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

My main problem is that I found a reference where it states that \begin{equation} \rho'(x,x')=\frac{x+\partial_x}{\sqrt{2}}\frac{x'+\partial_{x'}}{\sqrt{2}}\rho(x, x') \end{equation} 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?


1 Answer 1


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.

  • $\begingroup$ 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! $\endgroup$
    – Luthien
    Commented May 11, 2020 at 21:41
  • $\begingroup$ 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) $\endgroup$
    – Luthien
    Commented May 11, 2020 at 21:45
  • 1
    $\begingroup$ 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. $\endgroup$ Commented May 12, 2020 at 0:00
  • $\begingroup$ 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. $\endgroup$
    – Luthien
    Commented May 12, 2020 at 10:32

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