# Derivative with respect to perturbation in QM

The book "Introduction to Computational Chemistry" by Frank Jensen claims the following (Eq. 10.35): Let $H(\lambda) = H_0 + \lambda V$ be a Hamiltonian parametrized by a perturbation strength of $\lambda$ and $\Psi_i^{(0)}$ , a complete set of eigenfunctions of $H_0$. Then we have $$\left \lvert \frac{d\Psi_i(\lambda)}{d\lambda}\bigg|_{\lambda=0} \right\rangle = \sum_{k \neq i} \frac{\langle \Psi_k^{(0)} | V | \Psi_i^{(0)} \rangle}{E_i - E_k} |\Psi_k^{(0)}\rangle.$$ Now if $\Psi_i(\lambda)$ is analytical, then from the standard derivations of time-independent perturbation theory, the above is true. But I have heard that often perturbation theory is not convergent (e.g. quartic perturbation to harmonic oscillator). Does the above still make sense somehow (e.g. that the divergences only occur in higher-order terms) or is it plainly wrong in general? Maybe somebody will have an example where both sides of the equation can be determined exactly (note that both sides of the equation are well-defined quantities).

## 2 Answers

Both Sides of the Equation can be determined exactly for the harmonic oscillator, $V_h=w^2x^2$, with linear perturbation, $V_p = \lambda x$. You can solve it exactly and second order perturbation gets you the exact energy shift.

To solve it exactly complete the square of the total potential $V_T=w^2x^2+ \lambda x$. Doing this will get you $V_T = w^2(x+\lambda /2)^2 - w^2 \lambda^2/4$ which is just a shifted harmonic oscillator. To solve it perturbative, rewrite the perturbed potential with raising and lowering operators:

$$\lambda x = \lambda \sqrt{2 \hbar\over{mw}} (a + a^\dagger)$$

So \begin{align}\left \lvert \frac{d\Psi_i(\lambda)}{d\lambda}\bigg|_{\lambda=0} \right\rangle & = \sum_{k \neq i} \frac{\left\langle \Psi_k^{(0)} | \lambda \sqrt{2 \hbar\over{mw}} (a + a^\dagger) | \Psi_i^{(0)} \right\rangle}{E_i - E_k} \left|\Psi_k^{(0)}\right\rangle\\ &= \frac{\left\langle \Psi_{i-1}^{(0)} | \lambda \sqrt{2 \hbar\over{mw}} a | \Psi_i^{(0)} \right\rangle}{E_i - E_{i-1}} \left|\Psi_{i-1}^{(0)}\right\rangle +\frac{\left\langle \Psi_{i+1}^{(0)} | \lambda \sqrt{2 \hbar\over{mw}} a ^ {\dagger} | \Psi_i^{(0)} \right\rangle}{E_i - E_{i+1}} \left|\Psi_{i+1}^{(0)}\right\rangle\end{align}

At higher orders, the raising and lowering operators will mix more and more states, but there will only ever be a finite sum so at any given order there will be no divergence since $E_k$ isn't equal to $E_i$. If you calculate energy shifts, you should get 0 at first order and $-w^2 {\lambda}^2/4$ as suggested by the exact solution.

Only time you get divergences is if $\dfrac{\lambda}{E_k-E_i} >1$ and thus your series as a whole diverges(thus truly invalidating perturbation theory). Or if your eigenstates are degenerate (i.e. $E_i =E_k$) and this divergence can be avoided by a proper choice of basis for the degenerate states. The proper choice is one which diagonalizes V. This way the degenerate states that would have a 0 in the denominator have a 0 in the numerator since the numerator is the off diagonal term in the potential: $$\left\langle \Psi_{k}^{(0)} | V | \Psi_i^{(0)} \right\rangle=0$$

This is most likely what you hear when there are divergences. This is very often talked about in field theory, where this is a more serious problem because your eigenstates are infinitely degenerate, but we can still make sense of perturbation theory using tools like Renormalization Group.

So to answer your question: Yes, you can still make sense of it as long as the divergence is due to degeneracy, not from being outside the realm of perturbation theory.

It's true that perturbation theory is usually not convergent (a different issue from the question of whether the individual terms is the series are finite, which is what Shane's old answer adresses), but the derivatives of quantities at $$\lambda=0$$ are given correctly by the standard formulae.