Normalization to delta distribution

[Jano L.]

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

<br /> (\Phi_s,\Phi_s&#039;) = \int \Phi_s^*(q) \, \Phi_{s&#039;} (q)~ dq = \delta(s-s&#039;)~?<br />

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

<br /> \int \Phi_\epsilon^*(q) \, \Phi_{\epsilon&#039;} (q)~ dq <br />

with \Phi_\epsilon(q) = e^{i \sqrt{2m\epsilon} \,q/\hbar} does not equal to \delta(\epsilon - \epsilon&#039;).

 

[vanhees71]

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

Now we have
\langle \epsilon&#039;|\epsilon \rangle = \int_{\mathbb{R}} \mathrm{d} q \Phi_{\epsilon&#039;}^*(q) \Phi_{\epsilon}(q) = 2 \pi |N(\epsilon)|^2 \delta \left [\sqrt{2m}(\sqrt{\epsilon}-\sqrt{\epsilon&#039;} \right]=\sqrt{8m \epsilon} \pi |N(\epsilon)|^2 \delta(\epsilon-\epsilon&#039;).
Thus setting
N(\epsilon)=\frac{1}{\sqrt{\sqrt{8m \epsilon} \pi}}
normalizes your functions to a \delta distribution.

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

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
u_{\epsilon,P}(q)=\frac{1}{\sqrt{2}} [\Phi_{\epsilon}(q) + P \Psi_{\epsilon}(q)] \quad \text{with} \quad P \in \{-1,+1\}.

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

 

[Jano L.]

Thank you very much! It is much more clear now. I didn't realize one can normalize \Phi_\epsilon's by dividing by \sqrt{\epsilon}.

 

[Jano L.]

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