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.

 

[Entropia]

Hello,

Thank you for sharing your work and thoughts on estimating the ground state energy using Heisenberg's uncertainty relation. It is important to note that this method is just an approximation and is not exact. The substitution of (\Delta p) and (\Delta x) in the Hamiltonian is valid because it considers the minimum possible uncertainty in the momentum and position of the particle. This is why the energy obtained from this method is the lower bound or minimum possible energy. It is not the exact ground state energy, but rather an estimation.

As for your question about the validity of the substitution, it is important to understand that in quantum mechanics, the position and momentum of a particle cannot be precisely determined at the same time. This is a fundamental principle of quantum mechanics and is described by Heisenberg's uncertainty principle. The substitution takes into account this uncertainty and provides an estimation of the ground state energy.

It is also worth noting that this method is only applicable for certain potentials, such as the harmonic oscillator potential, and may not be accurate for other potentials. In general, it is best to use other methods, such as solving the Schrödinger equation, to obtain the exact ground state energy.

I hope this helps clarify the validity and limitations of using Heisenberg's uncertainty relation to estimate the ground state energy. Keep up the good work in exploring different methods in quantum mechanics!
Best regards,