# Eigenfunctions of Operators with Continuous Sprectra

Sferics
I'm self-studying Griffith's Intro to Quantum Mechanics, and on page 100 he makes the claim that the eigenfunctions of operators with continuous spectra are not normalizable. I can't see why this is necessarily true. Hopefully I am not missing something basic.

muppet
The problem comes when you try and work out the meaning of "orthogonality". Say x and y are position eigenstates. Then we want
$$\langle x | y\rangle =\delta(x-y)$$
For discrete quantities, the Kronecker delta is either one or zero. The Dirac delta function used here has to be something more like an infinite weight, in the handwaving way we tend to talk about these things. As it's neither a function, nor a number, there's no sense in which we can just divide by $\delta(0)$ to make the "norm" of such a state one.

Hope that helps.

Staff Emeritus
Gold Member
Momentum eigenfunctions have a similar problem. Consider the function g defined by g(x)=A exp(ipx) for all x. This is a momentum "eigenfunction". If we plug it into the usual formula for the norm of a wavefunction, we get
$$\|g\|^2=\int(Ae^{ipx})^*(Ae^{ipx})dx=|A|^2\int dx.$$ There's clearly no choice of A that makes the right-hand side =1, since ##\int dx=\infty##.

Note that the space of wavefunctions is the semi-inner product space of square-integrable ##\mathbb C##-valued functions on ##\mathbb R##. What I did above shows that a momentum "eigenfunction" isn't square-integrable. This means that it's not actually a member of that semi-inner product space. So it's actually more appropriate to say that the momentum operator doesn't have any eigenfunctions. These functions are a kind of generalized eigenfunctions.

Sferics
I understand that these two examples (which he provides) are not normalizable, but it almost seems as if it just happened that way. How do we know that, for any arbitrary operator with a continuous spectra, it's eigenfunctions are not normalizable?

muppet
For any continuous spectra, your "sum" over eigenstates will have to be an integral. This necessarily implies that the states will have a delta function norm.
If this isn't clear, think about some totally arbitrary continuum of values f, and think about how the identity operator acts on one specific eigenstate with eigenvalue $f_0$:
$$|f_0\rangle=\int df \langle f|f_0\rangle |f\rangle$$

Ilmrak