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.


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.

  • $\begingroup$ That clears it up. Thanks. So in such cases (again take the infinite potential well) the pure eigenstates are physically realisable , aren't they ? Also for clarification does the uncertainty principle hold only for such physically realisable solutions or for ANY solution to the Schrodinger equation? $\endgroup$ – Vijay Sep 16 '17 at 16:42
  • $\begingroup$ @Vijay.V.Nenmeli Whether these stationary states are physically realizable or not a) is irrelevant and b) depends on the situation. In the case of the infinite potential well they are realizable to the extent that an "infinite" potential well is realizable to begin with. $\endgroup$ – ACuriousMind Sep 16 '17 at 17:02

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.

  • $\begingroup$ I don't see how this answers the question at all. $\endgroup$ – ACuriousMind Sep 16 '17 at 12:42
  • $\begingroup$ Additionally, solutions to the problem of a particle in an infinite well are not improper wave functions; the solutions can be fully normalized in the usual sense. $\endgroup$ – ZeroTheHero Sep 16 '17 at 15:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.