# How do we take the limit of this quantum operation?

I am wondering how to take the following limit: \begin{align} L= \lim_{\tau \to \infty} \frac{1}{\tau} \int_{-\tau/2}^{\tau/2} dy \, \left(1 - \frac{1}{\sqrt{ \pi} \sigma } \int_{-\tau/2}^{\tau/2} d x \, e^{-\frac{(y-x)^2}{\sigma^2}}\right) e^{-i\hat{P} y} \rho e^{i\hat{P}y}, \end{align} where $$\rho = \int_{-\infty}^{\infty} dz \int_{-\infty}^{\infty} dz' \ \rho(z,z') \left|z\right\rangle\!\left\langle z'\right|$$ is a density matrix on the real line, $\hat{P}$ is the momentum operator which generates translations $$e^{-i\hat{P} y} \left|z\right\rangle = \left|z + y\right\rangle .$$

I expect the answer to be $0$, but I don't know how to show it rigorously. From the above expression it's clear the integrand vanishes in the limit $\tau \to \infty$, but I don't know if this implies the entire expression vanishes as well.

I've tried to apply L'Hopital's rule with no success: \begin{align} L &= \lim_{\tau \to \infty} \frac{1}{\tfrac{\partial}{\partial \tau} \tau}\frac{\partial}{\partial \tau } \int_{-\tau/2}^{\tau/2} dy \, \left(1 - \frac{1}{\sqrt{ \pi} \sigma } \int_{-\tau/2}^{\tau/2} dx \, e^{-\frac{(y-x)^2}{\sigma^2}}\right) e^{-i\hat{P}y} \rho e^{i\hat{P}y}\\ &= \lim_{\tau \to \infty} \frac{1}{2} \Bigg[ \left(1 - \frac{1}{\sqrt{ \pi} \sigma } \int_{-\tau/2}^{\tau/2} dx \, e^{-\frac{(\tau/2-x)^2}{\sigma^2}}\right) e^{-i\hat{P}\tau/2} \rho e^{i\hat{P}\tau/2} \\ &\qquad \qquad \qquad \qquad - \left(1 - \frac{1}{\sqrt{ \pi} \sigma } \int_{-\tau/2}^{\tau/2} dx \, e^{-\frac{(-\tau/2-x)^2}{\sigma^2}}\right) e^{i\hat{P}\tau/2} \rho e^{-i\hat{P}\tau/2} \Bigg] \\ &\qquad + \lim_{\tau \to \infty} \int_{-\tau/2}^{\tau/2} dy \, \frac{\partial}{\partial \tau} \left(1 - \frac{1}{\sqrt{ \pi} \sigma } \int_{-\tau/2}^{\tau/2} dx \, e^{-\frac{(y-x)^2}{\sigma^2}}\right) e^{-i\hat{P}y} \rho e^{i\hat{P}y}\\ &= \lim_{\tau \to \infty} \frac{1}{2} \Bigg[ \left(1 - \frac{1}{\sqrt{ \pi} \sigma } \int_{-\tau/2}^{\tau/2} dx \, e^{-\frac{(\tau/2-x)^2}{\sigma^2}}\right) e^{-i\hat{P}\tau/2} \rho e^{i\hat{P}\tau/2} \\ &\qquad \qquad \qquad \qquad - \left(1 - \frac{1}{\sqrt{ \pi} \sigma } \int_{-\tau/2}^{\tau/2} dx \, e^{-\frac{(-\tau/2-x)^2}{\sigma^2}}\right) e^{i\hat{P}\tau/2} \rho e^{-i\hat{P}\tau/2} \Bigg] \\ &\qquad - \frac{1}{\sqrt{ \pi} \sigma } \lim_{\tau \to \infty} \int_{-\tau/2}^{\tau/2} dy \, \left[ \frac{1}{2} e^{-\frac{(y-\tau/2)^2}{\sigma^2}} - \frac{1}{2} e^{-\frac{(y+\tau/2)^2}{\sigma^2}} \right]e^{-i\hat{P}y} \rho e^{i\hat{P}y}. \end{align} One might be able to argue that the first two terms in the square brackets in the last equality above vanish, by using the limit law: $$\lim_{x \to a} f(x) g(x) = \lim_{x \to a} f(x) \lim_{x \to a} g(x),$$ but that assumes that both the limits, $\lim_{x \to a} f(x)$ and $\lim_{x \to a} g(x)$; in this case it's not clear if $$\lim_{\tau \to \infty} e^{-i\hat{P}\tau/2} \rho e^{i\hat{P}\tau/2}$$ exists.

• From a quick glance, I'd say that you should Fourier transform the whole thing. This should turn the convolution into a multiplication, half of which is a Gaussian integral and thus well-known. Also, the Fourier transform of the momentum operator, which should allow you to tackle the other term. – Martin Nov 22 '15 at 12:05
• @Martin That’s a good idea! However, after taking the Fourier transform, to get a product of Fourier transforms of the two functions we would have to apply the convolution theorem. Doesn't the convolution theorem only hold in the limit $\tau \to \infty$? – e4alex Nov 22 '15 at 12:31
• This question lacks physical context, as it stands, it is a pure math question about how to evaluate a limit. – ACuriousMind Nov 22 '15 at 12:47
• @ACuriousMind I partially agree. The limit arises when trying to perform a G-twirl, that is, a weighted average of a density matrix over a group $G$, in this case the group of translations. The G-twirl is well defined for compact groups, however in this case we are averaging over the group of translations which is non-compact. $\tau$ is the range of possible translations, which I am trying to take to infinity. Physically, the G-twirl describes the a state when you lack a reference frame associated with $G$. – e4alex Nov 22 '15 at 13:14

Hint. ( I give absolutely no guarantee about not making calculational mistakes!)

\begin{eqnarray*} L(\tau ) &=&\frac{1}{\tau }\int_{-\tau /2}^{+\tau /2}dy\left( 1-\frac{1}{ \sqrt{\pi }\sigma }\int_{-\tau /2}^{+\tau /2}dx\exp [-\frac{(y-x)^{2}}{ \sigma ^{2}}]\right) e^{-iPy}\rho e^{+iPy}=L_{1}(\tau )-L_{2}(\tau ) \\ L_{1}(\tau ) &=&\frac{1}{\tau }\int_{-\tau /2}^{+\tau /2}dye^{-iPy}\rho e^{+iPy} \\ L_{2}(\tau ) &=&\frac{1}{\tau }\int_{-\tau /2}^{+\tau /2}dy\frac{1}{\sqrt{ \pi }\sigma }\int_{-\tau /2}^{+\tau /2}dx\exp [-\frac{(y-x)^{2}}{\sigma ^{2}} ]e^{-iPy}\rho e^{+iPy} \end{eqnarray*} Take matrix elements with momentum eigenstates of $P$, assuming that $<p_{1}|\rho |p_{2}>$ exists. Then \begin{eqnarray*} &<&p_{1}|L_{1}(\tau )|p_{2}>=\frac{1}{\tau }\int_{-\tau /2}^{+\tau /2}dye^{-i(p_{1}-p_{2})y}<p_{1}|\rho |p_{2}>=<p_{1}|\rho |p_{2}>\frac{1}{ \tau }\left[ \frac{e^{-i(p_{1}-p_{2})y}}{-i(p_{1}-p_{2})}\right] _{y=-\tau /2}^{y=+\tau /2} \\ &=&<p_{1}|\rho |p_{2}>\frac{1}{\tau }\frac{1}{-i(p_{1}-p_{2})} \{e^{-i(p_{1}-p_{2})\tau /2}-e^{+i(p_{1}-p_{2})\tau /2}\}=\frac{2}{\tau } \frac{1}{(p_{1}-p_{2})}\sin \{(p_{1}-p_{2})\tau /2\}<p_{1}|\rho |p_{2}> \\ &=&\frac{\sin \{(p_{1}-p_{2})\tau /2\}}{(p_{1}-p_{2})\tau /2}<p_{1}|\rho |p_{2}>\rightarrow \left\{ \begin{array}{cc} 0 & p_{1}\neq p_{2} \\ <p_{1}|\rho |p_{1}> & p_{1}=p_{2} \end{array} \right. \end{eqnarray*}

Next

\begin{eqnarray*} &<&p_{1}|L_{2}(\tau )|p_{2}>=-\frac{1}{\sqrt{\pi }\sigma }\frac{1}{\tau } \int_{-\tau /2}^{+\tau /2}dye^{-i(p_{1}-p_{2})y}\int_{-\tau /2}^{+\tau /2}dx\exp [-\frac{(y-x)^{2}}{\sigma ^{2}}]<p_{1}|\rho |p_{2}> \\ &=&-\frac{1}{\sqrt{\pi }\sigma }<p_{1}|\rho |p_{2}>\frac{1}{\tau } \int_{-\tau /2}^{+\tau /2}dye^{-i(p_{1}-p_{2})y}\int_{-\tau /2-y}^{+\tau /2-y}dx\exp [-\frac{x^{2}}{\sigma ^{2}}] \end{eqnarray*} and do a partial integration

\begin{eqnarray*} J(\tau ) &=&\int_{-\tau /2}^{+\tau /2}dye^{-i(p_{1}-p_{2})y}\int_{-\tau /2-y}^{+\tau /2-y}dx\exp [-\frac{x^{2}}{\sigma ^{2}}] \\ &=&\left[ \frac{e^{-i(p_{1}-p_{2})y}}{-i(p_{1}-p_{2})}\int_{-\tau /2-y}^{+\tau /2-y}dx\exp [-\frac{x^{2}}{\sigma ^{2}}]\right] _{y=-\tau /2}^{y=+\tau /2}+\int_{-\tau /2}^{+\tau /2}dy\frac{e^{-i(p_{1}-p_{2})y}}{ -i(p_{1}-p_{2})}\{-e^{-(y-\tau /2)^{2}}+e^{-(y+\tau /2)^{2}}\} \\ &=&\frac{e^{-i(p_{1}-p_{2})\tau /2}}{-i(p_{1}-p_{2})}\int_{-\tau }^{0}dx\exp [-\frac{x^{2}}{\sigma ^{2}}]-\frac{e^{+i(p_{1}-p_{2})\tau /2}}{ -i(p_{1}-p_{2})}\int_{-\tau }^{0}dx\exp [-\frac{x^{2}}{\sigma ^{2}}] \\ &&+\int_{-\tau /2}^{+\tau /2}dy\frac{e^{-i(p_{1}-p_{2})y}}{-i(p_{1}-p_{2})} \{-e^{-(y-\tau /2)^{2}}+e^{-(y+\tau /2)^{2}}\} \\ &=&\frac{1}{-i(p_{1}-p_{2})}\{e^{-i(p_{1}-p_{2})\tau /2}-e^{+i(p_{1}-p_{2})\tau /2}\}\int_{0}^{\tau }dx\exp [-\frac{x^{2}}{\sigma ^{2}}] \\ &&+\int_{-\tau /2}^{+\tau /2}dy\frac{e^{-i(p_{1}-p_{2})y}}{-i(p_{1}-p_{2})} \{-e^{-(y-\tau /2)^{2}}+e^{-(y+\tau /2)^{2}}\} \\ &=&\frac{2\sin (p_{1}-p_{2})\tau /2}{p_{1}-p_{2}}\int_{0}^{\tau }dx\exp [- \frac{x^{2}}{\sigma ^{2}}]+\int_{-\tau /2}^{\\+\tau /2}dy\frac{ e^{-i(p_{1}-p_{2})y}}{-i(p_{1}-p_{2})}\{-e^{-(y-\tau /2)^{2}}\\+e^{-(y+\tau /2)^{2}}\} \end{eqnarray*} Let

\begin{eqnarray*} X(\tau ) &=&\frac{1}{\tau }\frac{2\sin (p_{1}-p_{2})\tau /2}{p_{1}-p_{2}} \int_{0}^{\tau }dx\exp [-\frac{x^{2}}{\sigma ^{2}}]=\frac{2\sin (p_{1}-p_{2})\tau /2}{p_{1}-p_{2}}\frac{1}{\tau }\int_{0}^{\tau }dx\exp [- \frac{x^{2}}{\sigma ^{2}}] \\ p_{1} &=&p_{2}\Rightarrow X=\int_{0}^{\tau }dx\exp [-\frac{x^{2}}{\sigma ^{2} }]\rightarrow \int_{0}^{\infty }dx\exp [-\frac{x^{2}}{\sigma ^{2}}] \\ p_{1} &\neq &p_{2}\Rightarrow X\rightarrow 0 \end{eqnarray*} The remaining term \begin{equation*} Y(\tau )=\frac{1}{\tau }\int_{-\tau /2}^{+\tau /2}dy\frac{ e^{-i(p_{1}-p_{2})y}}{-i(p_{1}-p_{2})}\{-e^{-(y-\tau /2)^{2}}+e^{-(y+\tau /2)^{2}}\} \end{equation*} is harder. The two parts are

\begin{eqnarray*} \frac{1}{\tau }\int_{-\tau /2}^{+\tau /2}dy\frac{e^{-i(p_{1}-p_{2})y}}{ -i(p_{1}-p_{2})}e^{-(y-\tau /2)^{2}} &=&\frac{1}{\tau }\int_{-\tau }^{0}dy \frac{e^{-i(p_{1}-p_{2})(y+\tau /2}}{-i(p_{1}-p_{2})}e^{-y^{2}}\\=\frac{ e^{-i(p_{1}-p_{2})\tau /2}}{-i(p_{1}-p_{2})}\frac{1}{\tau }\int_{-\tau }^{0}dye^{-i(p_{1}-p_{2})y}e^{-y^{2}} \\ \frac{1}{\tau }\int_{-\tau /2}^{+\tau /2}dy\frac{e^{-i(p_{1}-p_{2})y}}{ -i(p_{1}-p_{2})}e^{-(y+\tau /2)^{2}} &=&\frac{1}{\tau }\int_{0}^{+\tau }dy \frac{e^{-i(p_{1}-p_{2})(y-\tau /2)}}{-i(p_{1}-p_{2})}e^{-y^{2}}\\=\frac{ e^{+i(p_{1}-p_{2})\tau /2}}{-i(p_{1}-p_{2})}\frac{1}{\tau }\int_{0}^{+\tau }dye^{-i(p_{1}-p_{2})y}e^{-y^{2}} \end{eqnarray*} Again we have to distinguish between $p_{1}=p_{2}$ and $p_{1}\neq p_{2}$. In the second case these objects tends to $0$. I leave the first case to you. It seems your guess is correct for $p_{1}\neq p_{2}$.