Group Representation: Understanding SO(3), SU(2), and the Clebsch-Gordan Theorem

[Calabi]

Good morning I'me french so excuse my bad language : so in this course : http://lapth.cnrs.fr/pg-nomin/salati/TQC_UJF_13.pdf take a look at page 16.
They say that all rotation auround a unitary vector \vec{u} of angle \theta in the conventionnal space
could be right like this with the matrix : e^{\theta(u_{x}A_{x} + u_{y}A_{y} + u_{z}A_{z})}, the 3 matrix A are given in the relation I-47 page 16.

So now take a look at those 3 matrix, we're going to right it X. They check the falowing commutation relation : [X_{a}, X_{b}] = \epsilon_{abc}X_{c}(1), the \epsilon are the Levi Civita symbol.

Now we're page 17 near relation I-55. We're going to associate at the matrix X_{x}, X_{y}, X_{z} 3 nxn matrix that we right A_{x}, A_{y}, A_{z} or like this A_{1}, A_{2}, A_{3}. Those 3 matrx nxn check the same commutation relation (1).
Actually we try to represent SO(3) and his Lie algebra so(3) in a \mathbb{C} vector space of dimension n. We call this space V.

Each matrix e^{\theta^{a}X_{a}} of SO(3) is represant by the unitary automorphism e^{\theta^{a}A_{a}} which proceed in V.

Now let's interrset us to the X_{z} matrox. She's represented in V by A_{z}. Now let's define H_{a} = iA_{a}. For exemple we have H_{3} = iA_{3}.

The matrix H_{3} could be considerate as an automorphism in V. Let's suppose there's a base in V where H_{3} is diagonalisable. Let's right the vector of this base(the eigenvectors of H_{3}.). like this : |m\rangle. We have H_{3}|m\rangle = m|m\rangle.

And now we define H_{+} = H_{1} + iH_{2} and H_{-} = H_{1} - iH_{2}.

We demonstrate that H_{+}|m\rangle = \beta_{m}|m+1\rangle. And H_{-}|m\rangle = \alpha_{m}|m-1\rangle.
The \alpha and \beta are to determine.

Now we can suppose that it exist in the base(form by the eigenvectors of H_{3}.). a eigenvector |j\rangle which is associated to a maximal eigenvalue j. There's also a minimal eigenvalue(which we demosntrate after is equats to -j.).

As you saw if you apply H_{+} on a eigenvector you obtein another eigenvector whose the eigenvalue was increase of 1.

In the same way if you apply H_{-} on a eigenvector you obtein another eigenvector whose the eigenvalue was decrease of 1.
Moreover you know that all the base of V contain n vectors.

So if you apply H_{+} on |j\rangle you obtein \vec{0}. In the same way if you apply H_{-} on |-j\rangle you obtein \vec{0}.
It's a sort of security.

So if you know |j\rangle you can obtein all the vectors of the base(form by the eigenvectors of H_{3}.). by applying n-1 time. on |j\rangle until you obtein |-j\rangle.

NOw we could demonstrate that j = \frac{n-1}{2}(2) and we also demosntrate that : \beta_{m} = \sqrt{j(j+1) - m(m+1)} and
\alpha_{m} = \sqrt{j(j+1) - m(m-1)}.

So now we're going to wright the vector of this base(the eigenvectors of H_{3}.). like this : |j, m\rangle. We have H_{3}|j, m\rangle = m|j, m\rangle.

Now with (2) you could see that : the eigenvalue are half integer(en français le mot est demi entière comme 1.5 3.5 4.5 5.5 enfin tu vois quoi.). if n is a peer number(2,4,4,8.). or whole number(1,2,3,4,5.). if n a odd number (1,3,5,7,9.).

An important things on that representation with H_{3} : we represent so(3) and in the same time su(2). Indeed those 2 Lie algebra are isomorph.

The representation I talk about from the beginning is irreducible. We wright it n.

For exemple 2 is the representation associated to the space whose on of base is (|\frac{1}{2},\frac{-1}{2}\rangle, |\frac{1}{2},\frac{1}{2}\rangle). In this base H_{3} is diagonalisable. And H_{3} = \frac{1}{2}\sigma_{z} = \begin{pmatrix} \frac{1}{2} 0 \\ 0 \frac{-1}{2} \end{pmatrix}.

And finally : we demonstrate that the representation 2 \otimes 2 = 3 \oplus 1 and the representation 3 \otimes 2 = 4 \oplus 2.

We have an illustration of the Clebsh Gordon theorem : a tensorial representation could be reduce at a addition of irreducible representation.

If you want I can explain more about this point. Just ask for it.

the quantity $\psi^a\psi_a$ invariant on a SU(2) transformation, Indeede I represant the SU(2) group whose de Lie algebra is isomorph to the Lie algebra of SO(3). At one element(rotation.). of SO(3) is associatted 2 element of SU(2). It come frome the fact that a spinor have to be turn in 4\pi to be equal as himself.

Whate ever : here is the probleme 30 translate in english :

"First check that the antysymetric part \psi_{[a \phi b]} could be wright like this : \psi_{[a \phi b]} = \epsilon_{ab}\psi_{m}\phi^{m}"

I've made it. It was easy.

Then they said : "show the antysymetric part is invariant on a SU(2) action"

I've made it to. Thene the course said : "In which representation is the antysyetric part" le cour dit : à quelle représentation la partie antysymetrique apartient t elle.

Then there's : "Show that the symetrics parts \psi_{(a \phi b)} has 3 components and comments".

I stump on those 2 question.

Could you help me to resolve it please?
Thank you very much and have a nice afternoon.

 

[Calabi]

I'll wright all of this on the course because it's in french. It's to avoid to you to translate and to presents the context.

Thank you in advance and have a nice afternoon.

 

[strangerep]

[...] here is the probleme 30 translate in english :

"First check that the antysymetric part \psi_{[a \phi b]} could be wright

FYI, the English word is spelled "write", not "right" nor "wright".

... like this : \psi_{[a \phi b]} = \epsilon_{ab}\psi_{m}\phi^{m}

I think you meant: $$\psi_{[a} \phi_{b]} = \epsilon_{ab}\psi_{m}\phi^{m} ~~~~~~~~ (1)$$

[...] Then they said : "show the antysymetric part is invariant on a $SU(2)$ action". I've made it to. Thene the course said : "In which representation is the antysyetric part"

Well, you've just shown that the expression (1) is invariant. And there's only 1 degree of freedom (i.e., $\psi_m\phi^m$). So... which representation has elements that don't change under any rotation?

Then there's : "Show that the symetrics parts \psi_{(a \phi b)} has 3 components and comments".

I guess you meant: $\psi_{(a} \phi_{b)}$ .

You've just got to count how many of the $\psi_{(a} \phi_{b)}$ products are independent. The $a,b$ indices take values in the set $\{1,2\}$, so try writing all possible combinations of "a" and "b" with those values, and don't worry about the order. How many different combinations do you get?

 

[Calabi]

Good morning strangerep and many thanks,

Well, you've just shown that the expression (1) is invariant. And there's only 1 degree of freedom (i.e., $\psi_m\phi^m$). So... which representation has elements that don't change under any rotation?

Euh there's 2 degree I think, this depend of \epsilon. Indeed : \epsilon_{ab} = -\epsilon_{ba}.

And it's not because : \psi_{a}\phi_{b} = \frac{1}{2}\psi_{[a} \phi_{b]}(= \epsilon_{ab}\psi_{m}\phi^{m}) + \frac{1}{2}\psi_{(a} \phi_{b)} and that there's 3 possibility for \psi_{(a} \phi_{b)} that you can said that there's 2 parts : one in the 1 representation and the 3 others in the 3 representation.

All I know for the moment is that ther's 4 possibility of \psi_{a}\phi_{b} which are the 4 components of the tensor |\psi\rangle \otimes |\phi\rangle in the base (|+\rangle \otimes |+\rangle, |-\rangle \otimes |-\rangle, |+\rangle \otimes |-\rangle, |-\rangle \otimes |+\rangle).

Could you be more clear please?

Maybe we should to develope |\psi\rangle \otimes |\phi\rangle in the base form by (|1,-1\rangle, |1,0\rangle, |1,1\rangle, |0,0\rangle) who's is described page 25 of this course : http://lapth.cnrs.fr/pg-nomin/salati/TQC_UJF_13.pdf.

Thank you in advance and have a nice morning.

 

[strangerep]

Euh

?? What does that mean?

there's 2 degree I think, this depend of \epsilon. Indeed : \epsilon_{ab} = -\epsilon_{ba}.

I should have said "... independent degrees of freedom". (If a degree of freedom is a constant multiple of another, then there's really only 1 independent degree of freedom, not 2.)

Could you be more clear please?

Since my French is much worse than your English, maybe I'll never be clear enough for you. But let us try one more thing...

Write out an arbitrary 2x2 matrix. Then write out the symmetric and antisymmetric parts of the matrix separately. How many independent degrees of freedom are there in each part?

 

[Calabi]

You've just got to count how many of the $\psi_{(a} \phi_{b)}$ products are independent. The $a,b$ indices take values in the set $\{1,2\}$, so try writing all possible combinations of "a" and "b" with those values, and don't worry about the order. How many different combinations do you get?

I get 3 combinations(because of the symettry.). but it's not because I see 3 that those 3 combinations are a part of the components of $|\psi\rangle \otimes |\phi\rangle$ in the parts of the base which is $(|1,-1\rangle, |1,0\rangle, |1,1\rangle)$.

Well, you've just shown that the expression (1) is invariant. And there's only 1 degree of freedom (i.e., $\psi_m\phi^m$). So... which representation has elements that don't change under any rotation?

It's the 1 representation(which is associated to the spin 0.). Indeed in this representation a rotation is : e^{0} = 1. It's the identity fonction so if you aply it you'll find the same components. The elements d'ont change under rotation.

But there's 3 posibility for : $\psi_{[a} \phi_{b]} = \epsilon_{ab}\psi_{m}\phi^{m}$.

If $a=b$ it's $0$, and $\psi_{[a} \phi_{b]} = \psi_{[b} \phi_{a]}$.

And even if they were one possibility in the same way, it's not because I see 1 that this 1 is a part of the components of $|\psi\rangle \otimes |\phi\rangle$ in the parts of the base which is $|0,0\rangle$. It's on this point I'm lock.

Indeed we try to made the Clebsh Gordon theorem but with the compnentns and not with the base this time(as we have made in the problem 25 page 25 of this course : http://lapth.cnrs.fr/pg-nomin/salati/TQC_UJF_13.pdf.).

Do you see what I mean please?

Thank you in advance and have a nice afternoon.

 

[strangerep]

[...] but it's not because I see 3 that those 3 combinations are a part of the components of $|\psi\rangle \otimes |\phi\rangle$ in the parts of the base which is $(|1,-1\rangle, |1,0\rangle, |1,1\rangle)$.

Yes, they are.

It's the 1 representation(which is associated to the spin 0.). Indeed in this representation a rotation is : e^{0} = 1. It's the identity fonction so if you aply it you'll find the same components. The elements d'ont change under rotation.

Yes.

[...] Do you see what I mean please?

Yes, I see exactly where you are confused.

I also see that you did not make any attempt to follow my suggestion at the end of my post #5.
If you do not at least try to work through my suggestions, then it's difficult for me to help you.
I will try one (but only one) more time...

Can you work out the following outer product, and then decompose the resulting matrix into symmetric and antisymmetric parts?
$$
\psi \phi^T ~=~
\begin{pmatrix} \psi_1 \\ \psi_2 \end{pmatrix} ~
\begin{pmatrix} \phi_1 & \phi_2 \end{pmatrix} ~=~ \cdots ~?
$$

 

[Calabi]

I get :
$$

\begin{pmatrix} \psi_1 \\ \psi_2 \end{pmatrix} \otimes
\begin{pmatrix} \phi_1 & \phi_2 \end{pmatrix} ~=~ \begin{pmatrix}\psi_1 \phi_1 & \psi_1 \phi_2 \\ \psi_2 \phi_1 & \psi_2 \phi_2\end{pmatrix} = \frac{1}{2}(\begin{pmatrix} \psi_1 \\ \psi_2 \end{pmatrix} \otimes
\begin{pmatrix} \phi_1 & \phi_2 \end{pmatrix} +
\begin{pmatrix} \phi_1 \\ \phi_2 \end{pmatrix} \otimes \begin{pmatrix} \psi_1 & \psi_2 \end{pmatrix} + \begin{pmatrix} \psi_1 \\ \psi_2 \end{pmatrix} \otimes
\begin{pmatrix} \phi_1 & \phi_2 \end{pmatrix} - \begin{pmatrix} \phi_1 \\ \phi_2 \end{pmatrix} \otimes \begin{pmatrix} \psi_1 & \psi_2 \end{pmatrix} ) = \frac{S+A}{2}
$$

Where $$S = \begin{pmatrix} \psi_1 \\ \psi_2 \end{pmatrix} \otimes
\begin{pmatrix} \phi_1 & \phi_2 \end{pmatrix} +
\begin{pmatrix} \phi_1 \\ \phi_2 \end{pmatrix} \otimes \begin{pmatrix} \psi_1 & \psi_2 \end{pmatrix}$$ is the symetric part and $$A = \begin{pmatrix} \psi_1 \\ \psi_2 \end{pmatrix} \otimes
\begin{pmatrix} \phi_1 & \phi_2 \end{pmatrix} - \begin{pmatrix} \phi_1 \\ \phi_2 \end{pmatrix} \otimes \begin{pmatrix} \psi_1 & \psi_2 \end{pmatrix}$$ is the antysymetric part.

Good afternoon.

 

[Calabi]

More : As we prevously show $$A = \begin{pmatrix} 0 & \psi_{1} \phi_{2} - \phi_{1} \psi_{2} \\ -(\psi_{1} \phi_{2} - \phi_{1} \psi_{2}) & 0\end{pmatrix} = \begin{pmatrix} 0 & \epsilon_{12}\psi_{m}\phi^{m} \\ \epsilon_{21}\psi_{m}\phi^{m} & 0\end{pmatrix}$$, since $$\epsilon_{12} = -\epsilon_{21}$$ this matrix could be considerate as to be in a 1 dimension matrix space. And we can also said that $$\psi_{m}\phi^{m} (1)$$ is the coponents of a vector in the base $$|0,0\rangle$$ which is in the irreducble 1 representation(which is a sub space of $$2 \otimes 2$$.). because (1) is invariant under a SU(2) action.

Then we can say that $$S = \begin{pmatrix} \psi_{1} \phi_{1} + \phi_{1} \psi_{1} & \psi_{1} \phi_{2} + \phi_{1} \psi_{2} \\ \phi_{1} \psi_{2} + \psi_{1} \phi_{2} & \psi_{2} \phi_{2} + \phi_{2} \psi_{2} \end{pmatrix} $$. I see 3 different comonents.

But I always can't see how to develope |\psi\rangle \otimes |\phi\rangle in the base form by (|1,-1\rangle, |1,0\rangle, |1,1\rangle, |0,0\rangle) who's is described page 25 of this course : http://lapth.cnrs.fr/pg-nomin/salati/TQC_UJF_13.pdf.

What can I do please?

Thank you in advance and have a nice afternoon.

 

[Calabi]

More : As we prevously show $$A = \begin{pmatrix} 0 & \psi_{1} \phi_{2} - \phi_{1} \psi_{2} \\ -(\psi_{1} \phi_{2} - \phi_{1} \psi_{2}) & 0\end{pmatrix} = \begin{pmatrix} 0 & \epsilon_{12}\psi_{m}\phi^{m} \\ \epsilon_{21}\psi_{m}\phi^{m} & 0\end{pmatrix}$$, since $$\epsilon_{12} = -\epsilon_{21}$$ this matrix could be considerate as to be in a 1 dimension matrix space.

THis matrix space is isomorph to the 1 representation.

Good afternoon.

 

[strangerep]

I get :
$$
\begin{pmatrix} \psi_1 \\ \psi_2 \end{pmatrix} \otimes
\begin{pmatrix} \phi_1 & \phi_2 \end{pmatrix} ~=~ [\cdots] ~=~ \frac{S+A}{2}
$$Where $$S = \begin{pmatrix} \psi_1 \\ \psi_2 \end{pmatrix} \otimes
\begin{pmatrix} \phi_1 & \phi_2 \end{pmatrix} +
\begin{pmatrix} \phi_1 \\ \phi_2 \end{pmatrix} \otimes \begin{pmatrix} \psi_1 & \psi_2 \end{pmatrix}$$ is the symetric part and $$A = \begin{pmatrix} \psi_1 \\ \psi_2 \end{pmatrix} \otimes
\begin{pmatrix} \phi_1 & \phi_2 \end{pmatrix} - \begin{pmatrix} \phi_1 \\ \phi_2 \end{pmatrix} \otimes \begin{pmatrix} \psi_1 & \psi_2 \end{pmatrix}$$ is the antysymetric part.

OK. Now what happens if you perform a rotation: $\psi \to \psi' = R\psi$ (and similarly for $\phi$). What happens on the right hand side of the 1st equation?

 

[Calabi]

Good morning,

OK. Now what happens if you perform a rotation: $\psi \to \psi' = R\psi$ (and similarly for $\phi$).

I transforme the components and the value of : \psi_{m}\phi^{m} is invariant under this rotation.

What happens on the right hand side of the 1st equation?

Whcih equation please?

If you talk about : \frac{S + A}{2} I know that A is invariant but I don't know what hapend to S.

What can I do please?

Thank you in advance and have a nice morning.

 

[strangerep]

If you talk about : \frac{S + A}{2} I know that A is invariant but I don't know what hapend to S.

Well, work it out. How did you figure out that A is invariant? Do a similar calculation for S and see what you get. (And if you still can't do it, show me how you figured out that A is invariant.)

 

[Calabi]

Good morning,

Well, work it out. How did you figure out that A is invariant? Do a similar calculation for S and see what you get. (And if you still can't do it, show me how you figured out that A is invariant.)

Well we've already said that A = \begin{pmatrix} 0 & \psi_{1} \phi_{2} - \phi_{1} \psi_{2} \\ -(\psi_{1} \phi_{2} - \phi_{1} \psi_{2}) & 0\end{pmatrix} = \begin{pmatrix} 0 & \epsilon_{12}\psi_{m}\phi^{m} \\ \epsilon_{21}\psi_{m}\phi^{m} & 0\end{pmatrix} so it's in a one dimension matrix space, so it's isomorph to 1. In 1 the components of the matrix are represented by \psi_{m}\phi^{m} which is invariant under a SU(2) action.

So we can conclude that A is invariant under a SU(2) action.

Good afternoon.

 

[Calabi]

So I'm going to start from the beginning. We have : $$\psi_{a}\phi_{b} = \frac{1}{2}\psi_{[a} \phi_{b]}(antisymetric) + \frac{1}{2}\psi_{(a} \phi_{b)}(symetric)$$.

First I've show that : $$\psi_{[a \phi b]} = \epsilon_{ab}\psi_{m}\phi^{m}$$.

More : we can say in matrix language that $$\begin{pmatrix} \psi_1 \\ \psi_2 \end{pmatrix} \otimes
\begin{pmatrix} \phi_1 & \phi_2 \end{pmatrix} ~=~ \begin{pmatrix}\psi_1 \phi_1 & \psi_1 \phi_2 \\ \psi_2 \phi_1 & \psi_2 \phi_2\end{pmatrix} = \frac{1}{2}(\begin{pmatrix} \psi_1 \\ \psi_2 \end{pmatrix} \otimes
\begin{pmatrix} \phi_1 & \phi_2 \end{pmatrix} +
\begin{pmatrix} \phi_1 \\ \phi_2 \end{pmatrix} \otimes \begin{pmatrix} \psi_1 & \psi_2 \end{pmatrix} + \begin{pmatrix} \psi_1 \\ \psi_2 \end{pmatrix} \otimes
\begin{pmatrix} \phi_1 & \phi_2 \end{pmatrix} - \begin{pmatrix} \phi_1 \\ \phi_2 \end{pmatrix} \otimes \begin{pmatrix} \psi_1 & \psi_2 \end{pmatrix} ) = \frac{S+A}{2}$$

And $$A = A = \begin{pmatrix} 0 & \psi_{m}\phi^{m} \\ -\psi_{m}\phi^{m} & 0\end{pmatrix}$$, so this matrix is in a 1 dimension matrix space with the components $$\psi_{m}\phi^{m}$$ in the base $$\begin{pmatrix} 0 & 1 \\ -1 & 0\end{pmatrix}$$.
So since we know that $$\psi_{m}\phi^{m}$$ is invariant under a $$SU(2)$$ rotation
we can say that this space matrix is a $$1$$ representation.

So the antysymetric part is a $$1$$ representation. The spin $$0$$. Indeed all rotation of $$SU(2)$$ are represented by the unity $$\mathbb{I}_{1} = 1$$ in the $$1$$ representation.The components are invariant under a $$SU(2)$$ rotation.

Now because of the symetry, there's 3 possibility(components.). for the symetric part : $$S = \begin{pmatrix} \psi_{1} \phi_{1} + \phi_{1} \psi_{1} & \psi_{1} \phi_{2} + \phi_{1} \psi_{2} \\ \phi_{1} \psi_{2} + \psi_{1} \phi_{2} & \psi_{2} \phi_{2} + \phi_{2} \psi_{2} \end{pmatrix}$$

So we can say I've resolve the problem. But they say in the course to comment.

What do you'll say please?

Thank you in advance and have a nice afternoon.

 

[strangerep]

So I'm going to start from the beginning. We have : $$\psi_{a}\phi_{b} = \frac{1}{2}\psi_{[a} \phi_{b]}(antisymetric) + \frac{1}{2}\psi_{(a} \phi_{b)}(symetric)$$.

Consider just the symmetric part. I'll call it $$\chi_{ab} ~:=~ \psi_{(a} \phi_{b)} ~.$$Now perform the rotation transformation on $\psi$ and $\phi$ that I mentioned in post #11. What is the explicit expression for the transformed $\chi$ ?

Then decompose the transformed $\chi$ into symmetric and antisymmetric parts. What do you find? And can you now conclude a bit more than you have so far?