# How to compute the measurement probabilities of $|\phi\rangle=\sum_k c_k |k\rangle$ in a rotated basis $V|k\rangle$?

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 $$|\langle \bar k|MV|\phi\rangle|^2$$?

• 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. Oct 11 at 8:42
• @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 :) Oct 11 at 8:44

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.

• 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 $V|\bar{k}\rangle$? Oct 12 at 6:17
• 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. Oct 12 at 6:37
• 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 $|\langle\phi|\bar{k}\rangle\langle\bar{k}|\phi\rangle|^2$ ? Oct 12 at 6:52
• yes, that's right. Oct 12 at 6:55
• 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 $\langle k|$? Oct 12 at 11:38