# Mathematical Reason for Uncertainty in Quantum Mechanics

Today I’d like to give a fairly simple account of why Uncertainty Principles exist in quantum mechanics. I thought I already did this post, but I can’t find it now. I often see in movies and sci-fi books (not to mention real-life discussions) a misunderstanding about what uncertainty means. Recall the classic form that says that we cannot know the exact momentum and position of a particle simultaneously.

First, I like this phrasing a bit better than a common alternative: we cannot measure perfectly the momentum and position simultaneously. Although, I guess this is technically true, it has a different flavor. This makes it sound like we don’t have good enough measuring equipment. Maybe in a hundred years our tools will get better, and we will be able to make more precise measurements to do both at once. This is wrong, and completely misunderstands the principle.

Even from a theoretical perspective, we cannot “know.” There are issues with that word as well. In some sense, the uncertainty principle should say that it makes no sense to ask for the momentum and position of a particle (although this again is misleading because we know the precise amount of uncertainty in attempting to do this).

It is like asking: Is blue hard or is blue soft? It doesn’t make sense to ask for the hardness property of a color. To drive the point home, it is even a mathematical impossibility, not just some physical one. You cannot ever write down an equation (a wavefunction for a particle) that has a precise momentum and position at the same time.

Here’s the formalism that lets this fall out easily. To each observable quantity (for example momentum and position) there exists a Hermition operator. If you haven’t seen this before, then don’t worry. The only fact we need about this is that “knowing” or “measuring” or “being in” a certain observable state corresponds to the wavefunction of the particle being an eigenfunction for this operator.

Suppose we have two operators ${A}$ and ${B}$ corresponding to observable quantities ${a}$ and ${b}$, and it makes sense to say that ${\Psi}$ we can simultaneously measure properties ${a}$ and ${b}$. This means there are two number ${\lambda_1}$ and ${\lambda_2}$ such that ${A\Psi = \lambda_1 \Psi}$ and ${B\Psi = \lambda_2 \Psi}$. That is the definition of being an eigenfunction.

This means that the commutator applied to ${\Psi}$ has the property

${[A,B] = AB\Psi - BA\Psi = A\lambda_2 \Psi - B \lambda_1 \Psi = \lambda_2\lambda_1 \Psi - \lambda_1\lambda_2 \Psi = 0}$.

Mathematically speaking, a particle that is in a state for which it makes sense to talk about having two definite observable quantities attached must be described by a wavefunction in the kernel of the commutator. Therefore, it never makes sense to ask for both if the commutator has no kernel. This is our proof. All we must do is compute the commutator of the momentum and position operator and see that it has no kernel (except for the 0 function which doesn’t correspond to a legitimate wavefunction).

You could check wikipedia or something, but the position operator is given by ${\widehat{x}f= xf}$ and the momentum is given by ${\widehat{p}f=-i\hbar f_x}$.

Thus,

$\displaystyle \begin{array}{rcl} [\widehat{x}, \widehat{p}]f & = & -ix\hbar f_x + i\hbar \frac{d}{dx}(xf) \\ & = & -i\hbar (xf_x - f - xf_x) \\ & = & i\hbar f \end{array}$

This shows that the commutator is a constant times the identity operator. It has no kernel, and therefore makes no sense to ask for a definite position and momentum of a particle simultaneously. There isn’t even some crazy, abstract purely theoretical construction that can have that property. This also shows that we can have all sorts of other uncertainty principles by checking other operators.