# Proof that (1/2,1/2) Lorentz group representation is a 4-vector

Taken from Quantum Field Theory in a Nutshell by Zee, problem II.3.1:

Show by explicit computation that $(\frac{1}{2},\frac{1}{2})$ is indeed the Lorentz vector.

This has been asked here:

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

but I can't really digest the formality of this answer with only a little knowledge of groups and representations.

By playing around with the Lorentz group generators it is possible to find the basis $J_{\pm i}$ that separately have the Lie algebra of $SU(2)$, and thus can be separately given spin representations.

My approach has been to write $$J_{+i}=\frac{1}{2}(J_{i}+iK_{i})=\frac{1}{2}\sigma_{i}$$ $$J_{-i}=\frac{1}{2}(J_{i}-iK_{i})=\frac{1}{2}\sigma_{i}$$ which implies that $$J_{i}=\sigma_{i}$$ $$K_{i}=0$$ However I don't really get where to go next.

• Related question by OP: physics.stackexchange.com/q/321276/2451 and links therein. – Qmechanic Mar 26 '17 at 10:53
• The isomorphism can be seen by realizing that the matrices $\sigma^\mu_{\alpha {\dot \beta}}$ forms a basis for all $2\times2$ matrices so an arbitrary matrix $A_{\alpha {\dot \beta}}$ can be written as $A_{\alpha {\dot \beta}} = A_\mu \sigma^\mu_{\alpha {\dot \beta}}$. The LHS of this transforms in the $(0,\frac{1}{2}) \otimes (\frac{1}{2} ,0) = ( \frac{1}{2} , \frac{1}{2})$. This equation tells you what is the change of the basis between the usual vector basis $A_\mu$ and the $(\frac{1}{2},\frac{1}{2})$ basis. – Prahar Sep 19 '17 at 2:51

First let's recall how to construct the finite dimensional irreducible representations of the Lorentz group. Say $$J_i$$ are the three rotation generators and $$K_i$$ are the three boost generators. \begin{align*} L_x = &\begin{pmatrix} 0&0&0&0 \\ 0&0&0&0 \\ 0&0&0&-1 \\ 0&0&1&0 \end{pmatrix}& L_y = &\begin{pmatrix} 0&0&0&0 \\ 0&0&0&1 \\ 0&0&0&0 \\ 0&-1&0&0 \end{pmatrix}& L_z = &\begin{pmatrix} 0&0&0&0 \\ 0&0&-1&0 \\ 0&1&0&0 \\ 0&0&0&0 \end{pmatrix}\\ K_x = &\begin{pmatrix} 0&1&0&0 \\ 1&0&0&0 \\ 0&0&0&0 \\ 0&0&0&0 \end{pmatrix}& K_y = &\begin{pmatrix} 0&0&1&0 \\ 0&0&0&0 \\ 1&0&0&0 \\ 0&0&0&0 \end{pmatrix}& K_z = &\begin{pmatrix} 0&0&0&1 \\ 0&0&0&0 \\ 0&0&0&0 \\ 1&0&0&0 \end{pmatrix}\\ \end{align*} They satisfy $$[J_i, J_j] = \varepsilon_{ijk} J_k \hspace{1 cm} [K_i, K_j] = -\varepsilon_{ijk} J_k \hspace{1 cm} [J_i, K_j] = \varepsilon_{ijk}K_k.$$ (Note that I am using the skew-adjoint convention for Lie algebra elements where I did not multiply by $$i$$.)

We then define $$A_i = \frac{1}{2} (J_i - i K_i) \hspace{2 cm} B_i = \frac{1}{2}(J_i + i K_i)$$ which satisfy the commutation relations $$[A_i, A_j] = \varepsilon_{ijk} A_k \hspace{2cm} [B_i, B_j] = \varepsilon_{ijk} B_k \hspace{2cm} [A_i, B_j] = 0.$$

Here is how you construct the representation of the Lorentz group: first, choose two non-negative half integers $$j_1$$ and $$j_2$$. These correspond to two spin $$j$$ representations of $$\mathfrak{su}(2)$$, which I will label $$\pi'_{j}.$$ Recall that that $$\mathfrak{su}(2) = \mathrm{span}_\mathbb{R} \{ -\tfrac{i}{2} \sigma_x, -\tfrac{i}{2} \sigma_y, -\tfrac{i}{2} \sigma_z \}$$ where $$[-\tfrac{i}{2} \sigma_i, -\tfrac{i}{2} \sigma_j] = -\tfrac{i}{2}\varepsilon_{ijk} \sigma_k.$$ For this question, we only need to know the spin $$1/2$$ representation of $$\mathfrak{su}(2)$$, which is given by $$\pi'_{\tfrac{1}{2}}( -\tfrac{i}{2}\sigma_i) = -\tfrac{i}{2} \sigma_i.$$

So okay, how do we construct the $$(j_1, j_2)$$ representation of the Lorentz group? Any Lie algebra element $$X \in \mathfrak{so}(1,3)$$ can written as a linear combination of $$A_i$$ and $$B_i$$: $$X = \sum_{i = 1}^3 (\alpha_i A_i + \beta_i B_i).$$ (Note that we are actually dealing with the complexified version of the Lie algebra $$\mathfrak{so}(1,3)$$ because our definitions of $$A_i$$ and $$B_i$$ have factors of $$i$$, so $$\alpha, \beta \in \mathbb{C}$$.)

$$A_i$$ and $$B_i$$ form their own independent $$\mathfrak{su}(2)$$ algebras.

The Lie algebra representation $$\pi'_{(j_1, j_2)}$$ is then given by \begin{align*} \pi'_{(j_1, j_2)}(X) &= \pi'_{(j_1, j_2)}(\alpha_i A_i + \beta_j B_i) \\ &\equiv \pi'_{j_1}(\alpha_i A_i) \otimes \big( \pi'_{j_2}(\beta_j B_i) \big)^* \end{align*} where the star denotes complex conjugation.

Sometimes people forget to mention that you have to include the complex conjugation, but it won't work otherwise!

If $$j_1 = 1/2$$ and $$j_2 = 1/2$$, we have $$\begin{equation*} \pi_{\frac{1}{2}}'(A_i) = -\frac{i}{2}\sigma_i \otimes I \hspace{2cm} \big(\pi_{\frac{1}{2}}' (B_i)\big)^* = \frac{i}{2} I \otimes\sigma_i^*. \end{equation*}$$ We can explicitly write out these tensor products in terms of a $$2 \times 2 = 4$$ dimensional basis. (Here I am using the so-called "Kronecker Product" to do this. That just a fancy name for multiplying all the elements of two $$2\times 2$$ cell-wise to get a $$4 \times 4$$ matrix.) \begin{align*} \pi_{(\frac{1}{2},\frac{1}{2})}'(A_x) &= -\frac{i}{2}\begin{pmatrix} 0&0&1&0 \\ 0&0&0&1 \\ 1&0&0&0 \\ 0&1&0&0 \end{pmatrix} & \big(\pi_{(\frac{1}{2},\frac{1}{2})}'(B_x)\big)^* &= \frac{i}{2}\begin{pmatrix} 0&1&0&0 \\ 1&0&0&0 \\ 0&0&0&1 \\ 0&0&1&0 \end{pmatrix} \\ \pi_{(\frac{1}{2},\frac{1}{2})}'(A_y) &= \frac{1}{2}\begin{pmatrix} 0&0&-1&0 \\ 0&0&0&-1 \\ 1&0&0&0 \\ 0&1&0&0 \end{pmatrix} & \big(\pi_{(\frac{1}{2},\frac{1}{2})}'(B_y)\big)^*&= \frac{1}{2}\begin{pmatrix} 0&-1&0&0 \\ 1&0&0&0 \\ 0&0&0&-1 \\ 0&0&1&0 \end{pmatrix} \\ \pi_{(\frac{1}{2},\frac{1}{2})}'(A_z) &= -\frac{i}{2}\begin{pmatrix} 1&0&0&0 \\ 0&1&0&0 \\ 0&0&-1&0 \\ 0&0&0&-1 \end{pmatrix} & \big(\pi_{(\frac{1}{2},\frac{1}{2})}'(B_z)\big)^* &= \frac{i}{2}\begin{pmatrix} 1&0&0&0 \\ 0&-1&0&0 \\ 0&0&1&0 \\ 0&0&0&-1 \end{pmatrix} \end{align*} We can then write out the matrices of the rotations and boosts $$J_i$$ and $$K_i$$ using $$J_i = A_i + B_i \hspace{2cm} K_i = i(A_i - B_i).$$ \begin{align*} \pi'_{(\frac{1}{2},\frac{1}{2})}(J_x) &= \frac{i}{2}\begin{pmatrix} 0&1&-1&0 \\ 1&0&0&-1 \\ -1&0&0&1 \\ 0&-1&1&0 \end{pmatrix} & \pi'_{(\frac{1}{2},\frac{1}{2})}(K_x) &= \frac{1}{2}\begin{pmatrix} 0&1&1&0 \\ 1&0&0&1 \\ 1&0&0&1 \\ 0&1&1&0 \end{pmatrix} \\ \pi'_{(\frac{1}{2},\frac{1}{2})}(J_y) &= \frac{1}{2}\begin{pmatrix} 0&-1&-1&0 \\ 1&0&0&-1 \\ 1&0&0&-1 \\ 0&1&1&0 \end{pmatrix} & \pi'_{(\frac{1}{2},\frac{1}{2})}(K_y) &= \frac{i}{2}\begin{pmatrix} 0&1&-1&0 \\ -1&0&0&-1 \\ 1&0&0&1 \\ 0&1&-1&0 \end{pmatrix} \\ \pi'_{(\frac{1}{2},\frac{1}{2})}(J_z) &= \begin{pmatrix} 0&0&0&0 \\ 0&-i&0&0 \\ 0&0&i&0 \\ 0&0&0&0 \end{pmatrix} & \pi'_{(\frac{1}{2},\frac{1}{2})}(K_z) &= \begin{pmatrix} 1&0&0&0 \\ 0&0&0&0 \\ 0&0&0&0 \\ 0&0&0&-1 \end{pmatrix} \\ \end{align*} These are strange matrices, although we can make them look much more suggestive in another basis. Define the matrix $$\begin{equation*} U = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & i & 0\\ 0 & 1 & -i & 0 \\ 1 & 0 & 0 &-1 \end{pmatrix}. \end{equation*}$$ Amazingly, $$\begin{equation*} U \big( \pi'_{(\frac{1}{2},\frac{1}{2})}(L_i) \big) U^{-1} = L_i \hspace{1cm}U \big( \pi'_{(\frac{1}{2},\frac{1}{2})}(K_i) \big) U^{-1} = K_i. \end{equation*}$$ Therefore, the $$(\tfrac{1}{2}, \tfrac{1}{2})$$ representation is equivalent to the regular "vector" representation of $$SO^+(1,3)$$. However, these "vectors" live in $$\mathbb{C}^4$$, not $$\mathbb{R}^4$$, which people usually don't mention.

Sorry this reply has come so long after your post. As this is a homework-type problem I won't perform all the calculations, but will spell out all the key points; it's probably simpler to follow this way anyway if you fill in the matrix algebra yourself!

To understand what's going on here, first of all think about a spin-1 system. In that case the action of $J_3$ on vectors is represented by the matrix $$\left(\begin{array}{ccc} 0 & -i & 0 \\ i & 0 & 0 \\ 0 & 0 & 0 \end{array} \right)$$ If you follow the standard construction of the spin-1 representation using ladder operators found in any QM textbook, you find instead that $J_3$ is represented by the matrix $$\left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{array} \right)$$ because it's acting on a basis of $J_3$ eigenstates. You can see that these matrices are unitarily equivalent by finding the eigenvalues of the first one to be $-1,0,1$; working out the corresponding eigenvectors gives you the matrix that changes basis (up to some ordering of that basis). (It's nice to notice that this matrix transforms linearly polarised transverse vector fields into circularly polarised ones!)

Now, in the given case where we have $(1/2,1/2) = SU(2) \otimes SU(2)$. $J_3$ acts on both the (1/2,0) and (0,1/2) representations by $\frac{1}{2} \sigma_3$. Because it's a generator of the group, and not a group element, the action on the tensor product of the two groups is given by $$J_3 = \frac{1}{2} \sigma_3 \otimes I + I \otimes \frac{1}{2} \sigma_3 = \left(\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \end{array} \right)$$ It's clear that this matrix gives us the action of $J_3$ on its eigenstates, and a change of basis gives us the 4-vector representation. The difference is that we now have two 0 eigenvalues; these have resulted from "adding the angular momentum" of our two spin-1/2 representations. One is from the spin-1 rep as before. The other is from the spin-0 representation; this is your time direction in the 4-vector, which is a scalar under rotations. So in this case the components of the matrix that changes basis acts on the 0-eigenvalue directions are the Clebsch-Gordon coefficients.

Hope that helps.

You have misunderstood the strategy. The idea is that we define new operators $A_i = \frac{J_i + iK_i}{2}$ and $B_i =\frac{J_i-iK_i}{2}$. Now compute $[A_i,A_j], [B_i,B_j]$ and $[A_i,B_j]$ The answer you should get is that the first two should give you the lie algebra of $SU(2)$ while the second should give you zero. This means we have at least locally $SU(2)\times SU(2)$

The lorentz boosts for a two component spinor is then $e^{i\vec{\sigma }\cdot \theta + \vec{\sigma} \cdot \phi}$ and the other is $e^{i\vec{\sigma }\cdot \theta - \vec{\sigma} \cdot \phi}$. Direct sum of this does the general job i.e $\begin{pmatrix}e^{i\vec{\sigma }\cdot \theta + \vec{\sigma} \cdot \phi}& 0 \\ 0 &e^{i\vec{\sigma }\cdot \theta - \vec{\sigma} \cdot \phi} \end{pmatrix}$

So from this you should be able to get the generators you want

• This is something that has been proven in the book. I think I need to explicitly calculate the generators of the Lorentz group by taking the tensor product of the two $SU(2)$ representations. How do I do this? – Watw Mar 26 '17 at 9:42
• I have added what the lorentz boosts look like so it should be clear what the generators are – Amara Mar 26 '17 at 13:00
• I believe you've actually misunderstood the question. Your answers relate to the Dirac spinor representation that's the direct sum (1/2, 0) +(0,1/2). The OP is asking for the equivalence of the (1/2, 1/2) representation to the fundamental 4-vector irrep. – rwold Apr 26 '17 at 11:43