What does it mean when the first order correction energy is 0?

Question

Suppose I have the following Hamiltonian to start

$$ H_0 = \begin{pmatrix} 0 & 0 & 0 & 0\\ 0 & 0 & 2 & 0\\ 0 & 2 & 0 & 0\\ 0 & 0 & 0 & 0 \end{pmatrix} $$

and now I add some pertubation $H_1$ to it to have the total Hamiltonian $H$ as

$$ H = H_0 + H_1 = \begin{pmatrix} 0 & 0 & 0 & 0\\ 0 & 0 & 2 & 0\\ 0 & 2 & 0 & 0\\ 0 & 0 & 0 & 0 \end{pmatrix} + \begin{pmatrix} -0.1 & 0 & 0 & 0\\ 0 & -0.1 & 0 & 0\\ 0 & 0 & 0.1 & 0\\ 0 & 0 & 0 & 0.1 \end{pmatrix} = \begin{pmatrix} -0.1 & 0 & 0 & 0\\ 0 & -0.1 & 2 & 0\\ 0 & 2 & 0.1 & 0\\ 0 & 0 & 0 & 0.1 \end{pmatrix} $$

Note that the lowest eigenvalue of $H$ is $\lambda_0^H = -2.00249844 $.

The lowest eigenvalue of $H_0$ is $\lambda_0^{H_0} = -2 $ with corresponding normalized eigenvector $|\lambda_0^{H_0} \rangle = \begin{pmatrix} 0 \\ -1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix}$.

Now to get the first order energy correction, $\delta E_1$, we can evaluate $\langle \lambda_0^{H_0} | H_1 | \lambda_0^{H_0} \rangle$ but this is always equal to 0 as one can see from the form of $H_1$ and $|\lambda_0^{H_0} \rangle $.

Question: How can you correct for this perturbation? That is to get back close $\lambda_0^H$ after obtaining $\lambda_0^{H_0} $ and $|\lambda_0^{H_0} \rangle $.

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As a side additional note: If $ H_0 = \begin{pmatrix} 0 & 0 & 1 & -1\\ 0 & 0 & 1 & 1\\ 1 & 1 & 0 & 0\\ -1 & 1 & 0 & 0 \end{pmatrix} $ and $H_1 = \begin{pmatrix} -0.1 & 0 & 0 & 0\\ 0 & 0.1 & 0 & 0\\ 0 & 0 & -0.1 & 0\\ 0 & 0 & 0 & 0.1 \end{pmatrix} $

then $H = H_0 + H_1$ has lowest eigenvalue of $\lambda_0^H = -1.48660687$. The lowest eigenvalue of $H_0$ is $\lambda_0^{H_0} = -1.4142135623 $ with corresponding normalized eigenvector $|\lambda_0^{H_0} \rangle = \begin{pmatrix} -1/\sqrt{2} \\ 0 \\ 1/2 \\ 1/2 \end{pmatrix}$. Then from here you can see that $$\delta E_1 = \langle \lambda_0^{H_0} | H_1 | \lambda_0^{H_0} \rangle = -0.05$$

Thus, adding this back bring me closer to the true value $\lambda_0^H = -1.48660687$.

Answer 1

The first-order correction $\langle \psi_0|H|\psi_0\rangle$ vanishes since $H|\psi_0 \rangle$ is orthogonal to $|\psi_0\rangle$. You need to go higher in perturbation theory to get a correction that differs from zero, or adjust your zeroth order Hamiltonian.

Such events often occur in physics. Take for instance the hydrogen atom in an electric field, ${\bf E}=(0,0,E)$, for which the potential is often chosen as $V=V(z)=Ez$. The hydrogen atom ground state is spherically symmetric, so the expectation value for the first-order term is zero.

Answer 2

There are already good answers to your immediate question, so I won't address it directly here. Instead, I'd like to delve slightly deeper into your question itself, in the hope that it might highlight the answer more clearly.

Whenever you have a perturbation, you should always have a parameter which represents the strength of that perturbation; let's call it $\alpha$. We now define not a single perturbed matrix $\mathrm{H}$, but a continuous family of matrices $\mathrm{H}(\alpha)$ such that,

$$ \mathrm{H}(\alpha) = \mathrm{H}_0 + \alpha \mathrm{H}_1. $$ Clearly, if $\alpha=0$ then we recover the original matrix, with its original eigenvalues and eigenvectors, which we assume are known already.

Let $\lambda_i(\alpha)$ be the $i$th eigenvalue of the perturbed matrix, with the corresponding eigenvector $\mathbf{u}_i(\alpha)$, and note that we already know $\lambda_i(0)$ and $\mathbf{u}_i(0)$ for all $i$.

The fundamental question perturbation theory addresses is: for small perturbations, how do $\lambda_i(\alpha)$ and $\mathbf{u}_i(\alpha)$ depend on $\alpha$?

When we refer to a "first-order change", we mean the change which is first-order in the strength of the perturbation, i.e. $\alpha$ in our nomenclature.

Now let us turn to your specific example. In this case,

$$ H(\alpha) = H_0 + \alpha H_1 = \begin{pmatrix} 0 & 0 & 0 & 0\\ 0 & 0 & 2 & 0\\ 0 & 2 & 0 & 0\\ 0 & 0 & 0 & 0 \end{pmatrix} + \alpha\begin{pmatrix} -0.1 & 0 & 0 & 0\\ 0 & -0.1 & 0 & 0\\ 0 & 0 & 0.1 & 0\\ 0 & 0 & 0 & 0.1 \end{pmatrix} $$ You ask about the change in the lowest eigenvalue, so let's actually examine it. Varying $\alpha$ from -1 to 1, we obtain:

The variation appears parabolic, and by symmetry we can see that the gradient of the curve is zero at $\alpha=0$, i.e. the first-order change is zero, which is what you had already calculated. In other words, the eigenvalue does not actually change appreciably for small perturbations of this form, so the correction is zero.

Another way of expressing this is to say that first-order perturbation theory tells you the linear change in the eigenvalue with respect to the perturbation. As you can see from the graph, the linear change is zero at $\alpha=0$ (your reference Hamiltonian). Clearly, this is not the same as saying that the eigenvalue doesn't change!

If you wish to correct the eigenvalue, therefore, you need to either go to a higher order of perturbation, as noted in the other answers, or to change your reference matrix. From the graph, you can see that any matrix with $\alpha>0$ would give a first-order change in the right direction (though second-order perturbation theory would still be more accurate, of course).