1
$\begingroup$

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?

$\endgroup$
1
  • 1
    $\begingroup$ For the opposite question, see this Phys.SE post. $\endgroup$
    – Qmechanic
    Feb 16 at 4:33
2
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Thank you very much for this. I would have answered sooner, but I am in Texas trying not to freeze. $\endgroup$ Feb 20 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 at 15:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.