Wave Function Collapse and the Dirac Delta Function

Question

When the wave function of a quantum system collapses, the probability of finding it at some specific point is given depends on $||\Psi||^2$: $$ \int_{\mathbb{R}^3}{d^3 \mathbf x \; |\Psi|^2} = 1 $$ Could this modulus square, the instant you measure, be thought as the Dirac Delta Function, because all the probability condensates to a single point, and its integral over all $\mathbb{R}^3$ gives 1. $$ |\Psi|^2 = \delta(\mathbf x)\\ \int_{\mathbb{R}^3}{d^3 \mathbf x \; \delta(\mathbf x)} = 1 $$

If yes, what are the initial conditions the wave equation must have the instant after being collapsed. The first one shall be this: $$ \Psi(\mathbf x, t) \\ |\Psi(\mathbf x, 0)|^2 = \delta(\mathbf x) $$ No? How would you plug this condition onto the Schrödinger Equation?

Answer 1

That is a good question; this exposes an idealization in the textbook formalism of Quantum Mechanics which is known not to apply exactly in real life. Notably, you realized that if the probability density is

$$|\Psi(\vec{x},0)|^2 = \delta(\vec{x})$$

Then what is the wave function itself? We don't have a square root of a delta function. Often the wave function might just be taken as a delta function itself, but this is of course not consistent because $\delta(\vec{x})^2 \neq \delta(\vec{x})$.

The answer, therefore, is that the wave function after measurement is not an exact delta function. This can also be seen by other means, and in fact there is no less than direct experimental evidence which is in contradiction with an exact delta function. Consider the way that momentum is measured by pixel detectors at the LHC:

The particle to detect traverses through a constant magnetic field, which induces circular motion. The volume of the detector contains many tiny pieces of silicon called pixel detectors. As the particle passes through these penetrable pieces of silicon, its position is recorded many times. These positions can be fit to a circle, and the radius of the circle determines the momentum through the simple relationship

$$|\vec{p}| = q|B|r$$

Now, what would we see if, after the first position was recorded, the wave function collapsed to an exact delta function in position? In this case, its momentum wave function would be

$$\psi(\vec{p}) = \langle \vec{p}|\psi \rangle = \langle p|x \rangle = e^{-i\vec{p} \cdot \vec{x}}$$

Which gives us

$$|\psi(\vec{p})|^2 = |e^{-i\vec{p}\cdot\vec{x}}|^2 = 1$$

Thus each momentum in $\mathbb{R^3}$ would be equally likely. We would expect a huge jolt in the motion of the particle, even to the extent that it could be found macroscopic distances away in fractions of a second. However, this is not what we see. Here are some reconstructed tracks from the LHC:

As you can see, we have particles moving in nicely-defined circles of relatively constant radii. Neutral particles are of course not bent by the magnetic field. This is not consistent with the infinite jolt that a particle which suddenly transitioned to an actual delta function would experience.

The delta function is just an approximation. If you want something "close enough", you can try

$$\psi(\vec{x},0) = \delta(\vec{x})$$

or else plug in a gaussian whose width is roughly equal to the experimental uncertainty, as a band-aid. But you are looking to know what is correct "in principle", you would have to look outside of textbook quantum mechanics.

Answer 2

When the wave function of a quantum system, collapses,

Collapse is a misleading word, the wavefunction is not a balloon. It is a mathematical solution of a specific wave equation with the specific potentials and boundary conditions of a specific problem. "Collapse" means that those boundary conditions and potentials are no longer valid, and an instance is "picked" (measured)from the original $Ψ^*Ψ$ probability distribution

what are the initial conditions the wave equation must have the instant after being collapsed.

a NEW $Ψ$ should be used dependent on the new boundary conditions after the measurement.

Take this one electron at at time double slit experiment.

Each dot is a point from the probability distribution of the wavefunction solution " electron of given four momentum scattering through double slits of given width and distance apart". Once it hits the screen it has "collapsed the original wavefunction". It is a different wavefunction solution that would describe the dot, that has nothing to do with the original wavefunction of the electron before the hit.

It is the accumulation of same boundary condition electrons that shows the probability distribution due to the original $Ψ^*Ψ$.

Delta functions have no meaning in this sequence, because interactions change the boundary conditions and the wavefunction solutions.

Answer 3

Could this modulus square, the instant you measure, be thought as the Dirac Delta Function, because all the probability condensates to a single point

It can't, because probability does not condensate to a single point, because result of measurement is not a single point, but at best a finite region of space, due to measurement having finite precision and accuracy. So instead of delta function, it is better for probability density to use function that is finite in some region $x_0-\Delta/2,x_0+\Delta/2$ with $\Delta$ appropriate for the precision of performed measurement.

Answer 4

I read this question as to be simple and precisely asked. Without knowing what a collapse of the wave function could be like, it understand: there is no longer a wave function when the object represented by the wave function is "measured".

But then the object exists, so it is just "there". It is all about "what actually is a wave function". It is a theoretical construct that can explain why atoms exist, why electrons can tunnel and we can produce EPROMS, and much more.

But the delta function is no function but a distribution and the only fact you know is that integral over defined range is 1.

Let there be a delta distribution. And do the integration from A to B. Will this be a timely process to integrate? Or spacial at no time? Whatever: one instant the integral value will jump from 0 to 1. But that doesn't allow you to stop the integration as you have to go to B to be sure, that the integral value will not change.

Answer 5

I suppose that you think that when a measurement is done then probability of a finding particle at a given point is certain, thus it can be equated with delta function which has value of one at a point and zero otherwise. Also this is after measurement, so wavefunction is collapsed.

In the case been asked, once a measurement has taken, state of a particle is changed. And how could particle is find at given position with certainity, when its probability of to be find there is a fraction smaller than one. So a wavefunction can't be equated with delta function even after collapse. Well, delta function is used to represent cause and not response, because such a confined response is not physical. A wavefunction is response to operator, in given conditions.