9
$\begingroup$

Say we had a Hadamard-like gate with the -1 in the first entry instead of the last. Let's call it $H^1$.

$$H = \begin{bmatrix}1&1\\1&-1\end{bmatrix}$$

$$H^1 = \begin{bmatrix}-1&1\\1&1\end{bmatrix}$$

From my maths it's a unitary matrix, so it's a valid quantum gate that does the following:

$$H^1|0\rangle = \left(\frac{-|0\rangle+|1\rangle}{\sqrt{2}}\right)$$

$$H^1|1\rangle = \left(\frac{|0\rangle+|1\rangle}{\sqrt{2}}\right)$$

These are similar to the true Hadamard gate, but with the sign flipped on the $|0\rangle$ instead of the $|1\rangle$.

Was the choice of the Hadamard as we know it an arbitrary decision like the right hand rule? Or is there a mathematical or historical reason for it?

$\endgroup$

2 Answers 2

8
$\begingroup$

The Hadamard gate has close ties to the discrete Fourier transform. Consider the DFT for an $N$-level system: $$\vert j \rangle = \frac{1}{\sqrt{N}} \sum\limits_{k=0}^{N-1} e^{\frac{i2 \pi j k}{N}} \vert k \rangle.$$ For $N=2$ this is simply $$\vert j \rangle = \frac{1}{\sqrt{2}}\begin{bmatrix}1 & 1 \\ 1 & -1 \end{bmatrix} \, \vert k \rangle = H \vert k \rangle.$$ For $H$ applied to $d$ qubits in parallel, this generalizes naturally to a $d$-dimensional DFT, $H^{\otimes d}$. In this sense, the sign convention of $H$ clearly aligns with its function as a unitary DFT matrix.

The choice of sign for $H$ also aligns, less directly, to the conventional $X$, $Y$, $Z$ basis of the Lie algebra $\mathfrak{su}_2$. For example, $H$ plays a part in the spectral decomposition of $X$ as $$X = H Z H,$$ noting that $H$ is both unitary and Hermitian. That $H$ contains both the eigenvectors and dual-eigenvectors of $X$, with eigenvalues in $Z$ can be a useful property. $H^1$, as defined in the question, would swap the position of the eigenvalues.

$\endgroup$
5
$\begingroup$

You can easily check that $H^1=XHX$, which is another way to say that $H^1$ is the same as $H$ modulo swapping $|0\rangle$ and $|1\rangle$.

This means that it is a different gate.

On the other hand, it is "equivalent" in the sense that, given an arbitrary circuit given as a sequence of gates $\prod_k U_k$, if you swap all the $|0\rangle$s and $|1\rangle$s, all the $H$ gates will become $H^1$ gates (and vice versa).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.