# Position vs momentum space calculation

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?

• $i \frac{d}{dp}$ is a Hermitian operator. 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, $$= \Big(i\hbar \frac{d}{dp} - \langle \hat x \rangle\Big)\Big(i\hbar \frac{d}{dp} - \langle \hat x \rangle\Big) = -\hbar^2\frac{d^2}{dp^2} - i\hbar \frac{d}{dp} + \langle \hat x \rangle^2.$$