In the Elementary gates for quantum computation paper by Barenco et al authors start their proofs by defining a generic form of 2x2 unitary matrix of $\mathbb{C}$ as follows:

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Can you help me with the basic arithmetic behind this statement? For unitary matrix $\begin{bmatrix}a & b \\ c & d\end{bmatrix}$ do we expect $ab + cd = 0$ ? Is this property really satisfied in this case?

Update Indeed! My confusion came from the incorrect treatment of inner products. As suggested below, the correct invariant is $\overline{a}b + \overline{c}d = 0$

  • 3
    $\begingroup$ No that is not a general property of a unitary matrix. Orthogonality of two vectors $(a,c)$ and $(b,d)$ with complex entries is defined as $ \overline{a} b + \overline{c} d = 0$ where $\overline{a}$ denotes the complex conjugate of $a$. Note you also need the vectors to be normalized. $\endgroup$
    – Rammus
    Commented Jul 23, 2023 at 14:05
  • $\begingroup$ Here: quantumcomputing.stackexchange.com/q/5199/9474 $\endgroup$ Commented Jul 23, 2023 at 15:53
  • $\begingroup$ Just make a conjugate transpose (i.e. transpose the matrix and replace all imaginary units with $-i$, denoted $A^\dagger$) and multiply it with the original matrix. If you get a unit matrix, both for $AA^\dagger$ and $A^\dagger A$, the matrix is unitary. $\endgroup$ Commented Jul 23, 2023 at 20:47

1 Answer 1


For a matrix to be unitary, you require $UU^\dagger=I$. Thus, $$ \begin{bmatrix} a & b \\ c & d \end{bmatrix}\begin{bmatrix} a^\star & c^\star \\ b^\star & d^\star \end{bmatrix}=\begin{bmatrix} |a|^2+|b|^2 & ac^\star+bd^\star \\ a^\star c+b^\star d& |c|^2+|d|^2 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} $$ The diagonal elements tell you that the rows must both have length 1. The off-diagonal elements both give the same information: that the two rows must be orthogonal to each other. Unlike what you stated in the question, that means you have to remember to take the Hermitian conjugate of one row.


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