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Q & A: What is the shape of a wavefunction?

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Most recent answer: 11/14/2013
Q:
What shape does a wave function of an electron, positron, and positronium have accordingly? And how do they look different? Could you show links to reliable graphical representations for each wave function?
- Anonymous
A:

Hi,

The wavefunction of any particle is determined by the potential it experiences. For example, the electron in a hydrogen atom experiences the Coulomb potential. If you solve the Schrodinger equation to find the electron's stable wavefunctions in this potential, you will find that they look like the usual hydrogen orbitals taught in basic chemistry classes:

However, if take the same electron away from the hydrogen nucleus, and put it inside a box, stable wave functions will will look like sine waves. So there isn't a single answer to your question: it depends on the situation.

The same applies to other types of particles, like positrons or muons.

Cheers,

David Schmid


(published on 11/02/2013)

Follow-Up #1: Shapes of wavefunctions and symmetries

Q:
Thank you, David, for your answer. "..put it inside a box, stable wave functions will will look like sine waves.." ->so this means, wave function (probability amplitude) is "not" spread out "evenly" inside the box but you will more like to find the electron in certain areas around the peaks(anti-nodes) of the sine wave of probability amplitude (almost? no electron is found at/around nodes)? correct? But what about "free electron in vacuum" without boundary(or very large volume of space/vacuum)? Does the wave function still look like sine wave? or something else?
- Anonymous
A:

Hi,

You are absolutely right. In a box, the stable wavefunctions of a system are not spread out evenly. If you look near a node of the wavefunction, you will almost never find the electron; it will most likely be found near the anti-nodes.

If you don't have a boundary, however, as in the case of a free electron, then the stable wavefunctions are in fact different. Specifically, they are complex exponentials, so their probability density is the same everywhere. This is what you would expect (even without calculating anything) by a simple symmetry argument: if the potential looks the same everywhere, why wouldn't the wavefunction look the same everywhere? 

[Note: This is a bit problematic unless the universe turns out to be finite. No non-zero probability density would give a probability adding  up to one. So in a simple flat, infinite universe, there are no stable single-particle wave functions./mw]

Side note: the specific wavefunction of a particle can actually be a function of your choice, if you are clever enough to find a way to shape it. The quantum potential that the particle is sitting in determines which wavefunctions are stable; the exact shape of a specific wavefunction in reality will be determined by its initial conditions and interactions with perturbing quantum potentials.

For any potential, you can find the set of stable wavefunctions, or "energy eigenstates." If you measure the energy of a particle, you force it into one of these stable wavefunctions (by some wavefunction collapse we don't yet understand), after which you know it has the wavefunction shapes I mentioned earlier (sine waves in a box, constant density in free space, etc).

If instead you force the particle into another state (for example, by measuring its position), then the particle's evolution will be completely specified by the quantum potential it is placed in.

[An exact  position measurement is not possible, since the resulting state would have infinite kinetic energy. Here David is referring to an approximate position measurement./mw]

Hope that makes sense,

David Schmid


(published on 11/14/2013)

Follow-up on this answer.