2
$\begingroup$

So I am studying non-degenerate time independent perturbation theory and I came across the derivation of the first order correction to the wavefunction. So the notes given to me affirm that the each "perturbed" part of the wavefunction can be expanded into: $$\psi^{(k)}_n=\sum_{k} c^{(k)}_m \psi^{(0)}_m$$ where $\psi^{0}_m$ are the unperturbed eigenfunctions. So for example the first order correction to the wavefunction would be: $$\psi^{(1)}_n=\sum_{m \neq n} c^{(1)}_m \psi^{(0)}_m$$

I don't understand why this can be done in the first place, what principle does it follow from? Why can the basis set of the unperturbed eigenfunctions be used to define the perturbed corrections? Also why does $m \neq n$ necessary in this definition (I am aware that the inner product $$\langle\psi^{(0)}_m|\psi^{(1)}_n \rangle = \frac{\langle\psi^{(0)}_m|H'|\psi^{(1)}_n \rangle}{E^{(0)}_n - E^{(0)}_m}$$ implies that this is valid only for $n \neq m$ as the denominator would be otherwise zero).

$\endgroup$

2 Answers 2

2
$\begingroup$

The eigenfunctions of the unperturbed Hamiltonian form a complete basis, so any function can be expressed as a linear combination of those eigenfunctions. This is how you obtain your first formula.

Your second formula is not a definition, it is a result of calculation in perturbation theory (see, e.g., here).

$\endgroup$
6
  • $\begingroup$ Just for a clarification what is meant by "complete basis" here (can there be incomplete basis)? $\endgroup$
    – user63248
    Commented Nov 25, 2018 at 17:54
  • 1
    $\begingroup$ @daljit97 "complete" means any function can be expanded in terms of this basis. An example of an incomplete set would be the unit vectors $\hat x$ and $\hat y$ in 3D space: vectors of the type $\vec A=\hat z A$ cannot be expanded on this set. It is a result from Sturm-Liouville theory that the eigenfunctions of any linear hermitian operators form a complete set (barring repeated eigenvalues and technical issues distinguishing hermitian from self-adjoint operators). $\endgroup$ Commented Nov 25, 2018 at 18:11
  • $\begingroup$ I think the completeness means that any function that solves the Schrodinger's equation with the unperturbed Hamiltonian can be expanded in that basis, it certainly does not mean any arbitrary function in the Hilbert space... So the expansion is really an assumption, and in some cases perturbation can introduce new phenomenon ill-described in the basis of the original problem (consider perturbing a harmonic oscillator by a cubic term to study anharmonicity, then we miss all these unbounded states) $\endgroup$
    – Macrophage
    Commented Dec 12, 2020 at 23:54
  • $\begingroup$ @Macrophage : My understanding is different: the functions of the basis do not depend on the time, and any function in 3D can be expanded in terms of the basis. Note that the perturbation theory considers the perturbation series both for the wave functions and for the energies. $\endgroup$
    – akhmeteli
    Commented Dec 13, 2020 at 1:29
  • $\begingroup$ @akhmeteli I wasn't talking about time dependence here. Also, perturbation theory certainly deals with both perturbed functions and energy corrections as you said, but I think that's again irrelevant to our discussion about what completeness means. In my example, the quantum harmonic oscillator only admits bound state solutions. If you perturb by adding in a cubic term, I think you will get unbounded states that cannot be expressed as a linear combination of the bounded basis functions. $\endgroup$
    – Macrophage
    Commented Dec 13, 2020 at 1:44
1
$\begingroup$

The unperturbed solutions form a complete set so that any function can be expanded in terms of functions in this set, just like any function on the sphere can be expanded in terms of $Y_{\ell,m}(\theta,\varphi)$ or any periodic function function on the line can be expanded in terms of $\cos(m x)$ and $\sin (n x)$.

The real issue is the practicality of using the unperturbed solutions: after all, to find some perturbed radial solution, you could choose to expand using the unperturbed radial functions for the spherical oscillator or the Coulomb problem. Physically however, it makes sense to use the unperturbed solutions: since the perturbed ones ought to be “close” to the unperturbed ones, one can expect the series to converge quickly.

Choosing the unperturbed solutions as a starting point is a bit like asking what is the best starting point to evaluate $\sqrt{17}$: you could start by writing $$ \sqrt{17}=\sqrt{16+1}=4\sqrt{1+1/16}\approx 4\left(1+\frac{1}{32}\right) =4+\frac{1}{8}=4.125\tag{1} $$ or you would write $$ \sqrt{17}=\sqrt{25-8}\approx 5\left(1-\frac{8}{50}\right)=5-\frac{8}{10}=4.2 \, .\tag{2} $$ Obviously (1) is a “better” zero’th order approximation (at least to first order) because $\sqrt{16}$ is clearly a better approximation to $\sqrt{17}$ than would be $\sqrt{25}$.

$\endgroup$
4
  • $\begingroup$ Hmm ok, could you explain what in $\psi^{(1)}_n=\sum_{m \neq n} c^{(1)}_m \psi^{(0)}_m$ is the significance of the $(1)$ as the superscript of the $c_m$? Does it mean that $c_m$ are unique for each order of correction? $\endgroup$
    – user63248
    Commented Nov 25, 2018 at 17:58
  • 1
    $\begingroup$ The first order correction $\psi_n^{(1)}$ to the state $n$ - which is a deviation from the unperturbed state $\psi_n^{(0)}$, will not be proportional to $\psi_n^{(0)}$ itself: the part proportional to $\psi_n^{(0)}$ itself can be absorbed in the normalization at the end. In $\psi_r^{(k)}$, $k$ denotes the $k$-order correction to the eigenstate number $r$, and yes the $c_m^{(1)}$ are unique to each order. $\endgroup$ Commented Nov 25, 2018 at 18:08
  • $\begingroup$ So are $c_m$ not numbers? Because if they are, shouldn't that imply proportionality? $\endgroup$
    – user63248
    Commented Nov 28, 2018 at 0:37
  • $\begingroup$ @daljit97 they are expansion coefficients so yes numbers. Your notation is a bit bad because you should have something like $c_{nm}^{(k)}$ to denote the coefficient of $\psi_m$ in $\psi_n$ at order $(k)$ but otherwise they are numbers. See en.m.wikipedia.org/wiki/Perturbation_theory_(quantum_mechanics) $\endgroup$ Commented Nov 28, 2018 at 0:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.