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When dealing with a Hamiltonian of the type:

$$H=\frac{p^2}{2m}+V(r)$$

I have a big problem in understanding why we factor the total eigenfunction of the Hamiltonian into a radial and an angular term.

I think I understand that since $H, L^2$ and $L_z$ are three operators which, in this context, commute, it is convenient to look for the simultaneous eigenfunctions of the three operators. Therefore, when we write the TISE

$$H\psi_n(r,\theta,\phi)=E\psi_n(r,\theta,\phi)$$

the $\psi_n(r,\theta,\phi)$ refers to the simultaneous eigenfunctions of the three operators.

The main problem is why we write

$\psi_n(r,\theta,\phi)=f(R)F(\theta,\phi)$.

Is this related to the fact that the Hamiltonian can be separated into a radial term $H_r$ and an angular term $H_{\theta}$ and so we apply the standard resolution for a differential equation of this type, namely: we introduce a "test" function which consists of the product of single function each of one depends only on one variable: $T(x,y)=X(x)Y(y)$ and then we divide the whole differential equations by $T(x,y)$ and we obtain two differents ordinary differential equation?

For example in this discussion Quantum Central Force Problem and Angular Momentum in the first answer in the "Factorization of eigenvalues" paragraph, I read

its obvious, this equation does not depend anymore on θ or ϕ and therefore the solution must simplify as: $\psi_{nlm}(r,\theta,\phi)=u_{nl}(r)Y_{lm}(\theta,\phi)$

I ask you: why the solution must simplify in that way? Is that answer related to mine?

If possible I would like an explanation that uses only time-independent Schrodinger equations and is not too difficult because I am on my first quantum mechanics course and am not very knowledgeable.

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One common way of solving a partial differential equation is by separation of variables. With a bit of experience it’s clear that, for potentials of the type $V(r)$, the angular part will not depend on $V$, and only the radial part will depend on $V$.

Separation of variables then allows you to transform the PDE into a set of ODEs linked by the separation constants.

It is not the only way to proceed but its advantages are numerous since you get out of that a complete set of functions, so that any solution can be written as a sum of products of separated functions. Also it simplifies the search for solution since you only need to “solve again” the radial part : the radial ODE is different for every potential but the angular part remains the same for all $V(r)$ so you only need to do that angular part once.

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  • $\begingroup$ Thank you, there is still something I don't understand though. Why with the separation of variables method we "get out of that a complete set of functions"? are you talking about the complete set of simultaneous eigenfunctions of the operators or in general, separating the variables, lead us to a "set of solution functions" which are complete by definition? $\endgroup$
    – Salmon
    Jan 31, 2022 at 22:56
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    $\begingroup$ All the resulting ODE’s are Sturm-Liouville problems… $\endgroup$ Jan 31, 2022 at 23:43
  • $\begingroup$ So why in a central field problem we say that $H,L^2,L_z$ commute with each other? Shouldn't we just show that the Hamiltonian is the sum of two parts that commute with each other and then we can factor the eigenfunction? Only $r$ and $L^2$ appear in the Hamiltonian, I should not be interested in $L_z$. $\endgroup$
    – Salmon
    Jan 31, 2022 at 23:51
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    $\begingroup$ You can approach this from several angles and you’re right that in a central field there are degeneracies. You need $L_z$ to identify the states with the same eigenvalue of $L^2$. And $H$,$L^2$ and $L_z$ do commute. $\endgroup$ Feb 1, 2022 at 0:08
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    $\begingroup$ Not sure what you mean by simplification. By construction separation of variables leads to $\psi(r,\theta,\phi)=R(r)\Theta(\theta,\phi)$ where $R(r)$ is a solution to an equation that depends only on $r$, and $\Theta(\theta,\phi)$ depends only on the angles. So in this sense the method precisely looks for solutions that simplify in this way. The lucky part is that all solutions are combos of functions of this type. $\endgroup$ Feb 1, 2022 at 0:26

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