Question

Suppose you’re given a two-level quantum system whose (time-independent) Hamiltonian ${\mathit{H}}^{\mathbf{0}}$admits just two Eigen states, ${\mathbf{\Psi }}_{\mathbf{a}}$ (with energy ${\mathit{E}}_{\mathbf{a}}$ ), and ${\mathit{\Psi }}_{\mathbf{b}}$(with energy ${\mathit{E}}_{\mathbf{b}}$ ). They are orthogonal, normalized, and non-degenerate (assume ${\mathit{E}}_{\mathit{a}}$ is the smaller of the two energies). Now we turn on a perturbation H′, with the following matrix elements:

(7.74).

where h is some specified constant.

(a) Find the exact Eigen values of the perturbed Hamiltonian.

(b) Estimate the energies of the perturbed system using second-order perturbation theory.

(c) Estimate the ground state energy of the perturbed system using the variation principle, with a trial function of the form

$\mathbf{\Psi }\mathbf{=}\left(\mathrm{cos\varphi }\right){\mathbf{\Psi }}_{\mathbf{a}}\mathbf{+}\left(\mathrm{sin\varphi }\right){\mathbf{\psi }}_{\mathbf{b}}$ (7.75).

where ϕ is an adjustable parameter. Note: Writing the linear combination in this way is just a neat way to guarantee that ψ is normalized.

(d) Compare your answers to (a), (b), and (c). Why is the variational principle so accurate, in this case?

Short answer

(a)$\Rightarrow E_{\pm }=\frac{1}{2}\left[E_a+E_b\pm \sqrt{{\left(E_a-E_b\right)}^2+4h^2}\right]$

(b)$E_{-}\approx E_a-\frac{h^2}{\left(E_b-E_a\right)};E_{+}\approx E_b+\frac{h^2}{\left(E_b-E_a\right)}$

(c)

(d) Using Taylor series (small h) to expand results in (a) we obtained results from

Step-by-step solution

(a) Finding the exact Eigen value

In order to find Eigen values of H, calculate:

$$\begin{gathered} det\left(H-\lambda .l\right)=0 \\ \left(E_a-\lambda \right)\left(E_b-\lambda \right)-h^2=0 \\ {\lambda }^2-\lambda \left(E_a+E_b\right)+E_a{E}_b-h^2=0 \end{gathered}$$

$$\begin{gathered} \lambda =\frac{1}{2}\left(E_a+E_b\pm \sqrt{E_a^2+2E_a{E}_b+E_b^2-4E_a{E}_b+4h^2}\right) \\ \Rightarrow E_{\pm }=\frac{1}{2}\left[E_a+E_b\pm \sqrt{{\left(E_a-E_b\right)}^2+4h^2}\right] \end{gathered}$$

Step 2:(b) Estimating the energies

Zero order: $E_a^0=E_a,E_b^0=E_b$

First order:

Second order:

 Step 3:(c) Estimating the ground state energy

Here estimating the ground state energy.

So,

$$\begin{gathered} =1-\frac{{\dot{o}}^2}{1+{\dot{o}}^2}=\frac{1}{1+{\dot{o}}^2}; \\ \cos 2\varphi =\frac{\overline{+}1}{\sqrt{1+{\dot{o}}^2}}\left(signdictatedby\tan 2\varphi =\frac{\sin 2\varphi }{\cos 2\varphi }=-\dot{o}\right).{\cos }^2\varphi \\ =\frac{1}{2}\left(1+\mathrm{co}\left(\right)s2\varphi \right) \\ =\frac{1}{2}\left(1\overline{+}\frac{1}{\sqrt{1+{\dot{o}}^2}}\right); \\ {\sin }^2\varphi =\frac{1}{2}\left(1-\cos 2\varphi \right) \\ =\frac{1}{2}\left(1\pm \frac{1}{\sqrt{1+{\dot{o}}^2}}\right) \end{gathered}$$

But

 Step 4:(d) Comparing (a),(b) and (c)

We can obtain results from task (b) if we expand in Taylor series exact energies from task (a). we use limit where h is very small, which has to be if we want to use perturbation theory:

$$\begin{gathered} E_{\pm }=\frac{1}{2}[\left(E_a+E_b\right)\pm \left(E_b+E_a\right)\sqrt{1+\frac{4h^2}{\left(E_b+E_a\right)}} \\ {E}_{\pm }\approx \frac{1}{2}\left\{E_a+E_b\pm \left(E_b+E_a\right)\left[1+\frac{2h^2}{{\left(E_b+E_a\right)}^2}\right]\right\} \\ =\frac{1}{2}\left[E_a+E_b\pm \left(E_b+E_a\right)\pm \frac{2h^2}{\left(E_b+E_a\right)}\right] \end{gathered}$$

so localid="1658396154675" $$\begin{gathered} E_{+}\approx E_b+\frac{h^2}{\left(E_b+E_a\right)}, \\ {E}_{-}\approx E_a-\frac{h^2}{\left(E_b+E_a\right)}, \end{gathered}$$

Confirming the perturbation theory results in (b). The variation principle (c) gets the ground state(E−) exactly right-not too surprising since the trial wave function Eq. 7.75 is almost the most general state (there could be a relative phase factor $e^{i\theta }$ .