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I was reading Griffith's book Introduction to Quantum mechanics and found that for the case of a free particle, we can diregard solutions of the form $e^{kx}$, where $k$ is real (positive or negative).

The book uses separation of variables to find possible solutions. I can see those solutions are not normalizable. I can even see their linear combination (Fourier transform) is not normalizable for all $f(x)$, where $f(x)$ is the initial wave function.

My question is, is that really enough to prove they are not solutions?

Suppose we perform a nonlinear mapping that transform $e^{kx}$ to $f(x)$,suppose that mapping exist. Wouldn't we need to prove that mapping does not exist in general in order to say they are not solutions ?

Why do we assume that only linear combinations works as general solutions ? I know Schrödinger equation is linear and it admits linear combinations, but that doesn't mean non-linear combinations are not allowed.

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closed as unclear what you're asking by Emilio Pisanty, Jon Custer, glS, AccidentalFourierTransform, ZeroTheHero Jul 18 '18 at 16:03

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    $\begingroup$ "but that doesn't mean non-linear combinations are not allowed" Yes it does. A 2nd order, linear, homogeneous ODE has two linearly independent solutions, and the most general possible solution is a linear combination of those two. $\endgroup$ – J. Murray Feb 23 '18 at 2:51
  • $\begingroup$ I understand the most general possible solution to the time independent schrodinger equation is the linear combination of those two. But in order to satisfy the initial value problem, you need to be able to somehow express the initial wave function in terms of those eigenfunctions. In this particular case there is an infinite not countable set of eigenfunctions. Suppose I can find a nonlinear transformation that acts over all the set of eigenfunctions and acts individually by producing a change of basis (meaning they would still generate the solutions to the second order ODE). Then what? $\endgroup$ – angel leonardo Feb 23 '18 at 4:41
  • $\begingroup$ I wan to clarify, the non-linear transformation I am talking about, would transform the set of eigen functions into f(x) (initial wave function) $\endgroup$ – angel leonardo Feb 23 '18 at 4:42
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The Schrödiger equation is not just a condition that wavefunctions of particles satisfy $H|\psi\rangle=i\hbar\partial_t|\psi\rangle$, but also that wavefunctions are normalizable, i.e.

$$\int\mathrm{d}^{d}x\,|\psi(x)|^2<\infty.$$

Since the Schrödinger equation is a linear eigenvalue problem, transforming the solutions nonlinearly will not result in a solution to the original problem.

I hope this helps!

TL;DR: If you apply a nonlinear transformation to the wavefunction, you have to apply the same nonlinear transformation to the equation of motion (the Schrödinger equation). No transformation you do will allow a non-solution to become a solution.

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  • $\begingroup$ So you are saying, that if you have a solution and a transformation is applied. The final result will be a solution if and only if the transformation you apply is linear $\endgroup$ – angel leonardo Feb 23 '18 at 20:51
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    $\begingroup$ The Schrödinger equation is the equation. We additionally require that the solution is normalised to unity. This requirement is a postulate and in no way - how ? - part of the equation. $\endgroup$ – my2cts Jul 16 '18 at 7:24
  • $\begingroup$ It's a part of the equation in the sense that this condition specifies the Hilbert space over which you're acting. $\endgroup$ – Bob Knighton Jul 16 '18 at 15:48
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I understand the most general possible solution to the time independent schrodinger equation is the linear combination of those two. But in order to satisfy the initial value problem, you need to be able to somehow express the initial wave function in terms of those eigenfunctions.

Yes, this is true.

In this particular case there is an infinite not countable set of eigenfunctions.

The eigenfunctions for the free particle take the form $$ \psi \sim e^{-ikx}$$ for some $k$. You want to take some (more or less) arbitrary initial function $f(x)$ and express it as a linear combination of those eigenfunctions. If there were a finite number of them, you would say

$$ f(x) = \sum \mathcal F_k e^{-ikx} $$

where $\mathcal F_k$ is the coefficient of the eigenfunction corresponding to $k$ - it shouldn't be too surprising that the generalization to a continuous set of eigenfunctions would simply be

$$ f(x) = \int_{-\infty}^\infty \mathcal F(k) e^{-ikx} dk$$

This should look quite familiar. It's not difficult to invert this to find the function $\mathcal F(k)$, so the problem is solved.


There are a few other mathematical points worth mentioning.

Suppose I can find a nonlinear transformation that acts over all the set of eigenfunctions and acts individually by producing a change of basis

A change of basis is always taken to be a linear transformation (in fact, a vector space isomorphism). A nonlinear change of basis is not impossible to construct, but there would be no point to it - there is always a linear transformation which connects any two bases.

Suppose we perform a nonlinear mapping that transform e^{kx} to f(x),suppose that mapping exist. Wouldn't we need to prove that mapping does not exist in general in order to say they are not solutions ?

No. If a function cannot be built out of linear combinations of the eigenfunctions of the Hamiltonian, then it is not a viable initial state for the system in the first place.

In most elementary applications of quantum mechanics, one assumes that the eigenfunctions of the Hamiltonian (or any other operator) span the entire relevant function space, which would mean that every function (with a few obvious limitations) can be built out of (possibly infinite) linear combinations of said eigenfunctions. This is very seldom proven to be true, but it holds well enough for introductory purposes.

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  • $\begingroup$ ok, I was reading about what you said in the last paragraph. It is mentioned as the completeness asumption. If you assume that, it means if you want to prove something is not a solution, then you only have to prove it is not complete. Which is the case of e^kx (with k being real), since the integral in the fourier transform or the "linear combination" would diverge. am I right thinking this? $\endgroup$ – angel leonardo Feb 23 '18 at 7:08
  • $\begingroup$ I don't understand. $e^{|k|x}$ is not a solution to the free particle Schrodinger equation, you can just plug it in and see. You don't need to do anything more complicated than that. $\endgroup$ – J. Murray Feb 23 '18 at 15:32

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