# Can the expectation value of the square of momentum be negative?

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?

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.$^1$

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$^1$ 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.

No. The expectation value of the square of the momentum operator cannot be negative.

The other answers address your particular problem on an integration level, but also notice that this can be easily shown in bra-ket notation.

Let $|\psi\rangle$ be any state in the Hilbert space, and let $\hat P$ be the momentum operator, then we have \begin{align} \langle \psi|\hat P^2|\psi\rangle = \langle \psi|\hat P^\dagger\hat P|\psi\rangle = \Big\|\hat P|\psi\rangle\Big\|^2 >0 \end{align} In the the first equality, we simply used the fact that $\hat P$ is hermitian; $\hat P^\dagger = \hat P$. In the second equality, we used the fact that the expression $\langle \psi|\hat P^\dagger\hat P|\psi\rangle$ is just the inner product of $\hat P|\psi\rangle$ with itself, and in the third equality we used the fact that the inner product of any vector with itself is just the square norm of that vector. The last inequality follows from positive definiteness of the inner product.

Let's set $\hbar$ and $\alpha$ to one. Then $\psi(x)=C\exp(-|x|)$. Let's compute the first derivative of $\psi(x)$ carefully. $d_x \psi(x) = -C \exp(-|x|)\mathrm{sgn}(x)$. Now let's carefully compute the second derivative of $\psi(x)$. \begin{aligned} d_x^2 \psi(x) &= -C d_x\exp(-|x|)\mathrm{sgn}(x) \\ &= C \exp(-|x|)\mathrm{sgn}(x)^2 -2C\exp(-|x|)\delta(x)\\ &= C\exp(-|x|)(1-2\delta(x)) \end{aligned} Now we want to compute $\int \psi(x) (-d_x^2) \psi(x)$ and see if we do really get something negative. Plugging in our expression for $d_x^2 \psi(x)$, we get the integral \begin{aligned} \int C\exp(-|x|)(- C\exp(-|x|)(1-2\delta(x))) \\ &= -C^2\int \exp(-2|x|)(1-2\delta(x))\\ &= -C^2\int \exp(-2|x|) + 2C^2 \int \exp(-2|x|)\delta(x)\\ &= -C^2 + 2C^2 \\ &= C^2. \end{aligned} Thus we get a positive answer. So your only mistake was that you forgot to do the chain rule when you took the derivative.