$\newcommand{\qr}[1]{|#1\rangle}$Say I begin with $4$ qubits $\qr{+}\qr{+}\qr{+}\qr{+}$ forming a register $B$. Name these qubits as $b_3, b_2, b_1, b_0$. Also, let $C$ be another register $\qr0\qr0\qr0\qr0$ whose qubit names are $c_3, c_2, c_1, c_0$.

Now apply $\operatorname{CNOT}(b_i, c_i)$, that is, entangle $b_3$ and $c_3$, $b_2$ and $c_2$ and so on. Suppose I measure $B$ and I obtain $\qr{0101} = \qr5$. What will I get if I measure $C$ now? I say I'll get precisely $\qr{5}$.

Does this happen a $100\%$ of the time?

  • $\begingroup$ @Blue Didn't know \operatorname. Thanks! $\endgroup$ – R. Chopin Oct 20 '18 at 13:25
  • $\begingroup$ No worries! You could equivalently also use \text{}. Works about the same. :) $\endgroup$ – Sanchayan Dutta Oct 20 '18 at 13:27

If you only operate on the $i$-th qubit, then the other qubits do not matter and can be ignored, so you effectively are considering the operation $$\lvert+\rangle\lvert0\rangle\to\frac{1}{\sqrt2}[\lvert0,0\rangle+\lvert1,1\rangle]$$ on the $i$-th qubit, with all other qubits not being affected. Then, measuring the first qubit always collapses the second one, like in the example you showed, which means that outcomes are always correlated.

More generally, if you apply CNOT operations to a subset of the qubits (say those with indices $i\in A\subseteq\{1,...,n\}$), then you are doing the following $$\left[\bigotimes_{k=1}^n \lvert+\rangle\right]\left[\bigotimes_{k=1}^n \lvert0\rangle\right] \rightarrow \left[\bigotimes_{i\notin A}\lvert+,0\rangle\right]\otimes 2^{-|A|/2}\left[\bigotimes_{i\in A} \big(\lvert0,0\rangle+\lvert 1,1\rangle\big)\right].$$ Measuring the first register does not do anything interesting on the indices $i\notin A$, while it completely collapses the qubits in the second register with indices $i\in A$. If the outcome measured on the first register is $\lvert\boldsymbol x\rangle\equiv\lvert x_1,...,x_n\rangle$, then the state of the second register becomes: $$\left[\bigotimes_{i\notin A}\lvert0\rangle\right]\otimes \left[\bigotimes_{i\in A}\lvert x_i\rangle\right].$$ Clearly, the state of the second register is entirely determined by what has been measured in the first register, so that the measurement results will always be correlated.

More precisely stated, the measurement results are "$100\%$" correlated in the sense that the conditional entropy $H(C|B)$ between the two measurement outcomes (when measuring in the computational basis) is zero.

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  • $\begingroup$ @gIs, I think you meant to write $\frac{1}{2^{|A|/2}}$ but you wrote $\frac{1}{2^{-|A|/2}}$. Do you confirm? Thanks! $\endgroup$ – R. Chopin Oct 21 '18 at 13:53

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