Register to reply 
What is Little Group ?by yicong2011
Tags: little group 
Share this thread: 
#1
Feb2012, 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. 


#2
Feb2112, 02:34 AM

Sci Advisor
P: 3,593

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 


#3
Feb2112, 07:56 AM

Sci Advisor
Thanks
P: 4,160

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


#4
Feb2112, 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. 


#5
Feb2112, 10:58 AM

P: 100

More grouptheoretically, 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.



#6
Feb2112, 11:15 AM

Sci Advisor
P: 3,593

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. 


#7
Feb2112, 07:09 PM

Mentor
P: 6,246




#8
Feb2212, 02:03 AM

Sci Advisor
P: 3,593




#9
Feb2212, 06:02 AM

Sci Advisor
HW Helper
Thanks
P: 26,148

hi yicong2011!
see it free online at http://books.google.co.uk/books?id=h...group%22&hl=en 


#10
Feb2212, 09:56 AM

PF Gold
P: 368

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



#11
Feb2212, 10:36 AM

Sci Advisor
P: 3,593




#12
Feb2212, 10:51 AM

Mentor
P: 6,246

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



#13
Feb2212, 11:50 AM

Mentor
P: 6,246

The universal cover of the restricted Lorentz group is [itex]SL\left( 2,\mathbb{C}\right)[/itex]. If a 4vector 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 4vector [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 twotoone 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] 


#14
Feb2312, 01:56 AM

Sci Advisor
P: 3,593

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. 


#15
Feb2312, 03:39 AM

Sci Advisor
P: 1,927

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] 


#16
Feb2612, 05:58 PM

Sci Advisor
P: 1,135

The A and B, acting on a momentum in the zdirection, 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 xy plane of the centrifugal acceleration. Small nitpick about his signs: In a righthanded 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,0,0k) 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 


#17
Feb2712, 12:48 AM

P: 76

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



#18
May1312, 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. 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 4vector [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]\leftu\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] \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]\leftu\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]\leftv\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. 


Register to reply 
Related Discussions  
In binary can we have a value with deci centi mili or more lower valued prefix?  Computers  14  
No prior geometry and the group field theory vehicle  Beyond the Standard Model  6  
Qs: How Group and Period affect semiconductors  Atomic, Solid State, Comp. Physics  6  
Question on Two theorems for the group velocity in dispersive media  Classical Physics  4  
Questions on Freidel's Group Field Theory (hepth/050516)  Beyond the Standard Model  3 