A problem I'm struggling with is following:

Consider the momentum space wave function

$$\Phi(p_x)= \begin{cases} 0, & \mbox{if }\vert\,p_x-p_0\vert > \gamma\\ C, & \mbox{if }\vert\,p_x-p_0\vert \le \gamma \end{cases}\,$$

where $C$ and $\gamma$ are constants.

Find the corresponding wave function $\Psi(x)$ in configuration space, and verify that $\Psi(x)$ is normalized to unity. Also, show that $\Delta x\Delta p_x \ge \hbar$ using reasonable definition of the widths $\Delta x$ and $\Delta p_x$.

I found that $C={1 \over \sqrt{2\gamma}}$ and the corresponding wave function is $\Psi(x)=\sqrt{\frac{\hbar}{\pi\gamma}}\,e^{\,ip_0x}{\sin{{(\gamma x/\hbar)}} \over x}$. And it was easy to verify that $\int_{-\infty}^{\infty}\vert \Psi(x)\vert^2\,dx=1$ using the fact that $\int_{-\infty}^{\infty}{\sin^2x \over x^2}\,dx=\pi$. For showing $\Delta x\Delta p_x \ge \hbar$, I chose $\gamma$ and $\frac{\hbar\pi}{\gamma}$(the first positive root of $\frac{\sin x}{x}$) for standard deviations of $p_x$ and $x$, respectively.

But I got stuck in calculating the expectation value of $x^2$. $$ \langle x^2\rangle=\int_{-\infty}^{\infty} \sin^2x\,dx=\,? $$ How can I deal with this awkward situation? Do I have to calculate the integration with a lower bound $-L$ and an upper bound $L$ and then take the limit $L \rightarrow \infty$? Are there any traits of the sinc wave function that make it special?

  • $\begingroup$ Is there any reason why you expect the variance $\langle x^2\rangle$ to be defined for this state in the first place? You've just shown that $x\Psi(x)$ isn't normalizable, so $\Psi(x)$ isn't (strictly speaking) in the domain of $\hat x$. $\endgroup$ Apr 12 '18 at 11:32
  • $\begingroup$ Thank you for your answer. If what I understand is right, you mean there are wave functions in the Hilbert space that are not in the domain of the operator $\hat{x}$. Then does this mean that wave functions not in the domain of $\hat{x}$ represent physically impossible states? $\endgroup$
    – asdf
    Apr 12 '18 at 11:47
  • 1
    $\begingroup$ Yep, that's correct. In the same way that there are wavefunctions (like the step function $\Psi(x)=\theta(a-|x|)$) that don't have derivatives and are thus not in the domain of $\hat p$. (And indeed, since $\hat x$ is a derivative in $p$-space, those two examples are one and the same.) The $1/x$ decay of the sinc function is too slow to be physical, as is the infinitely-sharp edge of the step on momentum space. $\endgroup$ Apr 12 '18 at 12:15

The fact that the integral $$ \langle x^2\rangle=\int_{-\infty}^{\infty} \sin^2x\,dx=\infty $$ diverges to infinity is the clear-cut indicator that your state is not in the domain of the position operator $\hat x: \mathcal D(x)\subset L_2(\mathbb R) \to L_2(\mathbb R)$. As such, the only meaningful way to assign the variance $\langle x^2\rangle$ is infinity, but really there is no pre-existing reason why that variance needed to exist in the first place.

This can be seen much more clearly in the original momentum-space formulation, where the position operator $\hat x = i\hbar \partial_p$ takes the form of a momentum derivative: your stepped wavefunction has a discontinuity, so asking about its derivatives is meaningless. This infinitely sharp step is as unphysical as the infinite span of the position-space wavefunction.


If a valid quantum wave function in the position basis must be continuous in both the wave function, $\psi(x)$, and it first derivative, $d \psi(x)/dx$, then the same must be true of the corresponding wave function in the momentum basis. A step function in the momentum basis is not a valid quantum wave function. Thus, I assume we cannot use $\sin(x)/x$ as a valid quantum wave function in the position basis, even though it is continuous in both the wave function and it first derivative.


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