# Solution of Time Independent Schrodinger Equation

In the book Modern Physics by Serway/Moses/Moyer (on page 200,chapter Quantum mechanics in one dimension) it is written that (or what i have understood)when there are no forces on a particle solution of schrodinger equation in separable form is identical to that of a plane wave. But the wavefunction associated with a particle has to be confined to a region. Then how it is possible?

The time-independent Schrödinger equation for a free particle (i.e. with a potential energy $$U(x)=0$$) $$-\frac{\hbar^2}{2m}\frac{d^2\psi(x)}{d x^2}=E\psi(x)$$ has indeed the solutions $$\psi(x) = Ae^{ikx} \quad\text{with any real }k$$ for the energy $$E=\frac{\hbar^2k^2}{2m}$$, as can easily be verified. Obviously these solutions are not localized in space. Instead they are evenly spread across the whole space from $$-\infty$$ to $$+\infty$$.
A stringent mathematician might complain, that these solutions cannot be normalized to $$\int_{-\infty}^{+\infty}|\psi(x)|^2 dx=1$$. But for a pragmatic physicist this is no problem. Such a solution just represents a particle moving with a perfectly certain momentum $$p=\hbar k$$ (thus $$\Delta p=0$$). Therefore, according to Heisenberg's uncertainty principle, it has a completely uncertain position $$x$$ (i.e. $$\Delta x=\infty$$).
• @naturallyInconsistent I disagree. In an (infinite) square well potential we have solutions like $A\cos(kx)$ or $A\sin(kx)$ (for certain $k$ values). But these are not momentum eigenstates, since they are superpositions of $e^{ikx}$ and $e^{−ikx}$ together. May 3, 2023 at 11:08