We know that the Schrödinger equation is $\hat H|\Psi\rangle=E|\Psi\rangle$. If we write the hamiltonian $\hat H$ as $\frac{\hat{p}^2}{2m}+V(x)$ and then use $\hat p=-i\hbar\partial _x $ we can write this equation in position basis as: $$\left [\nabla^2+\frac{2m}{\hbar^2}(E-V(x))\right ]\Psi(x)=0 \tag{1}$$ Now we can rearrange equation (1) to write it as:
$$\left [\nabla^2+\frac{2m}{\hbar^2}E\right ]\Psi(x)=\frac{2m}{\hbar^2}V(x)\Psi(x) \tag{2}$$ Equation (1) is a homogeneous equation that can be solved exactly analytically [for example for Hydrogen atom i.e. 1 electron in the potential field of 1 proton].
Equation (2) is just the rearranged version of equation (1). But it has the form of an inhomogenious equation. This can be solved by Green's function method and the solution looks as:
$$\Psi(x)=\Psi_0 (x)+\frac{2m}{\hbar^2}\int dx\, G_0(x)V(x)\Psi(x) \tag{3}$$ where $\Psi_0 (x)$ is the solution of the homogenious part "$\left [\nabla^2+\frac{2m}{\hslash^2}E\right ]\Psi_0(x)=0$" of equation (2) and $G_0(x)$ is the Green's function of the operator "$\left [\nabla^2+\frac{2m}{\hslash^2}E\right ]$". Now in order to find the expression of $\Psi(x)$ from equation (3) we have to use the Born approximation and then I dont know whether we can reach the exact solution that we obtain by analytically solving equation (1).

My questions are:

(A) Since equation (1) and equation (2) are the same equation but slightly rearranged, they should have same solution say for Hydrogen atom problem. Is it possible to reach the same solution form these 2 equations?

(B) Why we never solve inhomogeneous equation (eqn. 2) for Hydrogen atom problem.

(C) But why we solve inhomogeneous equation (eqn. 2) for scattering problem.

(D) Why we do not solve homogeneous equation (eqn. 1) for scattering problem.

(E) Generally speaking: what is the basic difference between a homogeneous and inhomogeneous differential equation and how to know which one to solve in which physical situation?


1 Answer 1


A linear differential equation is homogeneous when it can be written in a form $$ \hat{\mathcal{L}}\Psi(x,t)=0, $$ where $\hat{\mathcal{L}}$ is a differential operator, possibly involving partial derivatives and functions, but independent on $\Psi(x,t)$, since otherwise the equation would be non-linear. Inhomogeneous equation would have form $$ \hat{\mathcal{L}}\Psi(x,t)=f(x,t), $$ where $f(x,t)$ is some function, other than $\Psi(x,t)$.

Thus, Schrödinger equation (time dependent and time independent) is a homogeneous equation, since it can always be rearranged to have the homogeneous form.

A popularr method of solving Schrödinger equation, known as perturbation expansion, is based on treating this equation as an inhomogeneous one, by separating a part of this equation as if it were an external term $f(x,t)$. Solving the homogeneous part of this equation, it is converted into an integral equation, which then can be iterated infinitely many times to obtain a formally exact solution of the full equation. In practice this solution is either a) trancated at a certain order (perturbation theory) or b) formally used to obtain useful relations (scattering theory), or c) one can sum only sub-series of this solution (Feynmann-Dyson expansion and related methods).

There is no need to resort to any of these methods in case of the equation for a hydrogen atom, since it is exactly solvable. However one may use them if there is an additional perturbation, e.g., to study hydrogen atom in magnetic field. On could also try to treat Coulomb potential as a perturbation to a non-interacting particle and conceivably sum the perturbation series, but this is an unnecessarily hard way to do things.

Similarly, for some scattering problems the Schrödinger equation is exactly solvable and does not require using approximate methods. Scattering from a rectangular barrier dealt with in elementary quantum mechanics is just one example.

  • $\begingroup$ Thanks for the answer. When we do scattering from a central potential, we follow the inhomogeneous equation root. Why? Can it also be done in the homogeneous equation root too? (If possible please answer this question by editing your answer above.Your answer is already very appealing and the answer to this part will make it more complete.) $\endgroup$
    – user103515
    Commented Nov 19, 2020 at 10:19
  • $\begingroup$ As I tried to explain, we solve equation using perturbation series or other approximate methods when an exact solution is impossible or difficult to work with. Most differential equations cannot be solved exactly. I am not sure what you call "root"... It seems that you lack proper mathematical background in differential equations (particularly partial differential equations), and I strongly encourage you to learn more about them - it would answer most of your questions. $\endgroup$
    – Roger V.
    Commented Nov 19, 2020 at 10:26
  • 1
    $\begingroup$ You are right about my PDE familiarity. "root" was the wrong word i used. I wanted to say "why it was done by solving inhomogeneous equation". Now It is sufficiently clear from your comment. Thanks $\endgroup$
    – user103515
    Commented Nov 19, 2020 at 10:38

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