# Estimate ground state energy with HU

1. Jan 22, 2010

### saunderson

Hi,

i worked through several examples where the ground state energy of a particle in an arbitrary potential V(r) is estimated with Heisenberg's uncertainty relation.

In these examples they prepare the Hamiltonian for the particle. For example the Hamiltonian for a particle in a harmonic oscillator potential:

$$H = \frac{p^2}{2m} + \frac{1}{2} m \omega^2 x^2$$

Then they estimate the lower bound of the impulse

$$\Delta p = \frac{\hbar}{2\Delta x}$$

In the next step they substitute

$$p = \Delta p = \frac{\hbar}{2\Delta x} \qquad \mathrm{and} \qquad x = \Delta x$$
in the Hamiltonian and look for wich (\Delta x) the energy is minimal.

Of course they get the right result of the ground state energy, but for me it doesn't make sense. In my mind's eye i have always a distribution where (\Delta p) or (\Delta x) is the mean square deviation of the wave function in momentum space or position space. If, for example the mean value of p=0 then (\Delta p) says simply that the most measured impulses are in the range of (\Delta p)!? So, why is the above substitution valid?