# Experimental physics: $\langle\mathbf{P}\rangle\approx\Delta \mathbf{P}$?

In many of my experimental courses solving a problem in atomic or solid state physics comes down to using the Heisenberg Uncertainity relation $$\Delta \mathbf{X} \Delta \mathbf{P}\geq \frac{\hbar}{2}$$ as a mean to estimate the magnitude of the impulse of an particle $$\langle\mathbf{P}\rangle$$. I will give two examples:

1. Calculating the energy of an electron in an atom by saying that it's contained in a cube with bohr radius $$r$$ as side lengths.

2. Calculating the velocity of an electron bouncing up and down in a monoatomic layer with thickness $$d$$.

What all of these calculations have in common is that one estimates the spread of the position of a particle $$\Delta \mathbf{X}$$ and with that obtains a lower bound for the uncertainty in momentum $$\Delta \mathbf{P}$$ of this particle $$\Delta \mathbf{P}\geq \frac{\hbar}{2\Delta \mathbf{X} }$$ Until here this seems reasonable. Now two miracoulus things happen in all of these calculations

$$\langle\mathbf{P}\rangle\approx\Delta \mathbf{P}\approx \frac{\hbar}{2\Delta \mathbf{X} }$$

While the last approximation seems o.k. as a very conservative estimate for the uncertainty of the impulse, the first one makes no sense at all.

The expectation value and the uncertainty are not related at all. I can construct examples with very small expectation values but huge uncertainties.

After solving an entire worksheet with such exerices I got frustrated and asked my professor why we are assuming that this makes any sense at all. Many of the other students in the lecture felt the same way and anticipated an answer. My professor giggled and told me this is just how things are done around here, continuing to tell us that the results we achieved in the exercise can be measured to a good deal of accuracy.

I'm not willing to accept this, there must be a deeper explanation why this kind of approximation works.

• In the rest frame, the expectation value could be anything in the interval $[- \Delta p_x, \, \Delta p_x]$ (and yet this is an approximation, assuming an uniform probability distribution wihin the uncertainty). $-\, \Delta p_x \le p_x \le +\, \Delta p_x$.
– Cham
Nov 18, 2019 at 14:45
• I suspect part of the motivation in these exercises is the development of "back of the envelope" calculations. Such thinking is great for developing intuition. Notice these substitutions preserve units and are close in terms of orders of magnitude. As an aside: one early example is estimating the ground state of the QHO assuming the momentum is the uncertainty in momentum and the uncertainty is position is the distance between classical turning points. A lot of problems can be swapped out for a similar QHO problem. Nov 18, 2019 at 15:18

It's certainly not true in general that $$\langle P \rangle\approx \Delta P$$, or even that $$\langle P \rangle\sim \Delta P$$. We can change $$\langle P \rangle$$ to any value we like by changing frames of reference, but this will have no effect on $$\Delta P$$.

For an example like a neutron inside a nucleus, or other examples that are like a particle in a box, we typically talk about the center of mass frame, so $$\langle P \rangle=0$$, while it is certainly not true that $$\Delta P=0$$.

One such example would be calculating the energy of an electron in an atom by saying that it's contained in a cube with bohr radius r as side lengths. And assuming that Δx=r.

This is an order-of-magnitude estimate. To within this precision, we expect the kinetic and potential energies to have similar magnitudes but opposite signs (as in the virial theorem for classical mechanics, or just because there is nothing else to set an energy scale). So to order-of-magnitude precision, if we can estimate the kinetic energy, that's an estimate of the total energy.

The kinetic energy is $$P^2/2m$$, so we really want $$\langle P^2\rangle$$. But for a standing wave in one dimension, the wavefunction consists of two components, $$P=p_0$$ and $$P=-p_0$$. (This is in the rest frame of the box.) Therefore the uncertainty in momentum is on the order of $$p_0$$, and also $$\langle P^2\rangle=p_0^2$$. This is why the procedure works in this example.

Since $$\Delta p^2 \equiv \langle p^2 \rangle - \langle p \rangle^2$$, you could write this: $$\frac{p^2}{2m} \approx \frac{\langle p^2 \rangle}{2m} \equiv \frac{\Delta p^2}{2m} + \frac{\langle p \rangle^2}{2m}.$$ But then $$\langle p \rangle \approx 0$$ in the rest frame.

In using the Heisenberg Uncertainty Principle (HUP) one should keep in mind that one is not talking about accuracy of measurement . One is talking of the quantum mechanical uncertainty introduced by the probability base of quantum mechanical formulations.

The $$ΔxΔp>h/{2π}$$ means that if momentum is measured with very great accuracy,then the uncertainty in the x position is large given by the HUP. If the x position is measured with very great accuracy, then the uncertainty in the value of the momentum is large.

It means measurements, a fixed framework where measurements can be done in the lab.

Actually I have not seen the form with the expectation value before, and certainly it cannot be correct in general. The specific problem must be given.

I have seen the HUP used taking the de Broglie wavelength as an estimate .