0
$\begingroup$

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?

$\endgroup$
0

2 Answers 2

1
$\begingroup$

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 $$i\hbar\frac{d}{dt}\Psi=H\Psi.$$

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ 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. $\endgroup$ May 10, 2020 at 13:57
  • $\begingroup$ @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. $\endgroup$ May 10, 2020 at 14:33
  • $\begingroup$ 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. $\endgroup$ May 10, 2020 at 14:37
  • $\begingroup$ 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 $\langle n\rangle=|A_0|^2(|A_1|^2n_1+|A_2|^2n_2+|A_3|^2n_3+\cdots).$ $\endgroup$
    – Ruslan
    May 10, 2020 at 14:37

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.