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Old Jul 22nd 2017, 12:19 PM   #1
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Cool Discrete measurement operator quantum measurement theory

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.

Last edited by Jimmy; Jul 24th 2017 at 07:13 AM.
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discrete, measurement, measurements, operator, operators, quantum, quantum mechanics, quantum optics, theory



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