This comes from trying to understand the "Simon's algorithm".
So we have a set of $2^n$ kets $|x_i \rangle$ one each for $i \in \{0,1\}^n$. Each $x_j \in \{0,1\}^n$. And we have the further constraint that $x_i = x_j$ iff $i = j + s \pmod{2} = j \oplus s$ (bitwise addition modulo $2$) for a certain $s \in \{0,1\}^n$. (so every $|x_k\rangle$ is guaranteed to have a second copy)
Now one has this state given as $\psi_a = \frac{1}{\sqrt{2^n}}\sum_{i \in \{0,1\}^n} |i\rangle|x_i\rangle $ So this is a state of some 2 qubit system.
- Given this I want to understand why a measurement on the second qubit would necessarily collapse this to the state, $\psi_b = \frac{1}{\sqrt{2}} ( |j\rangle + |j \oplus s\rangle)|x_j\rangle$ for some $j \in \{0,1\}^n$?
- Is this measurement of the second qubit state trying to measure some observable/Hermitian operator whose eigenstates are precisely these $\psi_b$s? Then it would make some sense that the only things one sees are these $\psi_b$ states.
But this would still not clarify the weird probability addition that is happening here as shown below.
One notes that the state $\psi_b$ will be obtained with a probability of $(\bar{\psi_b} \psi_a)^2 = (\frac{ 2}{ \sqrt{2^{n+1} }})^2 = 1/2^{n-1}$. These $\psi_b$s obviously do not span the entire $2^{2n}$ dimensional Hilbert space available to these two qubits. They are not a basis.
But the probabilities don't add up rightly either $\sum_b ( \bar{\psi_b} \psi_a )^2 = 2$. Wonder is the interpretation of the fact that the probabilites over all these possibilities seem to add up to $2$!
Some explanations I found say that these states $\psi_b$ being the only possibilities is forced because of the fact that the second qubit is being measured in the "computational basis". This argument is not clear to me.
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