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Normalization to delta distribution

  1. Oct 13, 2012 #1

    Jano L.

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    Gold Member

    Do you know some example of an operator, other than momentum or position, that has (at least partially) continuous spectrum with eigenvalues [itex]s[/itex], and the corresponding eigenfunctions obey

    [tex]
    (\Phi_s,\Phi_s') = \int \Phi_s^*(q) \, \Phi_{s'} (q)~ dq = \delta(s-s')~?
    [/tex]

    EDIT
    For example, Hamiltonian of a free particle does not work with this, because the integral

    [tex]
    \int \Phi_\epsilon^*(q) \, \Phi_{\epsilon'} (q)~ dq
    [/tex]

    with [itex]\Phi_\epsilon(q) = e^{i \sqrt{2m\epsilon} \,q/\hbar}[/itex] does not equal to [itex]\delta(\epsilon - \epsilon')[/itex].
     
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  3. Oct 13, 2012 #2

    vanhees71

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    2016 Award

    Why not. Let's set
    [tex]\Phi_{\epsilon}(q)=N(\epsilon) \exp(\mathrm{i} \sqrt{2m \epsilon} q).[/tex]
    This is indeed the generalized eigenfunction of the free Hamiltonian
    [tex]\hat{H}=-\frac{1}{2m} \frac{\mathrm{d}^2}{\mathrm{d} q^2}[/tex]
    in position-space representation.

    Now we have
    [tex]\langle \epsilon'|\epsilon \rangle = \int_{\mathbb{R}} \mathrm{d} q \Phi_{\epsilon'}^*(q) \Phi_{\epsilon}(q) = 2 \pi |N(\epsilon)|^2 \delta \left [\sqrt{2m}(\sqrt{\epsilon}-\sqrt{\epsilon'} \right]=\sqrt{8m \epsilon} \pi |N(\epsilon)|^2 \delta(\epsilon-\epsilon').[/tex]
    Thus setting
    [tex]N(\epsilon)=\frac{1}{\sqrt{\sqrt{8m \epsilon} \pi}}[/tex]
    normalizes your functions to a [itex]\delta[/itex] distribution.

    Of course, this is not a complete set, because also
    [tex]\Psi_{\epsilon}(q)=N(\epsilon) \exp(-\mathrm{i} \sqrt{2m \epsilon} q)[/tex]
    are eigenfunctions of the Hamiltonian with the same eigenvalue [itex]\epsilon[/itex].

    Each energy eigenvalue is thus twofold degenerate. You can classify the energy eigenstates further by parity. That also explains the degeneracy: the parity operation commutes with the Hamiltonian and thus can be simultaneously diagonalized. The energy-parity eigenstates are given by
    [tex]u_{\epsilon,P}(q)=\frac{1}{\sqrt{2}} [\Phi_{\epsilon}(q) + P \Psi_{\epsilon}(q)] \quad \text{with} \quad P \in \{-1,+1\}.[/tex]

    The energy is degenerate, and the generalized basis built by[itex]\Phi_{\epsilon}[/itex] and [itex]\Psi_{\epsilon}[/itex] or [itex]u_{\epsilon,P}[/itex] is pretty inconvenient to work with. Much more convenient is the basis of momentum eigenfunctions,
    [tex]v_{p}(q)=\frac{1}{\sqrt{2 \pi}} \exp(-\mathrm{i} p q), \quad p \in \mathbb{R}[/tex]
    and has all nice properties of the above constructed energy-parity eigenbasis, because of course the [itex]v_{p}[/itex] are also generalized energy eigenstates.
     
  4. Oct 13, 2012 #3

    Jano L.

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    Gold Member

    Thank you very much! It is much more clear now. I didn't realize one can normalize [itex]\Phi_\epsilon[/itex]'s by dividing by [itex]\sqrt{\epsilon}[/itex].
     
  5. Oct 13, 2012 #4

    Jano L.

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    Gold Member

    Oh this somehow got to the classical physics forum, I apologize for that.
     
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