I was reading through a textbook, and the statement was made that the inner products are guaranteed to exist if the eigenvalue spectrum of the operator is discrete. I have come across no support for this claim, and the basis for this claim was not immediately apparent after quite a long time of consideration on my behalf. Furthermore, while trying to deduce the answer, I was lead to the complement of my first question: why does a continuous eigenvalue spectrum of an operator lead to non-normalizable eigenfunctions (i.e. the inner products don't exist)?

# [Physics] Why are the inner products of the eigenfunctions of an operator with a discrete eigenvalue spectrum guaranteed to exist

eigenvaluehilbert-spacemathematical physicsoperatorsquantum mechanics

#### Related Solutions

This could be a math question. But if you think about the physical aspect of the question, it is interesting to look at the Schrodinger equation:

For free fields (without potential), you have (in units $\bar h = m = \omega = 1$):

$$i\frac{\partial \Psi( k, t)}{\partial t} = \frac{ k^2}{2}\Psi( k, t)$$ or

$$ E \tilde \Psi( k, E) = \frac{ k^2}{2} \tilde \Psi( k, E)$$

Whose solution is :

$$\Psi( k, t) \sim e^{- i \frac{ k^2}{2} t}$$

or

$$ \tilde \Psi( k, E) \sim \delta \left(E - \frac{ k^2}{2}\right)$$

Here $ \tilde \Psi( k, E)$ is a Fourier transform of $\Psi( k, t)$.

It is clear, from the form of the equations, that there is no constraint about $E$. It is a continuous spectra, and this is clearly a non-normalizable solution.

However, with potentials, things appear differently, and you will have some differential equations, for instance, for the harmonic oscillator potential, you will have :

$$ E \Psi( k, E) = \frac{ k^2}{2} \Psi( k, E) - \frac{1}{2}\frac{\partial^2 \Psi( k, E)}{\partial k^2}$$

The solution for $\Psi$ involves a Hermite differential equation (multiply by some exponential $e^{-k^2}$).

If E is taking continuous, then the Hermite solution (with a real index) is not bounded at infinity, and so the solution is not normalizable.

If we want a normalizable solution, then we need a (positive) integer indexed solution $H_n$, whose name is Hermite polynomials. In this case, the spectrum of $E$ is discrete.

The choice of $E$ discrete (and so a normalizable solution) is then a physical choice. In the case of the harmonic oscillator, it is unphysical to suppose that the solution is not bounded at infinity.

The case of Hermite polynomials is a special case of orthogonal polynomials, which is very well suited to represent orthonormal states, corresponding to discrete eigenvalues of the hermitian operator Energy.

The thing of it all is this:

There are two kinds of eigenfunctions of hermitian operators. The ones which admit discrete spectra (eigenvalues are separated from one another) and the other, continuous spectra (eigenvalues fill out an entire range). If the spectra is continuous, then they **DO NOT** represent possible wavefunctions (only a linear combination of them.....yes a gaussian wavepacket kind of thing may be normalizable). In case of momentum operators, $$\frac{\hbar}{i}\frac{\partial f_p(x)}{\partial x} = pf_p(x)$$....$f_p(x)$ is some momentum eigenfunction....
solving the above gives $$f_p(x) = Ae^{\frac{ipx}{\hbar}}$$
which is not a square integrable one.

But since momentum is an observable, we only take real values of p and use dirac orthonormality by $$\int_{-\infty}^{\infty}f^*_{p^\prime}(x)f_p(p)dx = |A^2|\int_{-\infty}^{\infty}e^\frac{(p^\prime-p)x}{\hbar}dx = |A^2|2\pi\hbar\delta(p-p^\prime)$$, and then picking $A=\frac{1}{\sqrt{2\pi\hbar}}$, we have $$\langle f_{p^\prime}|f_p\rangle = \delta(p-p^\prime)$$......

Now, this means that the eigen functions of momentum are sinusoidal (this itself is unrealizable since any TRUE or PERFECT sine wave has to extend from $-\infty$ to $\infty$)

But there is **no** such thing as **a particle with definite momentum**, courtesy heisenberg uncertainty principle........also implies that measurement cannot collapse a wavefunction to an eigenstate with a perfectly defined momentum

This is why we make a normalizable wavepacket with a narrow range of momenta.....to make the whole thing physically realizable. None of the eigenfunctions of $\hat p$ live in hilbert space but those with real eigenvalues (wavepackets) and which are dirac normalizable do. They (eigenfuntions of $\hat p$) do not represent possible physical states but are very useful in problems like scattering from a potential hill or a barrier.

Reference :- Griffiths, Introduction to Quantum Mechanics

## EDIT

Please do not upvote my answer as it does not completely address the concerns raised by OP (take a look at the comment section below this answer), ie, a formal treatment of 'wavepacket'(not wavefunction) collapse.....i am terribly sorry if invoking such a statement is wrong. At best my answer is partially complete.

## Best Answer

The spectrum of a bounded linear operator $A$ is by definition the set of numbers $\lambda$ where $A-\lambda$ is not invertible. In the finite dimensional case, this means $A-\lambda$ is neither injective nor surjective, and the former statement is just a fancy way of saying that there exists an eigenvector; as this eigenvector is

by definitiona part of the Hilbert space, it is in partiular normalizable.However, in the infinite-dimensional case, injective and surjective are no longer equivalent and the spectrum decomposes into the point spectrum, the continuous spectrum and the residual spectrum.

The point spectrum is the part of the spectrum where the map is not injective (it might not be hard to prove that it is indeed discrete - too lazy to investigate right now). The residual spectrum is empty for normal operators (and thus in particular selfadjoint operators), which leaves the continuous spectrum where the map is injective (and thus there are no eigenvectors) but not surjective. The concept of eigenvector just doesn't make sense for the coninuous spectrum in this formalism.

The situation gets muddled (or, depending on your point of view, un-muddled) in the formalism of rigged Hilbert spaces, which is the proper framework for the treatment of unbounded operators: Rigged Hilbert spaces allow us to introduce 'eigenvectors' which aren't part of our Hilbert space. In particular, in case of the Hilbert space $L^2$, they can be non-normalizable functions (eg plane waves) or not functions at all (eg delta distributions).