SU_L x SU_R and SU_V x SU_A equivalence

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In summary, the group SU_L(N)\times SU_R(N) is the same as SU_V(N)\times SU_A(N) and it is possible to rewrite the transformation using N\times N SU(N) matrices as shown in the system:\begin{cases}\psi_L \to \psi'_L=V A\,\psi_L\\\psi_R \to \psi'_R=V A^\dagger\, \psi_R\end{cases}To prove this, one can use the fact that SU(N) matrices have a square root in SU(N), and consider infinitesimal elements of the group. However, it should be noted that the notation SU(2)V \times SU
  • #1
whitewolf91
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I want to show that the group [itex]SU_L(N)\times SU_R(N)[/itex] is the same as [itex]SU_V(N)\times SU_A(N)[/itex] - i.e. that it is possible to rewrite the transformation:

[tex]
\begin{cases}
\psi_L \to \psi'_L=V_L\,\psi_L\\
\psi_R \to \psi'_R=V_R\,\psi_R
\end{cases}
[/tex]

, where [itex]V_L[/itex] and [itex]V_R[/itex] are [itex]N\times N[/itex] [itex]SU(N)[/itex] matrices and [itex]\psi_L[/itex] and [itex]\psi_R[/itex] are [itex]N[/itex]-component vectors, as:

[tex]
\begin{cases}
\psi_L \to \psi'_L=V A\,\psi_L\\
\psi_R \to \psi'_R=V A^\dagger\, \psi_R
\end{cases}
[/tex]

, where again [itex]V[/itex] and [itex]A[/itex] are [itex]N\times N[/itex] [itex]SU(N)[/itex] matrices. The system:

[tex]
\begin{cases}
V_L=VA &\\
V_R=VA^\dagger
\end{cases}
[/tex]

must be solved for [itex]V[/itex] and [itex]A[/itex]:

[tex]
A^2=V_R^\dagger V_L
[/tex]

So, [itex]A[/itex] is the square root of the [itex]SU(N)[/itex] matrix [itex]V_R^\dagger V_L[/itex].

Problems:

1. The only way I can think of to give meaning to the square root of a matrix is a naive series expansion, but how to prove convergence?

2. Does the series expansion defines an [itex]SU(N)[/itex] matrix?


Random thoughts:

Probably 2. is the less problematic. I believe that it is true that if the square of a matrix equals an [itex]SU(N)[/itex] matrix, that matrix still lives in [itex]SU(N)[/itex]. It is not necessary to find the explicit expression of [itex] A[/itex] and [itex]V[/itex] so it would be enough to prove that the square root of an [itex]SU(N)[/itex] matrix is an [itex]SU(N)[/itex] matrix. Another approach is to consider infinitesimal elements of the group, then the previous system can be easily satisfied in the Lie algebra of the group. Is this enough to conclude that there are finite elements in the group that satisfy the same relationships?

Thank you for your time.
 
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  • #2
whitewolf91 said:
1. The only way I can think of to give meaning to the square root of a matrix is a naive series expansion, but how to prove convergence?

If a matrix A is similar to a diagonal matrix (there is some matrix ##P## and some diagonal matrix ##D## such that ##A = P^{-1} D P##) then you can define ##A^{1/2} \equiv P^{-1} D^{1/2} P##, where ##D^{1/2}## is defined by just taking the square root of the diagonal entries.
 
  • #3
An [itex]SU(N)[/itex] matrix is not necessarily hermitian, so it is not guaranteed that it can be diagonalized. If it were hermitian however then [itex]P[/itex] would be unitary and it would be automatic that the square root is in [itex]SU(N)[/itex].
 
  • #4
Unitary matrices are all diagonalizable with a unitary ##P##; see e.g. here. So all SU(N) matrices have a square root in SU(N).
 
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  • #8
@The_Duck: thank you, that solves the problem.

@samalkhaiat: I follow you until the very end, when you imply that:

[tex]
e^{i(\alpha+\epsilon \gamma_5)\,\cdot\, \tau/2}=e^{i\alpha\,\cdot\, \tau/2}e^{i\epsilon \gamma_5\,\cdot\, \tau/2}
[/tex]

Isn't this false since the exponents are non commuting?
 
  • #9
whitewolf91 said:
@The_Duck: thank you, that solves the problem.

@samalkhaiat: I follow you until the very end, when you imply that:

[tex]
e^{i(\alpha+\epsilon \gamma_5)\,\cdot\, \tau/2}=e^{i\alpha\,\cdot\, \tau/2}e^{i\epsilon \gamma_5\,\cdot\, \tau/2}
[/tex]

Isn't this false since the exponents are non commuting?

This should be understood as [itex]8 \times 8[/itex] matrix, when you use the BHC identity. Write it as
[tex]e^{ \frac{i}{2} ( \alpha \cdot \tau ) \otimes I } e^{ \frac{i}{2} ( \epsilon \cdot \tau ) \otimes \gamma_{ 5 } } .[/tex]
 
  • #10
My reasoning:

[itex]
[(\alpha\cdot \tau)\otimes I, (\epsilon\cdot \tau)\otimes\gamma_5]=
[/itex]

[itex]
=[\alpha\cdot\tau,\epsilon\cdot\tau]\otimes\gamma_5=
[/itex]

[itex]
=\alpha_a\epsilon_b[\tau_a,\tau_b]\otimes\gamma_5=
[/itex]

[itex]
=i\alpha_a\epsilon_b\epsilon_{abc}\tau_c\otimes\gamma_5=
[/itex]

[itex]
=i(\alpha \times \epsilon)\cdot \tau \otimes \gamma_5
[/itex]

, which is zero only if [itex]\alpha[/itex] and [itex]\epsilon[/itex] are parallel.
 
  • #11
One should stress that you shouldn't write [itex]\mathrm{SU}(2)_V \times \mathrm{SU}(2)_A[/itex] (BAD NOTATION!) since this the [itex]\times[/itex] symbol should be reserved for a direct product of groups. That's the case for [itex]\mathrm{SU}(2)_L \times \mathrm{SU}(2)_R[/itex], but [itex]\mathrm{SU}(2)_A[/itex] is not even a sub group but a coset.
 
  • #12
whitewolf91 said:
My reasoning:

[itex]
[(\alpha\cdot \tau)\otimes I, (\epsilon\cdot \tau)\otimes\gamma_5]=
[/itex]

[itex]
=[\alpha\cdot\tau,\epsilon\cdot\tau]\otimes\gamma_5=
[/itex]

[itex]
=\alpha_a\epsilon_b[\tau_a,\tau_b]\otimes\gamma_5=
[/itex]

[itex]
=i\alpha_a\epsilon_b\epsilon_{abc}\tau_c\otimes\gamma_5=
[/itex]

[itex]
=i(\alpha \times \epsilon)\cdot \tau \otimes \gamma_5
[/itex]

, which is zero only if [itex]\alpha[/itex] and [itex]\epsilon[/itex] are parallel.

I am very sorry, may be I was drunk. For some reason, I thought you were asking about eq(3) with the projection matrices. But you were asking about
[tex]U_{ L } U_{ R } = \exp ( i \frac{ \alpha \cdot \tau }{ 2 } ) \exp ( i \gamma_{ 5 } \frac{ \epsilon \cdot \tau }{ 2 } )[/tex]
This is not an algebraic equation. As explained in the paragraph bellows it, it means the following

For all [itex]g( \epsilon_{ L } , \epsilon_{ R } ) \in SU_{ L } \times SU_{ R }[/itex], a transformation with [itex]\epsilon_{ L } = \epsilon_{ R }[/itex], corresponds to the vector transformation [itex]g( \alpha )[/itex]. And, transformation with [itex]\epsilon_{ L } = - \epsilon_{ R }[/itex] corresponds to the axial transformation [itex]g( \epsilon )[/itex].

As pointed out by vanhees71 , the sign [itex]( \times ) [/itex] in [itex]SU \times SU_{ 5 }[/itex] is not a direct product , because the Lie bracket of [itex]SU_{ 5 }[/itex] does not closes on itself. But the group [itex]G = SU \times SU_{ 5 }[/itex] is a direct product of the two independent groups [itex]SU_{ L }[/itex] and [itex]SU_{ R }[/itex], with Lie algebra given as
[tex]\mathcal{ G } = su_{ L } \oplus su_{ R }[/tex]
Sorry for the confusion

Sam
 

1. What is SU_L x SU_R and SU_V x SU_A equivalence?

SU_L x SU_R and SU_V x SU_A equivalence is a concept in particle physics that refers to the symmetries between the strong nuclear force (SU_L x SU_R) and the weak nuclear force (SU_V x SU_A). It is a type of gauge symmetry that describes the interactions between subatomic particles.

2. How are SU_L x SU_R and SU_V x SU_A related?

SU_L x SU_R and SU_V x SU_A are related through the concept of chiral symmetry. Chiral symmetry refers to the symmetry between left-handed and right-handed particles, as well as between particles and their corresponding antiparticles. SU_L x SU_R and SU_V x SU_A are different aspects of chiral symmetry that describe different interactions between particles.

3. What is the significance of SU_L x SU_R and SU_V x SU_A equivalence?

The significance of SU_L x SU_R and SU_V x SU_A equivalence lies in its ability to explain the fundamental interactions between particles in the Standard Model of particle physics. It provides a mathematical framework for understanding the symmetries and interactions between subatomic particles, and has been extensively tested and confirmed through experiments.

4. How does SU_L x SU_R and SU_V x SU_A equivalence impact our understanding of the universe?

SU_L x SU_R and SU_V x SU_A equivalence plays a crucial role in our understanding of the universe by providing a way to unify the fundamental forces of nature. It is a key component of the Standard Model, which is the most comprehensive theory we have for describing the behavior of particles and their interactions. Additionally, the discovery of the Higgs boson, which was predicted by the Standard Model and confirmed in 2012, further supports the validity of SU_L x SU_R and SU_V x SU_A equivalence.

5. Are there any limitations to SU_L x SU_R and SU_V x SU_A equivalence?

While SU_L x SU_R and SU_V x SU_A equivalence has been successful in explaining the fundamental interactions between particles, it is not a complete theory. It does not incorporate gravity, and there are still unanswered questions and discrepancies in our understanding of the universe that require further research. However, it remains a crucial and widely accepted concept in particle physics.

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