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I came across the following question and have some conceptual questions.

Consider a general quantum state $|\phi\rangle$ of dimension $N$ spanned by some standard basis $\{|k\rangle,k=0,1,...N-1\}$. Suppose we wish to measure it in a different basis $\{|\bar{0},|\bar{1},...|\overline{N-1}\rangle\}$ such that $|\bar{k}\rangle=V|k\rangle$ for some unitary $V$ for each $k=0,1,...N-1$. Write down the probability of getting outcome $|\bar{k}\rangle$ in terms of $|\phi\rangle$ and $V$.

For a question like this,I would think of it as that we need to first apply V onto the state, and then do a measurement in the new basis, which would result in something like $ MV|\phi\rangle $ where $ M=\sum^{N-1}_i|\bar i\rangle \langle \bar i|$ and is the measurement operator, but how do we go on to find the probability? Do I simply insert $ \langle \bar k| $ in front and turn it into $ |\langle \bar k|MV|\phi\rangle|^2 $?

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    $\begingroup$ Hi and welcome to Quantum Computing SE. One of rule of this site is to ask one laser focussed question per a post. You can ask more questions of course but in serpareted posts. $\endgroup$ Oct 11 at 8:42
  • $\begingroup$ @MartinVesely if I break them up I will be posting 4 different questions at one go, was afraid that that would flood the forum, but sure I can do that later :) $\endgroup$
    – tangolin
    Oct 11 at 8:44
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Beware! If $M$ is an operator describing a measurement, it is not that the output after measurement is $M|\psi\rangle$ for initial state $|\psi\rangle$. Instead, let $\{P_i\}$ be projectors onto the different eigenspaces of $M$. The you get the outcome $i$ with probability $p_i=\langle\psi|P_i|\psi\rangle$ and the state after measurement, if you get that outcome, is $P_i|\psi\rangle/\sqrt{p_i}$.

So, in the case where $M$ only has unique eigenvalues, each of the $P_i=|\phi_i\rangle\langle\phi_i|$ and your probability is $p_i=|\langle\psi|\phi_i\rangle|^2$, giving the output state $|\phi_i\rangle$. This circles back around to your first question

I know that$ |\langle\psi|\phi\rangle|^2 $ gives a probability, but is it right to interpret it as the probability of the system in state $ > |\phi\rangle $ collapsing to state $ \langle\psi| $?

If you start with the state $|\phi\rangle$ and measure in some basis where one of the measurements is the state $|\psi\rangle$, then yes, this is the probabilty that you get that outcome.

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  • $\begingroup$ Just to check that my understanding is correct, to measure in a different basis in the case of the question above, it should then be $ |\bar k\rangle\langle\bar k|\phi\rangle $ since we are measuring in the $ |\bar k\rangle $ basis and the probability will be $ |\langle\phi|\bar k\rangle\langle\bar{k}|\phi\rangle|^2 $? We can then substitute $ |\bar k\rangle$ with $V|\bar{k}\rangle$? $\endgroup$
    – tangolin
    Oct 12 at 6:17
  • $\begingroup$ I'm not sure what you mean by the substitution. But everything up the that point is correct, where $\{|\bar k\rangle\}$ is any basis you want. $\endgroup$
    – DaftWullie
    Oct 12 at 6:37
  • $\begingroup$ Sorry, by subsitution I meant that in the picture of the question I posted, they mentioned that $|\bar{k}\rangle=V|k\rangle$, can we then replace the $|\bar{k}\rangle$ term in $|\langle\phi|\bar{k}\rangle\langle\bar{k}|\phi\rangle|^2$ ? $\endgroup$
    – tangolin
    Oct 12 at 6:52
  • $\begingroup$ yes, that's right. $\endgroup$
    – DaftWullie
    Oct 12 at 6:55
  • $\begingroup$ Since we are asked to express in $V$ and $|\phi\rangle$, I don't see how we can do it without throwing in $V^+$, $|k\rangle$ and $\langle k|$? $\endgroup$
    – tangolin
    Oct 12 at 11:38

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