Introduction to Perturbation Theory

1 Physics 342 Lecture 31 Introduction to Perturbation TheoryLecture 31 Physics 342 Quantum Mechanics IMonday, April 21st, 2008 The program of time-independent quantum mechanics is straightforward given a potentialV(x) (in one dimension, say), solve ~22m +V(x) =E ,( )for the eigenstates. These form a complete, orthogonal basis for all this adapted basis, generate generic initial configurations and timeevolve them according to (x,t) = j=1 j j(x)e iEj~t j= (x,0) j.( )At any given time, we have the probability density for the system, and cancalculate various physically measurable properties. If we actually performa measurement, the wavefunction takes on one of the eigenstates (of theoperator associated with the measurement), and returns the value of thephysical measurement for that state (part of our assumption).

2 The problem, as we have seen, is that solving ( ) for all but the simplestpotentials can be difficult. We turn now to the problem of approximatingsolutions our first (and only, at this stage) tool will be Perturbation technique is appropriate when we have a potentialV(x) that is closelyrelated to a simple (read solvable ) potential V(x). Perturbation PolynomialsBefore working on a full ODE like the time-independent Schr odinger equa-tion, let s get the basic arguments down for a polynomial equation, where1 of Perturbation POLYNOMIALSL ecture 31some of the issues are Distinct RootsConsider the roots of the polynomialax2+ x+c= 0,( )we know the solution here, just the quadratic formulax= 2 4ac2a.

3 ( )But suppose we didn t have/remember this. Further, suppose is itself asmall parameter, so that the form of ( ) is close to easily the solvableequation:ax2+c= 0( )which has roots:x= i idea behind Perturbation Theory is to attempt to solve ( ), given thesolution to ( ). Operationally, we take an ansatz forx:x=x0+ x1+ 2x2+..,( )and insert that into ( ). Note that an implicit assumption we are makinghere is that the coefficientsaandcare order one, and thatxitself is orderone (meaning that these quantities do not scale with ). Inputting our formgives:(ax20+c)+ (x0+ 2ax0x1) + 2(x1+ax21+ 2x0x2)+O( 3) = 0.

4 ( )Now for the simplifying trick we assume that the terms of different orderin do not talk to each other, that each order in must vanish separately 5 + = 0 implies that = 5 which is not small compared to 5. Using theindependence of order, the above gives us three equations that we can solveseparately to determine the set{x0,x1,x2}: 0:ax20+c= 0 1:x0+ 2ax0x1= 0 2:x1+ax21+ 2ax0x2= 0.( )2 of Perturbation POLYNOMIALSL ecture 31We can see how the = 0 equation ( ) plays a role here, it is the 0equation that starts off the process by allowing us to solve forx0. Noticethe cascade here, knowingx0= i ca, we can solve forx1(we don tactually needx0to findx1in the current case, but in general, we have ahierarchy of equations and perturbative dependence): i ca(1 + 2ax1) = 0 x1= 12a,( )and by knowingx0andx1, we can findx2: 12a+14a 2ia cax2= 0 x2= i8a2 ca.

5 ( )Putting it all together, we have, through order 2, the solution:x=x0+ x1+ 2x2= i ca 2a i 28a2 ca.( )Here it is easy to compare with ( ): Expanded in powers of via Taylorseries ( ) isx=12a( 4aci 1 e24ac) 12a( i 4ac(1 28ac))= i ca 2a i 28a2 ca.( )Here everything works out well. In some cases, we need to be more Degenerate RootsIn the preceding example, the roots of the unperturbed ( = 0) equationwhere separate, and we were effectively calculating corrections to each rootseparately. There is another case that has an analogue in our quantummechanical calculations suppose we had, for the unperturbed equation,the quadratic polynomial with degenerate roots:x2 2x+ 1 = 0,( )1 Think of what happens when we have x2+bx+c= 0, for of Perturbation FOR ODESL ecture 31for whichx= 1.

6 Now if we introduce a Perturbation , and define the per-turbed equation via:x2 2x+ (1 + ) = 0,( )and take, again,x=x0+ x1(this time, we will find only the first correc-tion), then:0 =(x20 2x0+ 1)+ (2x0x1 2x1 1) +O( 2)( )and our 0equation reproduces ( ), withx0= 1. But now, we haveno way to satisfy the equation, which becomes 1 = 0. What happened?Think of the assumption we ve made we wantxto be order unity, withcorrections coming at order , but this ignores the quadratic term, whichmakes no contribution to order, we have, apparently, gone out too far in without taking into account potential corrective terms.

7 Suppose we start,then, withx=x0+ x1.( )Then the perturbed equation becomes:(x20 2x0+ 1)+ x1(2x0 2) + (1 +x21)= 0( )and now the original solutionx0= 1 satisfies the equation, and we canmove on to the equation, where we learn thatx1= i, giving us twoperturbed solutions:x0= 1 i ,( )and splitting the degenerate root structure of the original equation ( ). Perturbation for ODEsThe same approach will work for ODEs, with similar caveats. Take theunperturbed equation: x(t) +x(t) = 0x(0) =A x(0) = 0,( )a harmonic oscillator that starts from rest. The solution isx(t) =Acos(t).Now suppose we want to solve x+x x= 0x(0) =A x(0) = 0,( )corresponding to a small frequency shift.

8 We make the usual ansatz:x(t) =x0(t) + x1(t)( )4 of Perturbation FOR EIGENVALUE PROBLEML ecture 31in order to generate the first order corrections. Then the ODE becomes:( x0+x0) + ( x1+x1 x0) +O( 2) = 0.( )The solution to the 0equation is justx(t) =Acos(t) as in the unperturbedcase. Our 1equation reads: x1+x1 x0= 0( )which looks like a driven harmonic oscillator, with driving forceAcos(t).We know how to solve this equation in general, but what should we doabout the boundary conditions? In this case, the full boundary conditionsare satisfied by thex0(t) solution, so we must have:x1(0) = 0 and x1(0) = the solution to the 1 ODE is:x1=12Ax0tsin(t)( )and our full solution isx(t) =Acos(t) +12A sin(t)t( )We can compare this with the Taylor expansion of the exact solution in thiscase:Acos( 1 t) Acos(t) +12 Asin(t)t.

9 ( )So to order , we have the correct Perturbation for Eigenvalue ProblemWe have seen how Perturbation Theory works, and what we need to doto get ODE solutions, the final element we need to consider to approachSchr odinger s equation perturbatively is to look at the Perturbation of theeigenvalue equation itself. The twist is that we are looking for both eigen-vectors and eigenvalues, and it is easiest to see how this will work out in thefinite matrix a symmetric real matrix,A=AT(so that we know the eigenvectorsare complete and can be made orthonormal) in IRN N. Suppose we knowthe eigenvectors and eigenvalues of this matrix:Axi= ixi,( )5 of Perturbation FOR EIGENVALUE PROBLEML ecture 31fori= 1 N, and we know that each eigenvalue has a distinct eigen-vector (no degeneracy).

10 We have constructed our eigenvectors so as to beorthonormal:xi xj= ij. Now we make a small Perturbation to the matrix,A A+ A, and we want to know how the eigenvalues and eigenvectorschange under this Perturbation . So introduce iand x, that are themselvesorder corrections:xi xi+ xi i i+ i,( )then the eigenvalue problem reads:(A+ A)(xi+ xi) =( i+ i)(xi+ xi).( )Expanding this, and keeping only those terms of order , we have:A xi+ Axi= i xi+ ixi.( )This seems like it will not be enough to determineboth iand xi. But wait,since xi IRN, it has a decomposition in terms of the set{xi}Ni=1: xi=N j=1 jxj,( )so that we can rewrite the first order equation asN j=1 j jxj+ Axi= iN j=1 jxj+ ixi.