# Time evolution Operator: write expression for $t>0$

I have a question related to time evolution operator. I was analyzing my teacher solution after solving the problem myself, but there is a detail I dont get. I have a hamiltonian that is represented by this matrix. The basis is $$|1\rangle$$ and $$|2\rangle$$.

$$\epsilon\begin{bmatrix}0&i\\-i&0\end{bmatrix}$$ where $$\epsilon$$ is positive and real number.

I know at $$t=0$$ the system is in the fundamental state, according tho the problem. The goal is to write the state for $$t>0$$. The final result is:

$$|\Psi(t)\rangle = e^{-i\hat Ht/\hbar }\,|\Psi(0)\rangle = e^{-i\hat Ht/\hbar }\,|F\rangle = \frac{e^{i\epsilon t/\hbar}}{\sqrt{2}}\,\left(|1\rangle - i |2\rangle\right) \quad .$$

I know how to get the eigenvalues and eigenvectors, and normalize them. My only doubt is in the signal in the exponential. When I apply the Hamiltonian why $$\epsilon$$ is positive, in the final expression? I know the state at $$t=0$$ is the state which the eigenvalue is $$+\epsilon$$, so why the final expression is $$e^{i\epsilon t/\hbar}$$ and not $$e^{-i\epsilon t/\hbar}$$ ?

• Well, the question itself, doesnt especify, but what I assumed and my teacher's solution confirmed is that you calculate the eigenvalues of the hamiltonian which are +ε and -ε, and the eigenvetor corresponding to +ε is the fundamental state. The result is 1/sqrt(2)*(|1⟩ - i |2⟩). Commented Jan 17, 2022 at 14:43
• I've edited the math symbols. Please try to use MathJax; you can find a tutorial here. Commented Jan 17, 2022 at 14:45
• Thank you, I will do my best to use Mathjax, I am still learning how to use it. Commented Jan 17, 2022 at 14:46
• In general If $H|F\rangle = \epsilon |F\rangle$, then $e^{-iHt}\,|F\rangle =e^{-i\epsilon t}\,|F\rangle$. Despite that, the Hamiltonian does not look hermitian, no? For example: $H_{12} \neq H_{21}^*$, where $*$ denotes complex conjugation. Is that a typo? Commented Jan 17, 2022 at 14:54
• I am sorry i made a mistake writing the matrix. This is the correct one, after editing. However the eigenvalues mentioned before are correct. Commented Jan 17, 2022 at 15:04

The "fundamental state" that you mention has the lowest energy possible. The two eigenvalues of the Hamiltonian are $$\pm \epsilon$$, so the fundamental state must have energy $$-\epsilon$$. Accounting for this mistake will immediately yield your teacher's solution.
• @AnaBranco "fundamental state" is not a uniquely defined term. If they mean "ground state" then it must be the state with the lowest energy and thus have eigenvalue $-\epsilon$. Perhaps check that you are using the same basis (ie is $|1\rangle\langle 1|$ the top left or bottom right part of the matrix?) Commented Jan 17, 2022 at 15:49
• @AnaBranco then there must be something inconsistent here in somebody's definition. Either $|F\rangle$ is the ground state, so the Hamiltonian must look like $H=-\epsilon i(|1\rangle\langle 2|-|2\rangle\langle 1|)$, or $|F\rangle$ is for some reason the excited state, in which case the sign that your teacher wrote in the exponential is indeed incorrect. The third possibility is that the Hamiltonian is correct but the fundamental state should be the Hamiltonian's ground state $|F\rangle\propto |1\rangle +i|2\rangle$ Commented Jan 17, 2022 at 17:32