So I am looking at non-degenerate perturbation theory. The idea is that the perturbing term in the Hamiltonian is small so you somehow expand the energies and wave functions in this small term and collect orders. Now I did an exercise in which you apply perturbation theory to a system, which is solvable. You then show by Taylor expanding the analytical result of the energies that the first order perturbation term is equal to the first order term in the Taylor expansion. Should this be obvious? I know that the first order perturbation theory was derived based on expanding the energies in the small perturbing term but somehow I cannot see that it is exactly equivalent to simply calculating the first order term in the energy.

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    $\begingroup$ If you Google this "Introduction to regular perturbation theory" you will be able to download a pdf which highlights the differences, my apologies but my tablet won't copy links for some frustrating reason. Scroll to page 3 for the methods used $\endgroup$ – user140606 Jan 22 '17 at 20:32

Yes. Suppose $$ H=H_0+\epsilon H_1=\left(\begin{array}{cc}A & 0 \\ 0 & B\end{array}\right)+ \epsilon\left(\begin{array}{cc}a & b \\ b & c\end{array}\right) $$ where $a, b$ and $c$ are real for simplicity. You can easily work out that the exact eigenvalues are $$ \lambda_\pm = \frac{1}{2}\left(A + B +\epsilon (a+ c) \pm \sqrt{(A-B)^2+\epsilon(A-B)(a-c)+\epsilon^2((a-c)^2+4b^2)}\right)\, . $$ Expanding - say - the first eigenvalue in powers of small $\epsilon$ and assuming that A-B is "large enough" to factor it from the square root gives $$ A+ a \epsilon +\epsilon ^2 \frac{b^2 }{A-B} $$ which is just $E_1^{0}+\epsilon \langle 1\vert H_1\vert 1\rangle + \epsilon^2 \frac{\vert \langle 1\vert H_1\vert 2\rangle\vert^2}{E^{0}_1-E^{0}_2} $ as "predicted" by perturbation theory.

I presume what you did must be basically equivalent to the above. This is a good exercise because you can also see why the unperturbed energies must be such that $A\ne B$: if $A=B$ then $H_0$ is basically the unit matrix and the eigenvectors will be those of $H_1$ since any similarity transformation will not affect a multiple of the unit matrix. In addition, the case $A=B$ results in a simplification in the discriminant, which no longer contains $A$ or $B$, making the assumption on the difference $A-B$ invalid.

This can be generalized to any hermitian matrix. The difficulty is that it's not so easy (and in fact in general impossible) to write down a closed form for the eigenvalues, but the $2\times 2$ example shows that, if you could, you must recover term by term the results of perturbation theory.

After all, the eigenvalues cannot depend on how you calculate them, i.e. exactly first and then expand, expand first and then add the terms.

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  • $\begingroup$ Good answer! I think that the key point is the last sentence. $\endgroup$ – AccidentalFourierTransform Jan 22 '17 at 20:03

I was thinking, if we write the matrix m=2 by 2 matrix. Then the $\lambda H'$ (which was treated as unknown), could be thought as the linear combination of $\lambda(H^1+\lambda H^2 + ...)$.

In a sense $\lambda$ here was $x$ in Taylor expansion, and the sequence of $H^1+\lambda H^2 + ...$, if you thought of the each of four indices, they were actually "independent" of each other. Thus you practically got four independent sequence for the taylor expansion of all possible function in the four places in the matrix. Where the combination of $\psi^j_n$ and $E^j_n$ were the resulted function and energy states for each $\lambda^j$'s power. (Notice $\psi^j_n$ and $E^j_n$ was the sum of all taylor expansion for the solution of $H^i$ in each $j$ states and then a double sum range from $n=0$ to infinity.)

In conclusion, there was two sets of taylor expansion. One for $H^i$ where represented the combination of taylor expansions for $m^2$ number of places. Two was the taylor expansion of $\psi^{ij}$ and $E^{ij}$ for each $H^i$ operator.

Notice the limits of taylr expansion, which was taken care of by the physical assumption that the system won't blow up(singularities) or discontinuity, and smooth meant you could always get one.

Also, just to be more clear, I suspect that $E^j_n$ in griffiths was actually the sum of all $j$ states of all $E^{ij}_n$ where $E^{0j}_n$ the base case was excluded, so did the $\psi_n^{j}$.

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  • $\begingroup$ Correction, the $\psi^{ij}_n$ was expand $j$ in term of the complete set $\psi_n^0$. $\endgroup$ – J C May 6 '18 at 18:13

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