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I'm trying to understand the terms "Operators" and "Commutators".

Operators / Variables helps us to derive a differential equation that our wave equation must satisfy. Ex. Momentum Operator $P = \frac{\hbar}{i} \frac{\partial}{\partial x}$, Energy Operator $E = i \hbar \frac{\partial}{\partial t}$ etc.

And Commutators are defined by: $[A,B] = AB - BA$

I know how to solve the problems regarding Operators and Commutators, but I still haven't understand the concept behind them. What are actually Operators and Commutators?

Will R
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rndflas
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    Operators are like machines that take in functions and output a new unique one. For example the derivative is an operator because we apply it to functions to get a new one. However the indefinite integral is not because the function outputed is not unique. The purpose of operators vary though. – Will Fisher Jun 16 '16 at 16:00

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The term operator is usually used to indicate a function that take an element of a vector space and gives an element of another (or the same) vector space. So the simpler example of an operator, that is also linear, is a square matrix that transforms a vector of a finite dimensional vector space in another vector of the same space. I suppose that you know that matrices, in general, does not commute, so that if we have two matrices $A$ and $B$ than $AB \vec x \ne BA \vec x$.

The commutator is the operator (a matrix) defined as: $[A,B]=AB-BA$.

In Quantum Mechanics the vector spaces used are spaces of functions and the (linear) operators take a function and give another function. If we consider a space of functions of one variable $f(x)$ ( with suitable conditions for differentiability), than the moment opertor $P=-i\hbar \frac{d}{dx}$ acts on a function as: $$ -i\hbar \frac{d}{dx}f(x)=-i\hbar f'(x) $$ and we can see that another operator as the position operator defined as: $$ Qf(x)=x\cdot f(x) $$ does not commute with $P$ and we have: $$ [QP-PQ]f(x)=-i\hbar x f'(x)+i\hbar f(x)+i\hbar xf'(x)=i\hbar f(x) $$ so that the commutator is a constant operator: $$ [QP-PQ]=i\hbar $$

Emilio Novati
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