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3.1.1 Simple examples of perturbation theory

Probably the simplest example we can think of is an infinite square well with a low step half way across, so that V (x) = 0 for 0 < x < a2, V 0 for a2 < x < a and infinite elsewhere. We treat this as a perturbation on the flat-bottomed well, so H(1) = V 0 for a2 < x < a and zero elsewhere.

The ground-state unperturbed wavefunction is ψ0 (0) = 2 a sin πx a , with unperturbed energy E0 (0) = π22(2ma2). A “low” step will mean V 0 E0(0). Then we have

E0(1) = ψ0(0) |H(1) |ψ0(0) = 2 aa2aV 0 sin 2πx a dx = V 0 2

This problem can be solved semi-analytically; in both regions the solutions are sinusoids, but with wavenumbers k = 2mE and k = 2m(E V 0) respectively; satisfying the boundary conditions and matching the wavefunctions and derivatives at x = a2 gives the condition k cot (ka2) = k cot(ka2) which can be solved numerically for E. (You ought to try this, it will be good practice for later sections of the course.) Below the exact solution (green, dotted) and E0(0) + E 0(1) (blue) are plotted; we can see that they start to diverge when V 0 = 5 (everything is in units of 2(2ma2)).

PIC

We can also plot the exact wavefunctions for different step size, and see that for V 0 = 10 (the middle picture, well beyond the validity of first-order perturbation theory) it is significantly different from a simple sinusoid.

PIC

Another example is the harmonic oscillator, with a perturbing potential H(1) = λx2 . The states of the unperturbed oscillator are denoted n(0) with energies E0(0) = (n + 1 2)ω.

Recalling that in terms of creation and annihilation operators (see section A.4), x ̂ = (2mω)(â + â), with [â , â ] = 1, and so

En(1) = n(0) |H(1) |n(0) = λ 2mωn(0) |(â )2 + â2 + 2â â + 1|n(0) = λ mω2ω(n + 1 2)

The first-order change in the wavefunction is also easy to compute, as m(0) |H(1) |n(0) = 0 unless m = n ± 2. Thus

n(1) = mnm(0)|Ĥ(1)|n(0) En(0) Em(0) m(0) = λ 2mω (n + 1)(n + 2) 2ω (n + 2)(0) + n(n 1) 2ω (n 2)(0) andEn(2) = n(0) |Ĥ(1) |n(1) = mn m(0)|Ĥ(1)|n(0)2 En(0) Em(0) = λ 2mω2 (n + 1)(n + 2) 2ω + n(n 1) 2ω = 1 2 λ mω2 2ω(n + 1 2)

We can see a pattern emerging, and of course this is actually a soluble problem, as all that the perturbation has done is change the frequency. Defining ω = ω1 + 2λ(mω2 ), we see that the exact solution is

En = (n + 1 2)ω = (n + 1 2)ω 1 + λ mω2 1 2 λ mω2 2 +

in agreement with the perturbative calculation.

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