We have the following wavefunction for the hydrogen atom:


where $a$ is the Bohr radius.

Question: How can I express the wavefunction above as a linear combination of the eigenstates of the hydrogen atom's Hamiltonian? In other words, we need to express the above in terms of a linear combination of

$$\psi_{nlm}=\sqrt{\left(\frac{2}{na}\right)^3 \frac{(n-l-1)!}{2n[(n+l)!]^3}}e^{-r/na}\left(\frac{2r}{na}\right)[L_{n-l-1}^{2l+1}(2r/na)]Y_l^m(\theta,\phi)$$

For some $n$, $l$, and $m$'s. How can I find such a linear combination without just a simple guess and check; i.e., what is a systematic way to find the linear combination?

  • $\begingroup$ Hint: orthonormalization relationships. $\endgroup$ – Jorge Lavín Oct 24 '13 at 17:59
  • $\begingroup$ Apparently you will have just one $\psi_{nlm}$ in your linear combination, and I guess with $n=2$, $m=1$ and you must see what is the correct value of $l$ for $\psi_{nlm}$ match your $\psi (r,\theta,\phi)$ $\endgroup$ – Ana S. H. Oct 24 '13 at 18:08

Suppose you have some wavefunction $\Psi$ that is a linear combination of eigenfunctions, $\psi$:

$$ \Psi = a_1\psi_1 + a_2\psi_2 + a_3\psi_3 + ... $$

You know the eigenfunctions are orthonormal, so $\langle\psi_i|\psi_j\rangle$ is zero if $i \ne j$ and 1 if $i = j$. Suppose you compute $\langle\psi_i|\Psi\rangle$:

$$\begin{align} \langle\psi_i|\Psi\rangle &= a_1\langle\psi_i|\psi_1\rangle + a_2\langle\psi_i|\psi_2\rangle + ... + a_i\langle\psi_i|\psi_i\rangle + ... \\ &= a_i \end{align}$$

because all the terms are zero except for $\langle\psi_i|\psi_i\rangle$.

So to calculate the coefficient for each eigenfunction just calculate $\langle\psi_i|\Psi\rangle$ for each eigenfunction.

This will always work, but computing the integrals can be a fussy business keeping track of all those prefactors. Physicists, being basically lazy, frequently look for easy solutions to problems such as looking at a table of hydrogen wavefunctions. At a quick glance I'd guess your wavefunction is a sum of (2,1,1) and (2,1,-1).


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