3
$\begingroup$

I know that, for a central potential $U(r,\theta, \varphi)=U(r)$, the Schrödinger equation takes the well-known form in spherical coordinates. I'm searching for separable solutions in the form: \begin{equation} \phi(r, \theta,\varphi)=a(r)b(\theta)c(\varphi) \end{equation} So, if I understood correctly the theory (I'm a beginner in theoretical physics so maybe I did some conceptual mistakes) my $\phi$ is the eigenfunction corresponding to some eigenvalue E. Solving the various steps this equation appears for $c(\varphi)$: \begin{gather} \frac{d^2c}{d\varphi^2}+m_l^2=0 \end{gather} And the solution should be: \begin{equation} c(\varphi)_{m_l}=Ae^{im_l\varphi}+Be^{-im_l\varphi} \end{equation} and after some considerations on the periodicity of $\varphi$, I got that $m_l$ is integer. My problem is the following:

My teacher, and also the majority of the documents I found online, states that the solution is \begin{equation} c(\varphi)_{m_l}=Ae^{im_l\varphi} \end{equation} saying "noting that $m_l$ and $-m_l$ yield the same general solution $c(\varphi)$". What does it mean? Every solution to the harmonic oscillator differential equation yield a solution of the same form, so what is the point? It seems to me that we are "throwing away" something (even if I'm sure it's not like this, but I don't understand why). This is because I think that the equation $$\phi(r, \theta,\varphi)_{m_l}=a(r)b(\theta)c(\varphi)_{m_l}$$ would be different if I use $c(\varphi)_{m_l}=Ae^{im_l\varphi}+Be^{-im_l\varphi}$ from the one obtained by using $c(\varphi)_{m_l}=Ae^{im_l\varphi}$.

I feel like I'm missing something really stupid here but I'm in a loop now and I can't see a solution.

$\endgroup$
  • $\begingroup$ Please provide a link to the pdf you are quoting from. $\endgroup$ – sammy gerbil Aug 22 '16 at 15:53
  • 1
    $\begingroup$ There are several, pdf and websites. One for example is physicspages.com/2011/03/25/… (in this one take a look at eqs (22)-(23)) $\endgroup$ – Luthien Aug 22 '16 at 15:58
  • $\begingroup$ With regards to the Angular Equation, you can find my own explanation here: sciencemadness.org/talk/…. With regards to the whole derivation for the particle in a central field problem, there are many excellent resources and Luthien's is a good one. $\endgroup$ – Gert Aug 22 '16 at 16:18
  • 1
    $\begingroup$ You can write either $c(\varphi)_{m_l}=Ae^{im_l\varphi}+Be^{-im_l\varphi}$ with integer $m_l>0$, or $c(\varphi)_{m_l}=Ae^{im_l\varphi}$ with arbitrary $m_l$ (either negative or positive). Because your final wave-function is a sum of such functions over $m_l$ (both negative or positive). $\endgroup$ – Hayk Hakobyan Aug 22 '16 at 16:18
  • $\begingroup$ Jack I like your answer because I was going in that direction after I posted this and I was going some thinking. You understood perfectly my struggle! Does this mean that I get the same eigenfunction $\phi$, either with, say, $m_l=2$ and $m_l=-2$? $\endgroup$ – Luthien Aug 22 '16 at 16:28
5
$\begingroup$

The overall theme is that you're not really looking for an arbitrary solution of the time-independent Schrödinger equation: instead, you're only looking for a basis of such solutions, which you can then use to build, via linear combinations of them, any arbitrary solution. This allows you a lot of freedom, because as long as you arrive at a complete set of solutions then you're free to restrict your search area as much as you need to.

In fact, you've already done precisely this sort of narrowing-down of the search area simply by the form of wavefunctions, $$\psi(r, \theta,\varphi)=R(r)\Theta(\theta)\Phi(\varphi), \tag1$$ that you're looking for. It's important to realize that these wavefunctions are intrinsically tied to a choice of coordinate system (and, particularly, to the choice of the direction for the $z$ axis), and that if you transform these wavefunctions to an arbitrary coordinate system then they will no longer be separable. Again, the reason you're allowed to make the Ansatz $(1)$ is that it will eventually lead you to a complete set of solutions that you can use to reconstruct any arbitrary solution.

Thus, when you're staring down at $$ \Phi_{m_l}(\varphi)=Ae^{im_l\varphi}+Be^{-im_l\varphi} $$ and you're trying to decide what to put in for $A$ and $B$, what you do is you realize that what you really have is simply a solution space of dimension two, and since we're looking for a basis anyways then it's perfectly OK to separate this into $$ \Phi_{m_l}(\varphi)=Ae^{im_l\varphi} \quad \text{and}\quad \Phi_{-m_l}(\varphi)=Be^{-im_l\varphi}, $$ since when taken together the two constitute a basis for the solution space. After that, it's just a bit of admin: you realize that the second solution is actually just the first solution with a different index, so you can simply keep $\Phi_{m_l}(\varphi)=Ae^{im_l\varphi}$ as your standard solution (while keeping in mind that $m_l$ might well be negative), and you will still be describing a basis for the entire space.

$\endgroup$
  • $\begingroup$ Emilio thank you very much for your detailed answer, now everything is more clear. Last thing: it seems to me that what matters in this case is the absolute value of $m_l$. Does this mean that I get the same $\phi$ either with, say, $m_l=+2$ and $m_l=-2$? $\endgroup$ – Luthien Aug 22 '16 at 17:26
  • $\begingroup$ No, you don't get the 'same' $\phi$ - you get two different, linearly independent functions of the overall basis. On the other hand, because the eigenvalue is $m_l^2$ and doesn't depend on the sign of $m_l$, the azimuthal and radial equations, on $\theta$ and $r$ resp., offer exactly the same choices for the $-m_l$ wavefunction as they do for the $+m_l$ one, and produce exactly the same azimuthal and radial solutions. Thus, you will get two almost-identical $\phi$s that only differ on the sign of $m_l$ (and, in fact, are complex conjugates of each other). $\endgroup$ – Emilio Pisanty Aug 22 '16 at 17:30
  • 1
    $\begingroup$ Yes sorry, I was referring to the second equation you wrote on your answer, the one with both A and B. With respect to this one my solution won't change If I use $m_l$ or $-m_l$. Then, if I choose to separate into $c(\varphi)=Ae^{im_l\varphi}$ and $c(\varphi)=Be^{-im_l\varphi}$, I can see my first equation as a linear combination of this two. Am I correct? $\endgroup$ – Luthien Aug 22 '16 at 17:40
  • $\begingroup$ Yes, exactly. You get the exact same solution, $e^{im_l\varphi}= e^{i(\mathrm{parameter}) \varphi}$, for both choices, so you simply embrace the parameter dependence, and then you build the 'general' solution (for the longitudinal equation in $\varphi$) as a linear combination of the same solution but with oppositely signed parameters. $\endgroup$ – Emilio Pisanty Aug 22 '16 at 17:46
1
$\begingroup$

A general solution is a linear combination of such separable solutions. Since Schrödinger equation is a linear equation, superposition of solutions is also a solution. You can take $A$ and $B$ as coefficients of two separable solutions one with $e^{im\phi}$ and another with $e^{-im\phi}$ as their $\phi$ dependent part.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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