Consider the Gaussian position measurement operators $$\hat{A}_y = \int_{-\infty}^{\infty}\frac{e^{-(x-y)^{2}/(4V)}}{(2 \pi V)^{1/4}}|x \rangle \langle x|dx$$ where \(\displaystyle |x \rangle\) are position eigenstates. Does anyone know how it can be shown that the required completion relation is satisfied: $$\int_{-\infty}^{\infty}A_{y}^{\dagger}A_{y}dy = 1$$ where \(\displaystyle 1\) is the identity operator.

I think I resolved it. I get $$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{e^{(x'-y)^2/(4V)}}{(2 \pi V)^{\frac{1}{4}}}\frac{e^{(x''-y)^2/(4V)}}{(2 \pi V)^{\frac{1}{4}}}\delta(x'-x'')|x' \rangle \langle x'' | dx' dx'' dy$$

So then you get $$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac{e^{(x''-y)^2/2V}}{(2 \pi V)^{1/2}}dy|x'' \rangle \langle x'' | dx'' = 1$$

Thanks.

I think I resolved it. I get $$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{e^{(x'-y)^2/(4V)}}{(2 \pi V)^{\frac{1}{4}}}\frac{e^{(x''-y)^2/(4V)}}{(2 \pi V)^{\frac{1}{4}}}\delta(x'-x'')|x' \rangle \langle x'' | dx' dx'' dy$$

So then you get $$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac{e^{(x''-y)^2/2V}}{(2 \pi V)^{1/2}}dy|x'' \rangle \langle x'' | dx'' = 1$$

Thanks.

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