Understanding 4-Vector Representations in the Lorentz Group

In summary, the irreducible representations of the proper orthochronous Lorentz group are labeled by 2 numbers (as it has rank 2). The 4-vector representation ##D^{(\frac{1}{2},\frac{1}{2})}## is not an IR.
  • #1
Silviu
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Hello! I am reading some notes on Lorentz group and at a point it is said that the irreducible representations (IR) of the proper orthochronous Lorentz group are labeled by 2 numbers (as it has rank 2). They describe the 4-vector representation ##D^{(\frac{1}{2},\frac{1}{2})}## and initially I thought this is an IR (also being a fundamental representation). However, further on they say that ##D^{(\frac{1}{2},\frac{1}{2})} = D^{(\frac{1}{2}, 0)} \oplus D^{(0,\frac{1}{2})}##, which implies that ##D^{(\frac{1}{2},\frac{1}{2})}## is not an IR. So I am confused, is it or is it not IR? The way I was thinking about it, is that the 4 dimensional vector representation (i.e. under Lorentz group) is an IR while the 4 dim spinor representation (i.e. under ##SL(2,C)##) is not IR. But wouldn't you need different notations for them? Then, ##D^{(0,\frac{3}{2})}## and ##D^{(\frac{3}{2},0)}## are also 4 dimensional, so what should I do with them? Are they IR, too? Can someone clarify this for me? Thank you!
 
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  • #2
For the first part, I will show that if a representation of a product group is constructed from an irreducible representation for each of the groups in it, than that representation is also irreducible.

For irreducibility: for a representation D(a) of a group, if a matrix X satisfies D(a).X = X.D(a), then X must be a multiple of the identity matrix.

For a group G that is a product G1 * G2, its elements are a = (a1,a2) where a1 ranges over all of G1 and a2 does so for G2. A representation D(a) = D1(a1)*D2(a2) where D1 is a rep of G1 and D2 of G2. In component form,

D(a)(i1j1,i2j2) = D1(a1)(i1,j1) * D2(a2)(i2,j2)

To test reducibility, we must find the possible matrices X(i1j1,i2j2) that satisfy D(a)(i1j1,i2j2) * X(j1k1,j2k2) = X(i1j1,i2j2) * D(a)(j1k1,j2k2) over all values of i1,i2,k1,k2 and summed over dummy indices j1,j2. Using the above decomposition of D(a) gives us

D1(a1)(i1,j1) * D2(a2)(i2,j2) * X(j1j2,k1k2) = X(i1i2,j1j2) * D1(a1)(j1,k1) * D2(a2)(j2,k2)

Setting a2 to the identity of G2 gives us D1(a1)(i1,j1) * X(j1i2,k1k2) = X(i1i2,j1k2) * D1(a1)(j1,k1) and since D1 is irreducible,

X(i1i2,j1j2) = δ(i1,j1) * X(i2,j2)

This in turn reduces the equation to D2(a2)(i2,j2) * X(j2,k2) = X(i2,j2) * D2(j2,k2) and that in turn gives us

X(i1i2,j1j2) = δ(i1,j1) * δ(i2,j2) * X

A multiple of the identity matrix. Thus, if component-group reps D1 and D2 are irreducible, then the product-group irrep D = D1 * D2 is also irreducible.

Thus, for the Lorentz group, every representation with form (j1,j2) is irreducible, since each part is irreducible.
 
  • #3
In general, a product representation of two irreducible representations will be reducible, but if one of the irreps is the unit representation D(a) = 1, then the product will also be an irrep. Thus, (j1,j2) = (j1,0) * (0,j2) where 0 is the unit representation.

The vector rep of the Lorentz group is (1/2,1/2), and is thus irreducible.

The spinor rep is (1/2,0) + (0,1/2) and is thus reducible to two smaller spinors, (1/2,0) and (0,1/2), both irreducible. Each one of them is a rep of SL(2,C), but each one is the complex conjugate of the other.

The reps (3/2,0) and (0,3/2) are both irreducible.

Though all of (1/2,1/2), (1/2,0) + (0,1/2), (3/2,0), and (0,3/2) have size 4, they are all inequivalent.
 

1. What is the Lorentz group?

The Lorentz group is a mathematical concept used in physics to describe the symmetry of space and time. It is a group of transformations that preserve the laws of physics, specifically the principles of special relativity.

2. What is a 4-vector?

A 4-vector is a mathematical construct used to represent physical quantities in four-dimensional spacetime. It consists of four components, three spatial coordinates and one time coordinate, and is often used in special relativity to describe the motion of objects.

3. How are 4-vectors used in the Lorentz group?

4-vectors are used in the Lorentz group to transform physical quantities between different frames of reference. This allows for the calculation of physical quantities, such as velocity and momentum, in different reference frames, taking into account the effects of special relativity.

4. How do you calculate the components of a 4-vector in the Lorentz group?

The components of a 4-vector can be calculated using the Lorentz transformation equations, which take into account the relative velocity between two frames of reference. These equations involve the speed of light and the transformation matrix of the Lorentz group.

5. Why is understanding 4-vector representations important in physics?

Understanding 4-vector representations is important in physics because it allows for the accurate description and prediction of physical phenomena, particularly in the realm of special relativity. It also helps to reconcile the differences between classical mechanics and the principles of special relativity.

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