What is "Little Group"?


by yicong2011
Tags: little group
yicong2011
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#1
Feb20-12, 08:58 PM
P: 76
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.
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DrDu
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#2
Feb21-12, 02:34 AM
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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
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#3
Feb21-12, 07:56 AM
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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
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#4
Feb21-12, 09:36 AM
P: 308

What is "Little Group"?


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
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#5
Feb21-12, 10:58 AM
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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
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#6
Feb21-12, 11:15 AM
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Quote Quote by Bill_K View Post
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.
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
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#7
Feb21-12, 07:09 PM
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Quote Quote by DrDu View Post
Are you sure that E2 is a sub-group of the Lorentz group?
Yes, in this context the (double cover) of E2 is a subgroup of the (cover of the) homogeneous Lorentz group.
Quote Quote by DrDu View Post
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.
Yes.
DrDu
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#8
Feb22-12, 02:03 AM
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Quote Quote by George Jones View Post
Yes, in this context the (double cover) of E2 is a subgroup of the (cover of the) homogeneous Lorentz group.
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
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#9
Feb22-12, 06:02 AM
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hi yicong2011!
Quote Quote by yicong2011 View Post
Or any good reference on it?
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
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#10
Feb22-12, 09:56 AM
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You can also find in Wigner little group details on E2 and T2
DrDu
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Feb22-12, 10:36 AM
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Quote Quote by naima View Post
You can also find in Wigner little group details on E2 and T2
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
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#12
Feb22-12, 10:51 AM
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Give me a bit of time, and I'll post the explicit construction.
George Jones
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Feb22-12, 11:50 AM
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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
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#14
Feb23-12, 01:56 AM
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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
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#15
Feb23-12, 03:39 AM
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Quote Quote by DrDu View Post
So the translation is a 50/50 mixture of rotation and boost.
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
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#16
Feb26-12, 05:58 PM
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Quote Quote by strangerep View Post
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]
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
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#17
Feb27-12, 12:48 AM
P: 76
Ah, Thanks a lot, guys. Thanks for glorious detail explanation.
sfn17
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#18
May13-12, 09:47 AM
P: 4
I'm preparing a little sketch on the irreps of the Poincaré group according to Wigner's classification. Unfortunately, this subject seems to be not too popular in the textbooks and in review articles. No author whom I found describes clearly and correctly the structure of the little group of the photon. The most part of them resort just to rescaling the angle (i.e. [itex]\phi\to\phi/2[/itex]), some give wrong statements on the semidirect structure.

I spent some time to puzzle everything together. Now, it's quite easy to explain. I try to make some comments. I am much indebted to the comments of George Jones.

According to Simms, I will denote the little group of the photon by Δ.

Quote Quote by George Jones View Post
The universal cover of the restricted Lorentz group is [itex]SL\left( 2,\mathbb{C}\right)[/itex].
That indicates, that Δ is also a sort of covering group. At least, its center will be [itex]\left\{ I_2,-I_2\right\}[/itex], [itex]I_2[/itex] being the identity element of [itex]SL\left( 2,\mathbb{C}\right)[/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.
This is the group Δ.

To get rid of using a chart of [itex]U(1)[/itex] (and thereby to avoid any problems with rescaling), I prefer the equivalent definition of the group Δ as follows:

The little group Δ 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}
u & u^{-1}b\\ 0 & u^{-1}
\end{pmatrix},
[/tex]
where [itex]u,b[/itex] are arbitrary complex numbers but with [itex]\left|u\right|=1[/itex].

The introduction of an additional factor [itex]u^{-1}[/itex] in the entry (1,2) simplifies the exhibition of the semidirect structure.

The key point is that the product in Δ resembles much to that in the Euclidean group E(2), but not fully. Namely we have
[tex]
\begin{pmatrix}
u_1 & u_1^{-1}b_1\\ 0 & u_1^{-1}
\end{pmatrix}\begin{pmatrix}
u_2 & u_2^{-1}b_2\\ 0 & u_2^{-1}
\end{pmatrix}
=
\begin{pmatrix}
u_1u_2 & u_2^{-1}u_1b_2+u_1^{-1}b_1\\ 0 & u_1^{-1}u_2^{-1}
\end{pmatrix}
=\begin{pmatrix}
u_1u_2 & (u_1u_2)^{-1}(u_1^2b_2+b_1)\\ 0 & (u_1u_2)^{-1}
\end{pmatrix}.
[/tex]
The difference lies in the [itex]u_1^2[/itex].

Let's describe the special Euclidean group SE(2) by real [itex]3\times3[/itex]-matrices: SE(2) consists of the elements
[tex]
\begin{pmatrix} R & b \\ 0 & 1 \end{pmatrix}
[/tex]
with [itex]R\in SO(2)[/itex], [itex]b\inℝ^2[/itex]. Note: Reflections should be excluded since they cannot be represented by any [itex]e^{i\theta}[/itex].

Ordinary matrix multiplication reproduces the structure as semidirect product of rotations and translations in two dimensions correctly:
[tex]
\begin{pmatrix} R_1 & b_1 \\ 0 & 1 \end{pmatrix}
\begin{pmatrix} R_2 & b_2 \\ 0 & 1 \end{pmatrix}
=
\begin{pmatrix} R_1R_2 & R_1b_2+b_1 \\ 0 & 1 \end{pmatrix}.
[/tex]
Here, the rotation [itex]R_1[/itex] does not appear as [itex]R_1^2[/itex].
To have then some sort of isomorphism to Δ, in the entry (1,2) should be a [itex]u_1[/itex] instead of [itex]u_1^2[/itex]. It seems that the elements [itex]u[/itex] and [itex]-u[/itex] have to be identified.

This identification is achieved by the following homomorphism

[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]
that means that I have got the homomorphism [itex]\phi[/itex]:
[tex]
\phi:\quad
\begin{pmatrix}
u & b \\ 0 & u^{-1}
\end{pmatrix}
\mapsto
\begin{pmatrix}
u^2 & b \\ 0 & 1
\end{pmatrix}.
[/tex]
(That I now suppress the additional factor [itex]u^{-1}[/itex] at entry (1,2) is of course of no significance.)

Note: I don't make use of any angle [itex]\theta[/itex] nor of some rescaling. Defining the group multiplication with elements written in the form [itex]g(\theta,b)[/itex] suffers from tackling the addition of two angles to a value exceeding [itex]2\pi[/itex] or even [itex]4\pi[/itex]. Somewhere in the literature, the mere extension of the interval [itex]\left[0,2\pi\right][/itex] to [itex]\left[0,4\pi\right][/itex] together with the replacement [itex]\theta[/itex] by [itex]\theta/2[/itex] seems to imitate the descent to a group covered by Δ.

The kernel of [itex]\phi[/itex] is obviously [itex]\left\{I_2,-I_2\right\}[/itex].
So, in taking the cosets of this kernel we get an isomorphism [itex]\iota[/itex]
[tex]
\iota:\quad
\left\{\begin{pmatrix}
u & b \\ 0 & u^{-1}
\end{pmatrix},
\begin{pmatrix}
-u & -b \\ 0 & -u^{-1}
\end{pmatrix}
\right\}
\mapsto
\begin{pmatrix}
u^2 & b \\ 0 & 1
\end{pmatrix}.
[/tex]
The group formed by the elements [itex]\begin{pmatrix}u^2 & b \\ 0 & 1\end{pmatrix}[/itex] with [itex]\left|u\right|=1[/itex] is of course the same as the group formed by the elements [itex]\begin{pmatrix}v & b \\ 0 & 1\end{pmatrix}[/itex] with [itex]\left|v\right|=1[/itex].

To prove that this group is isomorphic to the proper Euclidean group [itex]SE(2)[/itex], one needs now the identification [itex]v=e^{i\theta}[/itex] with [itex]\theta\in[0,2\pi[[/itex]. Then one can proceed in the manner as demonstrated by George Jones. No factor 2 is needed.

I conclude that the group Δ is a twofold covering group of the Euclidean group [itex]SE(2)[/itex]. The covering homomorphism (Δ onto an isomorphic copy of [itex]SE(2)[/itex]) is given by [itex]\phi[/itex] defined above.

I do not state that the little group Δ is the universal covering group of [itex]SE(2)[/itex] since I don't know whether Δ is simply connected. So, I refrain from writing [itex]\widetilde{SE(2)}[/itex] instead of Δ.


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