What is "Little Group"?

yicong2011

In my Quantum Field Theory class, I too often meet with the term "Little Group".
Unfortunately, I cannot find a good description of Little Group until now. I just know it is a subgroup of Lorentz Group.

Can anyone have any brief description of this concept? Or any good reference on it?

Thanks a lot.

DrDu

No, it is not a subgroup of the Lorentz group but rather of the Poincare group although it is also used in other context such as in crystallography.
It is the group which leaves a given k vector invariant. Hence different k vectors have different little groups.
See, e.g. Gordon Hamermesh, Group theory
or Eugene Wigners book on group theory as he introduced the concept in relativistic QM.
A more modern introduction is Sternberg, Group theory

Bill_K

No, in its usual application in quantum field theory, the "little group" is a subgroup of the Lorentz group. In general if you have a group G which acts on a space X, and an element x in X, the little group of x is the subgroup of G that leaves x invariant.

For example the Lorentz group acts on the space of 4-vectors. If x is taken to be a timelike vector, the little group is the SO(3) subgroup of the Lorentz group in the 3-space orthogonal to x. If x is spacelike or null, the little group will be SO(2,1) or E(2) respectively.

negru

Or if you're talking about massless particles, you can write their momentum as the product of two spinors \lambda \tilde{\lambda}

Then if you multiply lambda by t, and tilde lambda by 1/t, you leave the momentum unchanged.

chrispb

More group-theoretically, a little group is the group that leaves some particular state invariant. Poincare transformations act on good old quantum mechanical states; the little group of the state of one massive particle in its rest frame is therefore the SO(3) of rotations around it.

DrDu

Are you sure that E2 is a sub-group of the Lorentz group?
Anyhow, what I really wanted to say was that the little groups are used to construct representations of the Poincare group and not of the Lorentz group.

George Jones

Yes, in this context the (double cover) of E2 is a subgroup of the (cover of the) homogeneous Lorentz group.
Yes.

DrDu

Interesting! I am not much into the theory of the Lorentz Group. Could you give me an outline of how to see this? Where do the translations in the plane come from?

tiny-tim

hi yicong2011!
you could try Stephen Weinberg's "The Quantum Theory of Fields, Volume I" at page 64 …

see it free online at http://books.google.co.uk/books?id=h...group%22&hl=en

naima

You can also find in Wigner little group details on E2 and T2

DrDu

Thank you, that's an interesting link. But while it is clear that E2 can be obtained from O(3) by an Inonu Wigner transformation, this leads out of O(3) and maybe also out of the homogeneous Lorentz group?

George Jones

Give me a bit of time, and I'll post the explicit construction.

George Jones

The universal cover of the restricted Lorentz group is [itex]SL\left( 2,\mathbb{C}\right)[/itex]. If a 4-vector is written as [itex]X=x^{0}+x^{i}\sigma _{i}[/itex], then the action of [itex]A \in SL\left( 2,\mathbb{C}\right)[/itex] on [itex]X[/itex] is [itex]A X A^\dagger[/itex]. The little group of the lightlike 4-vector [itex]X=1+\sigma_3[/itex] is the subgroup of [itex]SL\left( 2,\mathbb{C}\right)[/itex] that consists of matrices of the form
[tex]
\begin{pmatrix}
e^{i\theta} & b\\
0 & e^{-i\theta}
\end{pmatrix},
[/tex]
where [itex]\theta[/itex] is an arbitrary real number and [itex]b[/itex] is an arbitrary complex number.

[tex]
\begin{pmatrix}
e^{i\theta} & b\\
0 & e^{-i\theta}
\end{pmatrix}
\rightarrow
\begin{pmatrix}
e^{i2\theta} & b\\
0 & 1
\end{pmatrix}
[/tex]
is a two-to-one homomorphism between groups of matrices. Write [itex]b = u+iv[/itex] for real [itex]u[/itex] and [itex]v[/itex], and [itex]\left( x , y \right) \in \mathbb{R}^2[/itex] as the column
[tex]
\begin{pmatrix}
x+iy\\
1
\end{pmatrix}.
[/tex]

[tex]
\begin{pmatrix}
e^{i2\theta} & u+iv)\\
0 & 1
\end{pmatrix}
\begin{pmatrix}
x+iy\\
1
\end{pmatrix}
=
\begin{pmatrix}
x\cos2\theta - y\sin2\theta + u + i\left( x\sin\theta + y\cos2\theta +v \right)\\
1
\end{pmatrix}
[/tex]

DrDu

Thank you George, very interesting.
So setting theta=0 yields the pure translations by b. The elements A are of the same form as the X, [itex]X=x^{0}+x^{i}\sigma _{i}[/itex]. If I remember correctly, imaginary parts of the x^i correspond to boosts. So a pure translation is of the form [itex]X=x^{0}+b/2\sigma _{x}+ib/2 \sigma_y[/itex]. So the translation is a 50/50 mixture of rotation and boost.

strangerep

There's some more info in Weinberg vol 1, pp69-74.

The explicit generators of the E2 little group are
[tex]
J_3 ~,~~~~ A ~:=~ J_2 + K_1 ~,~~~~ B ~:= -J_1 + K_2
[/tex]
so the commutation relations are
[tex]
[J_3, A] = iB ~,~~~~ [J_3,B] = -iA ~,~~~~ [A,B] = 0 ~.
[/tex]

Hans de Vries

Just to elaborate a bit:

The A and B, acting on a momentum in the z-direction, represent centrifugal
accelerations K compensated by counter rotations J. The total effect of A or
B is therefor zero. The general case is.
[tex]
A\cos \phi ~+~B\sin\phi
[/tex]
Where [itex]\phi[/itex] is the angle which determines the direction in the x-y plane of the
centrifugal acceleration. Small nitpick about his signs: In a right-handed
coordinate system they should be:
[tex]
J_3 ~,~~~~ A ~:=~ J_2 - K_1 ~,~~~~ B ~:= -J_1 - K_2
[/tex]
Note that if you reverse the direction of the momentum, (k,0,0-k) instead
of (k,0,0,k), that the signs of the generators also change. In this case you
get indeed.
[tex]
J_3 ~,~~~~ A ~:=~ J_2 + K_1 ~,~~~~ B ~:= -J_1 + K_2
[/tex]
One should expect this because under spatial inversion K behaves like a
vector and J like a pseudo vector.

Regards, Hans

yicong2011

Ah, Thanks a lot, guys. Thanks for glorious detail explanation.