# Quantum Phase estimation with $2\pi$ replaced with $2e$

The QPE on IBM platform finds the eigenvalue of a unitary operator, i.e $$U|\phi\rangle=e^{2\pi i\theta}|\phi\rangle$$ and uses the rotation operators as $$U(\theta)=\begin{bmatrix}0 & 1\\ 1 &e^{i\theta}\end{bmatrix}$$ My question is we can write $$U|\phi\rangle=e^{2\pi i\theta}|\phi\rangle$$ as $$U|\phi\rangle=e^{2e i\theta}|\phi\rangle$$ where, since even then the magnitude of the eigenvalue remains same. Of course that would require more number of measurement qubits. For instance for $$\theta=0.5$$, we had $$4$$, so this time we can increase the $$n$$ and get closer to $$\theta=0.5$$. My question is can this be done.

No - the key intuition of QPE is that $$e^{2 \pi i \theta}$$ holds for $$\theta \in [0, 1)$$, and the reading out the ancilla provides the binary representation of the fraction.
If you changed the operator to $$e^{2 e i \theta}$$, $$\theta \not \in [0, 1)$$ necessarily. Instead, you should just use the typical QPE approach, but then find $$\zeta$$ where $$2 e i \zeta = 2 \pi i \theta \implies \zeta = \frac{\pi}{e}\theta$$, and correct for symmetries afterwards.