Consider the following unitary defined for a system $A$ interacting with a bipartite system $BB^\prime$

$$U_{AB} = \Big[|0\rangle \langle 0|_{A} \otimes \mathbf{I}_{B} + |1\rangle \langle 1|_{A} \otimes \big(|1\rangle \langle 0|_{B} + |0\rangle \langle 1|_{B} \big) \Big] \otimes \mathbf{I}_{B^\prime},$$ with $\mathbf{I}_i$ being the identity.

My question:

  1. What is the physical meaning of operation $U_{AB}$?

  2. Can one represent $U_{AB}$ in terms of quantum logic gates ( a circuit diagram)?

  • 2
    $\begingroup$ The expression inside the square brackets is just a CNOT from A to B. It is usually considered to be an elementary gate, so it's representation is just itself. $\endgroup$ – Mateus Araújo Jul 31 at 12:49
  • $\begingroup$ I guess the "key" observation is that $| 1 \rangle \langle 0 |_{B} + | 0 \rangle \langle 1 |_{B} = \sigma^x_{B}$ $\endgroup$ – keisuke.akira Aug 1 at 1:58

The gate/operator in the brackets is the non-local CNOT gate, frequently used to create bipartite entanglement. Given it itself is a 2 qubit gate, then the tensor of this with the identity is simply a gate that acts on 3 qubits.

This gate will take a 3 qubit state and flip the second qubit of this state when the first is $|1\rangle$ It will take $|110\rangle \to|100\rangle, |111\rangle \to |101\rangle$ and vice-versa.

Edit: important to note is that $|1\rangle\langle 0|+|0\rangle\langle 1|$ is the Pauli X gate

| improve this answer | |
  • $\begingroup$ what you mean with "non-local CNOT gate"? It's just a normal CNOT $\endgroup$ – glS Aug 1 at 18:25
  • $\begingroup$ Yes, and a normal CNOT gate is non-local. It is used frequently to introduce non-local correlations. $\endgroup$ – GaussStrife Aug 1 at 18:35

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