# Could the Hadamard gate have been constructed differently with similar characteristics?

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?

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.
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 $$|1\rangle$$.
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).