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How do we evaluate the time-evolution operator of a perturbed system with time-independent perturbation ?

For example:

In a two state system acted up on by a time-independent perturbation, let's say $H' =\begin{pmatrix} 0 & V_{12} \\ V_{22} & 0 \\ \end{pmatrix}$ where $V_{21}=V^*_{12}$. So the total Hamiltonian, $$ H=H_{o}+H'=\begin{pmatrix} E_1 & 0 \\ 0 & E_2 \\ \end{pmatrix}+\begin{pmatrix} 0 & V_{12} \\ V_{22} & 0 \\ \end{pmatrix} $$

My understanding:

The eigenvalues of the perturbed system will be $$ E^{\pm}=\frac{1}{2}(E_{1}+E_{2})\pm\bigg[\frac{1}{4}(E_1-E_2)^2-V^2_{21}\bigg]^{1/2} $$

Given the perturbation is time-independent how do I approach this problem ?

Can I possibly do the following: $$ U(t)|+\rangle=exp\bigg(-\frac{iHt}{\hbar}\bigg)|+\rangle=exp\bigg(-\frac{i(H_o+H')t}{\hbar}\bigg)|+\rangle=exp\Bigg[-\frac{i\big(\frac{1}{2}(E_{1}+E_{2})+\big[\frac{1}{4}(E_1-E_2)^2-V^2_{21}\big]^{1/2}\big)t}{\hbar}\Bigg]|+\rangle $$ and $$ U(t)|-\rangle=exp\Bigg[-\frac{i\big(\frac{1}{2}(E_{1}+E_{2})-\big[\frac{1}{4}(E_1-E_2)^2-V^2_{21}\big]^{1/2}\big)t}{\hbar}\Bigg]|-\rangle $$

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2 Answers 2

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If you have the exact (constant) H matrix you can always solve exactly.
Find the Eigenvalues and Eigenvectors of $H$ - these are your true new Eigenenergies and Eigenstates. Rewrite your state in the basis og your new found eigenvectors and propagate each state with its appropriate energy. In your case - $$\text{$\unicode{f3b5}$H}=\unicode{f3b5} \left( \begin{array}{cc} \text{E1} & \text{V12} \\ \text{V12}^* & \text{E2} \\ \end{array} \right)$$,

$$\tilde{E1}=\frac{1}{2} \left(-\sqrt{4 \text{V12} \text{V12}^*+\text{E1}^2-2 \text{E1} \text{E2}+\text{E2}^2}+\text{E1}+\text{E2}\right)$$ $$\tilde{E2}=\frac{1}{2} \left(\sqrt{4 \text{V12} \text{V12}^*+\text{E1}^2-2 \text{E1} \text{E2}+\text{E2}^2}+\text{E1}+\text{E2}\right)$$ The appropriate eigenvectors (unnormalized) are -

$$|v1\rangle=\begin{pmatrix}-\frac{\sqrt{4 \text{V12} \text{V12}^*+\text{E1}^2-2 \text{E1} \text{E2}+\text{E2}^2}-\text{E1}+\text{E2}}{2 \text{V12}^*} \\ 1\end{pmatrix}=-\frac{\sqrt{4 \text{V12} \text{V12}^*+\text{E1}^2-2 \text{E1} \text{E2}+\text{E2}^2}-\text{E1}+\text{E2}}{2 \text{V12}^*}|+\rangle+1|-\rangle$$ $$|v2\rangle=\begin{pmatrix} -\frac{-\sqrt{4 \text{V12} \text{V12}^*+\text{E1}^2-2 \text{E1} \text{E2}+\text{E2}^2}-\text{E1}+\text{E2}}{2 \text{V12}^*} \\ 1\end{pmatrix}=-\frac{-\sqrt{4 \text{V12} \text{V12}^*+\text{E1}^2-2 \text{E1} \text{E2}+\text{E2}^2}-\text{E1}+\text{E2}}{2 \text{V12}^*}|+\rangle+1|-\rangle$$

After normalizing the eigenvectors, every initial condition can be rewritten as - $$|\psi \rangle=a|v1\rangle+b|v2\rangle$$ The evolution is given by - $$|\psi(t) \rangle=e^{-i\tilde{E1}t}a|v1\rangle+e^{-i\tilde{E2}t}b|v2\rangle$$ Only after evolving $|v_i\rangle$ with appropriate $ e^{-i \tilde{E_i} t }$ you should rewrite the solution with your original $|+\rangle$ and $|-\rangle$

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Yes, each state will have a different evolution operator whose Hamiltonian can be replaced by the eigenvalues associated with that state, so long as the states are stationary.

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  • $\begingroup$ pls check i have edited the original question. Is my approach correct ? $\endgroup$
    – Sooraj Soman
    Commented Mar 18, 2016 at 15:34
  • $\begingroup$ Yes, those values which you replaced the Hamiltonian with are the eigenvalues of the Hamiltonian. $\endgroup$
    – N. Carrara
    Commented Mar 18, 2016 at 15:46
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    $\begingroup$ Your answer is wrong (and doesn't suit even the pertubation theory). Exact solution - only the Eigenstates of H propagate with appropriate eigenenergies. After introducing the pertubation, the new eigenvectors are not the same as the old ones, and one should not propagate the old eigenvectors but the new eigenvectors with the new eigenenergies. Pertubation theory - the first order correction by pertubation theory in this case is the same as ignoring the pertubation matrix alltogether, since its diagonal is all zeros. $\endgroup$
    – Alexander
    Commented Mar 18, 2016 at 16:15
  • $\begingroup$ @Alexander so how do i approach the problem in that case? 'd be very helpful if u could provide proper instructions to deal with this problem $\endgroup$
    – Sooraj Soman
    Commented Mar 18, 2016 at 16:18
  • $\begingroup$ @ss1729 See the answer I posted $\endgroup$
    – Alexander
    Commented Mar 18, 2016 at 16:22

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