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I am interested in defining Kraus operators which allow you to define quantum measurements peaked at some basis state. To this end I am considering the Normal Distribution. Consider a finite set of basis states $\{ |x \rangle\}_x$ and a set of quantum measurement operators of the form $$A_C = \sum_x \sqrt{\mathrm{Pr}(x|C)} |x \rangle \langle x|.$$ I want to prove that $A_C$ defines valid Kraus operators. If I consider the Normal distribution $$\mathrm{Pr}(x|C) = \frac{1}{\sqrt{2 \pi \sigma^2}}e^{-\frac{(x-C)^2}{2 \sigma^2}}~~~~\text{where } x \in \{-M,-M+1,...,M \}$$ where $C$ is the mean, $x$ the random variable and $\sigma^2$ the variance, then
$$\int_{-\infty}^\infty \mathrm{Pr}(X=x|C)\,dC = \frac{1}{\sqrt{2 \pi \sigma^2}}\int_{-\infty}^{\infty}e^{-\frac{(x-C)^2}{2 \sigma^2}}dC = 1.$$ Thus it follows that

$$\int_{C}A_C^{\dagger}A_CdC = \int_{-\infty}^{\infty} \sum_{x}\mathrm{Pr}(x|C) |x \rangle \langle x | dC = \sum_x \bigg[\int_{-\infty}^{\infty}\mathrm{Pr}(x|C)dC\bigg]|x \rangle \langle x | = \sum_x |x \rangle \langle x | = 1 $$

hence we have shown $$\int_{C}A_{C}^{\dagger}A_C dC = 1$$ hence $A_C$ satisfies the conditions necessary for $\{A_C \}_C$ to be Kraus operators.

Question: Please advise if my working and conclusion that $\{A_c \}_C$ are valid Kraus operators is correct? Also, there are only a finite number of states $\{ |x \rangle \}_{x}$ yet the index $C$ is continuous and unbounded, so for some $C$ (the mean), the peak of $Pr(x|C)$ will be well outside the range of any $|x \rangle$. I don't think that this is a problem, but I'm not sure about this...

Thanks.

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Please advise if my working and conclusion that $\{A_c \}_C$ are valid Kraus operators is correct?

Your workings and your conclusion that $\{A_c \}_C$ are valid Kraus operators are correct.

Also, there are only a finite number of states $\{ |x \rangle \}_{x}$ yet the index $C$ is continuous and unbounded, so for some $C$ (the mean), the peak of $Pr(x|C)$ will be well outside the range of any $|x \rangle$. I don't think that this is a problem, but I'm not sure about this...

This is not a problem.

If all the $x$ in your finite set are bounded by some constant $|x|<M$, and you take some $C>M$, then you know that the norm of the corresponding Kraus operator is \begin{align} ||A_C||^2 & = \mathrm{Tr}(A_C^\dagger A_C) = \mathrm{Tr}\mathopen{}\left(\sum_x \mathrm{Pr}(x|C)|x \rangle \langle x|\right) = \sum_x \mathrm{Pr}(x|C) \\ & \leq \sum_x \mathrm{Pr}(M|C) = N \, \mathrm{Pr}(M|C) \end{align} if you have $N$ measurements, and this is exponentially small in $(C-M)^2/\sigma^2$. In other words, those Kraus operators are exponentially close to zero and they do not meaningfully contribute to anything.

I would like some reassurance about some minor technical aspect of my workings, and some validation of my status as a competent student of physics.

For the sorts of questions you've been asking, we can indeed provide that, but I would suggest that this is not necessarily the best use of your rep or of everyone's time. This kind of interaction is much more fruitful in person, and I would strongly encourage you to find a faculty member (or other such professional) to whom you can ask this kind of thing in person. It is quicker but, more importantly, the personal and interpersonal aspects of that kind of relationship are even more important to developing a healthy relationship with physics.

(Or, more specifically: this is obviously my interpretation, and I apologise if I'm overstepping here. However, please do take a good look at why exactly you're asking this style of question, where you've already solved the problem in full, and whether there are other ways to handle them that give you more in the end.)

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