# Wavefunction in the energy representation

## Homework Statement

$$\psi(x)=\frac{3}{5}\chi_{1}(x)+\frac{4}{5}\chi_{3}(x)$$

Both $$\chi_{1}(x) \chi_{3}(x)$$ 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

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

$$\left|\psi\right> = \frac{3}{5}\left|E_1\right>+\frac{4}{5}\left|E_3\right>$$

I'm not so sure. I think what they are after is $$\psi(E)$$ i.e. a function of energy. I suppose in analogy with $$\psi(x) = \left<x|\psi\right>$$ it would be $$\psi(E) = \left<E|\psi\right>$$. Completeness of states could be of help. I suspect you should get delta functions.

diazona
Homework Helper
I'm not so sure. I think what they are after is $$\psi(E)$$ i.e. a function of energy. I suppose in analogy with $$\psi(x) = \left<x|\psi\right>$$ it would be $$\psi(E) = \left<E|\psi\right>$$. 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.

Matterwave
Science Advisor
Gold Member
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:

$$\Psi (x) = \sum_{n=1}^{\infty} c_n\psi_n(x)$$

Where the $$\psi_n$$ represent energy eigenfunctions. The energy representation of the wave function would probably then just be the coefficients $$c_n$$ at different energies. The probabilities of finding a specific energy would be $$|c_n|^2$$. 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:

$$\Psi(E) = \frac{3}{5}\delta_{E, E_1} + \frac{4}{5}\delta_{E, E_3}$$

Not 100% on this though.

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