# What does motion mean in quantum mechanics?

• I
• sergiokapone
In summary, the particle does not have a trajectory, and it does not emit anything until it is detected.
sergiokapone
Consider the Schrödinger equation for a free particle:

-\frac{\hbar^2}{2m} \partial_i^2\psi = i\hbar\partial_t \psi.

Let us be interested in the motion of a free particle in quantum mechanics. We say ok, we have a solution to the Schrödinger equation for a free particle (1) with momentum ##\vec{p}## and energy ##E = \frac{p^2}{2m}##

\psi = Ae^{-\frac i\hbar(Et - p_ix_i)}.

But this solution is generally of little use, since it does not carry any useful information at all, since it gives the same probability of detecting a particle anywhere.

Second example. If we consider the problem of the state (I do not say motion) of a particle in a central field, then we must solve the Schrödinger equation:

-\frac{\hbar^2}{2m} \partial_i^2\psi - \frac{e}{r}\psi = i\hbar\partial_t \psi.

When we want to define a state with a well-defined energy (stationary states), we easily separate the phase factor ##e^{\frac{i}{\hbar} E t}##

\partial_i^2\chi - \frac{2m}{\hbar}\left(E + \frac{e}{r}\right)\chi = 0.

Let me remind you that for the ground state we have the solution:

\psi = \sqrt\frac{1}{\pi}\frac{1}{a^{3/2}}e^{r/a}e^{-\frac{i}{\hbar} E t},

where ##a## is the Bohr radius, ## E = - \frac {me ^ 2} {2 \hbar ^ 2} ##.

Again, did we get a motion? Of course not, we can only find out where the electron can be with one or another probability. Yes, the electron does not move, since it is in a stationary state. And naturally, the paradox with the emission of an electron is solved in quantum mechanics right away - since the electron does not move, it means that it does not emit.

In order for there to be movement, there must be a trajectory. It is obvious that classical trajectories are an emergent phenomenon of quantum mechanics. We need to understand how quantum mechanics, with its wave functions, manifests in the classics what we call motion and trajectory.

Is there a question here?

How does quantum mechanics manifest what we call motion in the classical sense?

sergiokapone said:
Let us be interested in the motion of a free particle in quantum mechanics.

There is no such thing. A particle in QM does not have a trajectory. It has a wave function. So it has no "motion". It just has probability amplitudes, given by the wave function, for various possible results of measurements of its position or momentum.

sergiokapone said:
Yes, the electron does not move, since it is in a stationary state.

This logic is not valid in QM. QM is not classical mechanics. A "stationary" state in QM is not a state in which the particle "does not move", since, as above, "motion" is not a valid concept for a particle in QM to begin with. A stationary state in QM is an eigenstate of the Hamiltonian.

sergiokapone said:
In order for there to be movement, there must be a trajectory.

This is classical reasoning, not QM reasoning.

sergiokapone said:
We need to understand how quantum mechanics, with its wave functions, manifests in the classics what we call motion and trajectory.

You are looking at the wrong things if that is what you want to understand. You need to look at, for example, the Ehrenfest theorem if you want to see how, under appropriate conditions, quantum systems can manifest classical behavior, such as (to a good approximation) classical trajectories with well-defined position and momentum.

Only in Bohmian mechanics one can assign ontologically a "does not move" description to stationary states.

Actually, "does not move" is already problematic due to Heisenberg UP.

physika and atyy
sergiokapone said:
How does quantum mechanics manifest what we call motion in the classical sense?
In your first example of the free particle there is the concept of a wave-packet.

In the case of the hydrogen atom, it doesn't - in the sense that there is no classical trajectory.

In many cases, the expected values follow classical laws. Look up the Ehrenfest Theorem and Quantum spin precession in a magnetic field, for example.

PeroK said:
In your first example of the free particle there is the concept of a wave-packet.

That in itself doesn't give a trajectory, even approximately, except possibly for a short time after a measurement that localizes the particle. Wave packets for a single particle spread very rapidly in position.

PeroK said:
In your first example of the free particle there is the concept of a wave-packet.

Yes, the concept of a wave packet is close to what might be called motion in the classical sense. And the velocity of movement can be related to the group velocity.

In order to see the movement, we need to build a world line on which we can see the cause-and-effect connection.

It can be argued that the Feynman picture - the path integral - better corresponds to the notion of "classical" motion?

PeterDonis said:
Wave packets for a single particle spread very rapidly in position.

Yes, but if the particle interacts with something, then it will again be collected in a packet.

sergiokapone said:
if the particle interacts with something, then it will again be collected in a packet.

That depends on the interaction. Not all interactions will re-localize the particle.

Dale
sergiokapone said:
It can be argued that the Feynman picture - the path integral - better corresponds to the notion of "classical" motion?

No. Why would you think so?

PeterDonis said:
No. Why would you think so?
The concept of motion just puzzled me when I finally saw formula (29.3). After all, it just connects two wave packets at different points and at different times. Is that right?

Here is ref on (29.3) (Sorry, in russuan)
http://alexandr4784.narod.ru/lev_2/lev2_05_gl_03_29.pdf

sergiokapone said:
formula (29.3). After all, it just connects two wave packets at different points and at different times. Is that right?

No. It relates the wave functions at different points and times. But there is no requirement that those wave functions be wave packets. They can be anything.

AlexCaledin
sergiokapone said:
It can be argued that the Feynman picture - the path integral - better corresponds to the notion of "classical" motion?

No, because it is a sum over many paths, whereas the classical particle takes one path. However, the classical trajectory can usually be recovered as the "saddle point" approximation to the Feynman integral.
http://fy.chalmers.se/~tfkhj/FPI.pdf

Another way to see the classical motion is to use the Ehrenfest theorem, as mentioned by other posters above.
https://en.wikipedia.org/wiki/Ehrenfest_theorem

Yet another way is to use the formalism of continuous measurement. which can be used to explain tracks in a cloud chamber.
https://arxiv.org/abs/math-ph/0512069
https://arxiv.org/abs/1209.2665v1
https://arxiv.org/abs/quant-ph/0611067 (see Fig. 3 Basic setup for imaging resonance fluorescence from a single atom as a continuous position measurement)

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sergiokapone
atyy said:
No, because it is a sum over many paths, whereas the classical particle takes one path. However, the classical trajectory can usually be recovered as the "saddle point" approximation to the Feynman integral.
It seems to me that this saddle point approximation is what one usually gets in QFT for S-matrix calculations in relativistic QFT. The very idea of "scattering" is quite classical.

sergiokapone said:
Consider the Schrödinger equation for a free particle:

-\frac{\hbar^2}{2m} \partial_i^2\psi = i\hbar\partial_t \psi.

Let us be interested in the motion of a free particle in quantum mechanics. We say ok, we have a solution to the Schrödinger equation for a free particle (1) with momentum ##\vec{p}## and energy ##E = \frac{p^2}{2m}##

\psi = Ae^{-\frac i\hbar(Et - p_ix_i)}.

But this solution is generally of little use, since it does not carry any useful information at all, since it gives the same probability of detecting a particle anywhere.

Second example. If we consider the problem of the state (I do not say motion) of a particle in a central field, then we must solve the Schrödinger equation:

-\frac{\hbar^2}{2m} \partial_i^2\psi - \frac{e}{r}\psi = i\hbar\partial_t \psi.

When we want to define a state with a well-defined energy (stationary states), we easily separate the phase factor ##e^{\frac{i}{\hbar} E t}##

\partial_i^2\chi - \frac{2m}{\hbar}\left(E + \frac{e}{r}\right)\chi = 0.

Let me remind you that for the ground state we have the solution:

\psi = \sqrt\frac{1}{\pi}\frac{1}{a^{3/2}}e^{r/a}e^{-\frac{i}{\hbar} E t},

where ##a## is the Bohr radius, ## E = - \frac {me ^ 2} {2 \hbar ^ 2} ##.

Again, did we get a motion? Of course not, we can only find out where the electron can be with one or another probability. Yes, the electron does not move, since it is in a stationary state. And naturally, the paradox with the emission of an electron is solved in quantum mechanics right away - since the electron does not move, it means that it does not emit.

In order for there to be movement, there must be a trajectory. It is obvious that classical trajectories are an emergent phenomenon of quantum mechanics. We need to understand how quantum mechanics, with its wave functions, manifests in the classics what we call motion and trajectory.
You are of course right in saying that the plane wave does not represent a state in quantum mechanics, because it's not a square-integrable (and thus normalizable) function of the Hilbert space ##\mathrm{L}^2(\mathbb{R}^3)##.

Further it's clear that any eigenstate of the Hamiltonian of a system does not describe motion but a static configuration. That's the important progress of quantum mechanics from classical mechanics if it comes to the description of atoms as a bound state of electrons around the atomic nucleus. In contradistinction to the inconsistent "old quantum theory" ("Bohr-Sommerfeld quantization") you have a static solution, which is consistent with the stability of the atom. In the classical picture there's no static stable solution due to the radiation of em. waves of accelerated charges moving around the nucleus.

The motion of a particle can of course be described in quantum mechanics by solutions of the time-dependent Schrödinger equation, e.g., a Gaussian wave packet for a free particle. There are of course no trajectories of point particles as in classical mechanics, but "only" probability distributions for the position of the particle. The probability distribution can be chosen as pretty localized initially (small standard deviation of the initial Gaussian wave function). For the free particle then the wave function stays a Gaussian forever with the expectation value of the position moving with constant velocity. Now according to the Heisenberg uncertainty relation a pretty narrow initial position distribution leads to a pretty broad momentum distribution, and this leads to an increase of the standard deviation of the position. So the wave packet gets broader and broader with time. Nevertheless it clearly describes the "motion" of the particle's position probability distribution and in this sense the dynamics of the particle.

AlexCaledin, sergiokapone, Dale and 1 other person
To avoid any confusion:

Quantum probability is not the probability of where a “particle“ is. It is the objective probability of where an observer will find it.

AlexCaledin
vanhees71 said:
the plane wave does not represent a state in quantum mechanics, because it's not a square-integrable

If one uses the rigged Hilbert space formalism, the plane wave actually can be treated as a state.

vanhees71 said:
it's clear that any eigenstate of the Hamiltonian of a system does not describe motion but a static configuration

This doesn't seem right, since, for example, there are eigenstates of the Hamiltonian for the hydrogen atom that have nonzero orbital angular momentum. Granted, the "orbits" in question are not classical orbits, but it doesn't seem appropriate to call them "static" either.

Tendex said:
It seems to me that this saddle point approximation is what one usually gets in QFT for S-matrix calculations in relativistic QFT. The very idea of "scattering" is quite classical.

No, the calculations include higher order terms beyond the saddle point approximation. The Feynman diagrams are a way to keep track of these higher order terms in the path integral.

atyy said:
the calculations include higher order terms beyond the saddle point approximation.
Why beyond? would you leave the higher order corrections outside of the approximation(that includes a deformation through analytic continuation to obtain them)? I at least was implying that the higher order terms were part of the saddle point method of approximation, at least when applied to Minkowski spacetime in relativistic QFT. Perhaps you were referring to the path integral saddle point approximation in quantum mechanics? I specified I was talking about realtivistic QFT.

PeterDonis said:
If one uses the rigged Hilbert space formalism, the plane wave actually can be treated as a state.
This doesn't seem right, since, for example, there are eigenstates of the Hamiltonian for the hydrogen atom that have nonzero orbital angular momentum. Granted, the "orbits" in question are not classical orbits, but it doesn't seem appropriate to call them "static" either.
In the rigged Hilbert-space formalism a vector representing a state must be in the Hilbert space. The plane wave is in the dual of the domain of the position and momentum operators. The latter is a dense proper subspace of the Hilbert space and thus its dual larger than the Hilbert space, i.e., it includes distributions (in the sense of generalized dual vectors or linear forms that are not continuous wrt. the Hilbert-space metric).

As any energy eigenstates also the energy eigenstates of the hydrogen atom describe a static situation, no matter whether ##\ell=0## or ##\ell \neq 0##. The expectation values of the position vector components as well as of all momentum components are all constant in time. It's, because the wave functions are standing rather than propagating waves, i.e., if prepared in an energy eigenstate at ##t=0## the wave function is
$$\psi(t,\vec{x})=\exp(-\mathrm{i} E_n t) u_{n \ell m}(\vec{x}),$$
and the probability distribution thus
$$P(\vec{x})=|\psi(t,\vec{x})|^2=|u_{n \ell m}(\vec{x})|^2$$
constant in time.

This is the solution of the paradox that the so described atom doesn't radiate when prepared in such a state: Because there's no time-varying charge distribution, there's no em. radiation.

This is of course valid only within this approximation, where the em. field is not quantized; if you quantize the em. field of course an energy eigenstate discussed above is only approximately and energy eigenstate, and there are non-zero transition probabilities from excited states of lower states due to spontaneous emission. The ground state of course still is a stable atom with no radiation possible (due to energy conservation).

vanhees71 said:
This is the solution of the paradox that the so described atom doesn't radiate when prepared in such a state: Because there's no time-varying charge distribution, there's no em. radiation.
M. Born Atomic Physics said:
From another point of view, the statistical interpretation of wave functions suggests how the radiation emitted by the atom may be calculated on wave-mechanical principles. In the classical theory this radiation is determined by the electric dipole moment ##\mathbf{p}## of the atom, or rather by its time-rate of variation. By the correspondence principle, this connexion must continue to subsist in the wave mechanics. Now the dipole moment ##\mathbf{p}## is easily calculated by wave mechanics; if we adhere to the analogy with classical atomic mechanics, it is given by

\mathbf{p} = e \int \mathbf{r} |\psi_n|^2 dV

where ##\mathbf{r}## stands for the radius vector from the nucleus to the point of integration, or field point. (As usual, the asterisk denotes replacement by the conjugate complex quantity.) The integral represents of course the position of the "electrical centroid of the electronic cloud ". Now, as is easily proved, this integral vanishes for all states of an atom, so that the derivative of the dipole moment vanishes, and accordingly the emitted radiation also; that is, a stationary state does not radiate. This gives an explanation of the fact — unintelligible from the standpoint of Bohr's theory —that an electron which is revolving about the nucleus, and according to the classical laws ought to emit radiation of the same frequency as the revolution, can continue to revolve in its orbit without radiating.

physika, atyy, mattt and 1 other person
Great quote from Born. From which paper/book is this?

atyy
vanhees71 said:
Great quote from Born. From which paper/book is this?

atyy and vanhees71
The quantum equivalent of "does not move" is "has zero probability current everywhere". This is true of e.g. atomic orbitals, even the ones with non-zero angular momentum; and of course they all are imputed "kinetic energy" by the quantum virial theorem. So in QM, having kinetic energy or angular momentum does not imply "moves".

PeroK
H_A_Landman said:
"has zero probability current everywhere".

Let's consider free particle with ##\vec{p'} = 0##. But, we have nonzeroth probability current ##\vec{j} = \frac{\sqrt{|\psi|^2}\vec{p}}{m} =\mathrm{const} \neq \vec{0}## in many other frames of reference. Does it "move" or not? In this case, we have ## \frac{\partial |\psi|^2}{\partial t} = -\mathrm{div}\vec{j} =0##

Isn't it more correct to say that "moves", it means that it is ##\frac{\partial |\psi|^2}{\partial t} \neq 0 \rightarrow \mathrm{div}\vec{j} \neq 0##?

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vanhees71
A stationary state doesn't imply a vanishing probability current, as shown nicely by the example in the previous post. For stationarity it's by definition sufficient that ##\rho=|\psi|^2## is time-independent and that's the case for energy eigenstates (of Hamiltonians non explicitly time dependent).

## 1. What is the definition of motion in quantum mechanics?

In quantum mechanics, motion refers to the change in position and momentum of a quantum system over time. It is described by the wave function, which gives the probability of finding the system at a particular position and momentum.

## 2. How does motion differ in classical mechanics and quantum mechanics?

In classical mechanics, motion is described by Newton's laws of motion and can be predicted with certainty. In quantum mechanics, motion is described by the probabilistic nature of the wave function and cannot be predicted with certainty.

## 3. Can particles in quantum mechanics have both position and momentum at the same time?

No, according to the Heisenberg uncertainty principle, it is impossible to know both the exact position and momentum of a particle at the same time. This is a fundamental principle in quantum mechanics.

## 4. How does the concept of motion relate to the uncertainty principle?

The uncertainty principle states that the more accurately we know the position of a particle, the less accurately we can know its momentum, and vice versa. This means that the concept of motion in quantum mechanics is inherently uncertain and probabilistic.

## 5. Are there any experiments that demonstrate the probabilistic nature of motion in quantum mechanics?

Yes, the double-slit experiment is a famous example that demonstrates the probabilistic nature of motion in quantum mechanics. It shows that particles can behave like waves and have a range of possible positions and momenta, rather than following a specific path like in classical mechanics.

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