# System where the energy is discreet

1. Mar 25, 2003

### GSXR750

I know when you have a system where the energy is discreet (i.e. a bound state), there is an discreet orthonormal base, and you can developpe(?) an arbitrary function in this base:

f(x)=Sum{a[n]*f[n](x)}. And you can find the a[n] by multipling with f[n'] and integrate over x. Then you get a[n']=Int{dx*f(x)*f[n'](x)}.

Now I have a continu(?) base and I can do:

f(x)=Int{dk*a(k)*f[k](x)}.

From now on I begin to doubt:

If I multiply with f[k'](x), the RHS becomes:

Int{dk*a(k)*delta(k-k'))=a(k') and thus:

a(k')=f(x)*f[k'](x)

where did I screw up??

Last edited by a moderator: Feb 4, 2013
2. Mar 25, 2003

### arcnets

Re: quantum

No, it doesn't. Instead, you get

Int(dk*a(k)*f[k](x)*f[k'](x))

Note the integration is over dk, not dx. So you can't simplify, since f[k'](x) does not depend on k.

I think you made the error of integrating over dx instead of dk.

3. Mar 25, 2003

### GSXR750

Then how can I find tha a(k) 's??

4. Mar 25, 2003

### arcnets

a(k') = Int{dx * f[k'](x) * f(x)}

Proof:
As you said,
f(x)=Int{dk*a(k)*f[k](x)}.
Multiply by f[k'](x):
f[k'](x) * f(x) = Int{dk*a(k)*f[k](x)* f[k'](x)}.
Integrate over dx:
Int{dx * f[k'](x) * f(x)}=Int{dx *Int{dk*a(k)*f[k](x)* f[k'](x)}}
(Here, some theorem about exchanging integrations is applied)
= Int{dk * a(k) *Int {dx * f[k](x) * f[k'](x)}}
(Now, execute the integration over dx, using orthonormality)
= Int{dk * a(k) * delta(k-k')}
(Now, execute the integration over dk)
= a(k')

Now, drop the ' and there you are. OK?

Last edited: Mar 25, 2003
5. Mar 25, 2003

OK Thanks