# Annihilation operator acting on density operator in the position representation

I have a silly question. I have a state $$\hat{\rho}$$ and I make the transformation $$\hat{\rho}'=\hat{a}\hat{\rho}\hat{a}^\dagger$$ (I want to subtract a photon).

I expand in the position basis the density operator: $$$$\hat{\rho}=\iint dx dx'|x\rangle \rho(x, x') \langle x'|$$$$ and the action of $$\hat{a}$$ is carried out by knowing that $$\hat{a}=(\hat{x}+i\hat{p})/\sqrt{2}$$. So $$$$\hat{\rho}'=\iint dx dx'\hat{a}|x\rangle \rho(x, x') \langle x'|\hat{a}^\dagger=\iint dx dx'|x\rangle \frac{x-\partial_x}{\sqrt{2}}\rho(x, x')\frac{x'+\partial_{x'}}{\sqrt{2}} \langle x'|\hat{a}^\dagger$$$$ where I used $$\hat{p}|x\rangle=i\partial_x$$ and $$\langle x|\hat{p}=-i\partial_x$$. So I conclude that $$$$\rho'(x,x')=\frac{x-\partial_x}{\sqrt{2}}\rho(x, x')\frac{x'+\partial_{x'}}{\sqrt{2}}$$$$

My main problem is that I found a reference where it states that $$$$\rho'(x,x')=\frac{x+\partial_x}{\sqrt{2}}\frac{x'+\partial_{x'}}{\sqrt{2}}\rho(x, x')$$$$ A part from the signs which are different, I NEED the differential operators to be on the left to carry out the proof which states $$W_{\hat{a}\hat{\rho}\hat{a}^\dagger}=\hat{D}W_{\hat{\rho}}$$, where $$\hat{D}$$ is a differential operator and $$W$$ is the Wigner quasi-probability distribution. How can I just take a differential operator and put it to the left when the function $$\rho(x,x')$$ is dependent on the variable $$x'$$?

Edit: I think I've always been kind of confused for what concerns differential operator being on the right of a function in QM. Are they acting from the right on $$\rho'(x,x')$$ (being $$\rho'(x,x')=\sum_i p_i \phi_i(x) \phi_i^*(x')$$ it would act from the right on $$\phi_i^*(x')$$). On what these operators are supposed to act?

Indeed, you have garbled your x-representation operators big-time. Here, derivatives always act on the right, and only integrations by part will get them to act on the left.

The important fact you could use is that $$\hat x = \int \!\! dx ~~|x\rangle x\langle x|, \qquad \hat p = \int \!\! dx ~~|x\rangle (-i) \partial_x \langle x|,$$ so that $$\bbox[yellow]{ \sqrt{2} \hat a = \int \!\! dx ~~|x\rangle (x +\partial_x ) \langle x|, \qquad \sqrt{2} \hat a ^\dagger = \int \!\! dx ~~|x\rangle (x -\partial_x ) \langle x| },$$ whence,
$$\hat{\rho}'=\hat{a}\hat{\rho}\hat{a}^\dagger=\frac{1}{2} \int \!\! dy ~~|y\rangle (y +\partial_y ) \langle y| \iint dx dx'~|x\rangle \rho(x, x') \langle x'| \int \!\! dz |z\rangle (z -\partial_z ) \langle z|\\ = \frac{1}{2} \iiiint dx dx' dy dz ~|y\rangle (y+\partial_y) \delta(y-x) \rho(x,x') \delta(x'-z) (z-\partial_z)\langle z| \\ = \frac{1}{2} \iiiint dx dx' dy dz ~|y\rangle \Bigl ( (y-\partial_x) \delta(y-x)\Bigr ) \rho(x,x') \Bigl ((z-\partial_{x'})\delta(x'-z)\Bigr ) \langle z| \\ = \frac{1}{2} \iint dx dx' |x\rangle \Bigl ((x+\partial_x) (x'+\partial_{x'}) \rho(x,x') \Bigr ) \langle x'| ,$$
after collapsing the δ-functions and integrating by parts their derivatives, mindful of symmetries such as $$(\partial_y+\partial_x)\delta(y-x)=0$$, etc.

N.B. You might also choose, with Segal and Bargmann (the magnificent), to change representation from $$|x\rangle$$ to Fock space, $$|x\rangle= \frac{e^{x^2/2}}{\pi^{1/4}} e^{-(a^\dagger-\sqrt{2} x)^2/2} |0\rangle,$$ but this is hardly recommended for your question.

• Thank you for the detailed answer Cosmas. I applied the annihilation operator directly on the kets as $\hat{a}|x>=\hat{x}+i\hat{p}/\sqrt{2}|x>$ $=x-i\partial_x/\sqrt{2}|x>$. I don't think it's a mistake (or is it?) but then of course I cannot find the right result or I would need to insert a delta anyway. Thank you very much! Commented May 11, 2020 at 21:41
• Thank you also for the Fock space representation tip. I will try to do the proof also in the Fock representation (it cannot hurt, for sure) Commented May 11, 2020 at 21:45
• Indeed, your expression above is off. The mnemonic is $\hat p \sim -i\partial_x$ only when acting on bras. When acting on kets it's $i\partial_x$. Check the expression above the yellow highlight. Commented May 12, 2020 at 0:00
• Yes, sorry it's a typo, I wanted to write 𝑥−∂𝑥 , I forgot to get rid of the 𝑖. Yes, this is clear to me, fortunately my QM professor made it very clear some years ago, even if sometimes out of disattention I forget. Commented May 12, 2020 at 10:32