Consider a Hermitian field operator $\phi(x)$ with eigenstates satisfying $$ \phi(x) |\alpha\rangle = \alpha(x) | \alpha \rangle $$ I'm trying to determine the inner product between the eigenstates. To do this, I consider $$ \langle\beta|\phi(x)|\alpha\rangle = \alpha(x)\langle\beta|\alpha\rangle = \beta(x)\langle\beta|\alpha\rangle $$ which implies $$ \left[ \alpha(x) - \beta(x) \right] \langle\beta|\alpha\rangle = 0\hspace{3cm} (1) $$ Q. What is the solution to this equation?

From the equation, I gather that $\langle\beta|\alpha\rangle = 0$ whenever $\alpha(x) \neq \beta(x)$ for any $x$ and therefore it has support only when $\alpha(x) = \beta(x)$. How can I represent this?

Is it obvious that this implies $$ \langle\beta|\alpha\rangle \propto \delta \left[ \alpha(x) - \beta(x) \right] $$ This solution seems weird since it seems to imply that the norm of the eigenstate is "infinite" (naively!), but this does not follow from $(1)$.

I know there are many subtleties here when dealing with infinite dimensional Hilbert spaces. The solution may lie in one of those subtleties. Any ideas?

  • $\begingroup$ I speculate that this question may not have a mathematically rigorous answer at all unless the field theory is extremely simple. Morally speaking, isn't $\langle \beta|\alpha\rangle$ the "$\beta$-$\alpha$ matrix element" of the partition function (in Euclidean signature), since written as a path integral the inner product would be $\int \mathscr D\phi e^{-S}$ with the appropriate boundary conditions? As another note, I would say that $\langle \beta|\alpha\rangle \propto \delta(\alpha-\beta)$ where $\delta$ is a "functional delta function," nonzero only if the functions are the same. $\endgroup$ – joshphysics Sep 4 '14 at 19:43
  • $\begingroup$ Formulated better in 312006. $\endgroup$ – Cosmas Zachos Feb 20 '17 at 1:33

The relation

$$\langle a|b\rangle\propto\delta(a-b)$$

is nothing unusual, it is simply an orthogonality condition. If the proportionality was an equality, and in addition we had completeness, the set of states would form an orthonormal basis. The reason why the delta function shows up is that you assume your operator to have a continuous spectrum of eigenvalues.

We are working with a vector space, it only seems natural that there should be some way to define an orthogonality condition. The delta distribution, as a linear functional on the Hilbert space, provides the appropriate structure for that. If you worry about infinities, there are two things to keep in mind: formally, the delta function is not really infinite, as it is strictly only defined under an integral. This is due to its nature as a distribution. The other thing is that it is not an observable quantity anyways: what is physically relevant are eigenvalues of operators and probabilities.


The steps that you wrote down till eq. 1 is in fact a simple proof of the following theorem (which can be looked up in elementary text books of quantum mechanics):

Eigen functions (of a Hermitian operator or more generally a symmetric operator on a separable Hilbert space) belonging to distinct eigenvalues are orthogonal.

This is always true for separable Hilbert spaces (having a countable basis), e.g. the Hilbert space $L^2$ of square integrable functions.

  • $\begingroup$ Don't we have to use some kind of rigged Hilbert space here? $\endgroup$ – Danu Sep 5 '14 at 8:05

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