Where's the rest of it? - Schrödinger equation for a free particle whose position is known

Question

Assumptions

• I will be working in one spatial dimension throughout. It should be possible to generalise to three dimension with relative ease but doing so adds little to the conversation, only serving to make calculation more difficult.

• Elementary particles are necessarily without any measure or internal structure whatsoever. This means that if an elementary particle ever has a position, then that particle exists only at a single point in space. From this, we may infer that if a particle's position at time $t$ is known to be $y$, then its wavefunction $\psi$ must be such that $|\psi(x,t)|^2 = \delta(x-y)$.

• The speed of a particle with positive rest mass, as measured in any reference frame, is always less than $c$, the speed of light. Thus, if a [massive] particle's position is $y$ at time $t$ and $y'$ at time $t'$, then $\left\vert\frac{y'-y}{t'-t}\right\vert< c$.

Notation

• Because the letter $\phi$ is commonly used for both phase and momentum, I would like to use $\varphi$ to represent phase and $\phi$ to represent momentum.

The Problem

Suppose that we measure the position of a free particle, assigning this position a value of $0$ and likewise marking the time as $0$. We then have the probability density function for the particle's position given by...

$$|\psi(x,0)|^2=\delta(x)$$

...from which we may infer...

$$\psi(x,0)=\delta(x)e^{i\varphi_0}\tag {eq.1}$$

where $\varphi_0$ is the initial phase of $\psi$.

From here, we wish to predict the particle's position at an arbitrary future time $t$. To do so, we must solve the time-dependent Schrödinger equation...

$$i\hbar\frac\partial{\partial t}\psi(x,t)=-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi(x,t)\tag {eq.2}$$

...with initial condition (eq.1). Fortunately, a solution to this problem already exists.

First Solution

The first solution is given by Emilio Pisanty here:

$$\psi(x,t)= \begin{cases} \delta(x) & t=0\\ \sqrt{\frac m{2\pi\hbar t}}\exp\left[i\left(\frac {mx^2}{2\hbar t}-\frac\pi4\right)\right] & t>0\end{cases}$$

Somewhat problematically, this solution is non-normalizable (we'll get into why this is problematic later). It does, however, indicate that 1) due to uncertainty, the position of the particle at time $t>0$ is always unknown and 2) as time goes on, the probability of finding the particle near its original location diminishes, as though it were "spreading out" into infinity. This second point provides some of the intuition behind the reasoning of this question.

Second "Solution"

The second solution - or rather, half solution - is the result of my feeble attempt at a normalizable solution. To begin with, we may consider our initial condition as the limiting case of a Gaussian distribution...

$$|\psi(x,t)|^2=\frac 1{\sigma(t) \sqrt {2\pi }} \exp\left(-\frac {x^2}{2\sigma(t)^2}\right)$$

...where $\sigma(0)=0$. From this, we have...

$$\psi(x,t)=\frac1{\sqrt{\sigma(t)\sqrt 2\pi}}\exp\left(-\frac{x^2}{4\sigma(t)^2}+i\varphi(t)\right)$$

We can feed this back into (eq.2), skipping the usual transform to and from momentum basis, to get...

$$x \hbar\sigma(t)-2 i m x^2 \frac\partial{\partial t}\sigma(t)+2 i m \sigma(t)^2\frac\partial{\partial t}\sigma(t)+4 m \sigma(t)^3 \frac\partial{\partial t}\varphi(t)=0\tag{eq.3}$$

...which we can then solve for pairs of functions $(\varphi,\sigma)$ which meet the desired conditions. Solving in the position basis is necessary because transforming a solution from the momentum basis will yield a non-normalizable solution again. Provided that solutions to {(eq.3), $\sigma(0)=0$} exist (pretty sure they don't, but I'm not writing the proof), we can use this to cheat uncertainty by avoiding momentum entirely.

Why both solutions (and almost all other solutions) are wrong

In both cases, we know that the exact location of the particle at time $t=0$ is $0$. This means that in order for the particle to be observed at some position $x'$ at time $t'$, it must possess, at some point, a speed of at least $|x'/t'|$.

Because this speed cannot exceed $c$, even if we have no idea where the particle is, we know that it isn't outside of the region $(-ct,ct)$. Hence, at any time $t$, $\psi(x,t)=0$ for all but countably many $x\notin(-ct,ct)$.

This is not the case if $\psi$ is non-normalizable (as in solution 1), and not necessarily the case even when $\psi$ is (as in solution 2, ostensibly), but strictly requires that the essential support of $\psi$ be compact. Specifically, we have the condition that $\operatorname{supp}(\psi(\cdot,t))\subseteq[-ct,ct]$. This condition is not imposed in any obvious way by (eq.2), which leads me to the following:

Either

1. (eq.2) is incomplete. There are additional terms and/or equations required which impose the restriction $\operatorname{supp}(\psi(\cdot,t))\subseteq[-ct,ct]$ on solutions whenever the initial position of the particle is known exactly. Additionaly, we have that $\psi(x,t)=0$ for all $x\notin(-ct,ct)$ when $m>0$, for all $x\notin[-ct,ct]$ when $m=0$, and possibly some other condition when $m<0$.

2. At least one of my assumptions is incorrect

   2a. Due to some quirk of the mathematics involved (e.g. a hidden cross product or anything involving quaternions), physically relevant solutions to the Schrödinger equation are only possible in 3+1 dimensions.

   2b. The notion of "position," in the classical sense, does not apply to quantum particles at all. Even measurement of a particle's position only yields a probability density function with a maximum located at the point it was observed. In other words, the initial condition $\psi(x,0)=\delta(x)e^{i\varphi_0}$ is invalid.

   This implies that it is possible to directly observe an "individual" particle simultaneously appearing at two separate points in space, and that particles can be detected at any distance (though the likelihood of detection decreases as a function of that distance).

   2c. Quantum mechanics permits violations to the cosmic speed limit. In this case, such violations ought to be ubiquitous over short distances and cosmic time scales. I would be very interested in any experimental evidence of this phenomenon.

3. Both of the above.

or

4. I have neglected something hugely fundamental to quantum mechanics, without which this question is ill-formed and irrelevant. Once I'm sure what it is (and no, it isn't the Fourier Transform), I can delete this question.

(There is of course, the possibility of literally anything else, but this hints at the terrifying truth that we don't actually know how the universe works; and since I don't particularly feel like questioning the entirety of modern physics at this moment, let's just assume that quantum mechanics is "mostly right," or at least a good starting point.)

Answer 1

Beware of using the word "measurement" to smuggle in an ill-defined or impossible physical process. The notion that there could be a position measurement resulting in a delta-function wavefunction is an example of an unphysical idea. Even if we replace the delta-function by a function of finite height and very small width, the whole calculation still goes wrong because now you have enough kinetic energy to make electron-positron pairs. So the calculation has to be done via quantum field theory.

One can generalise as follows. If we suppose a particle is confined to a region of width $a$ then the standard deviation of its momentum will be lower-bounded by $$ \Delta p \ge \frac{\hbar}{a} $$ (Heisenberg uncertainty principle). If the mean momentum is zero then it follows that the squared momentum is of order $(\hbar/a)^2$ and therefore the kinetic energy is of order $$ (\gamma - 1) m c^2 $$ where $$ \gamma \simeq \left(1 - \left(\frac{\hbar}{a m c}\right)^2 \right)^{-1/2} $$ where $m$ is the particle's rest mass. At the level of approximation involved we don't need to worry about the details; the Heisenberg uncertainty principle gives the right order of magnitude. This kinetic energy is enough to provide the energy to create a particle-anti-particle pair if $$ (\gamma - 1) m c^2 \simeq 2 m c^2 $$ i.e. $\gamma \simeq 3$. This happens when $\hbar \simeq a m c$, so the result is that particle confinement to a region smaller than $$ a \simeq \frac{\hbar}{m c} $$ is always questionable. And 'confinement' here does not need to mean 'permanent confinement'; it could include any process which temporally caused the particle's position wavefunction to be spread by $\hbar/mc$ or less. The combination $h / mc$ is also called the "Compton wavelength" because it comes up in the Compton effect, but its role here is perhaps its first importance. (See wiki for more info).

So, to repeat, once we are in this regime, non-relativistic quantum theory is no longer sufficient.

Similarly, to explore the notion of faster-than-light changes in wavefunctions, to be sure you are making no mistake, you need to go to a relativistic theory, and that is quantum field theory once again. And quantum field theory is different from non-relativistic quantum mechanics in profound ways.

Answer 2

Your answers 1 and 2b are roughly correct. The schrodinger equation you have written down is non relativistic. To capture relativity you must use quantum field theory (this is sort of your answer 4.

2b is also relevant because Dirac delta functions are generally not thought to be physical wave functions in the same way that pure sine waves (infinite in time, Dirac delta Fourier transform) are not physical signals. It is useful to include them In the mathematical formalism but only infinite sums of Dirac delta functions that are integrated over with some measure (eg a normal function like a Gaussian) are physical wave functions. These don’t have quite the same relativity problems that you raise but they may still have some.

But again, the question is only fully resolved when you mean to qft. There the mode function and some commutators are defined such that adherence to special relativity is built in.

Answer 3

The Schrödinger equation (with the hamiltonian used here) is a non-relativistic equation, so it does not obey the speed-of-light limit.

Starting with a particle in a position eigenstate, it contains all possible momentum values: the Fourier transform of the delta function is a constant, i.e. it is "equally likely" to take on any real number as a momentum, in some sense (although this is a non-normalizable momentum space wavefunction, so it's pretty unphysical). So, there is some amplitude for it to travel arbitrarily far in an arbitrarily short time period. Correspondingly, the position space wavefunction has infinite support for any $t > 0$.

So the solutions are not really wrong; they are correct within the confines of the mathematical framework you are using. To talk about relativistic particles you would have to try to switch to a relativistic hamiltonian, and then this rapidly leads you into the realm of quantum field theory.

In relativistic QFT one can indeed show that spacelike-separated field operators commute, which is QFT's way of expressing the idea that influences can't propagate faster than light through the field (so-called "microcausality").