I used to think $\lvert\langle p\lvert\psi\rangle\rvert^2$ had the meaning of some likelihood of the particle's momentum being $p$ (within some tolerance interval $\Delta p$). Now I'm just confused.

I'll use momentum $p$, wavenumber $k$, and velocity $v$ interchangeably here (i.e., I'll use the convention $m=1$, $\hbar=1$). I begin with the momentum distribution at the time $t=0$. Further, let's limit ourselves to wavefunctions that are initially real, and also describe a stationary state in some potential well configuration.

I want to compare the evolution of the above state, with the evolution of the state of a free particle. As soon as the clock moves forward just an instant beyond $t=0$, these two different systems evolve completely differently.

I believe a theory that allows the probability distribution of velocities to depend on $\frac{\partial \psi}{\partial t}$ at time $t=0$ would make more sense than a theory by which the above distribution depends only on the initial form of the wave-function $\psi_0$.

A) For exemplification of the evolution of a free particle consider the spreading of a wave-packet with the initial state $\psi_0(x)= \exp(-\lvert x\rvert)$,

$$\psi(x,t)\propto\int \mathrm{d}k \; \frac{\exp(i k x-i \omega_k t)}{k^2+1}, \tag{1}$$ where $\omega_k=k^2/2$.

B) However, if the same initial state would describe the particle in the well, the particle is in a bound, stationary state, and $\lvert\psi(x,t)\rvert^2$ just remains constant over time. Nothing similar to the evolution $\text {(1)}$.

Discussion : My point was to pick an initial state (real, no current) and show that there is a great amount of current within a very short time, i.e. it seems to me that the difference between $|\psi_A (x,t)|^2$ and $|\psi_B (x,t)|^2$ happens just about instantaneously.

In the state $\text {(1)}$ we have no current of probability, $j(x,t)=0$. (The current is given by the density of probability times the Bohmian velocity $j(x,t)=\lvert\psi\rvert^2 v_{\text{avg}}$, where $v_{\text{avg}}=\frac{\hbar}{m}\frac{\partial}{\partial x}\,\arg(\psi)$ is a quantity central in Bohmian interpretations of QM.) But the zero-current doesn't mean there isn't any movement, simply the rate of particles moving to the bright counter-balances the rate of leftward going particles. And I have a hard time believing the distribution of velocities in the cases A and B is exactly equal.

(There are similar well problems, one being the distribution $\psi_0=\exp(-x^2)$ and compared with a particle in a ground state harmonic oscillator well. This would be a better one to use if we want to talk about the velocity distribution and the kinetic energy. And, since there are higher bound state energy states, about the classical limit.)

C) An additional thought: in the classical limit (large number of nodes within some space $\Delta x$), one should be able to talk about a velocity distribution near a position $x$. It should be something like $0.5\delta(v-v_c)+0.5\delta(v+v_c)$ where $v_c=\sqrt{2T/m}$. How does this distribution compare to any meaning we can gather from the positionless distribution $\lvert\langle p\lvert\psi\rangle\rvert^2$? Thinking classically I believe that a factor $\delta t$ times the difference in acceleration between the two systems should explain the difference in evolution.

Would there be any way to obtain a position dependent velocity distribution, or even a quantity similar to this, that would agree in the classical limit?


What prevents us from talking about the proabibility to find a particle with momentum $\mathbf p$ at position $\mathbf x$ in quantum mechanics? It is of course the non-commutativity of $\hat x$ and $\hat p$, or in more physical terms, the Heisenberg uncertainty principle $$\Delta x \Delta p \ge \frac{\hbar}{2}.$$ This is why the Wigner function mentioned by Void is not a proper non-negative probability distribution.

But let us try to minimize the effect of Heisenberg's principle. Recall that $$I = \int \phi(x)^*\psi(x) \, dx$$ is a measure of the overlap of the states $\phi$ and $\psi$. In particular the probability to find the state $\phi$ is $P = |I|^2$. Now, consider the following state $$\phi(x; p,x') = C_1\exp\big[ -C_2(x-x')^2 + ipx] $$ where $C_1$ and $C_2$ are some unimportant normalization constants. It is not too hard to show that the expectation values of $\hat x$ and $\hat p$ in this state are $x'$ and $p$. Furthermore, for this state $\Delta x \Delta p = \hbar/2$, meaning that the uncertainty is minimal. Thus define the Husimi function $$Q(x',p) = \left | \int \phi(x; x', p)^* \psi(x) \, dx\right|^2$$ which then gives the probability to find a minimum uncertainty state with momentum $p$ at $x$.

The Husimi function can be related to the Wigner function and vice versa, and can be used to calculate anything the Wigner function can be used for.

  • $\begingroup$ You explanation is a thousand times easier to understand than what I could find on wiki (I never really quite wigner distros--wiki was my only source-- and hadn't heard of Husimi till today!). It looks like an inner product with a moving/modulated gaussian, and that should intuitively weigh the momentum more to what is near x'. (I'm not sure if there is a * missing on the $\phi$.) This Q(x',p) really looks like it may be what I'm looking for; thanks a billion $\endgroup$ – bert Feb 22 '15 at 6:13
  • 2
    $\begingroup$ +1, I did not think about this. It should be noted that this $Q(x',p)$ is non-unique. We have $C_2 = 1/(4(\Delta x)^2) = (\Delta p)^2$ ($\hbar=1$); choosing different $\Delta p$ you will use a different basis and get a different $Q(x',p)$. When you take $\Delta p \to 0$, you will have $Q(x',p) \to \rho(p)$ and when $\Delta p \to \infty$ you will get $Q(x',p) \to \rho(x')$. The Husimi function is useful if you have a well-defined basis into which to project your wavefunction and can be used as a tool to obtain the quasi-classical limit but one should beware of it's ambiguity. $\endgroup$ – Void Feb 22 '15 at 9:33

I used to think $\lvert\langle p\lvert\psi\rangle\rvert^2$ had the meaning of some likelihood of the particle's momentum being $p$ (within some tolerance interval $\Delta p$). Now I'm just confused.

$\rho(p)=\lvert\langle p|\psi\rangle\rvert^2$ is just another notation for $\lvert\tilde{\psi}(p)\rvert^2$ where

$$ \tilde{\psi}(p) = \frac{1}{\sqrt{2\pi\hbar}}\int \psi(q)e^{-i\frac{pq}{\hbar}}dq. $$

It is assumed (and mathematically it works well) that just as $\int_{q_1}^{q_2}|\psi(q)|^2dq$ gives probability that $q\in(q_1,q_2)$,

$$ \int_{p_1}^{p_2}\lvert\tilde{\psi}(p)\rvert^2dp $$

gives probability that $p\in(p_1,p_2)$-

However, I have a hard time believing the distribution of velocities of these two cases are exactly equal.

Take it is a prescription to calculate the distribution of velocities the best way we can. It does not necessarily reflect actual velocities.

In the classical limit (large number of nodes within some space $DX$), one should be able to talk about a velocity distribution near a position $x$. It should be something like $0.5\delta(v-v_c)+0.5\delta(v+v_c)$ where $v_c=\sqrt{2T/m}$. How does this distribution compare to any meaning we can gather from the positionless distribution $\lvert\langle p\lvert\psi\rangle\rvert^2$?

If you have $q$-dependent probability distribution for $p$, you can always calculate marginal probability distribution for $p$ (disregarding value of $q$) as

$$ \mu(p) = \int_{-\infty}^{\infty} \rho_q(p) dq $$ and compare it to $\rho(p)$. If $\mu(p)$ and $\rho(p)$ do not agree, you can then think why that is and which one you prefer as a better description.

  • $\begingroup$ You didn't answer him why the wave function evolves in the case A differently than in the case B. Both are quantum cases. He believes that one and the same wave-function can be initial wave-function (of which I am not sure, but let it be). And he asks why the velocity distribution will evolve differently. $\endgroup$ – Sofia Feb 21 '15 at 18:31
  • $\begingroup$ @Sofia, I did not see any such questions in the first post. $\endgroup$ – Ján Lalinský Feb 21 '15 at 19:38
  • $\begingroup$ It's O.K. I placed a comment that completes the issue. $\endgroup$ – Sofia Feb 21 '15 at 19:42
  • $\begingroup$ I think this helps, what I'm looking for is a $\rho_q(p)$ that would need to agree with $\rho(p)=|\langle p| \psi\rangle|$ when integrated over q as above. $\endgroup$ – bert Feb 21 '15 at 21:36

Interpreting the probability current $\vec{j}$ as an actual current determining a kind of Bohmian velocity of a stream of fluid (statistical ensemble) cannot be mixed with the usual interpretation of quantum mechanics, most importantly it cannot be mixed with the principle of superposition.

Consider the following example: we take a wavefunction $\sim e^{-ikx}$, this wavefunction would represent a streaming ensemble of velocity $k$ in the Bohmian picture ($\hbar=m=1$). Let us now superpose it with an ensemble of velocity $-k$, i.e. $\sim e^{ikx}$, so that we get counterstreaming particles. That means that from the Bohmian point of view we will get two counter-streaming ensembles which give a zero velocity and a flat spatial probability distribution, right? Not quite. Our resulting wave-function is $$\psi_{\pm k} \sim \cos(kx)$$ with the probability distribution $\sim 1+cos(2kx)$. The Bohmian trajectories are suddenly not peacefully streaming but frozen between the nodes of $\cos(kx)$. This is a picture very different from what a naive Bohmian-like interpretation would give! There is no way the particles could be streaming through the nodes of zero particle probability so we can only say particles are somehow oscillating between the nodes of $\cos(kx)$.

Either way, you can see that combining the concept of an $e^{-ikx}$ streaming ensemble with superposition does not give anything satisfactory. Superposition and a "beable" interpretation of the wavefunction just do not work together. "Beables" are non-linear and every "linearly" asked question will not provide any good answer.

Taking $\langle p|\psi \rangle$ with a Bohmian-like mindset is like asking "How much of a uniform $p$-stream ($\sim e^{-ipx}$) is linearly present in this ensemble $\psi$?" and you get a mathematical answer which has no good meaning in sense of the non-linear physics you believe to be true. You only get a useful Fourier transform of this peculiar function $\psi$ of your theory.

There is one "quasi-probability" function on the phase space which I find pretty neat, the Wigner quasi-probability distribution $P(p,x)$. It has all the nice properties which you would expect from such a distribution such as $\int P(p,x) dp = \rho(p), \int P(p,x) dx = \rho(x)$. It's really great, but... it is not positive definite. Nevertheles, it is sometimes used in the classical limit.

The truth is that the quantum state cannot be interpreted in terms of classical objects but it should be understood on it's own terms.

The true classical counter-part of the quantum state is the sharp position and momentum of the particle, because the quantum state is the basic building block of your theory, the sharpest you can get, the full characterization of the system and it's evolution. And the interpretation and convergence to the classical limit should be done only by checking the correspondence between $x \to \langle \hat{X}\rangle, \langle (\hat{X})^2\rangle - (\langle \hat{X}\rangle)^2 \to 0$ etc.

On the other hand, the quantum counter-part of a classical distribution is the density operator. In classical mechanics you have a distribution over the sharp objects of your theory, the position and momenta. In quantum theory also, the only difference being that you have a distribution over quantum-sharp objects, the states. As states converge to sharp positions and momenta, so does your density operator converge to a classical probability density.

Am I cursed or do I take too long to complete an answer so that another one gets accepted in the meantime? Anyways, enjoy an alternate viewpoint.

  • $\begingroup$ I think in any interpretation you will always have these interference effects; phase and interference effects seem like the main driver for quantum phenomena. For standing waves with no current, the nodes of the wavefn do give you lots of information on the phase; if you think of $\psi$ as superposition of two streams of equal magnitude and opposite direction, $\psi_{\hbox{Righwards}}$ and $\psi_{\hbox{Leftwards}}=\psi_{\hbox{Righwards}}^*$, it tells you that that the phase of these components are at $\pm \pi$ at the nodes (with the rare exception of case where the components have 0 magnitude. $\endgroup$ – bert Feb 22 '15 at 21:01

The time evolution can be analyzed by applying the time evolution operator, i.e.

$$ \hat U=e^{i\hat Ht} $$

Since the outcome of $\hat H\left|\psi\right>$ is different in the two cases you have stated (the Hamiltonian describing the problems are different) the states evolve differently. The expectation value for the momentum will reflect this situation, i.e.

$$ \left<\hat p(t)\right>=\left<\psi(t)\right|\hat p\left|\psi(t)\right> $$

will have different dependencies in the two cases and so the distribution of the momenta will change as the expectation value $ \left<\hat p(t)\right>$ changes.

  • $\begingroup$ OP knows how to use the evolution operator as is obvious from eq. (1). Furthermore, $\langle p\rangle$ is equal in this case which is kinda the point of the question. $\endgroup$ – Void Feb 21 '15 at 20:30
  • $\begingroup$ @Void I have to take back what I said about formula (1). I am sorry, it is correct. $\endgroup$ – Sofia Feb 22 '15 at 12:32

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