Are Photons and Electrons Indistinguishable Based on Their Wavefunction Alone?

[chrisphd]

I heard that a postulate of quantum mechanics is that a wave function tells you all you need to know about a particle. If a photon and an electron have the same probabilistic wavefunction, is the photon then indistinguishable to the electron or is there another property not in the wavefunction that specifies what type of particle exists(whether the particle is photon orelectron for example)?

 

[CompuChip]

I think that for a "real" proton and electron, the full quantum mechanical state would also encode quantum numbers like electric charge, lepton number, baryon number, etc which would differ for both particles and (most of which) we believe to be conserved.

Also (but I'm not really sure about this argument): note that a proton and an electron have very different masses, so they satisfy different Schroedinger equations (which contains m). Therefore, even if you put them in the same state, when you measure a property (like the energy) they cannot in general collapse to the same state because they don't share (all) eigenfunctions.

(Afterthought:) on the other hand, if two wavefunctions are exactly the same, any information we can extract from the wave function will also be the same so it's really no use to try and distinguish between two such particles. I.e. we have agreed that "by definition" an electron is the particle which, when I measure property such and so, I will get this and that value. So if a proton would give the same values, it wouldn't really be a proton, but an electron anyway.

 

[chrisphd]

Does this mean that any two particles can interfere with each other merely by superimposing their wavefunctions. For example, a photon can interefere with an electron etc.

 

[Davide86]

If you want to describe the interaction between a photon and an electron, you have to consider the potential that describe the interaction. I mean, if you are looking at the consequences of the interaction on the electron, you have to put the potential in the Schroedinger equation, then you have the new wavefunction of the electron.

But I think (not sure) that the right description of this interaction requires to use quantum field theory.

 

[CompuChip]

I think that if you superimpose the wave-functions, they will just pass right through each other.

As Davide says, you can either consider the electron to generate some potential which the proton feels (which would involve Coulomb force and possibly gravity) or you would have to solve the Schroedinger equation from the start for the combined proton/electron system.

 

[Phrak]

I heard that a postulate of quantum mechanics is that a wave function tells you all you need to know about a particle. If a photon and an electron have the same probabilistic wavefunction, is the photon then indistinguishable to the electron or is there another property not in the wavefunction that specifies what type of particle exists(whether the particle is photon orelectron for example)?

Some things are left out. Black holes, for instance, need only mass, angular momentum and charge to quantify them, right? 'Cause "black wholes have no hair", etc. But the triplet of descriptives is just cut-rate physics. Position in space, and in fact, the entire world-line is required--but no one will ever point that out...

Don't confuse the set with it's elements.

 

[vanesch]

No no, there is no possibility to have "the same wavefunction" for an electron and a photon for instance. The reason is that if you consider a system in which it is possible to have photons AND electrons, then this system is made up at least of two different quantum fields, and hence the state space is the tensor product of the state space of the "electron" quantum field, and the "photon" quantum field. Now, a specific electron state is nothing else but a particular state of the electron quantum field, and a specific photon state is nothing else but a particular state of the photon quantum field.

Symbolically, the first is: |electron state> x |empty_photon_state>

while the second is:
|empty_electron_state> (x) |photon state>

They are not the same states, although it is possible that there is a large ressemblance between say, the electron state of the first example and the photon state of the second example. But they play in different product spaces.

Only, people usually don't bother to write things that way, it would be way too cumbersome.