# Accuracy of Heisenberg Uncertainty Principle for estimating ground state energy of particle in potential well

I've understood the assumptions and logic behind the 'proof' that the ground state of a particle in an infinite potential well has a non-zero energy using the Heisenberg Uncertainty Principle. Essentially, it is proved from $$\left=\left/2m$$ and $$\Delta P\cdot\Delta X\geq \hbar/2$$ that $$\left\geq\frac{\hbar^2}{8m\left(\Delta X\right)^2}.$$

Hence the ground state's energy eigenstate $$E_1$$ is constrained by $$E_1=\left\geq\frac{\hbar^2}{2mL^2}$$where $$L$$ is the width of the well (i.e we're constrained to $$-L/2\leq x\leq L/2$$ for allowed values of $$x$$), if we consider an energy eigenstate with $$n=1$$. Obviously, $$n=0$$ is neglected because that suggests a zero-energy case with a useless wavefunction like $$\psi_0=0$$.

The canonical attempts to find actual energy eigenstates of a particle in a box gives $$E_n=\frac{\hbar^2\pi^2n^2}{2mL^2}.$$ Thus Shankar says in Principles of Quantum Mechanics "the uncertainty principle is often used in this fashion to provide a quick order-of-magnitude estimate for the ground-state energy".

But my question is why is this HUP method a good estimation? Why is this lower bound 'close' to the actual value? I don't see why it isn't a meaningless/trivial bound like saying "the lower bound for 1 mole is 2 particles because I can obviously divide 12 grams of carbon-12 more than twice without changing chemical properties".

Is it purely a coincidence that we're off by $$\pi^2$$ (and not some other 'huge' value or $$L$$-dependent factor) from the actual value when we use our estimate to find the lowest $$\left$$ which is not forbidden by HUP? Is it perhaps a bad estimation, since

1. with the appropriate value of $$L$$, the difference between HUP's lower bound and the actual $$E_1$$ is huge OR

2. $$\pi^2\approx9.86$$ is almost an order of magnitude, which means it isn't exactly an order-of-magnitude estimate any more?

Or is there a deeper concept which suggests that there is a low likelihood of $$\Delta P\cdot \Delta X \gg \frac{\hbar}{2}$$ being true, which would cause the expectation of the hamiltonian to be a whole lot larger than $$\hbar^2/(2ml)$$ for a particle in an infinite potential well?

• This isn't really an answer (hence a comment) but the HUP argument must produce a dimensionally correct result. What it predicts incorrectly is a dimensionless factor that as you note, isn't really that small. – jacob1729 Feb 7 at 15:57
• @jacob1729 That such an estimate/calculation of a lower bound must be off by a dimensionless constant (and not something like $L$ or $\hbar$) is actually an interesting point; as a parenthesized portion of my question indicates, I had not noticed that earlier! – Chair Feb 7 at 16:08
• However, now that I think of it, it should still be possible to create an $L$-dependent term which is dimensionless by adding some other constants and variables with a net dimension of $\left[L^{-1}\right]$. – Chair Feb 7 at 16:19
• It seems you (or Shankar) picked $\Delta X= L/2$, why so? – lcv Feb 7 at 17:46
• HUP only gives a lower bound for the ground state energy. – Qmechanic Feb 7 at 18:01