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Apologies for the over-broad question(s), but I'm having a hard time finding out where to look to answer these myself:

If a particle is a wavefunction describing a probability amplitude distributed through space, what happens when two wavefunctions meet? I imagine that their amplitudes begin to sum, so that as a photon's wavefunction approaches a wall (composed itself of many wavefunctions), eventually these two wavefunctions begin to overlap and sum (or multiply?), but what happens after this I'm still not clear on. Really I'm trying to figure out what is meant by an 'interaction'.

Is the better picture that it's all one wavefunction per field? In other words, is a light bulb is spraying trillions of 'individual' photon wavefunctions, or just one that describes all the potential individual photons?

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  • $\begingroup$ Re, a light bulb spraying [many] individual photons: IMO, it might be more interesting to ask that same question about a laser instead of a light bulb. $\endgroup$ – Solomon Slow Jun 17 '16 at 15:08
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Wavefunctions combine trough tensor products, which is not the addition that one would expect naively. The reason for this is that a wavefunction contains the description of all possible futures of the system at once, so if there are multiple subsystems, then the wavefuntion of the entire system has to describe all possible futures for each part independently of the other parts.

Let's say a system can have two possible futures |1> and |2>. Then two copies of the system can have the possible futures |11>, |12>, |21> and |22>. Three copies can have the eight possible outcomes |111>, |112>,...|222> etc.. This is greatly complicated by the fact that if the subsystems are indistinguishable, then the permutations of them only can occupy one (further indistinguishable) future state, so instead of |12> and |21>, we can only talk meaningfully about the symmetric linear combination (|12> + |21>)/$\sqrt 2$. This gets even more complicated when fermions are involved which require antisymmetric combinations of this kind.

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If a particle is a wavefunction describing a probability amplitude distributed through space, what happens when two wavefunctions meet?

The particle exists, the wavefunction is a solution of the quantum mechanical equation specifically defined for the boundary values of the observation. Its complex conjugate square gives the probability density of finding a particle in a specific location.

For two wavefunctions to meet, it means a different set up of the problem, possibly new potentials and boundary conditions will define a new probability distribution.

Many body systems are not analytically solvable in classical physics. The answer by CuriousOne describes the mathematics developed to describe the quantum mechanical solutions for many body systems .

is a light bulb is spraying trillions of 'individual' photon wavefunctions, or just one that describes all the potential individual photons?

The photons are the real thing. In principle one wavefunction describes a specific boundary value problem, giving the probability of a specific distribution. In practice since the photon photon interaction is very very small one can use an approximation of individual wavefunction describing each photon. A neighboring photon is not a boundary condition within errors. In general the specific boundary conditions dictate what type of solution will describe an experimental setup.

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It depends on which wavefunctions are meeting. There are actually several different scenarios in your question which will change how they are treated:

1) wavepackets, or portions of a wavefunction that describe the same single particle system add linearly, as you have guessed. This is because (most) field equations in free space are linear field operators: $\hat H \left(| \psi_1 \rangle + |\psi_2 \rangle\right) = \hat H | \psi_1 \rangle + \hat H |\psi_2 \rangle $. In this case, fields just add.

2) To describe more than one particle, a single wavefunction is used that treats each additional particle as a new argument (this is the situation that @curiousone is describing). This single wavefunction must not distinguish between particles; this is reflected mathematically by wavefunctions that are symmetric (or antisymmetric) under interchange of particles. This ties into the spin of the particle and can get complicated; the notion of a single wavefunction for each particle is false (so you can't talk about what happens when they collide); all particles with a single wavefunction. This is nuanced! You cannot separate out one photon from another - it's all wrapped up in a single field. This composite wavefunction obeys linear superposition.

3) Particles of different species can interact, typically via an interaction/coupling term in the Hamiltonian/Lagrangian. How these fields would interact depends on the nature of that term

4) as @AnnaV points out there can be nonlinear couplings in a field (ie, photon photon), but these can often be ignored (in vacuum). In the approximation these terms are negligible, linear superposition still holds.

Finally, when talking about a 'bath' of photons, such as the trillions streaming out from a lightbulb, my guess is that you could treat this as a continuous EM field (which also obeys linear superposition), and ignore most of the constraints of a 'many photon' wavefunction. I'm a little less certain on this last point.

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  • $\begingroup$ Thank you! No idea why your answer was downvoted. I upvoted it, in any case. $\endgroup$ – T3db0t Jun 17 '16 at 20:25
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If a particle is a wave function describing a probability amplitude distributed through space, what happens when two wave functions meet?

Well, I just try to think allowed and one can not portray a 'wave function' picture unless a physical system is at hand to use the picturization or description.

Let us think about a hole in the reactor and neutron beam coming out of its collimator and hitting paraffin- which has the proton target.

One knows the energy/momentum of the neutron and can depict its probability amplitudes through an expansion in a multitude of partial waves and those partial waves are the wave function description.

Those partial waves interact with protons i.e. their wave functions and the counters set up by the observer is catching /fixing the scattered neutrons-

Perhaps the above depiction gives you a tool to describe the particles by a suitable wave functions which are solutions of Schrodinger's equation and the measurable parameters that is the scattering cross sections demands further 'fixation/choices' of the theoretical functions depicting the event to be named as 'wave functions'.

so one may not get bogged down with billions of particles crossing the observational frame and always try to reduce the system to 'measurable' parameters and can do experiments with say 'single photon 'interference of photons and learn more about the superposition of 'mathematical' functions specially chosen with real boundary conditions.

I remember the 'single particle' shell model or 'Free electron theory of solids' which are QM systems and can be described using the reduced framework of superposition of wave function description.

I think this is only a comment to the honourable questioner rather than an answer.

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I think when two particles (with associated wavefunctions) with wavefunctions $\psi_1\left(x\right)$ and $\psi_2\left(x\right)$ overlap (i.e. they interact in vacuum) they will interfere and the resultant wavefunction will be $\psi\left(x\right)=\psi_1\left(x\right)+\psi_2\left(x\right)$ with some normallization constant. If there is constant phase difference between two wavefunctions, i.e. two wavefunctions are coherent with each other you will see time stationary fringes.

On the other hand when two particles interact in a medium then the medium's response can use both wavefunctions in a certain manner and you might see the multiplication of two in certain manner.

For example if the medium is linear then the final wavefunction will be addition $\psi\left(x\right)=\psi_1\left(x\right)+\psi_2\left(x\right)$

If the medium possesses second order nonlinearity i.e. $\psi\left(x\right)=\left(\psi_1\left(x\right)+\psi_2\left(x\right)\right)^2$, then you will see the final wavefunction having multiplication of the two wavefunctions. If two wavefunctions have frequencies $\omega_1$ and $\omega_2$ you will see the wavefunction with frequency component $\omega_1\pm\omega_2$, you will see higher powers of product if the medium has larger degree of nonlinearity.

If you still have some doubt please leave a comment.

EDIT:

@anon0909 : You have misunderstood the answer. If two particles occupy states $\psi_1\left(x\right)$ and $\psi_2\left(x\right)$ individually then it does not mean that you can distinguish between them when they interact.

Question is how you know about particles's state and the answer is via. measurement. When you measure the state of a particle you will get

$\left|\psi_1\left(x\right)\right|^2$ and $\left|\psi_2\left(x\right)\right|^2$,

when you make measurement after the interaction you will get

$\left|\psi\left(x\right)\right|^2=\left|\psi_1\left(x\right)+\psi_2\left(x\right)\right |^2$.

In this way both particle remain indistinguishable. Moreover if you make measurement before the interaction and try to interact these particles after the measurement was made you will just get $\left|\psi_1\left(x\right)\right|^2$ + $\left|\psi_2\left(x\right)\right|^2$ and no interference will be seen.

You may think my explanation as double slit interference experiment (although this can be applied to several other physical situations)

My effort was to put the tedious concepts in a simple manner rather than to stress over mathematical rigour.

Regards

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    $\begingroup$ this is not true. because particles of a given species are indistinguishable, there is a single field that takes multiple arguments that correspond to each particle, in some manner of speaking. This should be covered in most elementary text books. $\endgroup$ – anon01 Jun 17 '16 at 3:57
  • $\begingroup$ @anon0909 kindly see the edited answer $\endgroup$ – hsinghal Jun 17 '16 at 10:08
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    $\begingroup$ anon0909's complaint is completely correct, and your answer just false - the correct wavefunction for two particles is emphatically not $\psi(x) = \psi_1(x) + \psi_2(x)$, but instead rather something like $\psi(x_1,x_2) = \psi_1(x_1)\psi_2(x_2) + \psi_1(x_1)\psi_2(x_2)$. $\endgroup$ – ACuriousMind Jun 17 '16 at 10:12
  • $\begingroup$ @hsinghal Have a look here; I think it should make my objection clearer. Unfortunately what you have written is simply untrue in quantum theory. $\endgroup$ – anon01 Jun 17 '16 at 12:30
  • $\begingroup$ I can understand it from the theory of probability that if $P_1$ is the probability of instance 1 at any point P and $P_2$ is the probability of instance 2 at same point then for uncorrelated system the probability will be $P_1.P_2$ but now if you think that two waves have frequencies $\omega_1$ and $\omega_2$ then from this theory we should see sum and difference frequencies. I think this is not true. I am thinking over your comments and trying to revisit my answer $\endgroup$ – hsinghal Jun 17 '16 at 13:17

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