Calculate [D, r] and [D, p] with Dilatation Operator

[Goodwater]

Homework Statement

Consider the dilatation operator: D = r * p

Calculate [D , r] and [D , p]

Homework Equations

The Attempt at a Solution

 

[arkajad]

It is not clear what r stand for and what * stands for. Do you mean

$$D=\mathbf{r}\cdot\mathbf{p}=xp_x+yp_y+zp_z$$?

Do you mean $[D,\mathbf{r}]$ or $[D,r]$ where $r=\sqrt{\mathbf{r}^2}$. These details are important.

 

[Goodwater]

Hi.

Thank you for reply.

You are correct: D = xpx + ypy + zpz
I meant [D,r] where r is not square root.

I hope this is more understandable, as this is my first post I don't know how to get the notations right:)

 

[arkajad]

The you calculate using $[AB,C]=A[B,C]+[A,C]B$, for isntance

$$[xp_x,x]=x[p_x,x]+[x,x]p_x=\ldots$$

Mostly you will get zeros, but occasionally you will get a term that will contribute.

 

[paranormal]

To calculate [D, r], we can use the commutator property of operators, which states that [A, B] = AB - BA. Therefore, for [D, r], we have:

[D, r] = Dr - rD

Substituting in the given dilatation operator D = r * p, we get:

[D, r] = r * p * r - r * r * p

= r * (p * r - r * p)

= r * [p, r]

Similarly, for [D, p], we have:

[D, p] = Dp - pD

= r * p * p - p * r * p

= p * (r * p - r * p)

= 0

Therefore, [D, p] = 0, as the order of multiplication does not matter for scalars.

Overall, the dilatation operator [D, r] is equal to r * [p, r], while [D, p] is equal to 0. This shows that the dilatation operator does not commute with p, but it commutes with r. This information can be useful in studying the behavior of systems under dilatation transformations.