Estimate ground state energy with HU

[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 which (\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?


Thanks for your effort

and with best regards

 

[azntoon]

The substitution is valid because the Heisenberg uncertainty principle sets a limit on the precision with which we can measure both the position and momentum of a particle. This limit is expressed as the product of the uncertainties in position and momentum being greater than or equal to Planck's constant divided by 2 (ΔxΔp ≥ h/2). By substituting the uncertainty in position and momentum into the Hamiltonian, you are effectively setting this limit, and finding the minimum energy at which the particle can exist given this limit. In other words, the particle is confined to a certain region of space with a certain momentum, and the total energy of the particle is the lowest it can be when these conditions are met.