Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

This question is based on problem II.3.1 in Anthony Zee's book Quantum Field Theory in a Nutshell (I'm reading this for fun- it isn't a homework problem.)

Show, by explicit calculation, that $(1/2,1/2)$ is the Lorentz Vector.

I see that the generators of SU(2) are the Pauli Matrices and the generators of SO(3,1)is a matrix composed of two Pauli Matrices along the diagonal. Is it always the case that the Direct Product of two groups is formed from the generators like this?

I ask this because I'm trying to write a Lorentz boost as two simultaneous quatertion rotations [unit quaternions rotations are isomorphic to SU(2)] and tranform between the two methods. Is this possible?

In other words, How do I construct the SU(2) representation of the Lorentz Group using the fact that $SU(2)\times SU(2) \sim SO(3,1)$?

Here is some background information:

Zee has shown that the algebra of the Lorentz group is formed from two separate $SU(2)$ algebras [$SO(3,1)$ is isomorphic to $SU(2)\times SU(2)$] because the Lorentz algebra satisfies:

$$\begin{align}[J_{+i},J_{+j}] &= ie_{ijk}J_{k+} & [J_{-i},J_{-j}] &= ie_{ijk} J_{k-} & [J_{+i},J_{-j}] &= 0\end{align}$$

The representations of $SU(2)$ are labeled by $j=0,\frac{1}{2},1,\ldots$ so the $SU(2)\times SU(2)$ rep is labelled by $(j_+,j_-)$ with the $(1/2,1/2)$ being the Lorentz 4-vector because and each representation contains $(2j+1)$ elements so $(1/2,1/2)$ contains 4 elements.

share|improve this question
5  
It's really the basic problem one has to solve himself in order to understand how spinors work. Check a 3D subset of this problem at motls.blogspot.com/2012/04/why-are-there-spinors.html if you really need help. Just one correction: the complexifications of the $SU(2)\times SU(2)$ and $SO(3,1)$ algebras are the same. However, when the coefficients are real, they're different. $SU(2)\times SU(2)$ is $SO(4)$ while $SO(3,1)$, its pseudoorthogonal version, is the same Lie algebra as $SL(2,C)$. –  Luboš Motl May 17 '12 at 15:47
    
Is there a method taking the Direct Product of SU(2) and explicitly calculating the Lorentz Transform? I would like to see the Lorentz Transform actually come out of the calculation. –  MadScientist May 17 '12 at 16:26
1  
That is a way of computing the direct product of two groups. Perhaps you could edit that last comment into the question, since normally we don't allow homework-like questions (which this definitely is) without narrowing down the question to the specific concept you want to ask about. –  David Z May 17 '12 at 20:36

1 Answer 1

up vote 11 down vote accepted

Here is a mathematical derivation. We use the sign convention $(+,-,-,-)$ for the Minkowski metric $\eta_{\mu\nu}$.

I) First recall the fact that

$SL(2,\mathbb{C})$ is (the double cover of) the restricted Lorentz group $SO^+(1,3;\mathbb{R})$.

This follows partly because:

  1. There is a bijective isometry from the Minkowski space $(M(1,3;\mathbb{R}),||\cdot||^2)$ to the space of $2\times2 $ Hermitian matrices $(u(2),\det(\cdot))$, $$\mathbb{R}^4~=~M(1,3;\mathbb{R}) ~\cong ~ u(2) ~:=~\{\sigma\in {\rm Mat}_{2\times 2}(\mathbb{C}) \mid \sigma^{\dagger}=\sigma \} ~=~ {\rm span}_{\mathbb{R}} \{\sigma_{\mu} \mid \mu=0,1,2,3\}, $$ $$ M(1,3;\mathbb{R})~\ni~\tilde{x}~=~(x^0,x^1,x^2,x^3) \quad\mapsto \quad\sigma~=~x^{\mu}\sigma_{\mu}~\in~ u(2), $$ $$ ||\tilde{x}||^2 ~=~x^{\mu} \eta_{\mu\nu}x^{\nu} ~=~\det(\sigma), \qquad \sigma_{0}~:=~{\bf 1}_{2 \times 2}.$$

  2. There is a group action $\rho: SL(2,\mathbb{C})\times u(2) \to u(2)$ given by $$g\quad \mapsto\quad\rho(g)\sigma~:= ~g\sigma g^{\dagger}, \qquad g\in SL(2,\mathbb{C}),\qquad\sigma\in u(2), $$ which is length preserving, i.e. $g$ is a pseudo-orthogonal (or Lorentz) transformation. In other words, there is a Lie group homomorphism
    $$\rho: SL(2,\mathbb{C}) \quad\to\quad O(u(2),\mathbb{R})~\cong~ O(1,3;\mathbb{R}) , \qquad \rho(\pm {\bf 1}_{2 \times 2})~=~{\bf 1}_{u(2)}.$$

  3. The Lie group $SL(2,\mathbb{C})=\pm e^{sl(2,\mathbb{C})}$ has Lie algebra $$ sl(2,\mathbb{C}) ~=~ \{\tau\in{\rm Mat}_{2\times 2}(\mathbb{C}) \mid {\rm tr}(\tau)~=~0 \} ~=~{\rm span}_{\mathbb{C}} \{\sigma_{i} \mid i=1,2,3\}.$$

  4. The Lie group homomorphism $\rho: SL(2,\mathbb{C}) \to O(u(2),\mathbb{R})$ induces a Lie algebra homomorphism $$\rho: sl(2,\mathbb{C})\to o(u(2),\mathbb{R})$$ given by $$ \rho(\tau)\sigma ~=~ \tau \sigma +\sigma \tau^{\dagger}, \qquad \tau\in sl(2,\mathbb{C}),\qquad\sigma\in u(2), $$ $$ \rho(\tau) ~=~ L_{\tau} +R_{\tau^{\dagger}},$$ where we have defined left and right multiplication of $2\times 2$ matrices $$L_{\sigma}(\tau)~:=~\sigma \tau~=:~ R_{\tau}(\sigma), \qquad \sigma,\tau ~\in~ {\rm Mat}_{2\times 2}(\mathbb{C}).$$

II) Note that the Lorentz Lie algebra $o(1,3;\mathbb{R}) \cong sl(2,\mathbb{C})$ does not$^{\ddagger}$ contain two perpendicular copies of, say, the real Lie algebra $su(2)$ or $sl(2,\mathbb{R})$. For comparison and completeness, let us mention that for other signatures in $4$ dimensions, one has

$$o(4;\mathbb{R}) ~\cong~su(2) \oplus su(2), \qquad\text{(compact form)}$$ $$o(2,2;\mathbb{R}) ~\cong~sl(2,\mathbb{R}) \oplus sl(2,\mathbb{R}).\qquad\text{(split form)}$$

(The compact form has a nice proof using quaternions, see also this math.SE question.)

To decompose Minkowski space into left- and right-handed Weyl spinor representations, one must go to the complexification, i.e. one must use the fact that

$SL(2,\mathbb{C})\times SL(2,\mathbb{C})$ is (the double cover of) the complexified Lorentz group $SO(1,3;\mathbb{C})$.

Note that Ref. 1 does not discuss complexification$^{\ddagger}$. One can more or less repeat the construction from section I with the real numbers $\mathbb{R}$ replaced by complex numbers $\mathbb{C}$, however with some important caveats.

  1. There is a bijective isometry from the complexified Minkowski space $(M(1,3;\mathbb{C}),||\cdot||^2)$ to the space of $2\times2 $ matrices $({\rm Mat}_{2\times 2}(\mathbb{C}),\det(\cdot))$, $$\mathbb{C}^4~=~M(1,3;\mathbb{C}) ~\cong ~ {\rm Mat}_{2\times 2}(\mathbb{C}) ~=~ {\rm span}_{\mathbb{C}} \{\sigma_{\mu} \mid \mu=0,1,2,3\}, $$ $$ M(1,3;\mathbb{C})~\ni~\tilde{x}~=~(x^0,x^1,x^2,x^3) \quad\mapsto \quad\sigma~=~x^{\mu}\sigma_{\mu}~\in~ {\rm Mat}_{2\times 2}(\mathbb{C}) , $$ $$ ||\tilde{x}||^2 ~=~x^{\mu} \eta_{\mu\nu}x^{\nu} ~=~\det(\sigma).$$ Note that forms are taken to be bilinear rather than sesquilinear.

  2. There is a Lie group homomorphism
    $$\rho: SL(2,\mathbb{C}) \times SL(2,\mathbb{C}) \quad\to\quad O({\rm Mat}_{2\times 2}(\mathbb{C}),\mathbb{C})~\cong~ O(1,3;\mathbb{C})$$ given by $$(g_L, g_R)\quad \mapsto\quad\rho(g_L, g_R)\sigma~:= ~g_L\sigma g^{\dagger}_R, \qquad g_L, g_R\in SL(2,\mathbb{C}),\qquad\sigma~\in~ {\rm Mat}_{2\times 2}(\mathbb{C}). $$

  3. The Lie group $SL(2,\mathbb{C})\times SL(2,\mathbb{C})$ has Lie algebra $sl(2,\mathbb{C})\oplus sl(2,\mathbb{C})$.

  4. The Lie group homomorphism
    $$\rho: SL(2,\mathbb{C})\times SL(2,\mathbb{C}) \quad\to\quad O({\rm Mat}_{2\times 2}(\mathbb{C}),\mathbb{C})$$ induces a Lie algebra homomorphism $$\rho: sl(2,\mathbb{C})\oplus sl(2,\mathbb{C})\quad\to\quad o({\rm Mat}_{2\times 2}(\mathbb{C}),\mathbb{C})$$ given by $$ \rho(\tau_L\oplus\tau_R)\sigma ~=~ \tau_L \sigma +\sigma \tau^{\dagger}_R, \qquad \tau_L,\tau_R\in sl(2,\mathbb{C}),\qquad \sigma\in {\rm Mat}_{2\times 2}(\mathbb{C}), $$ $$ \rho(\tau_L\oplus\tau_R) ~=~ L_{\tau_L} +R_{\tau^{\dagger}_R}.$$

The left action (acting from left on a two-dimensional complex column vector) yields by definition the (left-handed Weyl) spinor representation $(\frac{1}{2},0)$, while the right action (acting from right on a two-dimensional complex row vector) yields by definition the right-handed Weyl/complex conjugate spinor representation $(0,\frac{1}{2})$. The above shows that

The complexified Minkowski space $M(1,3;\mathbb{C})$ is a $(\frac{1}{2},\frac{1}{2})$ representation of the Lie group $SL(2,\mathbb{C}) \times SL(2,\mathbb{C})$, whose action respects the Minkowski metric.

References:

  1. Anthony Zee, Quantum Field Theory in a Nutshell, 2003.

$^{\ddagger}$ For a laugh, check out the wrong second sentence on p.113 in Ref. 1. Nevertheless, let me rush to add that Ref. 1 is overall a very nice book.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.