# Estimate energy of infinite well (ground state)

thatguy14

## Homework Statement

We have to estimate the ground state energy of an infinite potential well (1d) using an argument based on the Heisenberg uncertainty principal. We then are supposed to compare it with the exact value from the eigenvalue equation.

Below

## The Attempt at a Solution

I am not really sure where to start on this one but this is what I tried:

ΔxΔp = h and since there is no potential where the particle is all the energy is kinetic so classically Ke = E = p^2/2m. then to look at change i just did

Δp = $\sqrt{ΔE2m}$

and subbed it into the above unvertainty principle. This gives

Δx$\sqrt{ΔE2m}$ = h
$Δx^{2}$ΔE2m = $h^{2}$
ΔE = $\frac{h^{2}}{2mΔx^2}$

The problem is that even if this is somewhat right I am not sure what I really did. Can anyone give me some hints?

Homework Helper
Gold Member
Usually Δ represents "change" in a quantity. But in the uncertainty principle, it indicates the uncertainty in a quantity. The uncertainty is loosely the "spread" in possible results of measuring that quantity.

You know the particle is somewhere in the box of length L. That's all the information you have to go on when trying to estimate the uncertainty in position. So, as a crude estimate, you can take the uncertainty in x to be L: Δx $\approx$ L. Using this estimate for Δx, the uncertainty principle gives an estimate for the uncertainty in momentum Δpx.

To estimate the ground state energy, you need an estimate of the momentum px since E = px2/(2m). You have an estimate for Δpx, but you need an estimate of px itself. Here's where the argument gets "shaky" in my opinion. It seems reasonable that px should be at least of the order of magnitude of Δpx. The usual assumption is to just estimate px by its uncertainty Δpx.

Now, it is easy to think of examples where the value of some quantity is orders of magnitude greater than the uncertainty in the quantity. So, you could worry that estimating px by Δpx might be a bad estimate. But that is what is done in this case.

You can sort of support this by considering the classical particle in a box. The classical particle bounces back and forth in the box with a fixed magnitude of momentum but with a changing direction. So, the momentum has two possible values $\pm$ px and each is equally likely. So, the "spread" in momentum is of the same order of magnitude as the momentum itself.

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