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Eigenfunction vs wave function

 
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Jun30-12, 09:11 PM   #1
 

Eigenfunction vs wave function


What is the difference between eigenfunction and wave function?

I'm always get confused when i am asked to write wave function and eigenfunction..
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Jun30-12, 09:15 PM   #2
 
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An eigenfunction is a type of wavefunction that has an eigenvalue when operated on. It is said to be "an eigenfunction of the operator".
Jun30-12, 10:40 PM   #3
 
Quote by Simon Bridge View Post
An eigenfunction is a type of wavefunction that has an eigenvalue when operated on. It is said to be "an eigenfunction of the operator".
is this right?

ψ(x,t) = ψ1(x) + ψ2(x)
wavefunction = eigenfunction1 + eigenfunction2
Jul1-12, 12:29 AM   #4
 
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Eigenfunction vs wave function


Quote by Flavia View Post
is this right?

ψ(x,t) = ψ1(x) + ψ2(x)
wavefunction = eigenfunction1 + eigenfunction2
Assuming ψ1,ψ2 are eigenfunctions, then yes... Just because the wavefunction is written as the sum of something doesn't mean those somethings are eigenfunctions!

Here's the full description of the situation: You have a quantum system which has only two possible states, 1 and 2. That means there are two quantum states, ψ1 and ψ2 describing the system in either state 1 or 2 (1 and 2 can be spin up or spin down, for example). In general, then, any arbitrary wavefunction can be written as a linear superposition of these two states, [itex] \Psi(x,t) = \alpha \Psi_1(x,t) + \beta \Psi_2(x,t) [/itex]. This is possible because the set of eigenstates (ψ1,ψ2) are complete and form a basis.
Jul1-12, 12:38 AM   #5
 
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Since the LHS is a function of time as well as position while the LHS is position only, not really. However ... lets say we have a set of wavefunctions [itex]\{ \psi_n \}[/itex] which has been selected so that [tex]\mathbf{H}\psi_n = E_n\psi_n[/tex] .... then each [itex]\psi_n[/itex] is said to be an eigenfunction of the Hamiltonian with eigenvalue [itex]E_n[/itex].

A system prepared in a superpostion state may have wavefunction [tex]\psi = \frac{1}{\sqrt{2}}\left ( \psi_1 + \psi_2\right )[/tex] (assuming each [itex]\psi_n[/itex] are already normalized.) In this case [itex]\psi[/itex] is not an eigenfunction of the Hamiltonian.

In general, the set of eigenfunctions of an operator can be used as a basis set. Any wavefunction can. Therefore, be represented in terms of a superposition of eigenfunctions ... including eigenfunctions of another operator. (Just in case someone infers that superpositions of eigenfunctions cannot be eigenfunctions.) It is also possible for a wavefunction to, simultaniously, be an eigenfunction of more than one operator.

Notice how careful I was in the way I phrased things above?
In QM it is very important to be careful about what exactly is being said about a system ... when you are starting out it is as well to get really pedantic about this.
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