Eigenvalues of $H=aX+bY+cZ+dI$ Suppose I have the hamiltonian $H=aX+bY+cZ+dI$, where $a,b,c,d$ are some real constants, and $X,Y,Z,I$ are Pauli matrices. I'm trying to figure out the range of possible energy eigenvalues. If I limit the range of $a,b,c,d$ to be $[-1,1]$, then I think the range of eigenvalues should be $[-4,4]$ (linear combination), since the eigenvalues of each Pauli matrix are $-1$ and $1$. is my assumption correct?
However, I tried to calculate the eigenvalues using python with such conditions, and it looks like the range of eigenvalues goes from $-\sqrt2-1$ to  $\sqrt2+1$. I don't know if my assumption is wrong or the calculation is not working.
Thanks!
 A: The way you are getting $\pm (a+b+c+d)$ is if all the eigenvectors lined up so that you would get $\pm a$, $\pm b$ etc from each of the summands. But the eigenvectors do not line up like that. That means the $[-4,4]$ bound is more loose than it has to be. The actual range of eigenvalues is a proper subset of that.
$$
H = \begin{pmatrix}
d + c & a - bi\\
a + bi & d - c\\
\end{pmatrix}\\
H - \lambda = \begin{pmatrix}
d + c - \lambda & a - bi\\
a + bi & d - c - \lambda\\
\end{pmatrix}\\
\det (H - \lambda) = (d-\lambda + c)(d-\lambda - c) - (a-bi)(a+bi)\\
= (d-\lambda)^2 - c^2 - a^2 - b^2 = 0\\
(d-\lambda)^2 = a^2 + b^2 + c^2\\
(\lambda_{\pm}-d) = \pm \sqrt{a^2 + b^2 + c^2}\\
\lambda_{\pm} = d \pm \sqrt{a^2 + b^2 + c^2}\\
$$
$\sqrt{a^2+b^2+c^2}$ ranges from $0$ to $\sqrt{3}$ as $a,b,c$ vary in $[-1,1]$. So the lowest $\lambda_-$ can be is if $d=-1$ and the square root gives $\sqrt{3}$. That is $-1 - \sqrt{3}$. The highest $\lambda_+$ can be is if $d=1$ and the square root gives $\sqrt{3}$. That gives $1 + \sqrt{3}$.
