# Expectation value in terms of Dirac function in Quantum Mechanics

Recently, I started with principles of Quantum Mechanics by R. Shankar, and I got stuck at a part in my calculation of the Expectation value of Position and Momentum operator in the case where the particle is in one of the states.
Starting with the Momentum operator, the eigenstates are given by:$$\psi_p(x)=\sqrt\frac{1}{2\pi}e^{i\frac{p}{\hbar}x}$$ Here $$\psi_p(x)$$ represents the eigenstate for the eigenvalue p, and the $$\sqrt\frac{1}{2\pi}$$ factor is the normalize the above factor.
In order to find the Expectation value of momentum when the system in in eigenstate $$|p\rangle(\psi_p(x)=\langle x|p\rangle)$$ where $$|p\rangle$$ refers to the eigenstate for the eigenvalue $$p$$.$$\langle P\rangle=\langle \psi|P|\psi\rangle$$ where $$|\psi\rangle$$ refers the the state of the system and P is the momentum operator(which is $$\frac{\hbar}{i}\frac{d}{dx}$$). Now If I assume that my particle is in one of the eigenstate $$|p\rangle(|\psi\rangle =|p\rangle)$$ then the equation for expectation value becomes:$$\langle P\rangle=\langle p|P|p\rangle$$ $$\langle P\rangle=\int_{-\infty}^{\infty}{\int_{-\infty}^{\infty}{\langle p|x\rangle\langle x|P|x'\rangle\langle x'|p\rangle}dxdx'}$$ and we know $$\langle P\rangle=\frac{\hbar}{i}\delta'(x-x')$$ and thus, $$\langle P\rangle=\frac{\hbar}{2\pi i}\int_{-\infty}^{\infty}{\int_{-\infty}^{\infty}{e^{-i\frac{p}{\hbar}x}\delta'(x-x')e^{i\frac{p}{\hbar}x'}}dx'dx}$$ $$\langle P\rangle=\frac{p}{2\pi}\int_{-\infty}^{\infty}{e^{-i\frac{p}{\hbar}x}e^{i\frac{p}{\hbar}x}dx}$$ and we can say that:$$\langle P\rangle=p\delta(0)$$ How can the expectation value have direac delta function? Isn't this meaning and infinite amount of momentum? Where am I wrong in my calculations? I tried doing the same with the $$X$$ operator and failed brutally(It will be helpful If someone can do this calculation.).

• You can shorten the calculation (completely analogous for the position operator, just replace $p\to x$ and $P\to X$). Indeed, formally, you have $\langle p|p^\prime\rangle=\delta(p-p^\prime)$ and $P|p\rangle=p|p\rangle$ and hence $\langle p|P|p\rangle=p\delta(0)$. This last expression is, as you noted, not well-defined. The problem is that objects like $|p\rangle$ are in a strict sense not vectors of the underlying Hilbert space, but they turn out as a useful tool in QM. But you have to follow certain rules to don't run into paradoxes and undefined expressions. This is one of such instances. Commented Dec 16, 2023 at 7:40
• @TobiasFünke So does this mean the we cannot determine the absolute value of momentum doing a measurement? Commented Dec 16, 2023 at 8:02
• I would like to point out that $\vert p \rangle$ is normalized only in the sense that it has $1$ as completion relation and that $\delta$ is a distribution meaning it cannot be evaluated at a specif point (only integrated on an integral). Commented Dec 16, 2023 at 8:53

The position and momentum observables are both unphysical idealisations of real measurements. In reality a measurement is a result of coupling a system $$S$$ with a measurement device $$M$$ in some environment $$E$$ and is described by some Hamiltonian $$H$$ that depends on the observables of $$S$$, $$M$$ and $$E$$. There are various kinds of models of such couplings and mathematical techniques for studying the kinds of states that they give rise to, for a survey see