# Why Heisenberg's Uncertainty Principle?

My teachers have explained this using the electromagnetic wave thing, where if the energy of the wave is high then the body which we are measuring has a change in velocity, and if the wavelength is high then the body's position is fuzzy.

However, is there no mathematics which can measure the effect which the high energy EM wave will have on the velocity of the object? Much like kinematics in classical physics where we can figure out the initial velocity from the acceleration and final velocity. Why are there no such formulations for the quantum world too?

For example, say I shine light with short wavelength on an object and get a highly accurate position information. Now say I shine a light ray with long wavelength to measure the velocity with high precision. So now I know the final velocity and I also have the mass of the particle and energy of the EM wave which is supposed to have caused a change in velocity of the particle. Why can't we just determine the initial velocity from calculating the change in energy? Why does it have to be uncertain?

• I'm not sure what sort of "why" you're looking for here. Have you seen the proof of the general uncertainty principle as a consequence of how quantum mechanical operators work? Your question sounds as if you might not have seen an actual derivation of the uncertainty principle. Commented Jul 25, 2022 at 15:29
• @ACuriousMind that's right I haven't, we are being taught only the basics of QM in school. I think it's my bad to ask this question without doing more research... should I delete my post then? Commented Jul 25, 2022 at 15:30
• @AlternatingCurrent Uncertainty is a property of our universe.
– user326901
Commented Jul 25, 2022 at 17:43
• Once one grants the correctness of basic modeling of quantum stuff by a bit of mathematics, the "uncertainty principle" becomes an easy (and well documented, and google-able) little fact about Fourier transform. Yes, seems weird to me, too. :) Commented Jul 26, 2022 at 0:01
• Well, $$e^{i\theta}=\cos(\theta)+i\sin(\theta)$$ This beautiful equation leads to many interesting things. :) Commented Jul 26, 2022 at 14:33

There are laws that allow the energy transfer between EM radiation and charged particles to be predicted when they collide; but it yields probabilities applicable to large numbers of interactions rather than precise answers for single interactions. That whole field is called quantum electrodynamics and Richard Feynmann wrote a useful book for beginners about it, which I highly recommend.

As alluded to by ACuriousMind, there are mathematical aspects of QM (noncommutativity) which mean that for quantum particles, if you measure position first and then momentum you'll get a different answer from when you measure momentum first and then position- and the more you try to make one measurement precise, the less precise will be the other.

Feynmann's take on this was to remember that quantum objects can behave like particles or like waves depending on their circumstances. So imagine that a train of EM waves passes by you. You can measure the period of the wave crests precisely but the question "where is the wave located?" has no answer because the wave train is continually expanding into 3-dimensional space: it is there and there and there all at the same time.

Regarding your last question, the flaw in the reasoning is this: the two measurements you propose are nonsimultaneous and nothing prevents you from making nonsimultaneous measurements of conjugate variables. So you can precisely determine the position of the particle at one particular instant in time and then its momentum at some later instant of time but the particle will be in two different quantum states at those two different times.

The reason for the uncertainty is that the map between momentum and position representation of the wavefunction is the Fourier transformation. The underlying math is a little subtle if you are new to the subject.

Naively: The eigenfunctions of the momentum operator for the eigenvalue $$k\in\mathbb{R}$$ are $$f(k)=e^{ik x}$$, while the eigenfunctions for the position operator satisfy $$xf(x)=\lambda f(x)$$. Neither of these are square-integrable functions, which makes them useless for any computation regarding a quantum system, where observables are described by self-adjoint operators.

The solution is that we define them on a dense subset $$D(A)$$ in $$L^2(\mathbb{R})$$. Usually this is the Schwartz space (the space of a smooth function approaching zero at infinity faster than any polynomial), or the space of smooth functions with compact support. Then the functions $$f$$ form a complete set of generalized eigenfunctions in that the satisfy:

$$T(Af)=\lambda T(f)$$

for a tempered distribution $$T$$ and $$A$$ a symmetric operator on the dense subset. The $$T$$ are elements of the dual space of the Schwartz space and form a ring with the Fourier transformation (which is an automorphism on Schwartz space) and the delta-distribution as the unit-element. The uncertainty principle follows now from the fact that delta-distributions are generalized eigenfunctions of the position operator, in the prescribed manner and by remembering that the momentum is $$p=\hbar$$k .

A bit of philosophy:

Physics is based on experiments. The assumption is that the physical world exists and that we can understand it by experiments.

Over time we have created models that very closely describe the behaviour we see in the experiments. We use these models to make predictions of the results we will get from other experiments. But we are always aware that if we can create one single experiment that breaks the model we have to change our model -- the real world experiment result is the quality marker for our model.

Notice that the models never really answer the question why. The physical world simply exists and does whatever it wants to do. When we use a model we make it to get a prediction of something, but this is not a description of why.

Over time we have created more and more complex models, often to describe the edge cases. Well known example is the progression from Newtons laws of motion, over special and general relativity to quantuum mechanics. Newtons laws are quite good for a lot of things and we still use them, but gives wrong predictions at speeds approaching the speed of light. Special relativity is very good at a lot of things but gives wrong predictions at very small sizes of thing where we instead use quantuum mechanics. Newton laws of motion does not predict Heisenbergs uncertainty in any way.

In every day usage and in the process of learning things we often use simpler models than the ones at the forefront of our current learning. The reason often is that the models are complicated or use specialized mathematical tools that the student does not know yet.

To get back to the case of the Heisenberg uncertainty principle it is verified by experiments to exist and is well described by our models. So from that point of view, it simply exists and hence the model is a good model in predicting this part of reality. Any model that does not predict this behaviour is, if you take it as an absolute, a false model but can still be useful. If we compare it to Newtons motion laws we know they are false but they are still useful in everyday life.

The question why the uncertainty exists then becomes a non-valid question -- it simply is how the world works. I believe a better question to ask is how does this arise (or not) from predictions in our current best model. You see answers to that question in other posts here.

So to conclude. Experiments show that Heisenbergs uncertainty principle exists. A model that does not include this experimental result is either wrong or possibly a simplification only allowed give certain assumptions.

NOTE: a google search for "experimental verification of Heisenberg uncertainty principle" gives some insight in how the model can be experimentally verified. Hint: not easy. Hint 2: if you could produce an experiment that shows that the model is wrong and the principle is shown to be false, I guess you might be in for a Nobel price.