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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.

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  • $\begingroup$ Please provide a link to the pdf you are quoting from. $\endgroup$
    – sammy gerbil
    Aug 22, 2016 at 15:53
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    $\begingroup$ 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, 2016 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, 2016 at 16:18
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    $\begingroup$ 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
    Aug 22, 2016 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, 2016 at 16:28

2 Answers 2

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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.

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  • $\begingroup$ 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, 2016 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, 2016 at 17:30
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    $\begingroup$ 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, 2016 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, 2016 at 17:46
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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.

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