Consider a gate $\mathcal{G}(\mu)=e^{-i \mu G}$ generated by a Hermitian operator G. If G has just two distinct eigenvalues(which can be repeated) we can, without loss of generality,
shift the eigenvalues to ±r, as the global phase is unobservable. Note that any single qubit gate is of this form.
I dont know How to shift the eigenvalues of a quantum Hermitian operator G to ±r ?
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2 Answers
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Consider a gate $\mathcal{G}(\mu)=e^{-i \mu G}$ generated by a Hermitian operator G. If G has just two distinct eigenvalues(which can be repeated) we can, without loss of generality, shift the eigenvalues to ±r, as the global phase is unobservable. Note that any single qubit gate is of this form.
I dont know How to shift the eigenvalues of a quantum Hermitian operator G to ±r ?
Let the distinct eigenvalues of $G$ be denoted as $g_1$ and $g_2$.
Consider the Hermitian operator $\tilde G$, defined as: $$ \tilde G \equiv G - \frac{1}{2}\left(g_1 + g_2\right)\mathbb{I} $$
The $\tilde G$ operator also has two eigenvalues, which can be denoted as: $$ r\equiv \frac{1}{2}\left(g_1 - g_2\right) $$ and $$ -r = \frac{1}{2}\left(g_2 - g_1\right) $$
So we can write $\mathcal{G}$ as: $$ \mathcal{G}(\mu)=e^{-i \mu G} =e^{-i \mu (\tilde G + \frac{1}{2}(g_1 + g_2)\mathbb{I})} $$ $$ =e^{-i\tilde \phi}e^{-i \mu \tilde G}\;, $$ where $\tilde \phi = \frac{\mu}{2}(g_1 + g_2)$. Note that the final equality above follows because $[\tilde G, \mathbb{I}]=0$, since every operator commutes with the identity operator.
More explicitly, we can write, in matrix form, in some basis that diagonalizes $G$ (and $\tilde G$), the result for $\mathcal{G}(\mu)$ as: $$ \mathcal{G}(\mu) = e^{-i\tilde \phi} \left( \begin{matrix} e^{-ir} & 0 & 0 & \ldots & 0 & 0 & 0 & \ldots \\ 0 & e^{-ir} & 0 & \ldots & 0 & 0 & 0 & \ldots \\ 0 & 0 & e^{-ir} & \ldots & 0 & 0 & 0 & \ldots \\ \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots\\ 0 & 0 & 0 & \ldots & e^{+ir} & 0 & 0 & \ldots \\ 0 & 0 & 0 & \ldots & 0 & e^{+ir} & 0 & \ldots \\ 0 & 0 & 0 & \ldots & 0 & 0 & e^{+ir} & \ldots \\ \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \end{matrix} \right) $$
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Say the eigenvalues of $G$ are $\lambda_1, \lambda_2$ with $\lambda_1 \neq \lambda_2$. Writing the $2\times 2$ Hermitian matrix $G$ in its eigenbasis (we can pick any basis, since unitary transformations will not affect the spectrum of our operator), we are looking for real numbers $a$ and $b$ such that \begin{equation} aG + b = \begin{pmatrix} a\lambda_1 + b & 0 \\ 0 & a \lambda_2 + b \end{pmatrix} = \begin{pmatrix} r & 0 \\ 0 & -r \end{pmatrix}. \end{equation} This gives a system of equations, \begin{align} a \lambda_1 + b &= r \\ a \lambda_2 + b &= -r, \end{align} for which you can easily verify the solution is \begin{align} a = \frac{2r}{\lambda_1 - \lambda_2}\\ b = -r \frac{(\lambda_1 + \lambda_2)}{(\lambda_1 - \lambda_2)}. \end{align} Then $aG + b$ has eigenvalues $\pm r$. Physically, this corresponds to running the modified gate \begin{equation} \mathcal{G}(\mu) = \exp(i (aG + b) \mu), \end{equation} which you can interpret as performing time evolution with respect to $G$ for a factor of $a$ longer (the $b$ introduces physically meaningless global phase, reflecting the fact that we have 'shifted' both of the original energies $\lambda_1,\lambda_2$ by some amount but only differences in energies really matter).
