I am not sure whether my question counts as homework and exercises or not, because I already know the answer. The problem is, I find Zettili answer rather unsatisfactory.

Problem 1.11 (a) Find the Fourier transform for $\phi(k)=A(a-|k|),~~|k|\leq a~$ where $a$ is a positive parameter and $A$ is a normalization factor to be found. (b) Calculate the uncertainties $\Delta x$ and $\Delta p$ and check whether they satisfy the uncertainty principle.

There are many ways to calculate $\Delta x$ from wave function. One can find it from $\phi(k)$ or $\psi(x)$ (wave function in position space) etc. The easiest way in my opinion is obtaining it from $\phi(k)$. Note that $\psi (x)=\frac{4}{x^2}\sin^2\left(\frac{ax}{2}\right)$ according to Zettili. So in momentum space we have: $$\hat{x}=i\frac{d}{dk}~~~\text{and}~~~~\hat{x}^2=-\frac{d^2}{dk^2}$$ $$\langle x\rangle=\int_{-a}^{a}\phi(k)\hat{x} \phi(k)dk=i\left(\int_{-a}^{0}A^2(a+k)dk-\int_{0}^{a}A^2(a-k)dk\right)=0$$ $$\langle x^2\rangle=\int_{-a}^{a}\phi(k)\hat{x}^2 \phi(k)dk=0$$ $$\rightarrow \Delta x=\sqrt{\langle x^2\rangle-\langle x\rangle^2}=0,$$ which obviously violates uncertainty principle. I asked about this my university professor, he said due to the discontinuity of wave function you should change $\hat{x}$ such that: $$\hat{x}=-i\frac{d}{dk}, ~ -a\leq x <0 $$ $$\hat{x}=i\frac{d}{dk}, ~~~~~~ 0\leq x <a $$ But it gives me a complex number for average of position! It does not make any sense. Besides it is possible to show to that average of position in position space is also zero

$$\langle x\rangle=\int_{-\infty}^{\infty}\psi(x)\hat{x} \psi(x)dx=0.$$

All wave functions are real, so I didn't use complex conjugate anywhere.

This is Zettili answer:

Now, let us find the width $\Delta x$ of $\psi(x)$ Since $\sin(a\pi/2a)=1$, $\psi (\pi/a)=4/\pi^2$ and $\psi(0)=a^2$ we can obtain $\psi (\pi/a)=4/\pi^2\psi (0)$ or $\frac{\psi (\pi/a)}{\psi(0)}=\frac{4}{\pi^2}$ This suggests that $\Delta x= \pi/a.$

Where did I go wrong?

  • 2
    $\begingroup$ Oof, that's a tricky thing to put into chapter 1 of a QM textbook! The problem is that the second derivative is singular, so you can't trust your evaluation of $\langle x^2 \rangle$. You can fix it by smoothing out the singularity, or by working with delta functions. $\endgroup$
    – knzhou
    Jul 10, 2019 at 14:16
  • $\begingroup$ @knzhou You are completely right. I'd have used Cauchy integral in the complex plane to find $\Delta x$ from $\psi (x)$. I'm not sure why I didn't see that. $\endgroup$
    – Paradoxy
    Jul 10, 2019 at 15:00

1 Answer 1


You can justify this more rigorously if you take the $|k|$ term in the initial wavefunction (which isn't differentiable at $k=0$) as a distribution, in terms of Heaviside step function:


After that, differentiating it will yield:

\begin{align}\frac{\mathrm d}{\mathrm dk}|k|&=2\operatorname{\theta}(k)-1, \\ \frac{\mathrm d^2}{\mathrm d^2k}|k|&=2\operatorname{\delta}(k), \end{align}

where $\operatorname{\delta}$ is the Dirac delta. After you take properties of $\operatorname{\delta}$ on integration you should be able to arrive at the correct result.


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