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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 $$\langle{\cal O}\rangle = \lim_{\epsilon \to 0}\dfrac{\langle \psi_\epsilon |{\cal O}|\psi_\epsilon\rangle}{\langle \psi_\epsilon|\psi_\epsilon\rangle}$$

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  • $\begingroup$ 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. $\endgroup$ – user1620696 Jun 7 at 2:15
  • $\begingroup$ That's a good clarification. Deleting my previous comment because it's no longer needed. $\endgroup$ – Chiral Anomaly Jun 7 at 13:11

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