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Perturbation Theory Help

  1. May 8, 2010 #1
    Perturbation Theory Help!!

    Hello physicsforums.com,

    The last two weeks of my nuclear engineering course covered a mathematical topic known as 'perturbation theory'. It was offered as a 'method to solve anything' with; the problem is, however, that nobody in my class understands it.

    Basic google searching has not yielded any great results, so I turn to the wise physicsforums.com community to perhaps help give new perspective or recommend some relatively easy to follow readings.

    Thanks in advance!
     
  2. jcsd
  3. May 9, 2010 #2

    HallsofIvy

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    Re: Perturbation Theory Help!!

    Well, it's a method to approximately solve anything! Specifically, it is a method to approximately solve non-linear equations, in particular functional equations like differential equations or integral equations. The "WKB" method used in quantum mechanics is a perturbation method.

    The basic idea of "perturbation" theory is to write the solution to your problem (typically, a differential equation or integral equation although it will work for other kinds of problems) as a power series, [itex]y= y_0+ \epsilon y_1+ \epsilon^2 y_2+ \cdot\cdot\cdot [/itex] where "[itex]\epsilon[/itex]" is some small number inherent in your problem. Write out both sides of your equation as power series in [itex]\epsilon[/itex] and set coefficients of the same powers of [itex]\epsilon[/itex] equal. The [itex]\epsilon^0[/itex] term gives the solution to the approximate linear problem, y_0, and the other equations will give solutions in terms of previous solutions- that is, [itex]y_1[/itex], [itex]y_2[/itex] in terms of [itex]y_0[/itex] and [itex]y_1[/itex], etc.

    Here's a trivial example: Imagine that we know how to solve equations of the form [itex]x^2= a[/itex] by just taking the square root but we don't know the "quadratic formula".

    Now, we want to solve the equation [itex]x^2+ \epsilon x- 4= 0[/itex] where [itex]\epsilon[/itex] is a very, very small (positive) number. We could argue that since [itex]\epsilon[/itex] is small, that equation is very nearly [itex]x^2= 4[/itex] which has solutions 2 and -2 and so our equation must have solutions very close to 2 and -2.

    Is that true? If it is, how could we prove it is true? And how could we use that information to get a better approximation to the true solution?

    Let [itex]x= x_0+ x_1\epsilon+ x_2\epsilon^2+ \cdot\cdot\cdot[/itex], a power series in [itex]\epsilon[/itex]. We will assume that [itex]\epsilon[/itex] is small enough that we can ignore [itex]\epsilon^3[/itex] (assuming that [itex]\epsilon[/itex] was small enough to ignore [itex]\epsilon^1[/itex] would give y_0 the linear solution).

    If [itex]x= x_0+ x_1\epsilon+ x_2\epsilon^2[/itex], then [itex]x^2= x_0^+ 2x_0x_1\epsilon+ 2x_0x_2\epsilon^2+ x_1^2\epsilon^2[/itex] where I have dropped the terms [itex]2x_1x_2\epsilon^3[/itex] and [itex]x_2^2\epsilon^4[/itex] since they are of higher than second dergee.

    [itex]x^2+\epsilon x- 4[/itex], then is [itex] x_0^2+ 2x_0x_1\epsilon+ 2x_0x_2\epsilon^2+ x_1^2\epsilon^2+ x_0\epsilon+ x_1\epsilon^2- 4[/itex] where I have dropped the term [itex]x_2\epsilon^3[/itex] from [itex]\epsilon x[/itex] as it is, again, of higher than degree 2.

    The equation becomes [itex](x_0^2- 4)+ (2x_0x_1+ x_0)\epsilon+ (2x_0x_2+ x_1^2)\epsilon^2= 0[/itex]. Equating corresponding components, we have [itex]x_0^2- 4= 0[/itex], [itex]2x_0x_1+ x_0= x_0(2x_1+ 1)= 0[/itex] and [itex]2x_0x_2+ x_2^2= 0[/itex].

    [itex]x_0^2- 4= 0[/itex] gives [itex]x_0= 2[/itex] or [itex]x_0= -2[/itex]. Since that is not 0, we can divide both sides of [itex]x_0(2x_1+ 1)= 0[/itex] by x_0 and get [itex]x_1= -\frac{1}{2}[/itex] for both values of [itex]x_0[/itex].

    If [itex]x_0= 2[/itex] and [itex]x_1= -1/2[/itex], then the third equation is [itex]4x_2+ \frac{1}{4}= 0[/itex] so [itex]x_2= -1/16[/itex].

    If [itex]x_0= -2[/itex] and [iterx]x_1= -1/2[/itex], then the third equation is [itex]-4x_2+ \frac{1}{4}= 0[/itex] so [itex]x_2= 1/16[/itex].

    That is, our two solutions are [itex]x= 2- (1/2)\epsilon- (1/16)\epsilon^2[/itex] and [itex]x= -2- (1/2)\epsilon+ (1/16)\epsilon^2[/itex].

    If, for example, [itex]\epsilon= .001[/itex], then those solutions are [itex]2- .0005- 0.0000000625= 1.9994999375[/itex] and [itex]-2- .0005+ 0.0000000625= -2.0004999375.

    We can use the quadratic formula (which we were pretending we did not know) to actually solve [itex]x^2+ .001x- 4= 0[/itex], getting [itex]x= (-.001\pm\sqrt{0.000001+ 16})/2[/quote] which gives x= 1.99950006 and -2.00050006 so what we got using "perturbation theory" was certainly better than 2 and -2 and also show that 2 and -2 are good first approximations.

    A equation of the form [itex]\epsilon x^2+ 2x- 4= 0[/itex] is a much harder problem. Here, just ignoring [itex]\epsilon[/itex], the equation becomes 2x- 4= 0 which has the single solution x= 2 while we expect a quadratic equation like this to have two solutions. For this problem we need "singular perturbation" which is a whole different story!
     
  4. May 14, 2010 #3
    Re: Perturbation Theory Help!!

    thanks for your help!

    I had an additional question to what you have explained. I have used the perturbation method for solving a set of non-linear differential equations where [tex]\epsilon[/tex] was not small, however the terms [tex]x_0,x_1,x_2[/tex]...etc. get progressively smaller.

    I have worked out the equations and they seem to match up quite accurately to the solution found by using ode45 in matlab (while having time-varying parameters in the function).

    I am trying to find some sort of reassurance that this is ok. Everything that I have found all suggests that epsilon must be small.
     
  5. Sep 9, 2011 #4
    Re: Perturbation Theory Help!!

    Many of the ab initio quantum chemistry methods use perturbation theory directly or are closely related methods. Møller–Plesset perturbation theory uses the difference between the Hartree–Fock Hamiltonian and the exact non-relativistic Hamiltonian as the perturbation. The zero-order energy is the sum of orbital energies. The first-order energy is the Hartree–Fock energy and electron correlation is included at second-order or higher. Calculations to second, third or fourth order are very common and the code is included in most ab initio quantum chemistry programs. A related but more accurate method is the coupled cluster method.
     
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