In a quantum measurement theory text, the following is stated. Please advise on the query regarding the calculation below.

Consider the following measurement operator: $$\hat{A}(\alpha) = \bigg(\frac{4k}{\pi dt}\bigg)^{\frac{1}{4}}\int_{-\infty}^{\infty}\text{exp}\bigg\{-2k\frac{(\alpha - xdt)^2}{dt} \bigg\} |x \rangle \langle x | dx = \bigg(\frac{4 k}{\pi dt}\bigg)^{\frac{1}{4}} \text{exp} \bigg\{ -2k \frac{(\alpha - \hat{x} dt)^2}{dt} \bigg\}$$

where $dt$, $\alpha$ and $k$ can be considered as real numbers, and $\hat{x}$ is some observable.

Consider the state $| \psi \rangle = \int \psi(x)| x\ \rangle dx$. Can anyone see how the following two equations are obtained:

$$\int_{-\infty}^{\infty} \alpha \text{Tr}\bigg[ A^2(\alpha)| \psi \rangle \langle \psi \bigg]d \alpha = \sqrt{\frac{4k}{\pi dt}}\int_{-\infty}^{\infty}\bigg[\int_{-\infty}^{\infty} \alpha e^{\frac{-4k(\alpha - xdt)^2}{dt}}d \alpha \bigg] | \psi(x)|^2 dx =\\ dt\int_{- \infty}^{\infty} x |\psi(x)|^2 dx$$

By the way, these equations are the expectation value $\langle \alpha \rangle$.

Thanks for any help.

  • $\begingroup$ which text are you talking about $\endgroup$ – Boltzee Jul 18 '17 at 14:57

First, $\hat{A}^2$ is still an operator. It is diagonal in position but you need to keep explicitly the eigenvector dependence. Your third equal in the first line of equation is wrong and you should instead write $$ \hat{A}^2 = \sqrt{\frac{4k}{\pi dt}} \int \exp \{-\frac{2k(\alpha - ydt)^2}{dt}\} |y\rangle\langle y| dy \int \exp \{-\frac{2k(\alpha - xdt)^2}{dt}\} |x\rangle\langle x| dx, $$ with $\langle y | x \rangle = \delta(x-y)$, you have $$ A^2 = \sqrt{\frac{4k}{\pi dt}} \int dx \ \exp \{-\frac{4k(\alpha - xdt)^2}{dt}\} | x \rangle \langle x |. $$

Second, the trace of an operator is given by summing the diagonal elements, that is to say $Tr[\hat{B}] = \int dy \ \langle y | \hat{B} | y \rangle$. With $\langle x | \psi \rangle = \psi(x)$ (or conjugate, not sure but doesn't matter here), you have $$ \int \alpha Tr[A^2 |\psi\rangle\langle \psi |] d\alpha = \sqrt{\frac{4k}{\pi dt}} \int dx \left[\int d\alpha \ \alpha \exp\{-\frac{4k(\alpha - x dt)^2}{dt}\}\right] \ \int dy \langle y | x \rangle\langle x | \psi \rangle \langle \psi | y \rangle $$ with in the last integral you change $\langle y | x \rangle = \delta(x-y)$, you get there $|\psi(x)|^2$ and the rest is the same. Finally, the integral over $\alpha$ is a gaussian function and can be integrated exactly and very easily.

Hope this helps!

  • $\begingroup$ Thanks for the excellent answer. The integral is not a Gaussian in the usual form as in the link, since it has a factor of $\alpha$ in front of the exponent but I will evaluate and confirm this integral computationally. Are you a physics student? $\endgroup$ – user101311 Jul 18 '17 at 16:46
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    $\begingroup$ For the integral, perhaps there is a subtility in this integral but wolfram alpha should do it! As for the personal question, I do study physics as a graduate student, more specifically I study condensed matter physics using numerical simulations. $\endgroup$ – gingras.ol Jul 18 '17 at 16:49

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