In QM: How to derive <x|f> from f_n(x)?

1. Oct 13, 2008

mishla

1. The problem statement, all variables and given/known data

If you have a function f_n(x), how do you get the equivalent representation <x|f>?

2. Relevant equations

I have a system with a given Hamiltonian (not in matrix-form), from which I derived the specter of energy eigenvalues E_n, and the corresponding energy eigenfunctions f_n(x). However, I am asked to derive the eigenstates in the form <x|f> also, how do I do that?

3. The attempt at a solution

2. Oct 13, 2008

gabbagabbahey

f_n(x) is equivalent to $\left< x | f_n \right>$

3. Oct 13, 2008

HallsofIvy

Staff Emeritus
By the way, it is "spectrum", not "specter". Although, with Halloween coming, it may be appropriate!

As gabbagabbahey said, f_n(x) IS $\left< x | f_n \right>$ . <x, f> is the sum of $\left< x | f_n \right>$ over all n.

4. Oct 13, 2008

mishla

Thanks for the answers! So what is then meant by

“When you have found the spectrum of energy-eigenvalues, find the corresponding energy-eigenstates, both the abstract number basis and the concrete position-representation <x|f>.”?

Is “the abstract number basis” the same as f_n(x) (the one I have found)? And is “the concrete position representation <x|f>” then the sum of f_n(x) over all n?

And thanks for the correction of my misspelling, English is not my mother tounge…

5. Oct 13, 2008

gabbagabbahey

Suppose one of the energy eigenstates in some abstract basis was

$$|f_1 \rangle =\frac{1}{\sqrt{2}} |x \rangle -\frac{i}{\sqrt{2}} |y \rangle$$

Then in the concrete basis it would be $f_1(x)=\langle x|f_1 \rangle=\frac{1}{\sqrt{2}}$