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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?

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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. $\Delta x=\infty$).

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  • $\begingroup$ Also, if it is confined to a region, that can just be taken to be the (infinite) square well potential, which then gives us momentum-like eigenstates in a box. $\endgroup$ May 3, 2023 at 5:41
  • $\begingroup$ @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. $\endgroup$ May 3, 2023 at 11:08
  • $\begingroup$ Hence the -like. $\endgroup$ May 3, 2023 at 13:01

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