving means that the state of a system is time-dependent.

That's how I understood you some posts up as well. No big point in arguing just about definitions.

In QT energy eigenstates are time-independent and thus the system is "not moving". It's the solution for the problem of instable atoms in the classical picture, where accelerating electrons crash into the nuclei within a very short time due to radiative energy loss. In QT that's not the case: Neglecting the quantization of the electromagnetic field you get electronic energy eigenstates which are the stationary states, and the atoms are thus stable in these states forever. '

Interestingly, in my preferred interpretation, the choice of encoding the information in the best way, also is what makes the agent fit which relates to stability as well. If the agent can find a transformation that transforms the pattern of incoming data strems into a stationary code, that must be a massive evolutioanry advantage. This is how i prefer to interpret this. The fourier transform is just one of many possible transforms.

I recall during my first introduction to QM, that I felt that the role of the "fourier transform" was suspicuously a key concept, but which was not really motivated in a deeper waya beyond the standard "wave duality stuff". I think I have come to a much deeper understanding of this over the years, and i think there is maybe yet deeper motivations for all this that is ahead of us.

/Fredrik