# A strange operator

In a previous post on using the Feynman’s trick for Discrete calculus, I used a very strange operator ( $\triangledown$ ). And whose function is the following :

$\triangledown n^{\underline{k}} = \frac{n^{\underline{k+1}}}{k+1}$

What is this operator? Well, to be quite frank I am not sure of the name, but I used it as an analogy to Integration. i.e

$\int x^{n} = \frac{x^{n+1}}{n+1} + C$

What are the properties of this operator ? Let’s use the known fact that $n^{\underline{k+1}} = (n-k) n^{\underline{k}}$

$\triangledown n^{\underline{k}} = \frac{n^{\underline{k+1}}}{k+1}$

$\triangledown n^{\underline{k}} = \frac{(n-k) n^{\underline{k}}}{k+1}$

And applying the operator twice yields:

$\triangledown^2 n^{\underline{k}} = \frac{n^{\underline{k+2}}}{(k+1)(k+2)}$

$\triangledown^2 n^{\underline{k}} = \frac{(n-k-1) n^{\underline{k+1}}}{(k+1)(k+2)}$

$\triangledown^2 n^{\underline{k}} = \frac{(n-k-1)(n-k) n^{\underline{k}}}{(k+1)(k+2)}$

We can clearly see a pattern emerging from this already, applying the operator once more :

$\triangledown^3 n^{\underline{k}} = \frac{(n-k-2)(n-k-1)(n-k) n^{\underline{k}}}{(k+1)(k+2)(k+3)}$

$\vdots$

Or in general, the operator that has the characteristic prescribed in the previous post is the following:

$\triangledown^m n^{\underline{k}} = \frac{n^{\underline{k+m}}}{(k+m)^{\underline{m}}} n^{\underline{k}}$

If you guys are aware of the name of this operator, do ping me !

# 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 )

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

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