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 $\left<H\right>=\left<P^2\right>/2m$ and $\Delta P\cdot\Delta X\geq \hbar/2$ that $$\left<H\right>\geq\frac{\hbar^2}{8m\left(\Delta X\right)^2}.$$

Hence the ground state's energy eigenstate $E_1$ is constrained by $$E_1=\left<H\right>\geq\frac{\hbar^2}{2mL^2}$$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 $\left<H\right>$ 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 not very 'small'?

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?

Related: Using the Heisenberg Uncertainty Relation to Estimate Ground State Energies

  • $\begingroup$ 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. $\endgroup$
    – jacob1729
    Feb 7, 2019 at 15:57
  • $\begingroup$ @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! $\endgroup$
    – user191954
    Feb 7, 2019 at 16:08
  • $\begingroup$ 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 $\left[L^{-1}\right]$. $\endgroup$
    – user191954
    Feb 7, 2019 at 16:19
  • 1
    $\begingroup$ It seems you (or Shankar) picked $ \Delta X= L/2$, why so? $\endgroup$
    – lcv
    Feb 7, 2019 at 17:46
  • 1
    $\begingroup$ HUP only gives a lower bound for the ground state energy. $\endgroup$
    – Qmechanic
    Feb 7, 2019 at 18:01

1 Answer 1


Firstly, if you consider another experiment you will get a similar "off by a factor of 10" result. I am thinking of the case where a light beam passes through a single slit and we look at the diffraction pattern on a screen. You can deduce a photon's position uncertainly from the width of the slit, and the momentum uncertainty from the width of the diffraction pattern.

The reason for the factor of 10 is hiding in the definition of the "delta". A wave function in position space is the fourier transform of its wave function in momentum space. Only in the case of a gaussian wave packet will it fourier transform into another gaussian. In that case you can define "delta" as some measure of the width of the gaussian and you will recover the uncertainly principle exactly.

However a gaussian extends over all space, even though its tails do fall off quickly. In the case of the infinite square well or the single slit experiment the walls are "hard" and one becomes a good deal more "certain" about the position. This extra certainty in position costs you in momentum - and thus drives up the energy which scales as momentum squared.

For comparison, consider a bound state in a box with only finitely high walls. As you relax the wall height you now find the particle may be outside the box, but you still pretty much know where it is. However, the energy will come way down and begin to approach the uncertainly principle that you know and love.

So, all in, I answer your question with: it is a so-so estimate, which is not so bad considering the artificial geometry.


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