# What is the wavefunction?

1. Nov 5, 2009

### gabrielh

In simple, non-mathematical terms, what is the wavefunction?

2. Nov 5, 2009

### Mosis

a summary of all the information you possess (and can possess, in principle) about a system

*ducks*

3. Nov 5, 2009

### sokrates

unfortunately it doesn't mean much in non-mathematical terms.

Last edited: Nov 5, 2009
4. Nov 6, 2009

### dmtr

It's a vector in the Hilbert space. State vector. Kind of. It is practically impossible to explain what it is, without using some basic math.

Fortunately QM can be described via the path-integral formulation, which can be explained geometrically, without any math :)
I would suggest watching: http://www.vega.org.uk/video/subseries/8" [Broken] which give very gentle introduction on the subject. All fun and no math.

-- Dmtr

Last edited by a moderator: May 4, 2017
5. Nov 6, 2009

### gabrielh

Thank you very much.

Last edited by a moderator: May 4, 2017
6. Nov 7, 2009

### damnedcat

As simple as that question sounds it is a really good question that I bet many people students can't really answer. I'm def not an expert at QM but i'll venture a try to explain without as little math as possible as best as I understand it.

The wavefunction is a statistical state function from which probability distributions for all observables can be calculated.

So this leads to the question of what is a 'state'? (this is a little more subtle but it deals with ensembles of identically prepared systems) Anyway to each 'state' there corresponds state operators and state vectors (how these states are assigned operators/matrices and state vectors deals with the topic of state determination which is another story). these vectors live in Hilbert space and like any vector can be expressed in different bases e.g position $$\Psi$$(x) (which is what you prob are most familiar with to get your pobabilities etc), momentum $$\Phi$$(p) etc.

Hope it helped. People feel free to amend or add to my answer as u see fit.

7. Nov 7, 2009

### blkqi

In my opinion the best physical (sort-of) interpretation of the wave function is by analogy with more corporeal wave functions, such as that of light or sound. It's not hard to imagine a 'waving' disturbance in a liquid or electromagnetic field. A quantum wave function describes a an excitation, or a particular state, in a continuous quantum field.

The energy density (energy per unit volume) of sound or radiation is proportional to the square of the corresponding wave function. In the case of radiation, energy is quantized into small corpuscles of energy--photons--so it appears the square wave function is telling us the number of photons per unit volume (photon density). But when only one photon is considered, and as we all know it's wave nature still exists, the square wave function translates into the probability density of that photon's location in space.

8. Nov 7, 2009

### I_am_learning

What helped me most was to think this way:
Light can be treated both as particle and wave. The particle Photon, can be thought to be associated with an Electromagnetic Waves. The Square of the Amplitude of the EM waves can be shown to be proportional to the photon Density there which in turn is proportional to probability of Finding a photon there.

So, just by analogy, De-broglie, Supposed that all matter also have their associated waves. This Wave is described by the Wave function. This wave function, by analogy, should also be consistent with the relation that --> square of its amplitude should be proportional of finding the particle there. The later mathematical procedure to find out the wave function, should work such a way that the above relation holds. Similarly, this wave (given by the wave function) should also show interference, diffraction and the like.
The only analogy thats disobeyed is that --> Unlike EM waves, in which the EM waves behaves directly as varying Electric and Magnetic Field , There is no direct physical significance of the wave function of a particle.

9. Nov 8, 2009

### ytuab

The particle Photons are said to be existing in the electromagnetic (EM) waves,
And the amplitude of the EM waves is related with the photon number.
But there is no concept like "the probability amplitude of photon particles at (x,y,z) in space".
So the photon particles are not distributed in space obeying the EM waves like
Schrodinger wave function. (Though it's difficult to imagine.)
Actually, the particle photon can not be found clearly like an electron and can not be imagined concretely.

First, the EM waves satisfies the Maxwell's equations (not Schrodinger equation).
When considering "the particle photon", the quantization of the EM fields (such as vector potenitial) is done.
In this procedure, we carry out a Fourier expansion of the EM field function, and introduce creation and annihilation operators for photons.

This vector potential need to show the property of a four-vector which transforms like x, y or z under Lorentz transformations,
and they must satisfy the Maxwell's equations which form are Lorentz invariant.

Due to these mathematical restrictions, we can not use the EM field functions freely as the functions which show the probability density of photons. (There is a concept of "density", but not a concept of "the probability density" by which we can find a photon at (x,y,z) like Schrodinger equation).
We represent N photons with the energy hv by integrating the functions in all space.

It is easy to consider the light as the EM waves, but when we incorporate the idea of "the particle photon",
the mathematical property becomes stronger, and it becomes difficult to actually imagine "the particle photon".

For electrons, it is more confusing.
Because there are non-relativistic Schrodinger equation and relativistic Dirac equation.
The Scrodinger wave function shows the probability density of an electron, but the relativistic wave function of Dirac equation doesn't.

Last edited: Nov 8, 2009