如何证明表达式不是一维薛定谔方程的解？关于自由粒子非归一化解的疑问

Hey there, let's break down your questions about Schrödinger equation solutions for free particles and how to verify if an expression qualifies as a valid solution.

一、仅“无法归一化”不足以否定$e^{kx}$是薛定谔方程的解 #

First, a critical distinction: the Schrödinger equation itself is a linear partial differential equation, and it only requires that mathematically consistent expressions satisfy the equation. Normalization is a physical constraint, not a hard requirement of the equation.

Take the expression $e^{kx}$ (where $k$ is a real number) as an example. Substitute it into the time-dependent Schrödinger equation for a free particle:
$$i\hbar\frac{\partial\psi}{\partial t}=-\frac{\hbar^{2}{2m}\frac{\partial}2\psi}{\partial x^2}$$
Using separation of variables $\psi(x,t)=\psi(x)e^{-iEt/\hbar}$, we get the time-independent equation:
$$-\frac{\hbar^2}{2m}\psi''(x)=E\psi(x)$$
Plugging $\psi(x)=e^{kx}$ into the left-hand side gives $-\frac{\hbar^2}{2m}k2e^{kx}$, and the right-hand side is $Ee^{kx}$. As long as we set $E=-\frac{\hbar^2k2}{2m}$, the equation holds perfectly. Mathematically, this makes $e^{kx}$ a valid solution to the Schrödinger equation.

Normalization is a requirement for wave functions to describe observable physical states (like a real particle), but this is a physical filter we apply to solutions, not a rule of the equation itself. The fact that $e^{kx}$ can't be normalized only means it doesn't correspond to a measurable single-particle state—not that it's not a solution. Also, note that linear combinations of such solutions (like Fourier integrals, e.g., Gaussian wave packets) can be normalized and describe physical states.

二、通用方法：证明某表达式不是一维薛定谔方程的解 #

To determine if an expression is not a solution, the core approach is direct substitution, with some auxiliary checks to speed up the process:

1. Direct Substitution (Most Fundamental Method) #

Plug the candidate expression $\psi(x,t)$ into both sides of the Schrödinger equation and verify if they are equal for all $x$ and $t$. For the time-dependent one-dimensional Schrödinger equation:
$$i\hbar\frac{\partial\psi}{\partial t}=-\frac{\hbar^{2}{2m}\frac{\partial}2\psi}{\partial x^2}+V(x)\psi$$

• Calculate the left-hand side: $i\hbar\frac{\partial\psi}{\partial t}$
• Calculate the right-hand side: $-\frac{\hbar^{2}{2m}\frac{\partial}2\psi}{\partial x^2}+V(x)\psi$
  If the two results don't match for all $x$ and $t$, the expression is not a solution.

2. Auxiliary Checks Using Solution Properties #

• Continuity: Unless there's an infinite potential jump (like the boundary of an infinite square well), $\psi(x,t)$ and its first spatial derivative $\frac{\partial\psi}{\partial x}$ must be continuous everywhere. If the candidate expression has discontinuities, it's not a physically valid solution (though it might still satisfy the equation mathematically).
• Asymptotic Behavior: Bound-state solutions must tend to 0 as $x\to\pm\infty$. Scattering-state solutions (like free-particle plane waves) don't tend to 0, but they also shouldn't diverge exponentially (e.g., $e^{kx}$ with positive $k$ diverges as $x\to+\infty$, corresponding to a negative energy state that we usually ignore in free-particle discussions—but it's still a mathematical solution).
• Linear Superposition: The Schrödinger equation is linear, so any linear combination of solutions is also a solution. If a candidate expression can't be written as a combination of known solutions and fails the substitution test, it's not a solution.

For example: Suppose someone claims $\psi(x,t)=e^{x2}e^{-iEt/\hbar}$ is a free-particle solution. Substituting gives the left-hand side as $E\psi(x,t)$, and the right-hand side as $-\frac{\hbar^2}{2m}(2+4x2)\psi(x,t)$. Since $E$ can't equal $-\frac{\hbar^2}{2m}(2+4x2)$ for all $x$, this expression is not a solution.

内容的提问来源于stack exchange，提问作者angel leonardo