# Looking for a very cheap heuristic demonstration (in a few lines) of Heisenberg inequality

While I perfectly know the true demonstration of Heisenberg uncertainty principle (from the full QM machinery), I'm looking for a very cheap heuristic way of getting $$\Delta x \, \Delta p \ge h, \tag{1}$$ without getting into the semi-classical "Heisenberg microscope" experiment, or any wave packet superposition. I need this for a very short and simple exposition to the subject for my very low level students.

I was thinking about using only the deBroglie relations (which my students are learning): $$p = \frac{h}{\lambda}, \tag{2}$$ $$E = h f, \tag{3}$$ and saying something like this: The particle's position at time $$t$$ is $$x(t) = x_0 + v_0 t$$. During a small amount of time $$\Delta t$$, its position cannot be measured with a better accuracy than $$\Delta x = v_0 \Delta t$$ (or maybe it should be $$\Delta x = \Delta v_0 \, t$$, or even $$\Delta x \approx \lambda$$ ?), thus ... after few lines of blabla and maths ... we get (1).

Any idea or suggestion on how I could get (1) by using (2) and (3), and a few basic classical kinematics formulas? Basic differential calculus is an option.

My goal is to show that the deBroglie relations (as a starting point) in one dimension only (ie on the $$x$$ axis) and some basic classical kinematics imply the Heisenberg uncertainty principle.

Maybe a way is to use the wave's frequency as a clock of resolution $$\Delta t = 1/f$$. I'll use $$E = p^2 / 2m$$. The particle's position cannot have a resolution better than $$\Delta x \ge v \, \Delta t = \frac{p \, \Delta t}{m} = \frac{h \, p}{m E} = \frac{2 h}{p}. \tag{4}$$ Then what? How to define the momentum's uncertainty with that clock? $$\Delta x \Delta p \ge \frac{h \, 2p \, \Delta p}{p^2} \approx h \, \frac{\Delta (p^2)}{p^2}.\tag{5}$$

I think the Heisenberg uncertainty principle, or at least the simple wave mechanics version of it (which may be the only one your students are equipped to understand at this point), really stems from the idea that most wavepackets don't have a definite frequency and wavelength. That is, in $$p = h/\lambda$$, there is not just one $$\lambda$$. Maybe this concept falls under the "wave packet superposition" idea that you wanted to avoid, but if that's the case I would argue that you cannot properly explain the HUP given your constraints, and any attempt to come up with a roundabout argument that finds its way to the right result without talking about superposition and wavepackets would be doing them a disservice.

That being said, I don't think wavepackets and superposition are that hard to figure out. If your students have access to graphing software it should be easy enough for them to try combining a few pure cosine waves and see what happens to the resulting function, along the lines of $$\frac{1}{N}\sum_{k=1}^{N}\cos(kx),\ N=\{1,2,3,\ldots\}$$ It's going to be quite obvious that the central peak shrinks as more terms are added to the sum. You could point them to Lagrange's identities, specifically $$\sum_{n=1}^{N}\cos(n\theta) = -\frac{1}{2} + \frac{\sin[(N+1/2)\theta]}{2\sin(\theta/2)}$$ and have them investigate how the width of that central peak - perhaps expressed by the position of the first zero - changes as a function of $$N$$.

Now, I don't know offhand if this is going to be enough to actually calculate the right inverse relationship between the spread in momentum (roughly, $$N$$) and the spread in position (the width of the central peak). If nothing else, they are working with discrete frequencies, which might mess up the principle a little bit. But it will hopefully get the main point across of what the uncertainties in HUP mean and why there is some kind of inverse relationship between them.

• I would prefer a route that take small variations in $\lambda$ : $\Delta p \approx h \, \Delta \lambda / \lambda^2$, and $\Delta x \approx \lambda$, so $\Delta x \, \Delta p = h \, \Delta \lambda / \lambda$. But then, how to justify $\Delta \lambda \ge \lambda$ ? – Cham May 21 at 3:17
• @Cham There is no justification for $\Delta x = \lambda$. It's just a rebranded version of the "observer effect". – Dvij D.C. May 21 at 3:20
• Then what about a clock to define an accuracy in measurements of position? Another "observer effect" I guess? – Cham May 21 at 3:21
• I now believe that it is true that we need to consider a superposition of waves, with some dispersion in wavelenghts: $\lambda_0 - \Delta \lambda < \lambda < \lambda_0 + \Delta \lambda$. In that case, $p = h / \lambda$ gives $\Delta p \approx h \, \Delta \lambda / \lambda_0^2$ (assuming $\Delta \lambda \ll \lambda_0$). But then, how should we define the uncertainty in position for a wave packet, without an integral calculation for an average? – Cham May 21 at 3:56
• If actually doing the integral to calculate the variance isn't a reasonable option, then you probably have to take some proxy like the width of the central peak, as I used in my answer. You do have the flexibility to choose to define that in the easiest possible way to calculate. – David Z May 21 at 4:01

For the level you seem to be aiming for, my advice is to stick with binary observables instead of continuous ones.

So here's a good exercise for your students:

Let $$A=(1,0)$$ and let $$B=(0,1)$$

Take an observable $$O_1$$ with eigenstates $$A,B$$.

Take an observable $$O_2$$ with eigenstates $$U,V$$ where $$U$$ is proportional to $$A+B$$ and $$V$$ is proportional to $$-A+B$$. (You can call these "horizontal spin" and "vertical spin" if you like.)

Now suppose the system is in the state

$$\xi=\sqrt{p}\cdot A+\sqrt{1-p}\cdot B$$

(Or you could start with a more general $$\xi$$, allowing non-real coefficients, but the real case is already probably sufficient to make the point.)

If you measure observable $$O_1$$, you'll get state $$A$$ with probability $$p$$ and state $$B$$ with probability $$1-p$$. It's reasonable to define the associated uncertainty as $$p(1-p)$$ (so that the uncertainty is zero when $$p=0$$ or $$p=1$$ --- i.e. the uncertainty is zero when you can predict the outcome of your measurement with certainty --- and the uncertainty is maximized when $$p=1/2$$).

Now use linear algebra to rewrite $$\xi$$ in terms of the basis $$U,V$$, and multiply the coefficients on $$U$$ and $$V$$ to get the uncertainty for a measurement of $$O_2$$.

Now you've got an expression for the uncertainty of $$O_1$$ and an expression for the uncertainty of $$O_2$$, both as functions of $$p$$. Multiply the two uncertainties and use calculus to find the $$p$$ that minimizes this product. Compute the corresponding minimum and discover that it is greater than zero.

• That answer is way too abstract and sophisticated for my students. They don't even know what is an eigenvalue! – Cham May 21 at 3:14