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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?

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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.

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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).

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