# How to shift the eigenvalues of a quantum Hermitian operator G to ±r?

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 ?

Schuld, Maria, et al. "Evaluating analytic gradients on quantum hardware." Physical Review A 99.3 (2019): 032331.

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: $$\mathcal{G}(\mu) = e^{-i\tilde \phi} \left( \begin{matrix} e^{-ir} & 0 & 0 & \ldots & 0 & 0 & 0 & \ldots \\ 0 & e^{-ir} & 0 & \ldots & 0 & 0 & 0 & \ldots \\ 0 & 0 & e^{-ir} & \ldots & 0 & 0 & 0 & \ldots \\ \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots\\ 0 & 0 & 0 & \ldots & e^{+ir} & 0 & 0 & \ldots \\ 0 & 0 & 0 & \ldots & 0 & e^{+ir} & 0 & \ldots \\ 0 & 0 & 0 & \ldots & 0 & 0 & e^{+ir} & \ldots \\ \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \end{matrix} \right)$$

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 $$$$aG + b = \begin{pmatrix} a\lambda_1 + b & 0 \\ 0 & a \lambda_2 + b \end{pmatrix} = \begin{pmatrix} r & 0 \\ 0 & -r \end{pmatrix}.$$$$ This gives a system of equations, \begin{align} a \lambda_1 + b &= r \\ a \lambda_2 + b &= -r, \end{align} for which you can easily verify the solution is \begin{align} a = \frac{2r}{\lambda_1 - \lambda_2}\\ b = -r \frac{(\lambda_1 + \lambda_2)}{(\lambda_1 - \lambda_2)}. \end{align} Then $$aG + b$$ has eigenvalues $$\pm r$$. Physically, this corresponds to running the modified gate $$$$\mathcal{G}(\mu) = \exp(i (aG + b) \mu),$$$$ which you can interpret as performing time evolution with respect to $$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).