How do I measure a single qubit in a two-qubit state?

Question

Let us suppose that I have the state
begin{equation} 
frac{1}{sqrt2}(alpha|0rangle|+rangle+beta|1rangle|-rangle) 
end{equation}
and I choose to measure the first qubit in the basis ${(1/sqrt{2})(|0ranglepm e^{iphi}|1rangle)}$.
How can I perform this measurement and how can I find the related outcomes (i.e. the eigenvalues) provided only the basis in which I have to measure the system? Being kinda new to quantum mechanics I am really confused about the algebraic aspects of this problem, but I have an intuitive idea of what performing a measurement is.

DaftWullie · Answer

Let $P_{pm}$ be the projectors onto the two orthonormal basis states of the measurement. So,
$$
P_+=frac{1}{2}(|0ranglelangle 0|+|1ranglelangle 1|+e^{-iphi}|0ranglelangle 1|+e^{iphi}|1ranglelangle 0|).
$$
Also, let $|psirangle$ be the state that you're measuring (note that this must be normalised. Yours might be depending on your constraints on $alpha$ and $beta$, but we wouldn't usually have a factor of $1/sqrt{2}$ present).
Since you are measuring the first qubit of two, the actual projectors should be written as $P_pmotimes I$. Now, we get the + result with probability
$$
p_+=langlepsi |P_+otimes I|psirangle,
$$
and if the result is $+$, the state after measurement is
$$
(P_+otimes I)|psirangle/sqrt{p_+}.
$$
In this case, $p_+=p_-=frac12$.

How do I measure a single qubit in a two-qubit state?

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