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

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.


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

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

  • $\begingroup$ I don't understand how you got the expression for $\Psi^*$. Can we simply take the conjugate within the integral sign? Does this mean that the integral of a conjugate is equal to the conjugate of the integral? $\endgroup$ – Joebevo Nov 22 '12 at 3:18
  • $\begingroup$ Also, I don't understand why you introduced the primed variables only to drop them in the last 2 expressions. $\endgroup$ – Joebevo Nov 22 '12 at 3:30
  • $\begingroup$ @Joebevo, As you expected, integral and complex conjugation can be interchanged. Physically, integral is just a sum of infinitesimally small quantities, and conjugation threads over sums. Thus there is no problem. $\endgroup$ – Sangchul Lee Nov 22 '12 at 5:49
  • $\begingroup$ @Joebevo, I introduced the primed $p$ just in order to avoid confusion by two same $p$'s. id you have two dummy variables with same label, the how can you distinguish one from another? That's the reason. $\endgroup$ – Sangchul Lee Nov 22 '12 at 5:56
  • $\begingroup$ Yes, but in the penultimate step, you have both $a(p)^*$ and $a(p)$ being integrated over $p$. This seems contrary to step 2, where $a(p)^*$ goes with $p$ and $a(p')$ with $p'$. So shouldn't your penultimate step read $\int_{-\infty}^{\infty} p' a(p)^*a(p') \, dp$? $\endgroup$ – Joebevo Nov 22 '12 at 6:10

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}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.