# Mean value of observable in non-normalizable state

If $$|\psi\rangle$$ is a normalizable state in the Hilbert space of a quantum system and if $${\cal O}$$ is some observable we can always evaluate the mean value of $${\cal O}$$ on the state $$|\psi\rangle$$ using the equation $$\langle {\cal O}\rangle=\dfrac{\langle \psi|{\cal O}|\psi\rangle}{\langle \psi|\psi\rangle}\tag{1}.$$

The point is that $$\langle \psi|\psi\rangle\in (0,+\infty)$$ and the operation is mathematically well-defined. Alternatively we can rescale $$|\psi\rangle$$ so that $$\langle \psi|\psi\rangle =1$$ and then the mean value becomes just $$\langle {\cal O}\rangle = \langle \psi|{\cal O}|\psi\rangle\tag{2}.$$

Now suppose $$|\psi\rangle$$ is a non-normalizable state, like a position eigenstate $$|x\rangle$$ or a momentum eigenstate $$|p\rangle$$. In that case the basis is normalized according to $$\langle x|y\rangle=\delta(x-y)\tag{3}.$$

There are two issues now: the first is that $$\langle x|x\rangle$$ is ill-defined. In that case it makes no sense to use (1) at all, nor to rescale $$|x\rangle$$ to use (2). The second is that recalling that $$\delta(x-y)$$ is a distribution, which is just a continuous linear functional on test functions, it makes no sense whatsoever to divide by a distribution.

On the other hand equation (1) formally works if the state is an improper eigenstate of $${\cal O}$$. In fact, suppose $${\cal O}$$ is the position operator $$X$$ and $$|\psi \rangle = |x\rangle$$, we get $$\langle X\rangle = \dfrac{\langle x|X|x\rangle}{\langle x|x\rangle} = x\dfrac{\langle x|x\rangle}{\langle x|x\rangle}=x\tag{4}$$

Obviously mathematically speaking this is non-sense since $$\langle x| x\rangle$$ is not even a well-defined number so that dividing it by itself and saying it equals one is not something we can do.

Question: given that we have some non-normalizable state, say $$|\psi\rangle =|x\rangle$$, how do we properly define mean values of observables in such a state? How do we deal with division by the ill-defined $$\langle \psi |\psi\rangle$$? Or should we regularize $$|x\rangle$$ taking a smooth function $$f_\epsilon(y)$$ so that $$|\psi_\epsilon\rangle = \int f_\epsilon(y)|y\rangle dy,\quad \lim_{\epsilon \to 0} f_\epsilon(y)=\delta(y-x)$$ and then define $$\langle{\cal O}\rangle = \lim_{\epsilon \to 0}\dfrac{\langle \psi_\epsilon |{\cal O}|\psi_\epsilon\rangle}{\langle \psi_\epsilon|\psi_\epsilon\rangle}$$

• The situation that led me to ask this was the evaluation of the expectation value an observable after a QFT scattering. The initial state is a momentum eigenstate and then working in the Heisenberg picture we would be in a situation like in the question. In the associated literature one semiclassical approximation is employed on which a single momentum eigenstate contributes to the outgoing state and the result ends a ratio $\langle f| {\cal O} |i\rangle /\langle f | i \rangle$. I wasn't able to understand how they got the final result and imagined it could be connected to this question. – user1620696 Jun 7 at 2:15
• That's a good clarification. Deleting my previous comment because it's no longer needed. – Chiral Anomaly Jun 7 at 13:11