I have noticed that for all the matrices representing quantum gates, if we read rows left-to-right and top to bottom, the read the same as columns top to bottom left to right.


\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1& 0 \end{pmatrix} (CNOT gate), if we read the values in from the first row, we see 1 0 0 0. Similarly if we read left-most column top to bottom. Second row reads 0 1 0 0, just like 2nd left column etc.

I believe this is a property of all matrices representing quantum gates, but cannot explain why exactly this is the case (and if it is, in fact, guaranteed for every quantum gate?)

  • $\begingroup$ This is because these gates are Hermitian, not because they're unitary! $\endgroup$ Jul 7, 2018 at 15:10
  • $\begingroup$ Thanks @YuvalFilmus! Would you consider wrapping it in an answer, and I shall change the question accordingly? $\endgroup$
    – 3yakuya
    Jul 7, 2018 at 20:02

1 Answer 1


A counter-example to your observation is the Pauli Y gate:

$$Y = \begin{bmatrix} 0&-i\\i&0\end{bmatrix}$$

The first row is 0 then -i, but the first column is 0 then +i.

Another example is the X->Y->Z->X gate (a 120 degree rotation around the X+Y+Z axis of the bloch sphere):

$$\frac{1-i}{2} \begin{bmatrix} 1&-i\\1&i\end{bmatrix}$$

The reason the rows often look so much like the columns, particularly for self-inverse operations, is because quantum gates are unitary and the inverse of a unitary matrix is its conjugate transpose. But you can easily break the symmetry by multiplying an individual row (or column) by a phase factor, which produces a different but valid quantum operation.


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