This is really confusing me, and I'm still only on time independent. Can anyone recommend a good text? Suppose I wanted to find the first order correction to the non degenerate energy level \(E_0\) when a Hamiltonian \(H_0\) is perturbed by the addition of \(\lambda H'\). I believe the way to go about it is to first assume that the eigenvalues and eigenfunctions of \(H = H_0 + \lambda H' \) can be expanded in terms of the perturbation parameter \(\lambda\). This seems like quite a big assumption, but if we run with it we get: \( E_n = \sum_{j=0}^{\infty} \lambda ^j E_n^{(j)} \) \( \mid \psi _n \rangle = \sum_{j=0}^{\infty} \lambda ^j \mid \psi_n ^{j} \rangle \) Subbing these into the time independent Schroedinger Equation we get \( (H_0 + \lambda H')(\mid \psi_n^{(0)} \rangle + \lambda \mid \psi_n^{(1)} \rangle + \ldots ) = (E_n^{(0)} + \lambda E_n^{(1)} + \ldots ) (\mid \psi_n^{(0)} \rangle + \lambda \mid \psi_n^{(1)} \rangle + \ldots ) \) Comparing coefficients of lambda we get: \( H_0 \mid \psi_n^{(1)} \rangle + H' \mid \psi_n^{(0)} \rangle = E_n^{(0)} \mid \psi_n^{(1)} \rangle + E_n^{(1)} \mid \psi_n^{(0)} \rangle \) Pre multiplying both sides by \( \langle \psi_n^{(0)} \mid \) and using orthonormality of our psi-s, we arrive at: \( E_n^{(1)} = \langle \psi_n^{(0)} \mid H' \mid psi_n^{(0)} \rangle \) Which is just the perturbation averaged over the corresponding unperturbed state of the system. Is this correct? Is this \( E_n^{(1)} \) what is known as the "First order correction to the energy" ? If so, how would I go about showing that if \( \phi = H' \psi_0\) is an eigenfunction of H_0 with energy not equal to \( E_0 \), then the first order energy correction vanishes, and that the first order correction to the wave function is proportional to \( \phi \)? Sorry if my notation is a bit off, any help would be much appreciated.
These lecture notes from when I did this stuff are pretty good, covering the various forms of this problem. Yes, its an assumption which isn't always true and this is why perturbation theory is only part of quantum mechanics but non-perturbative methods are pains for researchers, never mind QM students. Its the overlap of the perturbed state with the unperturbed state. There's no averaging over since you aren't doing any summations or integrals. I'm not sure how to explain it too much without just giving the answer. Eigenstates are orthogonal, a fact which follows from the orthogonality of the eigenvectors of any Hermitian matrix, which the representation of H would be. Thus if two eigenstates have different eigenvalues they cannot be the same eigenstate (up to proportionality etc) and thus are orthogonal. You need to put this argument into algebra.
Those lecture notes are awesome. They pretty much cover the first 3/4 of my course perfectly! I see how stupid I was being in that last part of my OP, it's all very obvious now. Do you know of any examples where the assumption isn't true? Thanks for the help, very much appreciated.
In quantum field theory you do perturbation theory in the coupling constant, so if the theory is strongly coupled then perturbation theory doesn't work. Two important QFT's are QED, the theory of the electromagnetic force coupled to charged particles and QCD, the theory of quarks and the strong interaction. QED is weakly coupled and PT works extremely well. QCD (as you may have guessed) is strongly coupled at low energies so PT fails. To get anything out of QCD at low energies you have to find non perturbative techniques and that typically means lattice QCD, where big computers get used to solve the equations of QCD numerically. Those sorts of techniques are no good for time dependent problems though.
Though I can't think of a case off the top of my head exponential functions appear all over the place in quantum mechanics and the function \(e^{-\frac{1}{x^{2}}}\) is non-perturbative, its Taylor expansion is zero so a Hamiltionian might give a Schrodinger equation with solution of that form where 'x' is some perturbative parameter. I can't be arsed to construct one, if it does exist, though....
I should add that there's plenty of non-perturbative things known in classical and quantum theories. Solitons and instantons are the prime examlpes. Solitons exist in wave mechanics too and are often such that the sum of two soliton solutions is not a soliton solution as they are governed by non-linear equations (such as the KdV equation). Unfortunately the majority of stuff on Wiki, particularly the QFT instanton section, is going to be meaningless to someone who hasn't done a fair chunk of QFT and gauge theory. They are not pleasant....
If you want to prove the last equation, one way to do it is to consider the following object: \(<\psi_n^0|H_0H'|\psi_n^0>\) Since \(H_0\) is Hermitian, it can act either to the left or to the right. Operating to the left, you get \(E_0<\psi_0|H'|\psi_0>\). Acting to the right, it follows from your initial assumptions that you get something slightly different. There's only one way this can be achieved, I'll leave it to you to fill in the details.
Thanks for the the help guys. I managed to complete the problem, seems I was overcomplicating things and it was rather simple in the end! My tutor pointed me to the book Lectures on Quantum Mechanics, G Baym for a small discussion on where perturbation theory will break down. It then also gives the example of the energy of the Cooper pair; after lots of horrible maths, the binding energy per electron is given by a highly non analytic function in V, which clearly cannot be expanded as a power series.
