I've been solving a problem in quantum mechanics, and I was deriving the standard deviation of $P$, knowing that $\langle P\rangle=0$. Because $\Delta P=\sqrt{\langle P^2 \rangle – \langle P \rangle ^2} = \sqrt{\langle P^2 \rangle}$, I was trying to calculate the expectation value of the square of the momentum. The wave function was given by $\psi(x)=\sqrt{\alpha}e^{-\alpha|x|}$ where $\alpha>0$.
Here is what I've done. $$\langle P^2\rangle = \int_{-\infty}^{\infty}\psi^*(x) \left(-\hbar^2 \frac{d^2}{dx^2}\psi(x)\right)dx = -\hbar^2\alpha^2$$
Now, we have negative expectation value of the square of the momentum, which I think is wierd, and we have to take square root of that value. That's impossible. I couldn't find out what's wrong with my idea. Can somebody help me with this?
Best Answer
It seems that OP already knows that the variance is a manifestly non-negative quantity, and he is struggling to explain a negative result that he got.
Hint: The wave function $\psi(x)=\sqrt{\alpha}e^{-\alpha|x|}$ is not differentiable in $x=0$. The generalized function $$\tag{1} \psi^{\prime\prime}(x)~=~\left(\alpha^{\frac{5}{2}}- 2\alpha^{\frac{3}{2}}\delta(x)\right)e^{-\alpha|x|}~=~\alpha^{\frac{5}{2}}e^{-\alpha|x|}- 2\alpha^{\frac{3}{2}}\delta(x)$$ will have contributions proportional to a Dirac delta distribution.$^1$
--
$^1$ In the last step we have ignored some mathematical subtleties concerning how to multiply a non-smooth function and a Dirac delta distribution. These become apparent if we try to differentiate the wave function (1) further.