# Scalar Operator

1. Jun 28, 2005

### mmgbmm

According to the definition, an operator T that commutes with all components of the angular momentum operator is a scalar, or rank zero, operator. What is the mathematical definition to that statement?
How can I prove that the four dimensional Laplacian is a scalar operator?

Regards,

Last edited: Jun 28, 2005
2. Jun 28, 2005

### mikeu

The mathematical statement is that if $$\hat{T}$$ is an operator such that $$[\hat{T},\hat{L}_x]=[\hat{T},\hat{L}_y]=[\hat{T},\hat{L}_z]=0$$ then $$\hat{T}$$ is a scalar operator.

So to show that an operator is a scalar operator, you just have to show that those three commutators are zero.

Mike

3. Jun 28, 2005

### dextercioby

$$\left[\hat{\vec{J}}\cdot\vec{n},\hat{T}^{(k)}_{q}\right]_{-} =\sum_{q'} T^{(k)}_{q'}\left\langle k,q'\left | \right \hat{\vec{J}}\cdot\vec{n}\left | \right k,q \rangle\right$$

,so it suffices to prove it for the "z" component.

HINT:Use the position representation in which

$$\nabla^{2} \sim \delta_{lm}\hat{p}_{l}\hat{p}_{m}$$

Daniel.

Last edited: Jun 28, 2005
4. Jun 28, 2005

### George Jones

Staff Emeritus

I used to wonder often about the relationship bewteen tensor operators in quantum mechanics and tensors for physical 3-dimensional space, where the latter are defined either as multilinear maps, or as "things" that satisfy certain transformation properties.

A short passage on page 118 of the book Group Theory and Physics by S. Sternberg showed me that there is a connection. Elements of the rotation group act both on vectors in physical space and on (state) vectors in the state space of a quantum system. In other words, both of these spaces are representation spaces for representations of the the rotation group.

Tensor operators are (related to) operators on state space that correspond to the standard tensors of physical tensors via an intertwining operator (or morphism) between the representations.

Sternberg is a very nice introduction to the use of group theory in physics from an abstract point of view. I've always meant to write a little blurb that shows the correspdence bewteen the terse, general, and abstract treatment in Sternberg, and more the more concrete expositions found in quantum mechanics books. Unfortunately, I don't have access to Sternberg. I do have a few lines of handwritten notes from page 118. Maybe that will be enough.

With respect to the Laplacian, I was going to mention something about partial derivatives commuting, but I see that while I was writing, dextercioby has already given an equivalent but better suggestion.

Regards,
George

5. Jun 28, 2005

### reilly

The zero commutators demonstrate that the "scalar" is independent of the two standard spherical coordinate angular coordinates. Thus the "scalar" is a function of r only, and does not change under rotations -- and that's what we mean by a scalar -under rotation.

For the 4D-Laplacian, the t part is trivial. The spatial part is, up to a constant, momentum squared, p**2, and this commutes, as it should, with all the angular momentum operators.

Edmunds book on angular momentum goes into spherical tensors, rotations and all that stuff in a very complete fashion. Well worth fighting your way through, sometime down the road. Note that spherical tensors are defined only for rotations, so they are somewhat analogous to multipoles in E&M, less so to the standard tensors of relativity.

Regards,
Reilly Atkinson