Irreducible representation of su(2)

Kara386
Messages
204
Reaction score
2

Homework Statement


Using the irreducible representation of ##su(2)##, with ##j=\frac{5}{2}##, calculate ##J_z##, ##exp(itJ_z)## and ##J_x##.

Homework Equations

The Attempt at a Solution


There seem to be loads of irreducible representations of ##su(2)## online, but no reference at all to a specific irreducible representation in my lecture notes. It will be a matrix representation, I suspect, involving something from physics because that's the context we're working in, so maybe the Pauli matrices? I'm completely stuck and any guidance or thoughts on what my lecturer might mean would be very much appreciated! :)
 
Last edited:
Physics news on Phys.org
Look into ladder operator methods. One defines ##J_\pm = J_x\pm i J_y## and gets commutation relations ##[J_z,J_\pm]=\pm J_\pm## where the signs I just wrote are likely all screwed up.
 
Paul Colby said:
Look into ladder operator methods. One defines ##J_\pm = J_x\pm i J_y## and gets commutation relations ##[J_z,J_\pm]=\pm J_\pm## where the signs I just wrote are likely all screwed up.
But the ladder operators aren't even elements of ##su(2)##... and they aren't generators. How can they be a representation?
 
Okay, best looked up in a book. In a nut shell one starts with ##J_-\vert -5/2\rangle = 0##. By applying ##J_+## to this "ground" state one generates all the eigen states in the rep. From these follow all operators in matrix form. It's work, that's why it's homework. It's also extremely elegant.
 
Kara386 said:
But the ladder operators aren't even elements of su(2)...

Ah, best looked up in a Physics book on quantum mechanics. Any intro text will do.
 
##\dim \mathfrak{su}(2) = 3## and therefore (both are simple) ##\mathfrak{su}(2) \cong \mathfrak{sl}(2,\mathbb{R})##.
Therefore you can get all representations as representations of ##\mathfrak{sl}(2,\mathbb{R})## which has a basis ##\{Y,H,X\}## with ##[H,X]=2X\, , \,[H,Y]=-2Y\, , \,[X,Y]=H## which can be represented by the matrices
$$ H=\begin{bmatrix}1&0\\0&-1\end{bmatrix}\, , \,X=\begin{bmatrix}0&1\\0&0\end{bmatrix}\, , \,Y=\begin{bmatrix}0&0\\1&0\end{bmatrix} $$
This makes it easier to find the representations as those of ##\mathfrak{sl}(2,\mathbb{R})## and easier to see the "ladder", as ##X=J_+\, , \,Y=J_-\, , \,H=J_z\,.## By the way, is ##j=\frac{5}{2}## meant to be the highest weight?

You can also look up the Wikipedia entry, which is not bad:
https://en.wikipedia.org/wiki/Representation_theory_of_SU(2)
 
fresh_42 said:
##\dim \mathfrak{su}(2) = 3## and therefore (both are simple) ##\mathfrak{su}(2) \cong \mathfrak{sl}(2,\mathbb{R})##.
Therefore you can get all representations as representations of ##\mathfrak{sl}(2,\mathbb{R})## which has a basis ##\{Y,H,X\}## with ##[H,X]=2X\, , \,[H,Y]=-2Y\, , \,[X,Y]=H## which can be represented by the matrices
$$ H=\begin{bmatrix}1&0\\0&-1\end{bmatrix}\, , \,X=\begin{bmatrix}0&1\\0&0\end{bmatrix}\, , \,Y=\begin{bmatrix}0&0\\1&0\end{bmatrix} $$
This makes it easier to find the representations as those of ##\mathfrak{sl}(2,\mathbb{R})## and easier to see the "ladder", as ##X=J_+\, , \,Y=J_-\, , \,H=J_z\,.## By the way, is ##j=\frac{5}{2}## meant to be the highest weight?

You can also look up the Wikipedia entry, which is not bad:
https://en.wikipedia.org/wiki/Representation_theory_of_SU(2)
The j values were defined as j=0 are scalars, ##j=\frac{1}{2}## are spinors, and so on. Every half integer increase in j seems to be associated with an increase in 1 of the dimension of the representation.
 
Back
Top