What's Bob's state after this quantum circuit? [closed]

Question

As shown in the picture, we know Alice's state will be intact after this circuit, but what about Bob's state, will it be $|0\rangle$ or $(|0\rangle+|1\rangle)/\sqrt{2}$ and why? I think it will be $(|0\rangle+|1\rangle)/\sqrt{2}$, but I can't get this from density martix, can anyone give me a detailed analysis of this? Thanks in advance.

Answer 1

You can use circuit simulators to help with understanding small circuits like this, and to check your work. For example, here's your circuit in Quirk:

The key things you need to understand for this circuit are:

1. How to apply an operation on paper to an entangled state (for the H after the CNOT).
   • Group by the uninvolved bits, and apply the operation independently to each group.
2. How to apply post-selection on paper to a state.
   • Throw out non-matching components of the state like $|11\rangle$, renormalize so things add up to 100% again.

Answer 2

You seem very confused. You can work this one out without using density matrices and density matrices would only add complexity in this case, so you should avoid them in this particular problem. To figure out the state, you consider what each gate does in turn. And in this case one of the gates, CNOT, involves an interaction between the two qubits, so you have to work out the joint state of the two qubits. The Hadamard gate $H = \tfrac{1}{\sqrt{2}}(|0\rangle\langle 0|+|0\rangle\langle 1|+|1\rangle\langle 0|-|1\rangle\langle 1|)\otimes I$. The CNOT gate is given by $C = |00\rangle\langle 00|+|01\rangle\langle 01|+|10\rangle\langle 11|+|11\rangle\langle 10|$. I would guess that the $|0\rangle$ means projection of the final state onto $|0\rangle$ for the first qubit. So the state will be $HCH|00\rangle$ projected onto $|0\rangle$ for the first qubit.