# Does a wave function not collapse upon detection?

In Sabine Hossenfelder’s YouTube video “The Delayed Choice Quantum Eraser, Debunked”, she states that even if you detect which slit the wave function goes through (in the double-slit experiment) then it’s still a wave function, only a single slit wave function.

But surely the wave function is smeared across both slits and the act of detecting which slit the photon could go through, if it were a particle, forces the wave function to collapse, and therefore it cannot continue as a wave.

Quantum mechanics is difficult to grasp, but this seems utterly contradictory! What am I misunderstanding?

• Quantum Mechanics is difficult to grasp but this seems utterly contradictory! I’m confused? - That means you are doing it right! Commented Mar 5, 2022 at 12:21
• Hi adlibber, whenever you refer to an external piece of media, it's always a good practice to provide a link to the same. I have edited your post and added a link to the said YouTube video. :)
– user87745
Commented Mar 5, 2022 at 12:33
• @Dvij D. C. Sorry and thank you very much! Commented Mar 5, 2022 at 18:07
• It is not necessary in a complex apparatus such as the DCQE that the wave function sees both slits. The addition of the beam splitter offers strong direct pathways to collapse. It may just be a case of entanglement of photons and electrons: 1) Source> double slit> crystal>splitter> detector or 2)Source> single slit> crystal> splitter> detector. All paths have probabilities. Commented Mar 5, 2022 at 21:32

I think this question arises from a simple misunderstanding of what a wave function is. The wave function of a particle doesn't need to be "wavy". The description of a system in quantum mechanics is always given via its state-vector in the Hilbert space and that can always be translated to the wave function of the said system in a basis of your choice, e.g., the position basis or the momentum basis.

A wave function $$\psi(x)$$ of a particle in position basis simply gives you the probability amplitude of the particle at position $$x$$ which is a complex number, i.e., it gives you two bits of information:

• The magnitude gives you the probability (density) that you would find the particle in the vicinity of $$x$$ if you measure its position.
• The phase gives you the information that you'd need on top of the probability (density) to construct the wave function in some other basis, e.g., the momentum basis, so that you can calculate the probabilities (probability densities) associated with the measurement of its momentum.

So, the point is that there is always a wave function of a particle -- regardless of whether it is very localized and point-like or not.

As to why wave functions are nonetheless called wave functions, I think it's a historic relic. There are two tangible historic reasons that resulted in this naming, I think:

• The position-basis wave function of a particle that has a definite momentum is $$\sim e^{ipx}$$ and it is actually wavy. These are the famous de Broglie matter waves.
• The time-evolution equation that all wave functions satisfy is called the Schrodinger wave-equation (because it was the equation that was followed by the de Broglie waves, I suppose). One should note that the Schrodinger equation is not exactly a wave-equation although it admits wave solutions. It's more like a diffusion equation with an imaginary diffusion coefficient.
• "simply gives you the probability amplitude of the particle at position 𝑥" but if it is "observed" doesn't that "collapse" the wave function form a probability to a dead certainty? Commented Mar 5, 2022 at 18:30
• ... and if it's now a dead certainty it can no longer be a probability wave or whatever the correct term is. Must say I much prefer Richard Feynman's view on QM very much a particle man! Commented Mar 5, 2022 at 18:32
• @adlibber The post-measurement wavefunction will give probability $1$ for the measured value of the measured observable and $0$ for other values of the measured observables. In other words, the wavefunction (in the basis associated with the measured observable) will be a delta function spiked at the measured value of the measured observable.
– user87745
Commented Mar 5, 2022 at 18:32
• @adlibber If your point is that why do we still need a wavefunction description when it only provides us trivial probabilities (0 or 1) then the answer is that even when the particle has a certain position (so no non-trivial probabilities associated with the position), the probabilities over the outcomes of its momentum measurement are not trivial.
– user87745
Commented Mar 5, 2022 at 18:38
• "there is always a wave function of a particle" A particle does not have a wavefunction. A wavefunction describes statistically where and when charge, energy-momentum etc can be found. Commented Mar 6, 2022 at 10:20

Wave function collapse is fiction. When a detector is part of the system, its wavefunction is entangled with that of the electron-double slit system. This will prohibit or reduce the interference if the detector reveals which-way information. Say interference occurs between two branches of the wave, $$\cal l$$ and $$\cal r$$, because these are not orthogonal but overlap. Assume a detector has two orthogonal states, $$L$$ and $$R$$. In a measurement setup these will be entangled with the electron wave function, giving new wave function branches $${\cal l}L$$ and $${\cal r}R$$. These are orthogonal and no interference occurs. The wave function has 'collapsed'.

A particle always has an associated wave function. When we talk of a wave function 'collapsing', what we mean is that the form of the wave function changes as a consequence of a physical interaction- it doesn't mean that the wave function disappears. Also, you should not suppose that a particle such as an electron actually is a wave when its position is undetermined and suddenly switches to becoming a particle as a consequence of the wave function collapsing.

If you take the case of an electron that has been passed through some form of two slits experiment and is detected by a photographic plate, say, (I like old fashioned physics!), the position of the electron is narrowed down to a very large extent, but it is not a 'dead certainty', to use the terminology in your comment, that the electron is in a position that can be specified with unlimited precision. If an electron's position could be more and more precisely confined, then its associated wave function would become more and more to resemble a spike.

You might like to google the 'Dirac delta function', which is an idealised from of a wave function which has an infinitesimal spatial spread, so it is just a single spike giving a probability of 1 (ie certainty) for the electron being at one point in space and zero for the electron being anywhere else. That is an idealised extreme case- in reality you could not design an experiment that would localise an electron to that extent.

The other idealised extreme is a wave function that is a plane wave with a specific frequency. In theory such a wave has to be spread throughout space, which is not a realistic proposition. So a wave function in reality will always be something in between those two idealised extreme- either very spread out with a reasonably well-defined frequency, or very localised with a poorly defined frequency, or anywhere in between.

So, if you think of the wave function as just being a mathematical function that doesn't have to look like our everyday conception of a nicely periodic wave, and if you take 'collapse' as meaning just 'change abruptly' you might find some of the ideas of quantum mechanics easier to digest.

Let’s go back to basics in mainstream physics.

What is a wave function? It is the solution of a wave equation. In quantum mechanics it is the solution of a specific equation where, in the double slit experiment, the boundary conditions and potentials give the function that describes an "electron scattering off double slits of given width and distance apart". $$Ψ$$, the wave function, a complex value function where $$Ψ^*Ψ$$ is the probability of finding the electron at (x,y,z,t). One needs an accumulation of electrons to be able to measure this probability.

Collapse is a bad description of what happens if the particle undergoes a second scattering, at the detecting screen for example. At the screen there is a new interaction with new boundary conditions that has to be taken if one wants to compute the wave function. It is not a balloon collapsing, but a new necessarily solution for a new interaction, the old wave function becoming mathematical history, recorded on the screen.

But surely the wave function is smeared across both slits

The correct view is a mathematical formula describing the probability of interaction across and through both slits.

and the act of detecting which slit the photon could go through,

It is a second interaction with boundary conditions different, i.e., the first wave function is invalid when a subsequent interaction happens, new boundary conditions are needed for the new solution

if it were a particle, forces the wave function to collapse therefore it cannot continue as a wave.

Let’s make this clear. The electron is always described by a wave function, and when the boundary conditions change, the wave function changes, as with all solutions of differential wave equations.

My answer here may help clear the experimental point. The particle nature appears in the detection of individual electrons. The wave nature in the probability distribution of many electrons.

In the theory of hidden variables (which is equivalent to objective collapse theory), the wave function is considered not a mathematical device, but a real feature of reality. A measurement, like any interaction, is seen as the cause of collapse of a (non-local) that physical wave.

So the photon wave (or electron wave when using electrons in the double slit) interacts with the slit before interacting with the screen, it will change the wave function accompanying the photon. So the wave function arriving at the screen will be a collapsed one and further collapse upon interaction with it.

So the problem you address, what do we make of it in this context? Can the photon wave function collapse and travel on like a wave from one slit? Before entering the two slits, it's a spherical wave impinging on the two slits. Some waves collapse beneath the slits and are absorbed.

But if the wave interacts with something in aperture of the slits, the photon is not absorbed, but its wave function collapses to the one slit wave function. When it impinges on the screen, they are annihilated at positions that reflect the wave amplitudes. The effect of these annihilations can be seen indirectly.

• You've got a photon wave and as well a wave function "accompanying the photon". I would agree but say that the first part (your photon wave) is virtual photons, i.e. forces from the excited electron, probably initially spherical but changing shape quickly to one that is path shaped ... along the direction of photon transit. But more importantly the calculated or observed pattern (Feynman path) is what is traditionally called the wave function and shows probabilities for all possible paths ... e.x. the interference pattern in the DSE. Commented Mar 5, 2022 at 21:25