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I'm a bit confused by an answer given on this question. In the answer with the animation of a moving free (chargeless) particle and a non-moving free particle (or a free particle with a non-zero momentum expectation value and one with a zero momentum expectation value), the wave function of the non-moving free particle stays stationary.

Now, I always thought that the (symmetric) wavefunction of a non-moving free particle isn't moving to the left or right, but that it does spread out in the passage of time. The expectation value of the momentum stays zero, but the wave function spreads out in space (while the spreading of the wave function in momentum space gets less (as reflected in the uncertainty principle). While the wavefunction in configuration space spreads out, the wavefunction in momentum space gets "thinner" and higher (approaching a Dirac delta function).

So my simple question is: What's going on?

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  • $\begingroup$ Maker of the GIF here: the first picture is not animated while the second one is. Figured it would be the clearest. $\endgroup$ – user3502079 Jun 14 at 21:06
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Clearly you are talking about a free particle. The most general Schrödinger wave function then is a linear combination of plane waves $ \Psi(\vec r, t)= N \int d\vec k\ \psi(\vec k) e^{i(\vec k \cdot \vec r -\frac{\hbar k^2}{2m} t + \varphi(\vec k ))} $ where the amplitude of each wave $\psi(\vec k)$ and its phase $\varphi(\vec k)$ are free to choose. $N$ is the normalisation factor. The wave function describes a particle at rest if the expectation value of $\vec p = -i\hbar \vec \nabla$ vanishes. Now $<\vec p > = N \int d\vec k\ |\psi(\vec k)|^2 \hbar \vec k$. It is clear that under this restriction $\Psi$ in general still depends on time. The probability density $ |\Psi(\vec r, t)|^2 $ will evolve with time. If $ |\Psi|^2 $ is a gaussian "wave packet" at $t=0$ then its standard deviation will increase indefinitely with time. For a detailed derivation of the spread see https://quantummechanics.ucsd.edu/ph130a/130_notes/node83.html .

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The question shouldn't be about a moving particle. Quantum mechanics has told use that trajectories aren't the right way to think about quantum particles. Indeed, the notion of a trajectory isn't a well defined concept in quantum mechanics.

The time evolution of the wave function is completely governed by the Hamiltonian (in particular the Potential $V(x)$).

Concerning your statement

Now, I always thought that the (symmetric) wavefunction of a non-moving object isn't moving to the left or right, but that it does spread out in the passage of time. The expectation value of the momentum stays zero, but the wave function spreads out in space (while the spreading of the wave function in momentum space gets less (as reflected in the uncertainty principle).

You seem to be generalising a very specific kind of application of schodinger's equation to all wave functions. You also seem to be combining facts about the wave function from two different potentials. Namely the notion of a wave function spreading out. This is true for a free particle, but in general this is not the case (e.g. the eigenstates of the infinite square well). Also, in the case of the free particle, the wave packet does have a group velocity and so we can regard it as “moving”, whereas you say that the wave packet will have no group velocity yet still spread out over time. Note: this does't have anything to do a "moving" particle, but rather a “free” particle.

Who said that the average momentum of a free particle has to be zero? In general, the expectation of momentum of a free particle is

$$\langle k\rangle= \int_{-\infty}^{+\infty} \mathrm dk\ k\ |\phi(k)|^2 \tag{1} $$

where $\phi(k)$ are the fourier coefficients of the plane wave expansion

$$\Psi=\frac{1}{\sqrt{2\pi}}\int \mathrm dk\ \phi(k)\;\mathrm e^{i(kx-\omega t)} \tag{2} $$

Indeed, in general the expectation of momentum of a free particle is in general nonzero.

Conclusion: It doesn't really make sense to talk about "moving/nonmoving particle" wave function. The closest we can get is the group velocity of the wave packet. Moreover, the time evolution of the wave function is not governed by anything have to do with “motion”, but rather the Hamiltonian. Different potentials will determine the different time evolutions of the wave function. Given how broad your question is "What's going on", it's a bit difficult to answer you further. I can only point out the conceptual errors that I mentioned above.

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  • $\begingroup$ I forgot to write down that I was indeed talking about a free particle. Sorry for that! And indeed, nobody says that the average momentum of a free particle has to be zero, but I assumed this in relation to the answer on a question about moving and non-moving objects. But anyhow, thanks for your answer. $\endgroup$ – descheleschilder Jan 3 at 8:18
  • $\begingroup$ It make sense to talk about a non-moving particle, in the sense of zero momentum expectation value. The increase of its position uncertainty with time should not be interpreted as motion. $\endgroup$ – my2cts Jan 3 at 22:34
  • $\begingroup$ That's fine and I agree. But it seemed that OP was thinking that a moving particle will have a particular kind of wave function, when in reality we may ascribe "motion" to any particle with a nonzero expectation of momentum (e.g. the harmonic oscillator), but this doesn't mean that it's positional uncertainty will decrease over time. $\endgroup$ – InertialObserver Jan 3 at 22:41
  • $\begingroup$ I was indeed writing about a free particle that has a zero momentum expectation value (like a particle with a normalized Gaussian wavefunction associated with it). And the increase in the position uncertainty (the spreading of the Gaussian) of course doesn't imply the particle is in motion. It's only the Gaussian that's in (symmetric) motion. $\endgroup$ – descheleschilder Jan 4 at 22:41
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I think you might be misunderstanding two points about the Uncertainty Principle, and the reason I think that lies in your following sentence:

The expectation value of the momentum stays zero, but the wave function spreads out in space (while the spreading of the wave function in momentum space gets less (as reflected in the uncertainty principle).

The two points are these: First, the Uncertainty Principle does not apply to the expectation values of the particle in configuration space and momentum space. It applies to the expected deviations in each space from the expectation values. The principle states that the product of the expected deviations can't be smaller than a certain amount.

And that leads me to the second point: The Uncertainty Principle isn't an equality, it's an inequality. While the deviations of the configuration and the momentum cannot both be arbitrarily small, the Uncertainty Principle gives us no reason why they can't both be arbitrarily large.

So to finally answer your question about what's going on, the expectation value of x remains some constant point over time, the expectation value of p remains zero over time, and the expected deviation of both from their expected values increases over time. And the Uncertainty Principle, being an inequality, likes this just fine.

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I'm interpreting your "non moving particle" as "zero momentum expectation value".

Now, I always thought that the (symmetric) wavefunction of a non-moving particle isn't moving to the left or right, but that it does spread out in the passage of time.

You're right. To be precise it's not necessarily so - the wave function could also be shrinking for some time, then begin to spread. But the initial w.f. must be cleverly arranged and corresponds to a state practically impossible to prepare.

The expectation value of the momentum stays zero, but the wave function spreads out in space (while the spreading of the wave function in momentum space gets less (as reflected in the uncertainty principle).

The last statement is wrong. For a free particle momentum is a constant of the motion, so that all average values like $\langle p \rangle$ and $\langle p^2 \rangle$ are constant.

Remember that the uncertainty relation is an inequality. Usually it's far from being an equality. In the present case just that is happening: $\Delta x$ increases whereas $\Delta p$ stays constant.

I could give further details, if you want them. But at a later time, perhaps tomorrow.

Edit

$\let\a=\alpha \let\b=\beta \let\dag=\dagger \def\ket#1{|#1\rangle} \def\bra#1{\langle#1|} \def\avg#1{\langle #1 \rangle} \def\D#1#2{{d#1 \over d#2}} \def\DD#1#2{{d^2#1 \over d#2^2}} \def\PD#1#2{{\partial#1 \over \partial#2}} \def\dx{\dot x} \def\ddx{\ddot x} \def\bt{\bar t}$ It's necessary to set notations. Many choices are possible, mainly a matter of taste. I'll adopt the following:

  • Operators are written as simple letters (lower-case or capital)

Kets and bras are labelled

  • by a generic label with no attached meaning (I'll use here greek letters): $\ket\a$
  • by a time variable: $\ket{\a,t}$.

Time evolution. We have $$\ket{\a,t} = T(t)\,\ket{\a,0} = e^{-i\,H\,t/\hbar}\,\ket{\a,0} \tag1$$ where for a free particle $$H = {p^2 \over 2m}.\tag2$$

We could compute the w.f. at a generic time $t$, in order to show how it gets spreading in time. The expression one finds is quite complicated - not a simple widening of its initial shape. But there is a much easier way to directly compute $\avg x$ and $\avg{x^2}$ as functions of time. We have only to switch to Heisenberg picture.

Consider a matrix element of some operator $F$, between time dependent states $\ket{\a,t}$, $\ket{\b,t}$. We have $$\bra{\a,t}\,F\,\ket{\b,t} = \bra{\a,0}\,T^\dag(t)\,F\,T(t)\,\ket{\b,0} = \bra{\a}\,F_t\,\ket{\b}$$ where $$F_t = T^\dag(t)\,F\,T(t) \tag3$$ is Heisenberg (time dependent) operator whereas $\ket\a$, $\ket\b$ are Heisenberg (time independent) state vectors. I dropped the label "0" now useless. (Note that $H_t=H$ and the same holds, for a free particle, if $F$ is a function of $p$.)

Differentiating (3) wrt $t$ and using (1), (2) we get $$i \hbar\>\D{}t\,F_t = [F_t,H] = {1 \over 2m}\,[F,p^2].$$ In particular $$i \hbar\>\D{}t\,x_t = {1 \over 2m}\,[x,p^2] = i\,{\hbar \over m}\,p.$$ $$\D{}t\,x_t = {p \over m}.\tag4$$

Taking the expectation value (EV) on $\ket\a$: $$\D{}t\,\avg{x_t}_\a = {1 \over m}\,\avg p_\a = 0$$ if $\ket\a$, as we assumed, is a "non moving" state ot the particle. Then $$\avg{x_t}_\a = \avg x_\a = 0.$$ (There is no special restriction in taking 0 as the EV for $t=0$; it's only a matter of choosing the coordinates' origin.)

Let's compute $\avg{x^2_t}_\a$. The first step is $$i\,\hbar\>\D{}t\,x^2_t = {1 \over 2m}\,[x^2_t, p^2] = i\,{\hbar \over m}\,(x_t\,p + p\,x_t)$$ $$\D{}t\,x^2_t = {1 \over m}\,(x_t\,p + p\,x_t) = x_t\,\dx_t + \dx_t\,x_t.$$

A second differentiation gives $$\DD{}t\,x^2_t = x_t\,\ddx_t + \ddx_t\,x_t + 2\,(\dx_t)^2.$$ But eq. (5) implies $\ddx_t = 0$ ($p$ is a constant of the motion) so that $$\DD{}t\,x^2_t = 2\,(\dx_t)^2$$ and all further derivatives vanish.

So we may write $$x^2_t = x^2 + (x\,\dx + \dx\,x)\,t + \dx^2\,t^2 = x^2 + {1 \over m}\,(x\,p + p\,x)\,t + {1 \over m^2}\,p^2\,t^2 \tag5$$ where $x$, $\dx$, $p$ are computed at $t=0$, i.e. coincide with the Schrödinger picture operators.

Now it's time to examine the choice of the initial w.f. We already know it has to satisfy $\avg p_\a = 0.$ But we are always free to require $\avg x_\a = 0$ too. This is because if a randomly chosen w.f. didn't satisfy that condition, we'd only to shift the $x$-origin to have it satisfied.

A subtler issue comes into play from $\avg{x\,p+p\,x}_\a$. Might we safely assume it also vanishes? To answer, remind that this operator has a constant derivative, positive definite (it's $2\,p^2/m$). Then if its EV doesn't vanish for $t=0$ it will certainly do at some other time $\bt$. And we have only to take as initial w.f. the one at $t=\bt$ to ensure $\avg{x\,p+p\,x}_\a=0$. But there's another way out. In order to fulfill our condition a real w.f. is enough. To see that, let's work in Schrödinger representation: $$\eqalign{\avg{x\,p+p\,x}_\a &= -i\hbar\!\int\!\!dx \left[\psi_\a^*\>x\,\PD{\psi_\a}x + \psi_\a^*\,\PD{}x (x\,\psi_a)\right] \cr &= -i\hbar\!\int\!\!dx \left[\psi_\a^*\>x\,\PD{\psi_\a}x - \PD{\psi_\a^*}x\>x\,\psi_a\right]\!.\cr}$$ The integral obviously vanishes if $\psi_\a$ is real.

Let's summarize. Taking EV's of eq. (5) $$\avg{x^2_t}_\a = \avg{x^2}_\a + {1 \over m^2}\,\avg{p^2}_\a\,t^2 \tag6$$ Eq. (6) shows that $\avg{x^2_t}_\a$ indefinitely increases with time, whereas $\avg{p^2}_\a$ stays constant.

I just have to prove my previous statement: the wave function could also be shrinking for some time, then begin to spread. To show this a short parenthesis is needed, i.e. a well-known theorem.

If $\psi(x,t)$ is a solution of TDSE for a free particle, so is $\psi^*(x,-t)$

The theorem holds for more general $H$ too, but this simple form is enough for our problem. To prove it you have only to take complex conjugate of TDSE and look at the result.

Now look at eq. (6). It shows that for all $t>0$ $$\avg{x^2_t}_\a > \avg{x^2}_\a.$$ Then take a $\bt>0$ at your pleasure and define $$\psi_\b(x,0) = \psi_\a^*(x,\bt).$$ The above theorem ensures us that TDSE solved with $\psi_\b(x,0)$ as initial condition starts with $$\avg{x^2_0}_\b = \avg{x^2_\bt}_\a$$ and at time $t=\bt$ $$\avg{x^2_\bt}_\b = \avg{x^2_0}_\a < \avg{x^2_0}_\b.$$ Only for $t>\bt$ does $\avg{x^2_t}_\b$ begin to increase.

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  • $\begingroup$ But ain't it so that if the wavefunction (for a particle with zero momentum expectation value) becomes, in the course of time, flat (zero) everywhere in space (but still being normalized), the momentum operator $\vec p = -i\hbar \vec \nabla$ operating on this function is zero everywhere, which means that the uncertainty in momentum becomes zero (while the uncertainty in (x,y,z) becomes infinite)? In other words, the wavefunction in momentum space becomes a delta function (distribution) at the origin? $\endgroup$ – descheleschilder Jan 4 at 22:59
  • $\begingroup$ But ain't it so that if the wavefunction (for a particle with zero momentum expectation value) becomes, in the course of time, flat (zero) everywhere in space (but still being normalized), the momentum operator $\vec p = -i\hbar \vec \nabla$ operating on this function is zero everywhere, which means that the uncertainty in momentum becomes zero (while the uncertainty in (x,y,z) becomes infinite)? In other words, the wavefunction in momentum space becomes a Dirac delta function (distribution) at the origin? $\endgroup$ – descheleschilder Jan 4 at 23:09

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