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Suppose I have a state expressed in its eigenbasis as follows. $\rho = \sum_i\lambda_i\vert i\rangle\langle i\vert$. It is now measured in some other basis $\{\vert x\rangle\}$ that is distinct from the eigenbasis. Let this measurement operation be

$$M: \sigma \rightarrow \sum_x \vert x\rangle\langle x\vert \sigma \vert x\rangle\langle x\vert$$

The outcome of the measurement on $\rho$ is

$$M(\rho) = \sum_x \omega_x\vert x\rangle\langle x\vert$$

I have come across a claim that $\forall x, \omega_x \leq\lambda_{\max}$. My questions are

  1. How can I prove this? Trying out some examples also suggests that $\forall x, \lambda_{\min}\leq \omega_x \leq\lambda_{\max}$.

  2. What is the physical meaning? Does measuring a state in different bases (and remembering the result) eventually give one a maximally mixed state?

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  • $\begingroup$ shouldn’t the measurement be a single projector? (or least not summed over $x$...) $\endgroup$ Commented May 27, 2020 at 3:25

1 Answer 1

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First of all, as @ZeroTheHero mentioned, a measurement operator is a single projection operator. There should be no sum over $x$.

What you have done here is express an operator in two different bases. Continuing in that sense, we have, in $\{|x\rangle\}$ basis, $$M(\rho) = \sum_x \omega_x\vert x\rangle\langle x\vert$$ $$ = \sum_x \langle x\vert \rho \vert x\rangle\vert x\rangle\langle x\vert$$ $$ = \sum_x \vert x\rangle\langle x\vert \rho \vert x\rangle\langle x\vert$$

Using the fact $\rho = \sum_i\lambda_i\vert i\rangle\langle i\vert$, we can write

$$M(\rho) = \sum_x \vert x\rangle\langle x\vert \left(\sum_i\lambda_i\vert i\rangle\langle i\vert\right) \vert x\rangle\langle x\vert$$ $$ = \sum_{x,i} \lambda_i \vert x\rangle\langle x\vert i\rangle\langle i\vert x\rangle\langle x\vert$$ $$ = \sum_{x,i} \lambda_i \vert x\rangle \left\Vert\langle x\vert i\rangle \right\Vert^2 \langle x\vert$$ $$ = \sum_{x} \left(\sum_i\lambda_i \left\Vert\langle x\vert i\rangle \right\Vert^2 \right) \vert x\rangle \langle x\vert$$ $$= \sum_x \omega_x\vert x\rangle\langle x\vert$$

So, we get $$\boxed{\omega_x = \sum_i\lambda_i \left\Vert\langle x\vert i\rangle \right\Vert^2}$$

I hope you will be able to conclude the validity of the claim in (1) using proper bounds on $\left\Vert\langle x\vert i\rangle \right\Vert^2$.

And as of now, your second question does not have a solid meaning, as this is not at all a measurement problem.

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