## Matrix Multiplication and Heisenberg Uncertainty Principle

We now understand that Matrix multiplication is not commutative (Why?). What has this have to do anything with Quantum Mechanics ?

Behold the commutator operator:
$[\hat{A}, \hat{B}] = \hat{A}\hat{B} - \hat{B}\hat{A}$

where $\hat{A},\hat{B}$ are operators that are acting on the wavefunction $\psi$. This is equal to 0 if they commute and something else if they don’t.

One of the most important formulations in Quantum mechanics is the Heisenberg’s Uncertainty principle and it can be written as the commutation of the momentum operator (p) and the position operator (x):

$[\hat{p}, \hat{x}] = \hat{p}\hat{x} - \hat{x}\hat{p} = i\hbar$

If you think of p and x as some Linear transformations. (just for the sake of simplicity).

This means that measuring distance and then momentum is not the same thing as measuring momentum and then distance. Those two operators do not commute! You can sort of visualize them in the same way as in the post.

But in Quantum Mechanics, the matrices that are associated with $\hat{p}$ and $\hat{x}$ are infinite dimensional. ( The harmonic oscillator being the simple example to this )

$\hat{x} = \sqrt{\frac{\hbar}{2m \omega}} \begin{bmatrix} 0 & \sqrt{1} & 0 & 0 & \hdots \\ \sqrt{1} & 0 &\sqrt{2} & 0 & \hdots \\ 0 & \sqrt{2} & 0 &\sqrt{3} & \hdots \\ 0 & 0 & \sqrt{3} & 0 & \hdots \\ \vdots & \vdots & \vdots & \vdots \end{bmatrix}$

$\hat{p} = \sqrt{\frac{\hbar m \omega}{2}} \begin{bmatrix} 0 & -i & 0 & 0 & \hdots \\ i & 0 & -i \sqrt{2} & 0 & \hdots \\ 0 & i\sqrt{2} & 0 &\-i \sqrt{3} & \hdots \\ 0 & 0 & i\sqrt{3} & 0 & \hdots \\ \vdots & \vdots & \vdots & \vdots \end{bmatrix}$