# Commutation of differential operators

• magicfountain
In summary, the commutator of the operators d/dx and x is equal to -2, or -1 if written in terms of [x,d/dx]. This can also be shown using properties of commutators such as [A+B,C]=[A,C]+[B,C] and [A,B]=-[B,A].
magicfountain

## Homework Statement

Evaluate the commutator $\left[\frac{d}{dx},x\right]$

## Homework Equations

$\left[A,B\right]=AB-BA$

## The Attempt at a Solution

$\left[\frac{d}{dx},x\right]=1-x\frac{d}{dx}$

I don't know how to figure out $x\frac{d}{dx}$.

Try using a test function here i.e. calculate [d/dx,x]f(x) and when you are done make sure you get some expression (...)f(x) again then what is between the brackets is your commutator.

I actually just figured this out.
Can somebody please just check this one I did for mistakes?

[A,B]=[(d/dx + x),(d/dx - x)]
(AB-BA)f=(d/dx + x)(d/dx - x) f -(d/dx - x)(d/dx + x) f
(d/dx + x)(df/dx - xf)- (d/dx - x)(df/dx + xf)
(df^2/dx^2 - f - x df/dx)+(x df/dx - x^2f)-(df^2/dx^2 + f +x df/dx)+(x df/dx + x^2f)
=-2f

--> [(d/dx + x),(d/dx - x)]=-2

If f is any function then (d/dx)(x f)=x df/dx+ f while x(d/dx)f= xdf/dx. The difference is f so the commutator, [d/dx, x] is the identity operator.

magicfountain said:
I actually just figured this out.
Can somebody please just check this one I did for mistakes?

[A,B]=[(d/dx + x),(d/dx - x)]
(AB-BA)f=(d/dx + x)(d/dx - x) f -(d/dx - x)(d/dx + x) f
(d/dx + x)(df/dx - xf)- (d/dx - x)(df/dx + xf)
(df^2/dx^2 - f - x df/dx)+(x df/dx - x^2f)-(df^2/dx^2 + f +x df/dx)+(x df/dx + x^2f)
=-2f

--> [(d/dx + x),(d/dx - x)]=-2
That's fine.

You could have also used a few properties of commutators to save yourself some work, namely [A+B,C]=[A,C]+[B,C] and [A,B]=-[B,A]. Using those, you should be able to show that [(d/dx + x), (d/dx - x)] = 2[x, d/dx], and you've probably already worked out that last one and found it to be equal to -1.

vela said:
That's fine.

You could have also used a few properties of commutators to save yourself some work, namely [A+B,C]=[A,C]+[B,C] and [A,B]=-[B,A].

## What is the definition of commutation of differential operators?

Commutation of differential operators refers to the process of determining whether two given differential operators commute, meaning that their order of application does not affect the final result of a mathematical operation. In other words, if two operators A and B commute, then applying A followed by B will yield the same result as applying B followed by A.

## Why is commutation of differential operators important?

Commutation of differential operators is important because it allows for the simplification of complex mathematical equations. By determining which operators commute, scientists and mathematicians can reduce the number of operations needed to solve a problem, making it more efficient and elegant.

## What are the methods used to determine commutation of differential operators?

There are several methods used to determine commutation of differential operators, including using commutator brackets, tensor calculus, and the Lie derivative. These methods involve manipulating the operators and comparing them to see if they commute or not.

## Can all differential operators commute with each other?

No, not all differential operators can commute with each other. Some operators have special properties that allow them to commute with specific types of operators, while others do not commute at all. It is important to understand the properties and behaviors of each operator in order to determine if they can commute.

## How is commutation of differential operators used in scientific research?

Commutation of differential operators is used in various fields of science, including physics, engineering, and mathematics. It allows scientists to simplify complex equations and better understand the relationships between different variables in a system. It is also used in the development of new mathematical models and theories, as well as in the analysis of data and experimental results.

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