Basic quantum mechanics- violation of the time-energy uncertainty principle by pure eigenstates of the infinite potential well?

Question

Just beginning to learn quantum mechanics and my textbook says that all pure eigenstates of the Schrodinger equation ( with a specified potential function) have a zero standard deviation for the expectation values of energy.So every measurement of the energy yields a fixed value. I've also read that pure eigenstates are physically realisable solutions in some situations ( like the infinite potential well). But wouldn't that violate the uncertainty principle? the standard deviation in expectation of energy is zero and hence $$ΔE~Δt=0.$$ Where am I going wrong? Also could someone also give me an insight as to what uncertainty in time even means? I haven't been able to grasp that concept clearly yet.

Answer 1

The correct interpretation of $$ \Delta E \Delta t \geq \hbar / 2,$$ is with $\Delta E$ as the standard deviation of the Hamiltonian and $$ \Delta t = \frac{\Delta \Omega}{\lvert \mathrm{d}\langle \Omega \rangle / \mathrm{d} t\rvert},$$ where $\Omega$ is any other observable. For more details, see this answer by joshphysics.

Now, in a stationary state we have be definition $\Delta E = 0$ since it is an eigenstate of the Hamiltonian, but we also have $\mathrm{d} \langle \Omega \rangle /\mathrm{d}t = 0$, since expectation values are constant in time for stationary states (hence the "stationary"). Therefore, $\Delta t$ is ill-defined due to division by zero for stationary states and likewise $\Delta E \Delta t$, so in particular it is not zero and hence does not violate the uncertainty principle.

Answer 2

In the continum such states with a definite energy are physically difficult to realise. They are termed improper wave functions and are not normalized in the usual sense to unity since the wavefunction is not localized.