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Wavefunction in the energy representation

  • Thread starter dave4000
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  • #1
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Homework Statement



[tex]\psi(x)=\frac{3}{5}\chi_{1}(x)+\frac{4}{5}\chi_{3}(x)[/tex]

Both [tex]\chi_{1}(x) \chi_{3}(x)[/tex] are normalized energy eigenfunctions of the ground and second excited states respectivley. I need to find the 'wavefunction in the energy representation'


The Attempt at a Solution



I can find the expectation value of the energy but what is the wavefunction in the energy representation?
 

Answers and Replies

  • #2
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It looks like it is given in the energy representation already. Except I'd write it as:

[tex]\left|\psi\right> = \frac{3}{5}\left|E_1\right>+\frac{4}{5}\left|E_3\right>[/tex]
 
  • #3
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I'm not so sure. I think what they are after is [tex]\psi(E)[/tex] i.e. a function of energy. I suppose in analogy with [tex]\psi(x) = \left<x|\psi\right>[/tex] it would be [tex]\psi(E) = \left<E|\psi\right>[/tex]. Completeness of states could be of help. I suspect you should get delta functions.
 
  • #4
diazona
Homework Helper
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I'm not so sure. I think what they are after is [tex]\psi(E)[/tex] i.e. a function of energy. I suppose in analogy with [tex]\psi(x) = \left<x|\psi\right>[/tex] it would be [tex]\psi(E) = \left<E|\psi\right>[/tex]. Completeness of states could be of help. I suspect you should get delta functions.
I wouldn't have thought of that, but yeah, that might be it.
 
  • #5
Matterwave
Science Advisor
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My guess at what they want: (I think it's the same as what Phsopher said, but possibly a bit more intuitive)

By completeness, you can represent any wave-function:

[tex]\Psi (x) = \sum_{n=1}^{\infty} c_n\psi_n(x)[/tex]

Where the [tex]\psi_n[/tex] represent energy eigenfunctions. The energy representation of the wave function would probably then just be the coefficients [tex]c_n[/tex] at different energies. The probabilities of finding a specific energy would be [tex]|c_n|^2[/tex]. This is a bit weird, because it's not any real function that I can think of (dirac deltas can't really be squared properly...as far as I know...so I don't think ).

I'm sort of stumped on this one too!

EDIT:

Oh, I just thought, maybe instead of using Dirac Deltas, you can use Kronecker deltas...

Perhaps:

[tex]\Psi(E) = \frac{3}{5}\delta_{E, E_1} + \frac{4}{5}\delta_{E, E_3}[/tex]

Not 100% on this though.
 
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