# Proof that momentum operator acting on a wavefunction gives the expectation value

I'm following a book where the author tries to prove that $$\langle p \rangle = \langle \psi \vert \hat{p} \vert \psi \rangle,\tag{0}$$ so he just computes the integral

$$\langle \psi \vert \hat{p} \vert \psi \rangle ~=~ \int_{\mathbb{R}} \psi^*(x) \left( -i\hbar \dfrac{\mathrm{d}}{\mathrm{d}x} \right) \psi(x) \, \mathrm{d}x \tag{1}$$

and starts by saying that

$$-i\hbar\dfrac{\mathrm{d}\psi(x)}{\mathrm{d}x} ~=~ -i\hbar\dfrac{\mathrm{d}}{\mathrm{d}x} \int_{\mathbb{R}} \tilde{\psi}(p) e^{ipx/\hbar} \, \mathrm{d}p ~=~ \int_{\mathbb{R}} p \, \tilde{\psi}(p) \, e^{ipx/\hbar} \, \mathrm{d}p \tag{2}$$

which is okay because we can expand the wavefunction on the position space as a sum of wavefunctions on the momentum space using the Fourier transform on $$\psi(x) .$$

Using this result, we have that

$$\langle \psi \vert \hat{p} \vert \psi \rangle ~=~ \int_{\mathbb{R}} \left( \int_{\mathbb{R}} \tilde{\psi}^*(p') \, e^{ip'x/\hbar} \mathrm{d}p' \right) \left( \int_{\mathbb{R}} p \, \tilde{\psi}(p) \, e^{ipx/\hbar} \mathrm{d}p \right) \mathrm{d}x \tag{3}$$

However, I don't get why the next two steps are mathematically correct and how he moves the integration signs:

\begin{align} \langle \psi \vert \hat{p} \vert \psi \rangle &= \int_{\mathbb{R}} \int_{\mathbb{R}} p\, \tilde{\psi}^*(p')\, \tilde{\psi}(p) \left( \int_{\mathbb{R}} e^{i(p-p')x/\hbar} dx\right) \, \mathrm{d}p' \, \mathrm{d}p \tag{4} \\[5px] &= \int_{\mathbb{R}} \int_{\mathbb{R}} \delta(p-p') \,p\, \tilde{\psi}^*(p')\, \tilde{\psi}(p) \, \mathrm{d}p \, \mathrm{d}p' \tag{5} \\[5px] &= \int_{\mathbb{R}} p \vert \tilde{\psi}(p) \vert^2 \, \mathrm{d}p \tag{6} \\[5px] &= \langle p \rangle \tag{7} \end{align}

Eq. (7) obviously follows from Eq. (6), because the expectation value of a random variable is equal to the integral of that variable times the probability density distribution, which is exactly what $$\vert\tilde{\psi}(p)\vert^2$$ is. But how does he get to Eq. (6)?

Question: How do we get Equations (5) and (6) starting from Equation (4)?

• I do not understand this question - $\langle p\rangle$ and $\langle \psi \vert \hat{p} \vert \psi\rangle$ are for me just two notations for the same thing. What exactly is the thing you want to prove here? – ACuriousMind Aug 18 '19 at 20:33
• Which textbook? – Qmechanic Aug 18 '19 at 21:06
• @ACuriousMind I suspect that Jamie_mc2 is using $\langle p\rangle$ to mean the statistical expectation value, i.e. $\langle X\rangle = E[X] = \int X P(X) \mathrm{d}X$, which is not a priori the same thing as a matrix element of an operator like $\langle\psi|\hat{p}|\psi\rangle$. – David Z Aug 18 '19 at 21:11
• Yes @DavidZ that was exactly what I meant. Thanks – Jaime_mc2 Aug 18 '19 at 21:14

From eq. (4) to eq. (5): The Fourier transform of $$1$$ is the $$\delta$$ function, i.e. $$\int \mathrm{e}^{\mathrm{i}px}\mathrm{d}p = \delta(x)$$, cf. e.g. Wikipedia.
From eq. (5) to eq. (6): This is just the characteristic property of the $$\delta$$-function, i.e. $$\int f(x) \delta(x - y) \mathrm{x} = f(y)$$.