# Measurement of energy apparently violating the position-momentum Uncertainty Principle in a potential that does not depend on distance?

I am taking a beginning course in QM and I have learnt that the measurement of energy collapses the wavefunction of a particle to one of its energy eigenstates. But some misconceptions regarding this is leading straightaway to the violation of the Uncertainty Principle in potentials that is independent of position of the particle. Consider the most elementary case of an infinite square potential well : $$V(x) = \begin{cases} 0, & 0 < x < L,\\ \infty, & \text{otherwise} \end{cases}$$ The energy eigenstates are : $$\psi_n(x) = \begin{cases} \sqrt{\frac{2 }{L}}\sin(\frac{n \pi x} { L}) & 0 < x < L,\\ 0, & \text{otherwise} \end{cases}$$ After, say, the measurement of energy returns a value $E_n$, the wavefunction becomes $\psi_n(x) e^{i \frac{2 \pi E_n } {h} t}$. Shouldn't then, the particle have a definite momentum $\sqrt{2 m E_n}$ ? The positional uncertainty is still (approximately) $L$ and hence the Heisenberg principle seems to get violated.

P.S. : I know that the actual momentum uncertainty calculated by the routine method $$\langle p \rangle = \int_{-\infty}^{\infty} p P_n(x)\,\mathrm{d}x$$ is $$\mathrm{Var}(p)=\left(\frac{\hbar n\pi}{L}\right)^2$$. But what is wrong with the above line of reasoning ?

• Why would the momentum be $\sqrt{2mE_n}$? $p=\sqrt{2mE}$ is only valid for free particles. – ACuriousMind Dec 3 '15 at 19:27
• Shouldn't the Hamiltonian be the total energy of the particle ? V = 0 in the well would imply the above relation. – Sagnik Dec 3 '15 at 19:30
• Oh, you're right. The issue here is indeed more subtle, I'll write an answer. – ACuriousMind Dec 3 '15 at 19:38
• Hi Sagy, there is a mistake above. The positional uncertainty by definition is the variance of x. What you have calculated is the expectation value of x (~L). – dergeophysiker Dec 3 '15 at 19:55
• The positional uncertainty is around L, too. – Sagnik Dec 3 '15 at 19:58

## 2 Answers

Your wavefunction does not have definite momentum.

Wavefunctions with definite momentum are infinite plane waves (i.e. $e^{-ipx/\hbar}$ everywhere). Your wavefunction looks like this within a finite region, and is zero outside that region. The only way to make it zero outside the region is by superposing other plane waves, which constructively interfere inside the well but destructively interfere outside it.

This is very general: localization requires destructive interference, and destructive interference requires multiple momenta present to interfere with each other. This is the basic idea behind the uncertainty principle.

You're claiming that, if $\psi$ is an eigenstate of the Hamiltonian $H = \frac{p^2}{2m}$, it should also be an eigenstate of $p$. This is just wrong, already in the finite-dimensional case:

Consider, for instance, a Pauli matrix. Its square is the identity, and every vector is an eigenvector of the identity - but a Pauli matrix has eigenvalues $1$ and $-1$, and the sum of the corresponding eigenvectors $v_1$ and $v_{-1}$ just isn't an eigenvector of the matrix, although it is certainly an eigenvector of its square.

Therefore, just because a state is an eigenvector of $H$ it does not follow that it must be one of $p$, so you cannot claim a state with definite energy has definite momentum.

• Doesn't it then follow that the Hamiltonian can never correspond to the total energy of a quantum particle ? – Sagnik Dec 3 '15 at 20:05
• @Sagy: Why would it? You just have to accept that just because something has a definite value for the square of a quantity, it doesn't have to have a definite value of the quantity itself. That's quantum physics for you. ;) – ACuriousMind Dec 3 '15 at 20:07
• But it does for a free quantum particle, right ? – Sagnik Dec 3 '15 at 20:17
• @Sagy: Not for all of the energy eigenstates, no (note that $\exp(\mathrm{i}px)+\exp(-\mathrm{i}px)$ is an energy-, but not a momentum eigenstate). And the "eigenstates" aren't even states ($\exp(\mathrm{i}px)$ is not square-integrable over all of $\mathbb{R}$). – ACuriousMind Dec 3 '15 at 20:23
• Yes, I think I see your point. But I am accepting Kevin Zhou's answer as it is easier to understand for beginners in QM like me. – Sagnik Dec 3 '15 at 20:59