I want to calculate $J^\dagger J$ with $J = x - \langle x\rangle$ , where $x$ is the position operator.

In position space: $J^\dagger J = (x - \langle x\rangle)^2 = x^2 - 2 x \langle x \rangle + \langle x \rangle^2$.

Now switching to momentum space with $ x = \mathrm{i} d/dp$ (setting $\hbar = 1$):

$J^\dagger J = - d^2/dp^2 - 2 \mathrm{i} d/dp \langle x \rangle + \langle x\rangle^2 $.

However, if I switch to momentum space right in the beginning, I get:

$J^\dagger J = d^2/dp^2 + \langle x \rangle^2$ (which is probably wrong).

Where is my mistake? I would guess that $(d/dp)^\dagger \neq (d/dp)$, since the derivative should act to the left ? Could someone point out where exactly I am wrong?

  • $\begingroup$ $i \frac{d}{dp}$ is a Hermitian operator. $\endgroup$
    – Prahar
    Apr 14 '20 at 17:06

You are right, that the problems is $d/dp$ not being hermitian. Since $\hat x$ is hermitian, however, we know that $\hat x = i\hbar \frac{d}{dp}$ is hermitian. Inserting the expression we find, $$J^\dagger J = \Big(i\hbar \frac{d}{dp} - \langle \hat x \rangle\Big)^\dagger\Big(i\hbar \frac{d}{dp} - \langle \hat x \rangle\Big).$$ In the first bracket we have $(i\hbar d/dp)^\dagger = i\hbar d/dp$ since it is hermitian and the problem is resolved, $$= \Big(i\hbar \frac{d}{dp} - \langle \hat x \rangle\Big)\Big(i\hbar \frac{d}{dp} - \langle \hat x \rangle\Big) = -\hbar^2\frac{d^2}{dp^2} - i\hbar \frac{d}{dp} + \langle \hat x \rangle^2. $$


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