I am reviewing basic quantum mechanics since I feel like I am struggling with the fundamentals. I am not sure exactly how much I am "taking for granted" or whether or not my logic is clear.

My goal is (in 1-D) to get to $[\hat{x},\hat{p}] = i$ where I set $\hbar = 1$. For now I am taking $\langle x' | x \rangle = \delta(x'-x)$, and $\langle p|x \rangle = \frac{1}{\sqrt{2\pi}}\exp(ipx)$. That last identity I am not sure whether I am taking for granted or not.

In any case, what I have is

$$ \langle \psi | \hat{x}\hat{p} | \psi \rangle = \int dx'dp' \langle\psi|\hat{x}|x'\rangle\langle x'|\hat{p}|p'\rangle\langle p'|\psi\rangle = \int dx' dp'dx'' x'p' \frac{1}{\sqrt{2\pi}}\exp(-ix'p')\langle\psi|x'\rangle\langle p'|x''\rangle\langle x''|\psi\rangle \\ = \frac{1}{2\pi}\int dx'dp'dx'' x'p'e^{-ip'(x'-x'')}\langle\psi|x'\rangle \langle x''|\psi\rangle. $$ Then we have $$ \langle \psi | \hat{p}\hat{x} | \psi \rangle = \frac{1}{2\pi}\int dx'dp'dx'' x'p'e^{ip'(x'-x'')}\langle\psi|x''\rangle \langle x'|\psi\rangle. $$ So we then have $$ \langle\psi|[\hat{x},\hat{p}]|\psi\rangle = \frac{1}{2\pi}\cdot2\text{Im}\Big[\int dx'dp'dx''x'p'e^{-ip'(x'-x'')}\langle\psi|x'\rangle\langle x''|\psi\rangle\Big]. $$ The only way I see the canonical commutation relation hold is if the imaginary part of the integrand evaluates to $\pi$. Am I making a mistake somewhere? If not, is there a neat trick I cannot see?

  • 1
    $\begingroup$ For the opposite question, see this Phys.SE post. $\endgroup$
    – Qmechanic
    Feb 16, 2021 at 4:33

1 Answer 1


Note that $\int dp'\ p' e^{-ip'(x'-x'')} = 2\pi i\delta'(x'-x'')$, where $\delta'$ is the distributional derivative of the delta function defined by

$$\int dx \ f(x) \delta'(x-a) := -f'(a)$$

Note also that for a complex number $a$, we have that $a-a^* = 2\color{red}{i}\mathrm{Im}(a)$, not just $2\mathrm{Im}(a)$. Those two pieces of information, plus the occasional integration by parts, should get you where you're trying to go.

  • $\begingroup$ Thank you very much for this. I would have answered sooner, but I am in Texas trying not to freeze. $\endgroup$ Feb 20, 2021 at 2:46
  • $\begingroup$ @user3166083 I'm glad I could help. I hope things get better down there soon, and that you can stay warm! $\endgroup$
    – J. Murray
    Feb 20, 2021 at 15:09

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