# Why is the second order perturbative correction to the ground state energy always down?

What is the physical/deeper reason for the second order shift of the ground state energy in time independent perturbation theory to be always down?

I know that it follows from the formula quite straightforwardly, but I could not find a deeper reason for it even if if I took up examples like the linear Stark effect.

Would you please tell me why?

• If I recall correctly, you can think of it as related to the variational principle in quantum mechanics. The zeroth-order guess for the ground state is not a very good guess, so its energy is too high. At higher order you have a better guess for what the ground state is, so its energy is lower. (It is not quite as simple as I'm describing ... there are details I am glossing over.) – Steve Byrnes Jul 14 '12 at 19:03

## 2 Answers

I) The lowering of the ground state energy is a special case of the more general phenomenon of level repulsion (because the excited energy levels by definition must be larger than the ground state energy).

II) Level repulsion is not just a quantum phenomenon. It also happens for purely classical systems, e.g. two coupled oscillators, as mentioned in the link.

III) Mathematically, the non-zero off-diagonal elements ("the interactions") of an $n\times n$ Hermitian matrix $(H_{ij})_{1\leq i,j \leq n}$ ("the Hamiltonian") cause the real eigenvalues $E_1$, $\ldots$, $E_n$, (counted with multiplicity) to, loosely speaking, spread out more than the distribution of diagonal elements $H_{11}$, $\ldots$, $H_{nn}$. This eigenvalue repulsion effect is e.g. encoded in the Schur-Horn Theorem.

IV) Perhaps a more physical understanding of level repulsion is as follows. When considering a perturbation problem $H=H^{(0)}+V$, we would like to find the true energy eigenstates $|1 \rangle$, $\ldots$, $|n \rangle$, with energy eigenvalues $E_1$, $\ldots$, $E_n$. A priori we only know the unperturbed energy eigenstates $|1^{(0)} \rangle$, $\ldots$, $|n^{(0)} \rangle$, with energy eigenvalues $H_{11}$, $\ldots$, $H_{nn}$. (We have for the sake of simplicity assumed that the interaction part $V$ has no diagonal part. This is always possible via reorganizing $H^{(0)} \leftrightarrow V$.) Thus an unperturbed energy eigenstates

$$|i^{(0)} \rangle ~=~ \sum_{j=1}^n |j \rangle~ \langle j |i^{(0)} \rangle$$

is really a linear combination of the true energy eigenstates $|1 \rangle$, $\ldots$, $|n \rangle$. The square of the overlaps $\langle j |i^{(0)} \rangle$ has a probabilistic interpretation

$$\sum_{j=1}^n |\langle j |i^{(0)}\rangle|^2 ~=~\langle i^{(0)}|i^{(0)}\rangle ~=~1.$$

Hence the unperturbed energy eigenvalue

$$H_{ii}~=~\langle i^{(0)}|H|i^{(0)}\rangle ~=~\sum_{j=1}^n E_j|\langle j |i^{(0)}\rangle|^2$$

is a quantum average of the true energy eigenvalues $E_1$, $\ldots$, $E_n$ of the system. Intuitively, the distribution of unperturbed energy eigenvalues $H_{11}$, $\ldots$, $H_{nn}$, must therefore, loosely speaking, be closer to each other than the distribution of true energy eigenvalues $E_1$, $\ldots$, $E_n$.

V) Finally, let us mention that eigenvalue repulsion plays an important role in random matrix theory. Integrating out "angular" d.o.f. leads to a Vandermonde measure factor

$$\prod_{1\leq i<j \leq n} |E_j-E_i|^{\beta},$$

so that the partition function ${\cal Z}$ favors that the eigenvalues $E_1$, $\ldots$, $E_n$ are different. Here the power $\beta$ is traditionally $1$, $2$, or $4$, depending on the random matrix ensemble. For Hermitian matrices $\beta=2$. See also this post.

• More or less related is that the (imaginary part of the) zeros of the Riemann zeta function display this repulsion phenomenon thus the conjecture that the zeros correspond to the eigenvalues of a Hermitian operator – Alfred Centauri Jul 16 '12 at 23:45
• For explanation IV), couldn't you invert the argument, pretending that you knew the eigenstates of the full Hamiltonian $H=H^{(0)}+V$ and treat $H^{(0)}=H-V$ perturbatively, and conclude that the eigenvalues of $H^{(0)}$ are more spread out that for $H$? – ChickenGod Mar 30 '13 at 1:32
• @ChickenGod: Nice question! Note that we have also used the Ansatz $\langle i^{(0)}|V|i^{(0)}\rangle=0$ to deduce that the eigenvalues of $H$ are more spread that the eigenvalues of $H^{(0)}$. Similarly, we could have used an "opposite" Ansatz $\langle i|V|i\rangle=0$ to deduce that the eigenvalues of $H^{(0)}$ are more spread that the eigenvalues of $H$. However, the latter "opposite" Ansatz is not natural, as it assume knowledge of the full system that we don't have. Thus the situation $H \leftrightarrow H^{(0)}$ is not symmetric, and no paradox arises. – Qmechanic Mar 31 '13 at 10:07

The true reason is the "perturbation" operator $\hat{V}$ being accidentally Hermitian (self-conjugate) in the linear space of non-perturbed Hamiltonian $\hat{H}_0$ in many practical cases: $V_{mn}=V_{nm}^*.$

Generally, a perturbation theory develops a function $f$ in a Taylor series in powers of a small parameter, let us say, $\varepsilon$, and this function is not obliged to have its second derivative of a negative sign at the expansion point $\varepsilon = 0$. The function $f(\varepsilon)$ may be concave or convex at the expansion point.

In terms of operators, the total Hamiltonian $\hat{H}=\hat{H}_0+\hat{V}$ should be Hermitian in the linear space of its exacts eigenfunctions $\varphi_n$, not in the linear space of the non perturbed Hamiltonian $\hat{H}_0$. So, the perturbation operator $\hat{V}$ may be non Hermitian in the space of non perturbed eigenfunctions $\varphi_n^{(0)}$. Then, the second-order correction may take a positive sign.

I encountered such an example in my practice. See it here, Chapter 2.1.