Perturbation Method#

What you need to know

  • Perturbation method, attempts to solve analytically intractable problem by idintifying an exactly sovable part and a “small” pertrubation. The method is similair in spirit to taylor expansion of functions familiar from calculus and is the single most important method of solving problems in quantum mechanics.

  • Application of pertrubation theory proceeds in two steps. Step one identify solvable part and perturbation. Part two expand energy and eigenfunctions as series of corrections of increasing order. In particle first and second order corrections to energy suffice.

Time independent perturbations#

We have hamiltonian H^0 for some exactly solvable problem think particle in a box, harmonic oscilator, etc:

H^0n0=En0n0
  • Note the 0 superscript: It indicates exactly solvable hamitlonian, eigenfunctions and eigenvalues. The n0 is the eigenfunction corepsonding to the nth eigenvalue En0. Now we perturb the hamitlonian by adding a “small” pertrubation H1^. Where λ turns perturbation on λ=1 and off λ=0.

H^=H^0+λH^1

  • The objective of perturbation theory is to solve the following problem with new hamiltonian expressed entirely in terms of eigenvalues and eigenfunctions of exactly solvable problems.

H^n=Enn

It’s just like Taylor expansions!#

We assume that eigenvalues and eigenfunctions can be expanded in power series in the parameter λ to be set to 1 in the end.

En=En0+λEn1+λ2En2+...
n=∣n0+λn1+λ2n2...
(H^0+λH^1)(n0+λn1)=(En0+λEn1+λ2En2)(n0+λn1)
H^0n0+λ(H^1n0+H^0n1)+λ2H^1n1==En0n0+λ(En1n0+En0n1)+λ2(En1n1+En2n0)

Petrubation equations of order 0, 1 and 2#

Opening the brackets and collecting different orders of λ we have 0, 1 and 2nd order perturbation equations:

H^0n0=En0n0
H^0n1+H1^n0=En0n1+En1n0
H1^n1=En1n1+En2n0
  • Note how the sum of upstairs index determines the order of perturbation expansion

  • Note that 0 order is just the exact solution.

  • Note that hamitonian only has first order expansion while eigenfunctions and eigenvalues are expanded to infinite terms. Usually going to second order is enough for most problems.

Fixing the normalization#

If we have normalization the zero order eigenfunctions, then unperturbed eigenfunctions will be orthogonal to all higher order eigenfunctions:

n0n0=1
n0n=n0n0+λn0n1=1+0+...
n0nk=0k=1,2,..

1st order correction to energy En1#

We multiply first order pertrubation equation by n0. The first terms on the right is zero becasue of rothogonality. The first term on left is alos zero because of orhtogonality and hermitian property : n0H^0n1=n1H^0n0=En0n1n0=0

H^0n1+H1^n0=En0n1+En1n0
n0H^0n1+n0H1^n0=En0n0n1+En1n0n0
En1=n0H^1n0

  • We have obtained the central result of pertrubation theory: The 1st order correction to energy En1

En=En0+En1=En0+n0H^1n0
  • Note how this expression is different from expectation expressions we have seen before. Here the eigenfunctions of H^0 sandwich the the pertrubation hamitlonian H^1 . The two hamitlonians in general do not share eigenfunctions!

1st order correction to eigenfunction n1#

We express unknown first order eigenfunctions n1 in terms of known eigenfunctions k0 which form complete basis set. Another conseuqence of hermitian operators.

n1=knckk0
  • The coefficients are ck=k0n1 .

  • Becasue of orthogonality cn=n0n1=0 therefore we have indicates kn condition in the sum.


Inserting the expanstion of n1 and this taking dot product with bra k0 we find the coefficients of expansion:

H^0n1+H1^n0=En0n1+En1n0
H^0knckk0+H1^n0=En0knckk0+En1n0
ckEk0+k0H^1n0=ckEn0
ck=k0H^1n0En0Ek0=HnkEn0Ek0

n1=knckk0=knHnkEn0Ek0k0
  • We we have intduced convenient notaation for matrix elements of pertrubed hamitlonian:

2nd order correction. Obtaining expression for En2#

H0^n2+H1^n1=En0n2+En1n1+En2n0
n0H0^n2+n0H1^n1=E0n0n2+En1n0n1+En2n0n0
En2=n0H^1n1
  • Once again wetook dot product with bra n0. First term on the left. is zero (hermitian+orthogonality). The firs two terms on the right are zero due to orthogonality.

  • We are not done yet, the expression contains eigenfunction n1 which we need to expres in terms of known solutions n0

2nd order correction. Obtaining expression for n1#

En2=n0H^1n1=knckn0H^1k0=knckHnk
En2=knHnk2En0Ek0
E=En0+n0H1n0+knHnk2En0Ek0
  • The energy in the denominator is the difference between energy of a given state En from all other states Ek with k being the summation index.

  • If the matrix elements of H^1 are of comparable magnitude the neighbouring levels make larger contributions that distance levels.

Ground state energy perturbations#

Let us write second order correction explicitely for the ground state for some exactly solvable hamiltonian H0^ pertrubed by H1^

En=En0+Hnn+knHnk2En0Ek0
E0=E00+H00+H012E00E10+H022E00E20+H032E00E30+...
  • Notice that for the ground state the second order correction thereofre will always be negative because ΔE0k=E0Ek<0

Applications#

Example-1: Magnetic field#

Hydrogen atom in magnetic field problem can be seen as as a hamitonian of H atom to which we have added a small pertrubation in the form of interation with magnetic field.

H^=H^0+e2meBL^z=H^0+H^1
  • Using 1st order pertrubation expression we can calculate for instance how ground state energy will be perturbed. Where on right hand side we define RH as Rydberg’s and βB as Bohr’s magneton, both constants.

E0=E00+0H^10=RHn2+mlβBB

  • In a similiar way the effect of spin orbit coupling (LS)

H^=H^0+ASOL^S^
E=E0+ASO0L^S^0

Example-2: Particle in a box#

Estimate the energy of the ground-state and first excited-state wavefunction within first-order perturbation theory of a system with the following potential energy:

V(x)=V00x
V(x)=+x, x

This problem can be seen as a particle in a box pertrubed by the presence of a potential energy V0

En1=nV0n=V02L0Lsin2nπxLdx=V0

Thus we find that energy level of PIB are pertrubed by a constant shift up term:

En=En0+En1n2h28mL2+V0

Exaple-3 Unharmonic oscillator#

Unharmonic oscillator problem can be seen as a problem fo harmonic oscillator + pertrubation in the form of unharmonic term:

H^=K^+kx22+γx3=H^0+γx3
  • Using first order pertrubation we find an interesting result after evaluating the integral to by using simple symmetry arguments.

En1=0γx3n=even/oddoddeven/odd=0

  • But energy levels surely must experience change since we added a new term to hamitlonian. To see the change we must therefore turn to second order and use ground state as an example

E02=k00γx3k2E00Ek0=0γx312E00E10+nγx332E00E30+...
E0=E0+E1+E2=ω2+0+H012ω+H0322ω+...

Thus we see that only terms odd terms of the sum contribute. The matrix elements need to be evaluated explicitly using Hermite polynomials.