I'm studying perturbation theory in the context of quantum mechanics. My lecture notes say that in order to calculate the first-order correction of eigenfunction $\psi_n$, that is $\psi_n^{(1)}$, I shall use the formula:

\begin{equation} \psi_n^{(1)}=\sum_{m,m\ne n} \frac{V_{mn}}{E_n^{(0)} - E_m^{(0)}}\psi_m^{(0)} \end{equation}

where $E_n^{(0)}$ is the zeroth-order correction (i.e., the unperturbed), and $V_{mn}$ are given by:

\begin{equation} V_{mn} = \left(\psi_m^{(0)}, V\psi_n^{(0)}\right) \end{equation}

and V is the "weak" potential that represents the physical disturbance to our system. $(x,y)$ represents inner product.

My question is: What values does $m$ take in the above summation ? Say I'd like to calculate the first-order corrected $\psi_1$, that is to calculate $\psi_1^{(1)}$. What values would $m$ assume ? Infinite values except for $1$ ? Values $0,\ldots,n-1$ ?


1 Answer 1


Yes, the summation is taking over all possible integer value of $m=0,1,2,...$ except $m=n$. It can be easily seen by following the derivation of the first order perturbation theory.

In your example $\psi_1^{(1)}$, it is sum over $m=0,2,3,4,...$. Note that sometimes the index start from 1 instead of 0 such as infinite square well, then you should skip 0.

  • $\begingroup$ Thanks @hwlau! So is it an infinite sum, right ? Due to the fact that one writes $\psi_n^{(1)}$ in the basis of unperturbed eigenfunctions $\psi_n^{(0)}$ which happen to be infinite ? $\endgroup$
    – stathisk
    Commented Dec 17, 2013 at 1:41
  • $\begingroup$ Yes, it is inifinite sum, and it could converge. If people want to mention it is not infinite like your writing above, they will put m<n instead $\endgroup$
    – unsym
    Commented Dec 17, 2013 at 1:42

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