Editing Particle in a box (section)

=== Momentum wave function ===

The momentum wave function is proportional to the [[Fourier transform]] of the position wave function. With <math>k = p / \hbar</math> (note that the parameter {{mvar|k}} describing the momentum wave function below is not exactly the special {{math|''k<sub>n</sub>''}} above, linked to the energy eigenvalues), the momentum wave function is given by
<math display="block">\phi_n(p,t)
= \frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^\infty \psi_n(x,t)e^{-ikx}\,dx 
= \sqrt{\frac{L}{\pi \hbar}} \left(\frac{n\pi}{n\pi+k L}\right)\,\operatorname{sinc}\left(\tfrac{1}{2}(n\pi-k L)\right)e^{-i k x_c}e^{i (n-1) \tfrac{\pi}{2}}e^{-i\omega_n t} ,
</math>
where sinc is the cardinal sine [[sinc function]], {{math|1=sinc(''x'') = sin(''x'')/''x''}}. For the centered box ({{math|1=''x<sub>c</sub>'' = 0}}), the solution is real and particularly simple, since the phase factor on the right reduces to unity. (With care, it can be written as an even function of {{mvar|p}}.)

It can be seen that the momentum spectrum in this wave packet is continuous, and one may conclude that for the energy state described by the wave number {{math|''k<sub>n</sub>''}}, the momentum can, when measured, also attain ''other values'' beyond <math> p = \pm \hbar k_n </math>.

Hence, it also appears that, since the energy is <math display="inline"> E_n = \frac{\hbar^2 k_n^2}{2m} </math> for the ''n''th eigenstate, the relation <math display="inline"> E = \frac{p^2}{2m} </math> does not strictly hold for the measured momentum {{mvar|p}}; the energy eigenstate <math> \psi_n </math> is not a momentum eigenstate, and, in fact, not even a superposition of two momentum eigenstates, as one might be tempted to imagine from equation ({{EquationNote|1}}) above: peculiarly, it has no well-defined momentum before measurement!