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System where the energy is discreet

  1. Mar 25, 2003 #1
    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:


    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:


    where did I screw up??
    Last edited by a moderator: Feb 4, 2013
  2. jcsd
  3. Mar 25, 2003 #2
    Re: quantum

    No, it doesn't. Instead, you get


    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.
  4. Mar 25, 2003 #3
    Then how can I find tha a(k) 's??
  5. Mar 25, 2003 #4
    a(k') = Int{dx * f[k'](x) * f(x)}

    As you said,
    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
  6. Mar 25, 2003 #5
    OK Thanks
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