# How is this circuit equivalent to a Bell measurement? The circuit elements are :

1. CNOT gate, first qubit is the control bit
2. Hadamard gate (H)
3. The dials represent measurements made in the computational basis

Therefore, the circuit (without the measurements) is equivalent to $\text{CNOT}\;(H \otimes I)$.

My question is, how is this circuit equivalent to a Bell measurement?

In this paper (pages 14-15) the author mentions that the initial part of this circuit (without the measurements) is also equivalent to a change of basis from the computational basis to the Bell basis.

But at the same time, Nielsen and Chuang mention in their book (Quantum Computation and Information, Page 188) that the initial part is a change of basis from the Bell basis to the computational basis!! Which one is true and why is it true?

Edit :- By comparing the matrix of the net operation $\text{CNOT}\;(H \otimes I)$ and the actual change of basis matrix (from Bell to computational), as given in the paper, I could figure out that the circuit does represent a change of basis from the computational basis into the Bell basis. But why should both be true?

• Both is true. What did you try to verify either of the statements? – Norbert Schuch Aug 16 '15 at 20:47
• edited the question in response. Could you please answer it? – transistor Aug 16 '15 at 21:14
• Did you try to write down the matrix for the opposite basis transformation? Did you compare the two? What did you find? (Note on the side: The circuit implements $(H\otimes I)\mathrm{CNOT}$.) – Norbert Schuch Aug 17 '15 at 11:35

## 1 Answer

The reverse of the circuit in your picture prepares the Bell states:
\begin{align} C_X (H\otimes I) |00\rangle &=(|00\rangle +|11\rangle)/\sqrt{2}\\ C_X (H\otimes I) |01\rangle &=(|01\rangle +|10\rangle)/\sqrt{2}\\ C_X (H\otimes I) |10\rangle &=(|00\rangle -|11\rangle)/\sqrt{2}\\ C_X (H\otimes I) |11\rangle &=(|01\rangle -|10\rangle)/\sqrt{2} \end{align} (The order of $C_X$ and $H\otimes I$ in the picture is correct.)

Just multiply all equations with $(H\otimes I)C_X$ from the left to get \begin{align} |00\rangle &=(H\otimes I)C_X (|00\rangle +|11\rangle)/\sqrt{2}\\ |01\rangle &=(H\otimes I)C_X (|01\rangle +|10\rangle)/\sqrt{2}\\ |10\rangle &=(H\otimes I)C_X (|00\rangle -|11\rangle)/\sqrt{2}\\ |11\rangle &=(H\otimes I)C_X (|01\rangle -|10\rangle)/\sqrt{2} \end{align} and these equations describe how your measurement works: read it from right to left: the incoming Bell states are processed by the controlled-Not gate $C_X$ and then $H$ acts on the first qubit. The final state is measured in the computational basis.