# Is the ground state closest to the uncertainty relation? [duplicate]

For simplicity, suppose we are only talking about discrete energy levels, ie, bound state case. The energy levels are $E_1, E_2\cdots$, and the corresponding wave functions are $\psi_1, \psi_2 \cdots$.

My question is, is it true that $\sigma_x \sigma_p$ is minimum when $n=1$ for the eigenstates.

I came across question because I found harmonic oscillator and infinite potential well problems satisfy this statement, so I want to know if this is a general case.

I think this may be true because for ground state, there is no node ground state wave function. Thus the $\sigma_p$ may be small compare to other eigenstates.

• Possibly duplicate of physics.stackexchange.com/q/14181 – Ruslan Sep 24 '14 at 8:01
• No. The minimum in uncertainty is obtained for so-called coherent states: en.wikipedia.org/wiki/Coherent_states – Bubble Sep 24 '14 at 9:50
• @Bubble Yes, you are right. However I am asking which stationary state is nearest to the minimum. – an offer can't refuse Sep 24 '14 at 10:08
• – user10851 Sep 24 '14 at 21:16

The ground state of a system is by definition the state of minimal energy, i.e. the system is located at the minimum point of the potential.

Now, if we were in classical mechanics, this would mean that the system is at a stable fixed point.

Of course in QM that is not possible since we have to satisfy the Heisenberg uncertainty.

And so, I would say yes, in general. there might be some configurations where we might be in a false minimum (or a local minimum), which might also satisfy the uncertainty minimum, alternatively there might be a need to transform coordinates and redefine displacement and momentum, but if we are working with canonical conjugates, it is supposed to be true.

Take it as you want, but this is the way I interpret the necessary existence of a ground state (at finite energy) for any bound system in quantum mechanics. The idea, in my view, consists in finding the minimum energy value of an hamiltonian of the form $H = \frac{\vec{p}^2}{2m} + V(\vec{x})$ under the statistical constraint that, say, $\Delta x \Delta p_x \geq \frac{\hbar}{2}$.

A rapid account of this strategy (I have seen a more rigorous reference somewhere else but I can't retrieve it for now) can be found here.

A Gaussian wave packet $\psi(r)=e^{-r^2/2a}$ actually satisfies $\triangle x \triangle p_x = \frac{\hbar}{2}$. As @luming points out, this is the ground state of the harmonic oscillator.

In general a Gaussian isn't going to be an eigenstate of the Hamiltonian, but the lowest energy eigenstate should have the fewest number of oscillations, and so has the best chance of being a Gaussian-like lump. I'll try to update with a more rigorous answer, as @doetoe has pointed out that solving for the potential given a wave function is actually more straightforward solving for the wave function.

• Finding a Hamiltonian for which a given function and number are an eigenstate and energy eigenvalue only involves differentiation: $-{1\over\psi}\left({\hbar^2\over 2m}\nabla^2\psi\right) = E - V(x,y,z)$. The only potential giving a Gaussian is a quadratic potential (harmonic oscillator). Since a twice differentiable potential looks like a harmonic oscillator potential at a minimum, up to second order, it could be expected that the associated eigenstate also looks like a Gaussian up to a certain approximation, like you suggest. – doetoe Sep 24 '14 at 21:44
• The ground state of Harmonic Oscillator is a Gaussian. You saying "Over time, a free particle Gaussian will spread out in configuration space and contract in momentum space, maintaining the minimal uncertainty relation." is not correct. The uncertainty will enlarged with time – an offer can't refuse Sep 25 '14 at 3:52
• Thank you for the corrections. My original answer was incredibly wrong! I've updated it. – adipy Sep 25 '14 at 13:07