I understand that it is possible to write an uncertainty relation between the Hamiltonian of a system and time, where the time uncertainity is defined by the amount of time it takes an arbitrary operator to change by its standard deviation.

$\frac{ΔAΔH}{|\frac{d<A>}{dt}|} ≥ ℏ/2$

(Reference: University of California - Riverside)

Does this result in all quantum systems having a zero point energy, as ΔH must be non-zero ? This contradicts what I have learnt about certain model problems (particle in infinite free space, particle in a box, particle on a ring of defined radius), so I am pretty sure it is not correct.

  • $\begingroup$ In cases where you have $H=0$, doesn't the denominator go to $0$ as well? $\endgroup$ – Scott Lawrence Apr 15 '15 at 21:32
  • $\begingroup$ If the operator itself was zero, yes, surely, by the TDSE. $\endgroup$ – J. LS Apr 15 '15 at 21:35

The zero-point energy is the energy eigenvalue $\langle H \rangle_\Omega$ of the lowest lying energy eigenstate $\Omega$.

$\Delta H_\Omega = 0$, so this has nothing to do with the zero-point energy. This does also not contradict the uncertainty principle, since $\frac{\mathrm{d}\langle A \rangle}{\mathrm{d}t} = 0$ for a energy eigenstate (which are the stationary states, after all!), so the LHS of the uncertainty relation is undefined.

  • $\begingroup$ What is the link between zero-point energies and the uncertainity principle then ? I often see them connected in a handwavey way. $\endgroup$ – J. LS Apr 15 '15 at 22:01
  • $\begingroup$ @J.LS: Oh, that is a different question from what you asked, and answered by Wikipedia. $\endgroup$ – ACuriousMind Apr 15 '15 at 22:03
  • $\begingroup$ It's not really answered by Wikipedia though. The article says "All quantum mechanical systems undergo fluctuations even in their ground state and have an associated zero-point energy, a consequence of their wave-like nature." But then it only accounts for the relationship between ZPE and the uncertainity principle for the case of a harmonic oscillator. $\endgroup$ – J. LS Apr 15 '15 at 22:09
  • $\begingroup$ @J.LS: It says (for all systems, not just the oscillator): "In particular, there cannot be a state in which the system sits motionless at the bottom of its potential well, for then its position and momentum would both be completely determined to arbitrarily great precision. Therefore, the lowest-energy state (the ground state) of the system must have a distribution in position and momentum that satisfies the uncertainty principle, which implies its energy must be greater than the minimum of the potential well." $\endgroup$ – ACuriousMind Apr 15 '15 at 22:12

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.