After a perturbation $V(x)$ added to the system, a matrix element $H_{nn}$ calculated in unperturbed Eigenstates for one-dimensional harmonic oscillator is given as:

$$\epsilon \hbar \omega_0\begin{bmatrix} 1 & 0 &- \sqrt \frac{1}{2} & 0 & \sqrt \frac{3}{8}\\ 0 & 0 & 0 & 0 & 0 \\ - \sqrt \frac{1}{2} & 0 & 1/2 & 0 & - \sqrt \frac{3}{16} \\ 0 & 0 & 0 & 0 & 0 \\ \sqrt \frac{3}{8} & 0 & - \sqrt \frac{3}{16} & 0 & \frac{3}{8} \\ \end{bmatrix}$$

Where $n = 0$ is the lowest. And it can easily be seen that the odd no of energy states don't have any energy corrections.

My confusion arises when I'm reading the the energy corrections from the matrices.

In the author's notes: The energy for the second level is written as: $$E_0^\prime = (\frac{1}{2} + \epsilon ) \hbar \omega $$ $$E_2^\prime = (\frac{5}{2} +\color{red}{\frac{1}{2}} \epsilon ) \hbar \omega $$

I want to read the new energies for the first five energy levels to first order in perturbation theory. My problem is that why did not they take they square root of the matrix element in the energy level of $E_2^\prime$?

Should it not be like this? $$E_2^\prime = (\frac{5}{2} -\color{green}{\sqrt \frac{1}{2}} \epsilon ) \hbar \omega $$


Suppose that you have a perturbed Hamiltonian of the form

$$H_{mn} = E^0_n \delta_{mn} + g V_{mn}$$

where $g$ is a small coupling. Then famously the $n$-th energy level is of the form

$$E_n = E^0_n + g V_{nn} + O(g^2)$$

meaning that you just add the diagonal matrix element of the perturbation $V$. In this case, for $n=2$ that matrix element is just $$g V_{22} = \epsilon \cdot \frac{1}{2}$$ in units where $\hbar = \omega = 1$ etc. Off-diagonal matrix elements of $V$, like $V_{02} = V_{20}$ etc. start playing a role at the next order in perturbation theory.

  • $\begingroup$ Thanks. I was worrying about the off diagonal elements in the first order corrections. I understand now. $\endgroup$
    – user193422
    Aug 19 '19 at 22:05

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