Editing Quantum mechanics (section)

=== Free particle ===
{{Main|Free particle}}
[[File:Guassian Dispersion.gif|360 px|thumb|right|Position space probability density of a Gaussian [[wave packet]] moving in one dimension in free space]]
The simplest example of a quantum system with a position degree of freedom is a free particle in a single spatial dimension. A free particle is one which is not subject to external influences, so that its Hamiltonian consists only of its kinetic energy:
<math display=block>H = \frac{1}{2m}P^2 = - \frac {\hbar ^2}{2m} \frac {d ^2}{dx^2}. </math>
The general solution of the Schrödinger equation is given by
<math display=block>\psi (x,t)=\frac {1}{\sqrt {2\pi }}\int _{-\infty}^\infty{\hat {\psi }}(k,0)e^{i(kx -\frac{\hbar k^2}{2m} t)}\mathrm{d}k,</math>
which is a superposition of all possible [[plane wave]]s <math>e^{i(kx -\frac{\hbar k^2}{2m} t)}</math>, which are eigenstates of the momentum operator with momentum <math>p = \hbar k </math>. The coefficients of the superposition are <math> \hat {\psi }(k,0) </math>, which is the Fourier transform of the initial quantum state <math>\psi(x,0)</math>.

It is not possible for the solution to be a single momentum eigenstate, or a single position eigenstate, as these are not normalizable quantum states.{{refn|group=note|A momentum eigenstate would be a perfectly monochromatic wave of infinite extent, which is not square-integrable. Likewise, a position eigenstate would be a [[Dirac delta distribution]], not square-integrable and technically not a function at all. Consequently, neither can belong to the particle's Hilbert space. Physicists sometimes introduce fictitious "bases" for a Hilbert space comprising elements outside that space. These are invented for calculational convenience and do not represent physical states.<ref name="Cohen-Tannoudji" />{{rp|100–105}}}} Instead, we can consider a Gaussian [[wave packet]]:
<math display=block>\psi(x,0) = \frac{1}{\sqrt[4]{\pi a}}e^{-\frac{x^2}{2a}} </math>
which has Fourier transform, and therefore momentum distribution
<math display=block>\hat \psi(k,0) = \sqrt[4]{\frac{a}{\pi}}e^{-\frac{a k^2}{2}}. </math>
We see that as we make <math>a</math> smaller the spread in position gets smaller, but the spread in momentum gets larger. Conversely, by making <math>a</math> larger we make the spread in momentum smaller, but the spread in position gets larger. This illustrates the uncertainty principle.

As we let the Gaussian wave packet evolve in time, we see that its center moves through space at a constant velocity (like a classical particle with no forces acting on it). However, the wave packet will also spread out as time progresses, which means that the position becomes more and more uncertain. The uncertainty in momentum, however, stays constant.<ref>{{cite book |title=A Textbook of Quantum Mechanics |first1=Piravonu Mathews |last1=Mathews |first2=K. |last2=Venkatesan |publisher=Tata McGraw-Hill |year=1976 |isbn=978-0-07-096510-2 |page=[https://books.google.com/books?id=_qzs1DD3TcsC&pg=PA36 36] |chapter=The Schrödinger Equation and Stationary States |chapter-url=https://books.google.com/books?id=_qzs1DD3TcsC&pg=PA36}}</ref>