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While learning time-independent perturbation theory, there was something that I couldn't understand, and it has to do something with order of $\lambda\\$

For example, when deriving the equation for time-independent perturbation of energy eigenstates, we assume that the hamiltonian can be expressed as a combination of the unperturbed and perturbed hamiltonians. i.e. $H=H_0 +\lambda\ H_1 $. Along with that equation, there comes this condition that $H'=\lambda\ H_1$ and that $\lambda\ <<1$ (which does seem to make sense. I mean, you would want the perturbing part to be small)

However, some problems arise when I also did the similar things to eigen states. The author assumes that

$|n>=|n^0 > +\lambda\ |n^1> + \lambda\ ^2 |n^2> + \cdots$

this (somewhat) seems to make sense. I mean, if we can expand almost any function using the Taylor expansion, why not do it to states too?

But what I can't understand is when we put those two together in the Energy Eigenvalue equation.

$H|n>=E_n |n>$

$(H_0 +\lambda\ H_1)(|n^0 > +\lambda\ |n^1> + \lambda\ ^2 |n^2> + \cdots)=(E_n^1+\lambda E_n^2 +\cdots )(|n^0 > +\lambda\ |n^1> + \lambda\ ^2 |n^2> + \cdots)$

Then, we collect the terms in powers of $\lambda $ and then say that since the equation should be true no matter what value of $\lambda$ we give, the left side and the right side are equivalent, and therefore the coefficient of the $\lambda$ on the right side equals the coeffient of the same thing on the left side etc etc.

However, why does the equation have to hold for every $\lambda$? I can't seem to intuitively understand why.

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First of all, the series expansion of new wave function in terms of unperturbed eigenstates is not a taylor expansion. It's simply telling that, for small values of $\lambda$ perturbed wave function belongs to the same hilbert space of unperturbed one , hence it can be written in the basis of old eigenstate vectors.

Then, you just compare both sides of the equation at each order of $\lambda$. This does not mean $\lambda$ takes a large value so that perturbation becomes invalid. Often $\lambda$ is somewhat like undetermined coupling (can be determined from experiment) between some operators. In that spirit, it is mentioned that, it can be anything.

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  • $\begingroup$ Thank you for your answer. I'm sorry but I still don't get it. I don't know what undetermined coupling is.. Could you use another example or method to explain it to me? $\endgroup$ – Danny Han Dec 3 '19 at 14:16
  • $\begingroup$ Take perturbation due to some electric field, H' = e E x . Here $\lambda$ is electric charge 'e', which is only fixed by doing experiment. You would do perturbation in the order of e,e^2, so on. Then you have to compare perturbation series result up to some order of 'e' with experiment to get the value of 'e'. $\endgroup$ – DEEP GHOSH Dec 4 '19 at 4:34
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When the operators are matrices in finite dimensional spaces, it can be shown that the eigenvalues and eigenvectors vary analytically with $\lambda$ (near $\lambda = 0$ in the complex plane). This does not generally hold in infinite dimensional spaces when the potential is unbounded (so there is no ground state when $\lambda \le 0$). Physics textbooks pretend that it works the same way in general. In practice it does work for "small" $\lambda$ in many situations.

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