# More questions about Lorentz representation

1. Jan 23, 2015

### Mishra

Hello,

First of all, sorry if the question has been asked. I tryied to find some answers but my ignorance goes too deep for any of the previous topics I could find.

I'm completly lost when it comes to the Lorentz/Pointcarré groups representations.
From what I manage to understand, a representation is a way to associate a matrix to an abstract object (element of a group). The element of the group is now defined as a matrix which will act on a vector space.

In the case of QFT, this is done to be able to apply Lorentz transformations to mathematical objects such as, vectors, and mostly fields.

Each representations you create must obey the Lorentz Algebra. So far I've been able to find the Lorentz algebra using its representation acting on 4-vectors. In the way we introduced the exponential map, and generators. And some cool theorems.

Since we also wanted to study how fields transform we created the representation for scalar fields, which don't "really" transform (only the coordinates do) so that was easy.

The part that I don't understand is the spinors representation, (1/2,0), (0,1/2) and (1/2,1/2).
I understand we want to create a representation acting on something that would have a spin. So those representations act on Weyl field (left/right handed spinors). I just don't understand what they are and what this representation is.

-(1/2,0) seem to be the eigenvalues of the spin operator, but I've never seen a 1/2 spin transforming into a 0spin
-(1/2,1/2), here does it mean a combination of two 1/2 spins (like a 2 electrons system)?
I guess my problem comes from the fact that I can't understand what the vector space, on which the Lorentz transformation will act, really is.

I hope my question makes sense,
thank you!

2. Jan 23, 2015

### Mishra

Just to make a little more sense into my question:

If I only consider the Pointcarré transformations: $x^{\mu} \rightarrow x'^{\mu}=\Lambda^{\mu}_{\nu}x^{\nu}+a^{\mu}$.
In all generality, a field $\phi_a(x)$ will transforms as such: $\phi(x)_a \rightarrow \phi'(x')_a=D(\Lambda)^{b}_{a}\phi(x)_b$
By "canceling" the coordinates change (unpriming) one gets: $\phi'(x)_a=D(\Lambda)^{b}_{a}\phi(\Lambda^{-1}(x-a))_b$
Where $D(\Lambda)^{b}_{a}$ is the representative of the Pointcarré transformation in the representation corresponding to $\phi(x)_a$.

So far,

-Scalar field: $\phi(x)_a \rightarrow \phi'(x')_a=\phi(x)_a$, so $D(\Lambda)^{b}_{a}=1$ it is the trivial representation.
-Vector field: $\phi(x)_a \rightarrow \phi'(x')_a=\Lambda^{b}_{a}\phi(x)_b$, so $D(\Lambda)^{b}_{a}=\Lambda^{b}_{a}$ since the field is a vector, it transforms like one.
-Spinors: what are they? how do they transform ? why the Pauli matrices are the generators ?

3. Jan 23, 2015

### samalkhaiat

Do you know the relation between Lorentz group and the group of angular momentum, i.e. SU(2)?

4. Jan 24, 2015

### Mishra

Well, I know that the Spin operators $J_x,J_y,J_z$ obey the same commutation relations than the Lorentz generators.
After a little work you can show that SU(2)xSU(2)=SO(1,3) (only true for algebras, not groups).

Are you trying to tell me that if you try to find a representation of the Lorentz group acting on spins (meaning |j,m> Hilbert space) you find that the generators are the Spin operators $J_x,J_y,J_z$ ?
I still don't get it, what are fields doing here in this case ?...

5. Jan 24, 2015

### samalkhaiat

As well as the generators $J_{ i }$ (of rotations), there are also 3 boosts generators $K_{ i }$. The complex combinations $J^{ L ( R ) }_{ i } = J_{ i } \pm i K_{ i }$, split the Lorentz algebra into 2 conjugate, commuting $SU(2)$ algebras. So, on the level of algebra, you can write $SO(1,3)_{ \mathbb{ C } } \sim SU_{ L } ( 2 ) \times SU_{ R } ( 2 )$.
To be exact, what you have written is not true even for the algebra. The correct relations are $SO(4) \sim SU(2) \times SU(2) ,$ and $SO(1,3)_{ \mathbb{ R } } \sim SL( 2 , \mathbb{ C } ) / \mathbb{ Z }_{ 2 }$.
The above complex combinations means that the finite dimensional irreducible representation of the Lorentz algebra can be characterized by pair $( j_{ L } , j_{ R } )$ of positive half-integers. The eigen-values of $J_{ L ( R ) }^{ 2 }$ are given by $j_{ L ( R ) } ( j_{ L ( R ) } + 1 )$, the dimension of the representation space $D^{ ( j_{ L } , j_{ R } ) }$ is given by $$\mbox{ dim } ( D^{ ( j_{ L } , j_{ R } ) } ) = ( 2 j_{ L } + 1 ) ( 2 j_{ R } + 1 ) ,$$ and the spin of the representation is given by $( j_{ L } + j_{ R } )$.
The problem of going from the abstract representation space $D^{ ( j_{ L } , j_{ R } ) }$ to finite component fields on Minkowski spacetime is a very complicated problem in mathematical physics. It turns out that it is always possible to identify $( j_{ L } , j_{ R } )$ with multi-component field $\Psi_{ A } ( x )$ with $A = 1 , 2 , \cdots , ( 2 j_{ L } + 1 ) ( 2 j_{ R } + 1 )$. See posts 6 & 7 in

Sam

Last edited: Jan 24, 2015
6. Jan 25, 2015

### Mishra

So, all we did was identifying the Lorentz algebra to the product of two SU(2) algebras (we did that by separating the boosts and the rotations).
And then we defined left/right handed spinors to be objects that have eingenvalues 1/2 and 0 under the lorentz transformations $J^+, J^-$ ?

In other words, the spinors are just an abstract construction based on their eigenvalues ?

7. Jan 25, 2015

### vanhees71

Last edited by a moderator: May 7, 2017
8. Jan 25, 2015

### dextercioby

It's customary to denote Lie algebras with lower case (even Fraktur), so that one should write $su(2)$ or $\mathfrak{su}(2)$, that's why putting $SO(4) \simeq SU(2)\times SU(2)$ is wrong, one should write this for their Lie algebras: $\mathfrak{so}(4) \simeq \mathfrak {su}(2)\oplus \mathfrak{su}(2)$.

While Sexl & Urbantke is a standard text, one may also use Wu Ki Tung's Group Theory for Physicists text.

9. Jan 25, 2015

### Mishra

I could not get my hands on the book you mentionned, but your notes are very good and I just noticed I already had them in my dropbox!
I'm starting to get it now. I was trying to find what this representation was; thinking it was "built" from the ground like we did for the scalar field. In fact, it emerges from the algebra.

10. Jan 26, 2015

### samalkhaiat

If this is directed at me, then I stand by everything I said. Regarding the group $SO(4)$, the symbol $\sim$ (which is an equivalence relation) represents groups homomorphism. Here is my justification for that: It is an easy exercise to show that, for any $g( \vec{ \alpha } , \vec{ \beta } ) \in SO( 4 )$, there exists a non-singular matrix $S$ such that $$g( \vec{ \alpha } , \vec{ \beta } ) \sim S \ g( \vec{ \alpha } , \vec{ \beta } ) \ S^{ - 1 } = \exp ( - i \vec{ \alpha } \cdot \frac{ \vec{ \sigma } }{ 2 } ) \otimes \exp (- i \vec{ \beta } \cdot \frac{ \vec{ \sigma } }{ 2 } ) . \ \ (1)$$ This means that each $g ( \vec{ \alpha } , \vec{ \beta } ) \in SO(4)$ is equivalent ($\sim$) to the direct product of two group elements $U( \vec{ \alpha } ) = \exp( - i \vec{ \alpha } \cdot \vec{ \sigma } / 2 )$ and $U( \vec{ \beta } ) = \exp( - i \vec{ \beta } \cdot \vec{ \sigma } / 2 )$ of $SU(2)$. That is to say that there exists a homomorphism (not isomorphism) of $SU(2) \times SU(2)$ onto $SO(4)$. Indeed, these are two different groups [*]. However, if the group $SU(2) \times SU(2)$ is quotient by the kernel of the homomorphism, an exact isomorphism is obtained. To do that, first observe that $$\exp ( - i \vec{ \theta } \cdot \frac{ \vec{ \sigma } }{ 2 } ) = D^{ ( 1 / 2 ) } \left( U ( \vec{ \theta } ) \right) , \ \ \ (2)$$ is the 2-dimensional irreducible representation of $SU(2)$, and for any group element $$\left( U ( \vec{ \alpha } ) , U ( \vec{ \beta } ) \right) \in SU(2) \times SU(2) ,$$ we have $$D^{ ( 1 / 2 ) } ( U ( \vec{ \alpha } ) ) \ \otimes \ D^{ ( 1 / 2 ) } ( U ( \vec{ \beta } ) ) = D^{ ( 1 / 2 \ , \ 1 / 2 ) } \left( U ( \vec{ \alpha } ) , U ( \vec{ \beta } ) \right) . \ \ \ (3)$$ Using (2) and (3) in (1), we obtain $$g ( \vec{ \alpha } , \vec{ \beta } ) \sim D^{ ( 1 / 2 \ , \ 1 / 2 ) } \left( U ( \vec{ \alpha } ) , U ( \vec{ \beta } ) \right) . \ \ \ \ (4)$$ This shows that every $g ( \vec{ \alpha } , \vec{ \beta } ) \in SO(4)$ is equivalent to the 4-dimensional irreducible representation, $( 1 / 2 , 1 / 2 )$, of $SU(2) \times SU(2)$. But the relation $$D^{ ( \frac{ 1 }{ 2 } , \frac{ 1 }{ 2 } ) } \left( U ( \vec{ \alpha } ) , U ( \vec{ \beta } ) \right) = [ - U ( \vec{ \alpha } ) ] \otimes [ - U ( \vec{ \beta } ) ] = D^{ ( \frac{ 1 }{ 2 } , \frac{ 1 }{ 2 } ) } \left( - U ( \vec{ \alpha } ) , - U ( \vec{ \beta } ) \right) ,$$ implies that each $g \in SO(4)$ is equivalent to two different group elements of $SU(2) \times SU(2)$. Thus, $\{ \mathbb{ E }, - \mathbb{ E } \} = \{ ( I , I ) , ( - I , - I ) \}$ is the kernel of the homomorphism of $SU(2) \times SU(2)$ onto $SO(4)$. This forms the group $\mathbb{ Z }_{ 2 }$ of order 2 which is the centre of both groups. Now, we can happily write down the following isomorphism $$SO(4) \cong \frac{ SU(2) \times SU(2) }{ \{ \mathbb{ E } , - \mathbb{ E } \} } .$$ This shows that $SU(2) \times SU(2)$ is the universal (double-)covering group of $SO(4)$.
[*]For example, $SU(2) \times SU(2)$ has, among its inequivalent irreducible representations, representations of dimensions $(2 + 4k), \ k = 0 , 1 , 2 , 3 , \ …$, while $SO(4)$ has no such representations. This explains the absence of the $(2 + 4k)- \mbox{fold}$ degeneracy in the H-atom. In fact, all inequivalen irreducible representations, $( j_{ 1 } , j_{ 2 } )$, of $SO(4)$ satisfy $j_{ 1 } + j_{ 2 } = \mbox{ integer }$, i.e. either $j_{ 1 }$ and $j_{ 2 }$ are both integer or both half-integer.
Finally, I leave you with the following exercise.
Show that $( 0 , \frac{ 1 }{ 2 } )$
(i) is an irreducible representation of $SU(2) \times SU(2)$,
(ii) is not faithfull representation of $SU(2) \times SU(2)$,
(iii) is not a representation of $SO(4)$.
Sam

Last edited: Jan 26, 2015
11. Jan 26, 2015

### vanhees71

Let's just add, that in quantum theory what we need are not representations but unitary ray representations, which are "representations up to a phase factor". At the end the space-time symmetry group (Poincare for special-relativistic, Galilei for non-relativistic physics) is substituted by the representation of a central extension of its covering group. For the Poincare group there are no non-trivial central charges, and thus it's simply the covering group of the proper orthochronous Poincare group, where the SO(1,3) is substituted by SL(2,C). The corresponding analysis of the unitary representations leads to all possible realizations of the symmetry. So far one could make physical sense only for the massive and massless representations. In addition also only socalled "local realizations" in terms of a local microcausal QFT lead to sensible models, most prominently the standard model of elementary particle physics.

The non-relativistic case is a bit more complicated, because (fortunately) there is a non-trivial central charge. The classical space-time symmetry, the proper orthochronous Galilei group, is thus represented by the corresponding central extensions of the covering of the classical Galilei group with mass as the central charge. The unitary representations of the classical Galilei group do not lead to a physically relevant realization of QT, i.e., massless particles do not make sense in non-relativistic physics.

12. Jan 26, 2015

### Mishra

If anyone is still interested, I've found a very good book which explains this with words: "A mordern introduction to quantum field theory. M Maggiore" on page 24,25.
I've seen the demonstrations over and over, but group theory, Lie Algebra, and manifolds being new to me they did not bring any clarity.

I think this book will help others and is by far the most comprehensive approach (to newbies) I could find.

13. Jan 26, 2015

### vanhees71

Interesting. How is this done without group and representation theory in this book?

14. Jan 27, 2015

### Mishra

It is done with group representation. I was just saying the authors is very understandable for someone like me.
The book is more a recap of QFT to be honest, not much computations. That's probably why it is so clear, you do not get lost in the computations which you have to do by yourself.