What is the matrix representation of a generic two-qubit controlled unitary operation?

Question

I have been taught that an arbitrary two qubit controlled unitary (first qubit control, second qubit target) can be represented as

\begin{matrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & a & b\\ 0 & 0 & c & d \end{matrix}

I can write down some controlled unitary gates (CZ, CNOT, etc) to convince myself of this but I haven't been able to find a way to derive this from scratch.

I would like to find a similar matrix representation for the case where the first qubit is now target and the second qubit is control.

Answer 1

Note that if you have the first qubit as the control qubit, and the second qubit as the target, then you can write $CU$ gate as follow:

$$ CU_{12} = |0\rangle \langle 0| \otimes I + |1 \rangle \langle 1| \otimes U $$

If you work this out then this is equivalent to the matrix representation of

$$ CU_{12} = \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \\ 0 & 0 & u_{11} & u_{12} \\ 0 & 0 & u_{21} & u_{22} \end{pmatrix} \hspace{1 cm} \textrm{where} \ \ U = \begin{pmatrix} u_{11} & u_{12} \\ u_{21} & u_{22} \end{pmatrix} $$

So for instance, if $U = X$ then you have the familarity $CNOT_{12}$ gate (with the first qubit being controlled and the second being the target)

$$ CNOT_{12} = \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{pmatrix} $$

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As for the case where the first qubit is the target and the second qubit as the controlled, then you can write it as

$$ CU_{21} = I \otimes |0\rangle \langle 0| + U\otimes |1\rangle \langle1|$$

I will let you work out the matrix representation here.

A final note that might help while you working out all the details yourself: $$ |0\rangle \langle 0 | = \begin{pmatrix}1 \\ 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \end{pmatrix} = \begin{pmatrix}1 & 0 \\ 0 & 0 \end{pmatrix} \hspace{1 cm} |1\rangle \langle 1 | = \begin{pmatrix}0 \\ 1\end{pmatrix} \begin{pmatrix} 0 & 1 \end{pmatrix} = \begin{pmatrix}0 & 0 \\ 0 & 1 \end{pmatrix} $$

Answer 2

For your first question, you are right, so here are 2 examples:

And for the second question, since operation happen only in case second qubit is 1, and nothing is changing when second qubit is 0:

Edit: you should look seperatly, what the gate is doing seperatly to each state in the $|x_1 x_2\rangle$ state. In cases that the control is zero, the state is not changing, therfore multiplied by one, without adding any other states (the rest of the row is 3 zeros).

In cases that the control is 1, the qubit changes as it was changes is it was only 1 qubit gate on the target alone, while you ignore the 1 in the control.

Another way to see, in case first is control, for any unitary $U$:

$$CU(|00\rangle+|01\rangle+|10\rangle+|11\rangle)=(|00\rangle+|01\rangle)+|1\rangle \otimes U(|0\rangle+|1\rangle)$$