# Measure two qubits, one from entangled pair, in Bell basis?

In quantum teleportation we require both sender (Alice) and receiver (Bob) to share an entangled pair of qubits in the $|\beta_{00}\rangle$ Bell state. Then, if Alice wants to teleport a quibt $|v\rangle$ to Bob, she needs to measure $|v\rangle$ and her qubit from an entangled pair in the Bell basis. Such measurment requires her to perform a CNOT on the two qubits with $|v\rangle$ being the control qubit and then apply Hadamard transformation to $|v\rangle$.

I can't understand how can she perform the above. If we represent quantum gates as matrices, what vector would represent the input to the CNOT matrix (the pair $|v\rangle$ and Alice's qubit from an entangled pair?) In other words, how can we represent a pair that Alice wants to apply CNOT to (when performing measurement in the Bell Basis?)

I know that Bell measurement does not require CNOT in general, but this is one of the ways it could be implemented (assuming we just want to get the classical 2-bit value).

My intuition is that because we need to operate on a tensor product of $|v\rangle$ and the entangled pair, we need to create a matrix representing an operation performing CNOT on $v\rangle$ and first qubit from the pair, and I on the second qubit from the pair, so it would be like $CNOT \otimes I$. Is this correct?

• the realization that bell type experiments/ setups are connected with CNOT is a rather newer discovery. here is some discussion of it, maybe helpful? – vzn Dec 17 '14 at 5:13
• Will be a good read to understand better, thanks. – 3yakuya Dec 17 '14 at 23:22

I figured out that as we are operating on 3 qubits ($|v\rangle$ and $\beta_{00}\rangle$ pair) and we want to perform CNOT on the first two (if we enumerate like 1- $|v\rangle$, 2- first from pair, 3 - second from pair) it is just like we want to perform $CNOT \otimes I$ on $|v\rangle \otimes |\beta_{00}\rangle$.
After performing the CNOT described above we need to perform H on the $|v\rangle$ qubit, which is now entangled (as CNOT is an entangling operation), so it is like we want to perform $H \otimes I$ on the three (where $I$ is 3 x 3 identity matrix).
Beaing able to perform both steps described above we are able to perform a measurement of qubit $|v\rangle$ and the first qubit from the entangled pair $|\beta_{00}\rangle$ in the Bell basis.
• $I$ is probably 4 on 4 in the second case, or better, just write $H\otimes I\otimes I$. – Ran G. May 26 '15 at 2:07