# Do all quantum systems have zero point energy ?

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.

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