# Normalization to delta distribution

1. Oct 13, 2012

### 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

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

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

$$\int \Phi_\epsilon^*(q) \, \Phi_{\epsilon'} (q)~ dq$$

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

2. Oct 13, 2012

### 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
$$\langle \epsilon'|\epsilon \rangle = \int_{\mathbb{R}} \mathrm{d} q \Phi_{\epsilon'}^*(q) \Phi_{\epsilon}(q) = 2 \pi |N(\epsilon)|^2 \delta \left [\sqrt{2m}(\sqrt{\epsilon}-\sqrt{\epsilon'} \right]=\sqrt{8m \epsilon} \pi |N(\epsilon)|^2 \delta(\epsilon-\epsilon').$$
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.

3. Oct 13, 2012

### 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 $\sqrt{\epsilon}$.

4. Oct 13, 2012

### Jano L.

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