VI.4. σ Models | 341
freedom are the two nucleons and the three pions. The proton and the neutron transform as
a spinor ψ ≡
p
n
under the SU(2) of isospin. Consider the kinetic energy term
¯
ψiγ∂ψ =
¯
ψ
L
iγ∂ψ
L
+
¯
ψ
R
iγ∂ψ
R
. We note that this term has the larger symmetry SU(2)
L
×SU(2)
R
,
with the left handed field ψ
L
and the right handed field ψ
R
transforming as a doublet
under SU(2)
L
and SU(2)
R
, respectively. [The SU(2) of isospin is the diagonal subgroup
of SU(2)
L
× SU(2)
R
.] We can write ψ
L
∼ (
1
2
,0) and ψ
R
∼ (0,
1
2
).
Now we see a problem immediately: The mass term m
¯
ψψ = m(
¯
ψ
L
ψ
R
+ h.c.) is not
allowed since
¯
ψ
L
ψ
R
∼ (
1
2
,
1
2
), a 4-dimensional representation of SU(2)
L
× SU(2)
R
,a
group locally isomorphic to SO(4).
At this point, lesser physicists would have said, what is the problem, we knew all along
that the strong interaction is invariant only under the SU(2) of isospin, which we will
write as SU(2)
I
. Under SU(2)
I
the bilinears constructed out of
¯
ψ
L
and ψ
R
transform as
1
2
×
1
2
=0 + 1, the singlet being
¯
ψψ and the triplet
¯
ψiγ
5
τ
a
ψ. With only SU(2)
I
symmetry,
we can certainly include the mass term
¯
ψψ.
To say it somewhat differently, to fully couple to the four bilinears we can construct out
of
¯
ψ
L
and ψ
R
, namely
¯
ψψ and
¯
ψiγ
5
τ
a
ψ , we need four meson fields transforming as the
vector representation under SO(4). But only the three pion fields are known. It seems
clear that we only have SU(2)
I
symmetry.
Nevertheless, Gell-Mann and L
´
evy boldly insisted on the larger symmetry SU(2)
L
×
SU(2)
R
SO(4) and simply postulated an additional meson field, which they called σ ,
so that (σ , π) form the 4-dimensional representation. I leave it to you to verify that
¯
ψ
L
(σ + i τ
.
π)ψ
R
+ h.c. =
¯
ψ(σ + i τ
.
πγ
5
)ψ is invariant. Hence, we can write down the
invariant Lagrangian
L =
¯
ψ[iγ∂ + g(σ + i τ
.
πγ
5
)]ψ + L(σ , π) (1)
where the part not involving the nucleons reads
L(σ , π) =
1
2
(∂σ )
2
+ (∂ π)
2
+
μ
2
2
(σ
2
+π
2
) −
λ
4
(σ
2
+π
2
)
2
(2)
This is known as the linear σ model.
The σ model would have struck most physicists as rather strange at the time it was
introduced: The nucleon does not have a mass and there is an extra meson field. Aha, but
you would recognize (2) as precisely the Lagrangian (IV.1.2) (for N =4) that we studied,
which exhibits spontaneous symmetry breaking. The four scalar fields (ϕ
4
, ϕ
1
, ϕ
2
, ϕ
3
)
in (IV.1.2) correspond to (σ , π). With no loss of generality, we can choose the vacuum
expectation value of ϕ to point in the 4th direction, namely the vacuum in which 0|σ |0=
μ
2
/λ ≡ v and 0|π |0=0. Expanding σ = v + σ
we see immediately that the nucleon
has a mass M = gv. You should not be surprised that the pion comes out massless. The
meson associated with the field σ
, which we will call the σ meson, has no reason to be
massless and indeed is not.
Can the all-important parameter v be related to a measurable quantity? Indeed. From
chapter I.10 you will recall that the axial current is given by Noether’s theorem as
J
a
μ5
=
¯
ψγ
μ
γ
5
(τ
a
/2)ψ + π
a
∂
μ
σ − σ∂
μ
π
a
. After σ acquires a vacuum expectation value,