General solution to initial value problem of free particle for 1D Schrodinger equation

Question

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.

Answer 1

"Not normalizable" is sufficient to rule a solution to Schrodinger's equation as not MEANINGFUL. The purpose of the wave function is to act as a probability distribution for the particle. A non-normalizable function may satisfy the equation itself, but it will not be regarded as having any physical relevance.

Answer 2

The free-space particle solution is not normalizable because $e^{kx}$ is not absolutely square integrable over all $\mathbb{R}$, regardless of the choice of $k\in\mathbb{C}$. The conclusion is that the free space particle is not consistent within the framework of Quantum Mechanics. However, it is approximately meaningful. For example, you can have a distribution of such solutions with nearly the same values of $k$, such as $$ \psi(x)=\frac{1}{\sqrt{2\pi}}\int_{p-\epsilon}^{p+\epsilon} A(k)e^{ikx}dk, $$ where $A(k)$ is a smooth function that vanishes outside $[p-\epsilon,p+\epsilon]$. In that case, the Plancherel Theorem for the Fourier transform gives $$ \int_{-\infty}^{\infty}\psi^*(x)\psi(x)dx = \int_{p-\epsilon}^{p+\epsilon}A(k)^*A(k)dk. $$ If you normalize the right side involving $A$ to be $1$, then $\psi$ is a valid possible free-space state vector.

It is valid to use an integral to build up a state vector from the $e^{ikx}$ in this way. The resulting function $\psi$ is a valid Quantum state vector with a momentum that is almost certain, and given by $$ \int_{p-\epsilon}^{p+\epsilon}kA(k)^*A(k)dk \approx p. $$ A state vector where the momentum is perfectly known to be $p$ has a position that is completely uncertain, and is given by $e^{ipx}$ in the limit as the uncertainly in momentum shrinks to $0$. So these idealized states do not exist, but approximations to them are realizable, to arbitrary accuracy.

In this case, $e^{ikx}$ is essentially given meaning through approximations such as those mentioned above, but only for real $k$. Complex $k$ would give complex values of momentum. The eigenvalues of operators in question must always be real, which implies that idealize states can only be given meaning when they are those corresponding to real eigenvalues of the operators in question. The eigenvalues of momentum, for example, must always be real. So, $$ -i\hbar\frac{\partial}{\partial x} e^{kx}=-i\hbar k e^{kx} $$ makes sense only when $-i k$ is real, which means $k$ must be imaginary. So $e^{ikx}$ only makes sense in this context if $k$ is real.

The "observables" of Quantum are selfadjoint linear operators acting on the state vectors. That you can take as an axiom, which it is.