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SU(2) axial and vector subgroups

Tags: axial, subgroups, vector
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Mar15-12, 08:48 AM
P: 411

Could someone explain to me how to split SU(2) into its axial and vector subgroups, what does this mean?

(The context I'm trying to understand this in is the U(2)_L x U(2)_R global flavour sym of chiral Lagrangian)

A related question: I know that the three axial generators of SU(2)_L x SU(2)_R get broken, and this leads to the three (pesudo)-goldstone bosons; the three pions. But why are there three axial generators of this group?

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Mar16-12, 11:00 AM
Sci Advisor
P: 779
The group SU(2)_L x SU(2)_R has 6 generators [itex]T^a_L, T^a_R[/itex] (a=1,2,3).

We can also use generators [itex]T^a_L + T^a_R[/itex] and [itex]T^a_L - T^a_R[/itex]. The first one generates SU(2)_V and the second one generates SU(2)_A. So you can think of vector and axial symmetries as a change of basis of SU(2)^2.

The axial generators are the broken ones in QCD, giving rise to the 3 pions.

Note that your question as phrased: "how to split SU(2) into its axial and vector subgroups?" makes no sense, as a SINGLE SU(2) doesn't have such a decomposition.
Mar16-12, 01:37 PM
PF Gold
P: 365
Hi Blechman,

Could you tell me why with addition of these generators the group is "vectorial" and "axial" with subtraction ?


Mar16-12, 01:49 PM
Sci Advisor
P: 779
SU(2) axial and vector subgroups

"Vector" means L and R fermions transform the same way:

[tex]\psi_L\longrightarrow e^{i\alpha^aT_L^a}\psi_L,\qquad\psi_
R\longrightarrow e^{i\alpha^aT_R^a}\psi_R[/tex]

Hence + sign.

"Axial" means L and R fermions transform oppositely:

[tex]\psi_L\longrightarrow e^{i\alpha^aT_L^a}\psi_L,\qquad\psi_
R\longrightarrow e^{-i\alpha^aT_R^a}\psi_R[/tex]

Hence - sign.
Mar16-12, 05:32 PM
PF Gold
P: 365
I accept your answer but what is the relationship with the usual definition of vectors and pseudovectors.
I'm sorry if this is obvious.
Mar16-12, 08:11 PM
Sci Advisor
P: 908
The word “Axial” indicates the presence of [itex]\gamma_{5}[/itex] in the transformation law.
Consider the following axial transformation of a fermion field with TWO FLAVOURS;
[tex]\Psi \rightarrow U_{5}\Psi \equiv \exp (i \frac{\epsilon . \tau}{2}\gamma_{5}) \Psi . \ \ (1)[/tex]
In this, [itex]\gamma_{5}[/itex] operates on the Dirac components of [itex]\Psi[/itex], while Pauli’s matrices [itex]\tau^{i}[/itex] operate on the internal 2-dimensional flavour space of the fermions. We may call this group [itex]SU(2)_{5}[/itex]. This transformation takes on a simple form when expressed in terms of the chiral components of [itex]\Psi[/itex]. That is
\Psi_{R}\rightarrow V \Psi_{R} \equiv \exp( i \frac{\epsilon . \tau}{2}) \Psi_{R},
\Psi_{L}\rightarrow V^{\dagger}\Psi_{L} \equiv \exp( - i \frac{\epsilon . \tau}{2}) \Psi_{L}.
This can be shown by expanding eq(1) in a power series, and using the following
[tex]\Psi = \Psi_{R} + \Psi_{L},[/tex]
[tex]\gamma_{5}\Psi = \Psi_{R} - \Psi_{L}.[/tex]
So, the [itex]8 \times 8[/itex] transformation matrix [itex]U_{5}[/itex] can be written as
[tex]U_{5} = VP^{+} + V^{\dagger}P^{-}\ \ \ (2),[/tex]
[tex]P^{\pm} = \frac{1}{2}(1 \pm \gamma_{5}),[/tex]
are projection operators.
From eq(2), it follows that
[tex]\gamma^{\mu}U_{5}= U^{\dagger}_{5}\gamma^{\mu}.[/tex]
This means that [itex]\Psi[/itex] and [itex]\bar{\Psi}[/itex] transform in the same way,
[tex]\Psi \rightarrow U_{5}\Psi,[/tex]
[tex]\bar{\Psi}\rightarrow \bar{\Psi}U_{5}.[/tex]
Thus, chirally invariant Lagrangian must be constructed out of massless fermions; the presence of a small fermion mass term provides a mechanism for breaking chiral symmetry.
Since our fermion field has two flavours, the theory must also be invariant under the global group [itex]SU(2)[/itex]. So, the total symmetry group is [itex]SU(2) \times SU(2)_{5}[/itex]. This group is equivalent to [itex]SU(2)_{L}\times SU(2)_{R}[/itex] with element
U_{L}U_{R}= \exp(i\frac{\epsilon_{L}.\tau}{2}P^{-}) \exp( i \frac{\epsilon_{R}.\tau}{2}P^{+}),
where [itex]\epsilon_{L}[/itex] and [itex]\epsilon_{R}[/itex] are the independent parameters. To see the equivalence, let [itex]\Psi_{L}= P^{-}\Psi[/itex], and [itex]\Psi_{R}= P^{+}\Psi[/itex] be associated with the (finite dimensional) irreducible representations of Lorentz group [itex](1/2,0) \oplus (1/2,0)[/itex] and [itex](0,1/2) \oplus (0,1/2)[/itex], respectively. The transformations under two commuting [itex]SU(2)[/itex] groups are
[tex](2,1) \in SU(2)_{L}:[/tex]
[tex]\Psi_{L}\rightarrow U_{L}\Psi_{L}\equiv \exp( i\frac{\epsilon_{L}.\tau}{2})\Psi_{L},[/tex]
[tex]\Psi_{R}\rightarrow \Psi_{R},[/tex]
which can be combined into
\Psi \rightarrow e^{i\frac{\epsilon_{L}.\tau}{2}} \frac{1}{2}(1 - \gamma_{5})\Psi + \frac{1}{2}(1 + \gamma_{5})\Psi,
[tex]\Psi \rightarrow \exp(i\frac{\epsilon_{L}.\tau}{2}P_{-})\Psi .[/tex]

[tex](1,2) \in SU(2)_{R}:[/tex]

[tex]\Psi_{L}\rightarrow \Psi_{L},[/tex]
[tex]\Psi_{R}\rightarrow U_{R}\Psi_{R} = \exp( i\frac{\epsilon_{R}.\tau}{2})\Psi_{R},[/tex]
which we can write as
[tex]\Psi \rightarrow \exp(i\frac{\epsilon_{R}.\tau}{2}P^{+})\Psi[/tex]
Therefore, the combined [itex]SU(2)_{L}\times SU(2)_{R}[/itex] transformation is given by
\Psi \rightarrow U_{L}U_{R}\Psi = \exp[i(\epsilon_{L}P^{-} + \epsilon_{R}P^{+}) . \frac{\tau}{2}] \Psi . \ \ \ (3)
If we define
[tex]\alpha = \frac{1}{2}(\epsilon_{R} + \epsilon_{L}),[/tex]
[tex]\epsilon = \frac{1}{2}(\epsilon_{R} - \epsilon_{L}),[/tex]
then, the [itex]SU(2)_{L}\times SU(2)_{R}[/itex] element [itex]U_{L}U_{R}[/itex] becomes
U_{L}U_{R} = \exp(i\frac{\alpha . \tau}{2}) \exp(i\frac{\epsilon . \tau}{2}\gamma_{5}),
which belongs to the original symmetry group [itex]SU(2) \times SU(2)_{5}[/itex].
The meaning of chiral symmetry is, according to eq(3), the statement that an [itex]SU(2)[/itex] symmetry can be INDEPENDENTLY realized on the two subspaces projected out by the [itex]P^{\pm}[/itex] operators, i.e., the transformations on these two spaces can have different parameters.

Hans de Vries
Mar17-12, 10:50 AM
Sci Advisor
P: 1,135
Quote Quote by naima View Post
Could you tell me why with addition of these generators the group is "vectorial" and "axial" with subtraction ?

In the essence it's simple: The two chiral components [itex]\psi_L[/itex] and [itex]\psi_R[/itex]
have some similarities with left and right circular polarized photons.
Independently they propagate at the speed of light, however if you
couple them together then the combined momentum is the sum of
the two individual momenta and the resulting propagation speed of
the particle can be anywhere between +c and -c.

For the combined momentum one first calculates the individual momenta.

[itex]\psi_L\tilde{\sigma}^\mu\psi_L[/itex] and [itex]\psi_R\sigma^\mu\psi_R[/itex]

The sum of these two is the combined momentum, a vector.
However [itex]\psi_L\tilde{\sigma}^\mu\psi_L[/itex] is also a measure for left handedness while
[itex]\psi_R\sigma^\mu\psi_R[/itex] is a measure for right handedness.

This means that when you subtract the two you get their combined
"handedness", which is an axial vector.

Regards, Hans
Mar17-12, 07:00 PM
PF Gold
P: 365
Thank you very much for your answers.
It just remains for me to make a synthesis of these aspects!
Oct16-12, 08:33 PM
P: 1
Citation from Weinberg´s second Volume on Quantum Field Theory Page 184 (botton of the page):

"It is not true that an unbroken chiral
symmetry necessarily implies a zero nucleon mass, unless we make further assumptions
about the matrix elements of the axial-vector current ."

A good and lucid treatment on chiral spontaneous symmetry breaking can be found in my opinion in Chapter 9 of "Stefan Pokorski-Gauge Field Theories".
Feb7-14, 09:43 AM
P: 1
I have another question that maybe fits here:

In a hadron physics lecture some ratios of hadron masses where given to proof that SU(2)_V is conserved, while SU(2)_A is spontaneously broken, e.g.

[tex] SU(2)_V \hspace{20mm} \frac{m_{\pi^+} - m_{\pi^-}}{m_{\pi^+} + m_{\pi^-}} = 0.1\% [/tex]
[tex] SU(3)_V \hspace{20mm} \frac{m_{f_0} - m_{\pi}}{m_{f_0} + m_{\pi}} = 56\% [/tex]

While the first one kind of makes sence to me the second does not.

How does the vector and axial charge act on hadrons?


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