# How to get effective quantum numbers of a linear combination of $\rm H$-atom wavefunctions?

The convention for the Hydrogen atom's interpretation subject to the laws of quantum mechanics is that you can prove the quantization of $$|L|$$, $$L_z$$, and Energy through quantum numbers $$\ell$$, $$m_\ell$$, and $$n$$ respectively. You can check the wavefunction with some parameters as ($$n$$, $$\ell$$, $$m_\ell$$) based on the appropriate spherical harmonics (based on $$\ell$$ and $$m_\ell$$) and the radial solutions (based on $$n$$ and $$\ell$$). You can get $$|L|$$, $$L_z$$, and Energy as follows:

$$L_z=m_{\ell}\hbar$$ $$|L|=\sqrt{\ell(\ell+1)}\hbar$$ $$E=\frac{-13.6 eV}{n^2}$$

If you add different wavefunctions, say with form $$A_0(A_1 \Psi_{n_1,\ell_1,m_\ell,1} + A_2 \Psi_{n_2,\ell_2,m_\ell,2} + A_3 \Psi_{n_3,\ell_3,m_\ell,3} + ...),$$ I know that this should be a solution to the equation since it is just a linear combination of different solutions to the Hydrogen atom. How would I go about solving for the "effective quantum numbers" that this linear combination has? Does such a thing exist?

## 2 Answers

No, assuming the individual $$\Psi_{nlm}$$ terms are time-independent, then a linear combination like $$A_0(A_1 \Psi_{n_1,l_1,m_l,1} + A_2 \Psi_{n_2,l_2,m_{l,2}} + A_3 \Psi_{n_3,l_3,m_{l,3}} + ...)$$ is not a solution of the Schrödinger equation.

Since terms with different $$n$$ have different energies $$E_n$$, you need to account for the time-dependent phase factors $$e^{-iE_n t/\hbar}$$.

So for example, the linear combination $$A_0(A_1 \Psi_{n_1,l_1,m_{l,1}}\ e^{-iE_{n_1}t/\hbar} + A_2 \Psi_{n_2,l_2,m_{l,2}}\ e^{-iE_{n_2}t/\hbar} + A_3 \Psi_{n_3,l_3,m_{l,3}}\ e^{-iE_{n_3}t/\hbar} + ...)$$ would be a solution of the time-dependent Schrödinger equation $$i\hbar\frac{d}{dt}\Psi=H\Psi.$$

The quantum numbers $$n,l,m$$ label the different possible energy eigenstates. Moreover, $$n$$ labels pertains to the radial wavefunction, $$l$$ labels the magnitude of orbital angular momentum where $$m$$ labels one of its components.

In a general superposition, the state may no longer be an eigenstate of the Hamiltonian, angular momentum operator (and one projection operator of it). Thus there are no $$n,l,m$$. One can still calculate expectation values of these operators and get an “effective” $$n,l,m$$ just as with any probability distribution.

• I'm sorry if I come off as inexperienced (this is our current topic in class), but how would one go about calculating the expectation values of these quantum numbers? As far as I know, to get say <n> you'd calculate it with $\int^{+\infty}_{-\infty} \Psi^*n\Psi$ but I don't know if n is equivalent to some operator, or if what I've written makes sense at all. I'm again sorry for my inexperience. May 10, 2020 at 13:57
• @IanAngeloAragoza there’s no need to apologise when learning! And, yes. That in general is the calculate the expectation value. Now say $\psi(r)$ goes as $e^{ir/n}$ (this is the case with hydrogen) then we may differentiate it with respect to $ir$ to get $1/n$ out of it and evaluate the integral. This gives us an expectation that we can invert to get <n>. However, you know that for hydrogen the energy goes as $1/n^2$ so you can extend this definition and evaluate the energy expectation to get <n> out of that. May 10, 2020 at 14:33
• However, in the example superposition you’ve given, the evaluation of these expectations are more straightforward than that. Because you’ve already expressed them in an orthonormal basis. So <n> $=\sum_i {A_i}^2 n_i$ How useful these definitions are depends on the context. May 10, 2020 at 14:37
• Simple single differentiation is unlikely to give you good results. Don't forget about the factors such as the Laguerre polynomials for $n>1$ and $r^\ell$ for $\ell>0$. A better way would be to expand in the eigenbasis and simply get that $\langle n\rangle=|A_0|^2(|A_1|^2n_1+|A_2|^2n_2+|A_3|^2n_3+\cdots).$ May 10, 2020 at 14:37