Consider a particle in the ground state of a infinite well with sides at ##x=0## and ##x=a##. The ground-state energy ##E_1=\frac{\hbar^2\pi^2}{2ma^2}##. The energy of the particle is entirely kinetic. Hence, ##E_1=\frac{p^2}{2m}##. Solving for ##p##, we get ##p=\pm\frac{\hbar\pi}{a}##. So the momentum ##p## of the particle can only take on one of these two values. In other words, we expect the wave function ##\phi(p)## in momentum space to peak at these two values: ##\phi(p)=\frac{1}{2}\delta(p+\frac{\hbar\pi}{a})+\frac{1}{2}\delta(p-\frac{\hbar\pi}{a})##, where ##\delta## is the Dirac delta function. But ##\phi(p)## is calculated to be ##\phi(p)=e^{-ipa/2\hbar}\frac{2\pi/a}{\sqrt{\pi\hbar a}}\frac{cos(pa/2\hbar)}{(\pi/a)^2-(p/\hbar)^2)^2}## (see 3rd equation in the picture(adsbygoogle = window.adsbygoogle || []).push({}); ^{1}).

We are led to conclude that ##E\neq\frac{p^2}{2m}##, or equivalently, the energy of the particle is not entirely kinetic. But why?

^{1}There is a typo in the text: an ##i## is missing in the exponent in the expression for ##\phi(p)## and there shouldn't be a negative sign. Reference: http://en.wikipedia.org/wiki/Uncertainty_principle#Particle_in_a_box

Source: Quantum Physics, 3rd Edition by Gasiorowicz, p. 57.

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# I Why is E not equal to p^2/2m for a particle in a box?

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