Expectation value of $e^{frac{i f(t)}{2} sigma^z_1 sigma^z_2 }$

My intuition is usually to find the eigenstates of the operator, express the state in terms of that eigenbasis, then take the appropriate linear combination of the eigenvalues weighted by the probabilities of the state in the eigenbasis.

This leads to another question: is there an intuitive way to find the eigenvalues and eigenstates of this operator? Yes! Any state $|psirangle$ that is a joint eigenstate of both $sigma_1^z$ and $sigma_2^z$, satisfying some equations $sigma_i^z|psirangle=lambda^{(i)}|psirangle$ will be an eigenstate of your operator, with eigenvalue $e^{frac{i f(t)}{2}lambda^{(1)}lambda^{(2)}}$. So if we label these eigenstates by their eigenvalues, where $sigma_1^zotimes sigma_2^z|lambda^{(1)}_i,lambda^{(2)}_irangle=lambda^{(1)}_i lambda^{(2)}_i|lambda^{(1)}_i,lambda^{(2)}_irangle$, then the expectation value of your operator in the state $$|Psirangle=sum_i psi_i |lambda^{(1)}_i,lambda^{(2)}_irangle$$ will be $$langlePsi|e^{frac{if(t)}{2}sigma_1^zotimessigma_2^z}|Psirangle=sum_i |psi_i|^2e^{frac{i f(t)}{2}lambda^{(1)}_ilambda^{(2)}_i}.$$

How do we justify the above? We expand our operator using the definition of an exponential: $$e^{A}=sum_{n=0}^infty frac{A^n}{n!}.$$ When acting on an eigenstate $|arangle$, we obtain the desired eigenvalue equation $$e^{A}|arangle=sum_{n=0}^infty frac{A^n}{n!}|arangle=sum_{n=0}^infty frac{a^n}{n!}|arangle=e^a|arangle.$$