Is there a 4x4 unitary matrix with 0 diagonal and non-zero off-diagonal?

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This question is homework from Scott Aaronson's 2017 Quantum Information course.

a) Give an example of a 2x2 unitary matrix where the diagonal entries are 0 but the off-diagonal entries are nonzero.

b) Give an example for a 4x4 unitary matrix.

c) Is it possible to have a 3x3 unitary matrix with this condition? If no, prove it!

(a) is easy.

$$ \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} $$

(c) is not possible. Since for

$$ \begin{pmatrix} 0 & a & b \\ c & 0 & d \\ e & f & 0 \end{pmatrix} $$

to be unitary the inner product $0\cdot a + c \cdot 0 + e \cdot f$ needs to be zero, but then at least one of $e$ or $f$ needs to be 0.

For (b) I assume he means a 4x4 matrix where the diagonal elements are 0 but all other elements are non-zero (not just any unitary matrix).

I tried without luck:

  • Building the 4x4 unitary matrix from 2x2 unitary block matrices. The closest I got so far is

    $$ U = \begin{pmatrix} X & H \\ H & -X \end{pmatrix} $$

    with

    $$ X = \begin{pmatrix} 0 & 1 \\ 1 & -0 \end{pmatrix} $$

    and

    $$ H = \frac{1}{\sqrt 2} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} $$

    Unfortunately $U^\dagger U$ leaves some non-diagonal elements.

    $$ U^\dagger U = \begin{pmatrix} 2 & 0 & \sqrt 2 & 0 \\ 0 & 2 & 0 & \sqrt 2 \\ \sqrt 2 & 0 & 2 & 0 \\ 0 & \sqrt 2 & 0 & 2 \\ \end{pmatrix} $$

  • Starting with an arbitary first column and finding orthogonal column vectors with the desired 0 elements, but this seems tricky.

Any hints?

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Example: $$ \frac{1}{\sqrt{3}} \begin{bmatrix} 0 & 1 & 1 & 1 \\ 1 & 0 & 1 & -1 \\ 1 & -1 & 0 & 1 \\ 1 & 1 & -1 & 0 \end{bmatrix}. $$