Showing that $\langle p\rangle=\int\limits_{-\infty}^{+\infty}p |a(p)|^2 dp$

Question

How do I show that $$\int \limits_{-\infty}^{+\infty} \Psi^* \left(-i\hbar\frac{\partial \Psi}{\partial x} \right)dx=\int \limits_{-\infty}^{+\infty} p \left|a(p)\right|^2dp\tag1$$

given that $$\Psi(x)=\frac{1}{\sqrt{2 \pi \hbar}}\int \limits_{-\infty}^{+\infty} a(p) \exp\left(\frac{i}{\hbar} px\right)dp\tag2$$

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My attempt: $$\frac {\partial \Psi(x)}{\partial x} = \frac{1}{\sqrt{2\pi \hbar}} \int\limits_{-\infty}^{+\infty} \frac{\partial}{\partial x} \left(a(p)\exp\left(\frac{i}{\hbar} px\right)\right)dp\tag3$$

$$=\frac{1}{\sqrt{2\pi \hbar}} \int\limits_{-\infty}^{+\infty} a(p) \cdot \exp\left(\frac{i}{\hbar} px\right)\frac{i}{\hbar}p \cdot dp\tag4$$

Multiplying by $-i\hbar$: $$-i\hbar \frac {\partial \Psi}{\partial x}=\frac{1}{\sqrt{2\pi \hbar}} \int\limits_{-\infty}^{+\infty} a(p) \cdot \exp\left(\frac{i}{\hbar} px\right)p \cdot dp\tag5$$

At this point I'm stuck because I don't know how to evaluate the integral without knowing $a(p)$. And yet, the right hand side of equation (1) doesn't have $a(p)$ substituted in.

Answer 1

The conclusion follows from the Fourier inversion formula (in distribution sense):

$$\begin{align*} &\int_{-\infty}^{\infty} \Psi^{*} \left( -i\hbar \frac{\partial \Psi}{\partial x} \right) \, dx \\ &= \int_{-\infty}^{\infty} \left( \frac{1}{\sqrt{2\pi\hbar}} \int_{-\infty}^{\infty} a(p)^{*}e^{-ipx/\hbar} \, dp \right) \left( \frac{1}{\sqrt{2\pi\hbar}} \int_{-\infty}^{\infty} p' a(p')e^{ip'x/\hbar} \, dp' \right) \, dx \\ &= \frac{1}{2\pi\hbar} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} p' a(p)^{*}a(p') e^{i(p'-p)x/\hbar} \, dp'dp dx \\ &= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} p' a(p)^{*}a(p') \left( \frac{1}{2\pi\hbar} \int_{-\infty}^{\infty} e^{i(p'-p)x/\hbar} \, dx \right) \, dp' dp \\ &= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} p' a(p)^{*}a(p') \delta(p-p') \, dp' dp \\ &= \int_{-\infty}^{\infty} p a(p)^{*}a(p) \, dp = \int_{-\infty}^{\infty} p \left| a(p) \right|^2 \, dp. \end{align*}$$

Answer 2

You almost got it, you just have to plug your result (5) up in the left hnd side of (1) and also use

\begin{equation} \Psi^*(x)=\frac{1}{2\pi}\int\thinspace dq\thinspace a^*(q)\exp(-iqx) \end{equation}

where I have set $\hbar=1$. The left hand side of (1) then reads

\begin{eqnarray} \int dx \thinspace \Psi^*(-i\frac{\partial}{\partial x})\Psi&=&\frac{1}{2\pi}\int dp\int dq\int dx\thinspace a^*(q)\thinspace p \thinspace a(p)\exp\Big(i(p-q)x\Big)\\\\\\&=&\frac{1}{2\pi}\int dp\int dq\thinspace a^*(q)\thinspace p \thinspace a(p)\Big[\int dx\exp\Big(i(p-q)x\Big)\Big]\\\\\\\end{eqnarray}

The integral in $x$ is simply the plane wave represeantation of a Dirac delta function $\int dx\exp\Big(i(p-q)x\Big)=2\pi\thinspace\delta(p-q)$. Pluging this above and integrating in $q$ gives you what you want, i.e.

\begin{eqnarray} \int dx \thinspace \Psi^*(-i\frac{\partial}{\partial x})\Psi&=&\int dp\int dq\thinspace a^*(q)\thinspace p \thinspace a(p)\delta(p-q)\\ &=&\int dp\thinspace p\thinspace a^*(p)\thinspace a(p)\\ &=&\int dp\thinspace p\thinspace |a(p)|^2 \end{eqnarray}