The state at wave function collapse

Question

In the double slit experiment, why does the wave function collapse into the coordinate base? Why not into something else? How does the particle know that its position is measured (hence, it gets "localized" at a point)? What happens on the screen exactly? (The same question could arise in other measurements as well.)

Answer 1

(Disclaimer: This answer is written in the context of non-relativistic quantum mechanics. For a paper which presents a perspective based on more advanced concepts, which is compatible with special relativity, see the end of my answer.)

The difficulty is that you are thinking that the wavefunction describe “the system” whereas better wording would be that it describes “the system as we know it”.

Thus the particle doesn’t “know” it’s detected somewhere, the measurement detect the particle somewhere, and from this we adjust the wavefunction to reflect this new knowledge of the system.

It is conceptually simpler to discuss the case of discrete outcomes - of say energy - rather than continuous outcomes like position, so let’s start there. Assume that you have good reasons to describe your system by a linear combination $$ \vert \Psi_0(t)\rangle = e^{-i H (t-t_0)/\hbar} \left(\alpha \vert E_0\rangle + \beta \vert E_1\rangle\right) \tag{1} $$ of two states with energy $E_0$ and $E_1$, and measure the energy to be $E_1$ at some time $t=t_0$. Then surely you have gained information on the system and can anticipate that the energy at a time $t$ very shortly after $t_0$ will again very probably be $E_1$. So having gained information about the system at $t_0$, the state of Eq.(1) no longer describes your knowledge of the system at that time (the probabilities of getting $E_0$ and $E_1$ are no longer $\vert \alpha\vert^2$ and $\vert \beta\vert^2$) and you need to “collapse” the old state $\vert \Psi_0\rangle$ to a new state $$ \vert\Psi (t_0) \rangle \to \vert\Psi_1(t_0)\rangle=\vert E_1\rangle $$ with subsequent evolution $$ \vert\Psi_1(t)\rangle=e^{-i H (t-t_0)/\hbar}\vert E_1\rangle\, ,\qquad t\ge t_0. $$

Note that the new state $\vert\Psi_1(t)\rangle$ at $t_0$ must be “part of” the linear combination that describes the old state for $t<t_0$, i.e. $\langle \Psi_1(t_0)\vert \Psi_0(t_0)\rangle\ne 0$, hence the name “collapse” in the sense that the vector $\vec r=x\hat x+y\hat y$ “collapses” to $x\hat x$ upon projection to the $\hat x$ axis.

If $\langle \Psi_1(t_0)\vert \Psi_0(t_0)\rangle=0$, the probability $\vert\langle\Psi_1(t_0)\vert\Psi_0(t_0)\rangle\vert^2$ of getting the outcome $E_1$ when you measure the energy is $0$, so your measurement could not have yielded $E_1$ as the energy of the system at that time.

Notice now quantities such as average values etc are different for $\vert\Psi_0(t)\rangle$ and $\vert \Psi_1(t)\rangle$. This is of course fully expected because $\vert\Psi_1(t)\rangle$ reflects additional knowledge gained at $t_0$.

Note also that, since the outcome of the measurement is probabilitics, the state $\vert\Psi_0\rangle$ will collapse to $\vert E_1\rangle$ $\vert\beta\vert^2$ fraction of the time, and to $\vert E_0\rangle$ $\vert\alpha\vert^2$ fraction of the time. However, whatever the outcome of the measurement, you need to time-evolve either $\vert E_0\rangle$ OR $\vert E_1\rangle$ after the measurement, but not their linear combination $\vert\Psi_0\rangle$. Computing an average for $e^{-i H (t-t_0)/\hbar}\vert E_0\rangle$ and for $e^{-i H (t-t_0)/\hbar}\vert E_1\rangle$ will give different value from computing average for $\vert\Psi_0\rangle$.

It’s broadly the same arguments with the 2-slit experiment. Once you “measure” your particle to be at a particular point, you need to “reset” the wavefunction of your system to reflect this new knowledge and, in this specific case, the time evolution can no longer take the system through 2 slits so you cannot get interference between the slits

I’ve glossed over a number of concepts here, such as “what is a measurement”, and this is really a cartoon view of the difficulties with collapse and measurements.

There is a partial discussion of this in

Peres A. When is a quantum measurement?. American Journal of Physics. 1986 Aug;54(8):688-92

where the author emphasizes the irreversible nature of measurements.

There are also discussions elsewhere in the literature of the “boundary” between the apparatus and the system, since the apparatus, after all, is made of atoms subject to the same laws of quantum mechanics as the quantum system under observation, and needs to interact with the system in order to “make a measurement”.

---

Nota bene: There is also a very interesting discussion by Peres in

Peres A. What is a state vector?. American Journal of Physics. 1984 Jul;52(7):644-50,

where the author argues there is no collapse or continuous evolution between measurements. The argument is compatible with special relativity and based on $S$-matrix theory and path integration, which are more advanced notions which I took to be outside the scope of the question.

Answer 2

First of all, the concept of "wave function collapse" is completely useless. A wave function is the description of the quantum mechanical ensemble for all possible preparations and emissions. It is similar to a probability distribution. The probability distribution for dice tells us nothing about an individual outcome when we throw dice, it only tells us what happens on average for an infinite number of repetitions of that throw. We also don't say that the probability distribution collapses whenever dice come to rest. The probability distribution always stays the same. A single dice throw can not change what would take an infinite number of throws to measure in the first place. One throw of dice is simply not the same experiment as the infinite number of throws that the statistical ensemble describes.

We have the same situation in quantum mechanics, except that we aren't throwing dice. Instead we are transferring energy into a quantum system. We call that "preparation" or "emission" from a "source". The energy then flows around in that quantum system in some (complicated) way and eventually we take that energy out again. That's called "absorption" or "measurement". Every state change of the quantum system is therefor an irreversible transfer of energy. What happened to the energy while it was in the quantum system we can't tell because we can't perform measurements on it without removing the energy once and for all from the system.

If you pay attention to the details of the dynamics of dice you will notice the very same processes... our hand adds kinetic energy to the dice and they lose that energy to friction while they are bouncing around and come to rest. We do not care in probability theory what happened to that energy while the dice were moving. The difference is that in a dice throw we merely don't care, but in the case of quantum mechanical systems we are fundamentally prevented from knowing.

In an optical experiment the energy loss (the mechanical equivalent of friction) occurs wherever we place a detector or another absorbing surface. We chose where that irreversible energy loss can happen. So, technically, what we are measuring in a detection process is simply the location of the detector. The quantum system merely supplies the energy to make that measurement.

What quantum mechanics tells us is that the quantum system does not "know" where that energy deposition will take place. Nature is just as uncertain about it as we are. It is not that we are just not very good at measuring energy quanta, quanta are fundamentally not localized. They can be absorbed wherever an irreversible absorption process can take place and the wave function of the system isn't zero. Energy, momentum and angular momentum conservation then guarantee that we can never absorb the same quantum twice... once we have taken the energy our of the system, it's lost forever. That is also the reason why we can't determine a "path" of quantum energy: it can't be measured more than once and there is no way to make a consecutive sampling of locations of it.

I would also point you to an often overlooked detail about the double slit experiment: it never was a quantum experiment in the first place. One can predict the brightness on the screen fairly well with scalar wave theory (Young did that in 1801) and perfectly with Maxwell's equations. Planck's constant never even enters the stage in these experiments. The probability to "find a photon" is simply proportional to the brightness of the classical wave diffraction pattern.

Technically even the "measurement at the screen" is a false explanation. A screen in an optical experiment is not an absorber or detector. It is a near perfect scattering surface. The actual "measurement" takes place in your eye after the scattered light from individual points of the screen has been re-focused onto your retina.

If you do the experiment with e.g. electrons, then the screen is the actual absorber. The quantum of kinetic energy of the electron is now converted into light and heat inside the material of the screen.

In other words, the double slit experiment is a horrible example to teach quantum mechanics with. It is a real shame that it is being overused so badly because it teaches all the wrong lessons and none of the right ones.