Particle Superposition

by karkas
Tags: particle, superposition
karkas is offline
Mar17-09, 03:31 PM
P: 125
Hello, 've been progressing through my self-studying of the Schrodinger Equation in both its time-dependent and independent forms, and I have come across an unknown term.

Super Position ( in my book it's translated in greek literally superposition = υπέρθεση)

My guess so far is that a superposition is when a particle is described by two wavefunctions, which happen to be two eigenfunctions [latex]\psi_n[/latex]with the same (perhaps with different, {not sure there} ) eigenvalues En. Am I correct? If not, please enlighten me :)
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malawi_glenn is offline
Mar17-09, 03:46 PM
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well it does not have to be a particle either! :-)

But consider deuterium, a bound proton - neutron state. It's state function is a linear combination of two terms:
karkas is offline
Mar17-09, 04:21 PM
P: 125
I think "Linear Combination of Wavefunctions" was the term I was looking for, eh?

confinement is offline
Mar17-09, 04:35 PM
P: 192

Particle Superposition

Quote Quote by karkas View Post
I think "Linear Combination of Wavefunctions" was the term I was looking for, eh?
The term you are looking for is "linear combination of Energy eigenstates." For example, take the case of a particle in a 1D box of width L. The energy eigenfunctions are:

[tex]\phi_n = A sin(\frac{n \pi x}{L}) [/tex]

Where A is a normalization factor. Just because a particle is in this box, does not mean that it is one of the states, those are only the states with definite well defined energy. A particle could be in a super position of energy eigenstates:

[tex]\psi= B sin(\frac{\pi x}{L}) + C sin(\frac{2 \pi x}{L})[/tex]

where a condition on B and C is to normalize the wavefunction, as usual. Notice that the two states which are involved are the n = 1 state (the B term) and the n = 2 state (the C term). Now when we measure the energy of a particle in this state we do not know whether
you will get n = 1 or n = 2 but we can calculuate the probabiity of either!
karkas is offline
Mar18-09, 03:41 AM
P: 125
Thank you! Really nice explanation there!

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