# Why does the unperturbed stationary state energy level could be the same as the perturbed?

In the book principle of quantum mechanics by Dirac, when analyzing the scattering problem, it assumes that unperturbed stationary state energy level is the same as the perturbed. It doesn't agree with my instinct, after perturbation, the energy level should a little higher. So why this assumption could be applicable in this situation?

Here I quote the original text in the book:

We shall now consider the calculation of scattering coefficients, taking first the case when there is no absorption and emission, which means that our unperturbed system has no closed stationary states. We may conveniently take this unperturbed system to be that for which there is no interaction between the scatterer and particle. Its Hamiltonian will thus be of the form $$E = H_s+W$$ where $$H_s$$ is that for the scatterer alone and W that for the particle alone, namely, with neglect of relativistic mechanics, $$W = \frac12m(p_x^2+p_y^2+p_z^2;)$$ The perturbing energy $$V$$ assumed small, will now be a function of the Cartesian coordinates of the particle x, y, z, and also, perhaps, of its momenta px, py , pz, together with dynamical variables describing the scatterer. Since we are now interested only in stationary states of the whole system, we use a perturbation method like that of§ 43. Our unperturbed system now' necessarily has a continuous range of energylevels, since it contains a free particle, and this gives rise to certain modifications in the perturbation method. The question of the change in the energy-levels caused by the perturbation, which was the main question of§ 43, no longer has a meaning, and the convention in § 43 of using the same number of primes to denote nearly equal eigenvalues of E and H now drops out. Again, the splitting of energylevels which we had in§ 43 when the unperturbed system is degenerate cannot now arise, since if the unperturbed system is degenerate the perturbed one, which must also have a continuous range of energy levels will also be degenerate to exactly the same extent. We again use the general scheme of equations developed at the beginning of§ 43, equations (1) to (4) there, but we now take our unperturbed stationary state forming the zero-order approximation to belong to an energy-level E' just equal to the energy-level H' of our perturbed stationary state.

For one, consider the ground state of the 1D harmonic oscillator, $$\phi_0(x)=Ae^{\frac{-m\omega x^2}{2\hbar}}$$ and a perturbing electric field $$V(x)=Fx$$. Then the first-order correction to the ground state is $$E_0^{(1)} = \langle\phi_0|V|\phi_0\rangle = F|A|^2 \int\limits_{-\infty}^\infty dr \; x e^{-\frac{m\omega x^2}{\hbar}} = 0$$ where the integral is $$0$$ because it is an odd function over symmetric limits of integration. Therefore, the lowest-order nonzero correction to the energy comes in at second order (at least) in perturbation theory.