A whole lot of doubts on Lorentz representation

Question

Can someone tell me in layman's language how the $(1/2,1/2)$ represents a vector field and $(0,1/2)$ or $(1/2,0)$ represents spinors and $(0,0)$ represents scalar field. Please don't be pedantic on mathematics part. I didn't take a course on group theory yet. Give me physical arguments why this is true? I have come across this in QFT course that I am currently enrolled in.

Answer 1

There is a definition that $\left( \frac{m}{2}, \frac{n}{2}\right)$ representation is equal to spinor tensor $$ \psi_{a_{1}...a_{m}\dot{b}_{1}...\dot{b}_{n}}, $$ where $\psi_{\dot{b}}$ transforms as complex conjugation of $\psi_{b}$. Why do we assume that $\left( \frac{1}{2}, 0\right)$ and $\left( 0, \frac{1}{2}\right)$ represent spinors? You can think about it (without lots of the group theory) by the next way.

We may introduce, by the closest analogy with complex number (which can describe rotation in a plane), sets of 4 hypercomplex numbers (quaternions) from which we can construct 3-rotations and Lorentz boosts matrices in space of some 2-component vectors, which we can call spinors. From two spinors then we can construct 2*2 matrix which behaves as 4-vector under quaternion transformations.

This shows that the most "elementary" non-invariant representation of the Lorentz group is spinors (by the definition they are marked as $\left( \frac{1}{2}, 0\right)$ and $\left( 0, \frac{1}{2}\right)$, where the second one transforms as complex conjugated first one). The invariant representation is, of course, scalar representation, which is marked as $\left( 0 , 0\right)$, because it doesn't have spinor indices so it is scalar.

As for $\left( \frac{1}{2}, \frac{1}{2}\right)$, there is a connection between 4-tensor and corresponding spinor tensor: $$ \psi_{\mu_{1}...\mu_{n}} \to \psi_{a_{1}...a_{n}\dot{b}_{1}...\dot{b}_{n}} = \sigma^{\mu_{1}}_{a_{1}\dot {b}_{1}}...\sigma^{\mu_{n}}_{a_{n}\dot {b}_{n}}\psi_{\mu_{1}...\mu_{n}}, $$ or $$ \psi_{a_{1}...a_{n}\dot{b}_{1}...\dot{b}_{n}} \to \psi_{\mu_{1}...\mu_{n}} = \frac{1}{2^{n}}Tr\left( \tilde{\sigma}_{\mu_{1}}^{\dot{b}_{1}a_{1}}...\tilde{\sigma}_{\mu_{n}}^{\dot{b}_{n}a_{n}}\psi_{a_{1}...a_{n}\dot{b}_{1}...\dot{b}_{n}}\right). $$ So $\left( \frac{1}{2}, \frac{1}{2}\right)$ represents 4-vector.

Little addition - correspondence between $\left(\frac{1}{2} , \frac{1}{2} \right)$ and 4-vector

The representation $\left(\frac{1}{2} , \frac{1}{2} \right)$ is constructed as $$ \left( \frac{1}{2} , 0 \right) \otimes \left( 0, \frac{1}{2} \right) = \left(\frac{1}{2}, \frac{1}{2} \right) . $$ Since both of $\psi_{a}, \psi_{\dot {b}}$ have 2 components, an object $\psi_{a \dot{b}}$ has 4 parameters. It can be given as hermitean $2 \times 2$ matrix. Each $2 \times 2$ hermitean matrix can be given in form $$ \tag 1 \psi_{a \dot {b}} = \begin{pmatrix} A_{0} + A_{3} & A_{1} - iA_{2} \\ A_{1} + iA_{2} & A_{0} - A_{3} \end{pmatrix} = A^{\mu}\sigma_{\mu}, \quad \sigma_{\mu} = (\hat{E} , \sigma )_{\mu}, \quad det (\psi ) = A_{0}^{2} - \mathbf A^{2}. $$ An object $(1)$ transforms under specific transformations (matrix $\hat{S}$ isn't arbitrary) $$ \psi {'} = \hat{S} \psi \hat{S}^{\dagger}, \quad det \hat{S} = 1 \Rightarrow det (\psi {'}) = det (\psi ) = inv $$ the same way as 4-vector.

So it's not hard to conсlude that $\left( \frac{1}{2}, \frac{1}{2} \right)$ in a form $(1)$ is homomorphic to usual 4-vector representation $A_{\mu}$.

It's not hard to see that 4-vector $A_{\mu}$ can be extracted from $(1)$ by the relation $$ A_{\mu} = \frac{1}{2}Tr(\sigma_{\mu}\psi ). $$

Answer 2

Recall that when classifying representations of the Lorentz group, we consider \begin{equation} \textbf{N}_{\pm} = \frac{\textbf{J} \pm i\textbf{K}}{2}, \tag{1} \end{equation} where $\textbf{J}$ is the angular momentum (rotation generator) and $\textbf{K}$ is the boost generator. The generators $\textbf{J}$ and $\textbf{K}$ satisfy \begin{equation} [J^{i},J^{j}] = i\varepsilon^{ijk} J^{k}, \tag{2} \end{equation} \begin{equation} [J^{i},K^{j}] = i\varepsilon^{ijk} K^{k}, \tag{3} \end{equation} and \begin{equation} [K^{i},K^{j}] = -i\varepsilon^{ijk} J^{k}. \tag{4} \end{equation} From the above relations, one can show that each of $\textbf{N}_{\pm}$ satisfies the "angular momentum commutation relation", and $\textbf{N}_{+}$ and $\textbf{N}_{-}$ commute, viz., \begin{equation} [N_{\pm}^{i},N_{\pm}^{j}] = i\varepsilon^{ijk} N_{\pm}^{k} \tag{5} \end{equation} and \begin{equation} [N_{\pm}^{i},N_{\mp}^{j}] = 0. \tag{6} \end{equation} By the $(A,B)$ representation of the Lorentz group, we mean that "angular momentum quantum numbers" corresponding to $\textbf{N}_{+}$ and $\textbf{N}_{-}$ are $A$ and $B$.

Now let's focus our attention on the rotation generator $\textbf{J} = \textbf{N}_{+}+\textbf{N}_{-}$. Possible "angular momentum quantum numbers" for $\textbf{J}$ are given by the usual angular momentum addition rule, i.e., \begin{equation} j = |A-B|, \ldots, A+B. \tag{7} \end{equation} We see that for the $(1/2,0)$ and $(0,1/2)$ representations, the only possible value of $j$ is $1/2$, which means that they behave like a spinor under rotation.

On the other hand, for the $(1/2,1/2)$ representation, we have $j=0,1$. This is exactly how a Lorentz vector would behave under rotation: the spatial components (a 3-vector) corresponds to $j=1$, whereas the time component is invariant under rotation, i.e, $j=0$.

I hope that by above, I have persuaded you that it is plausible to identify the $(1/2,1/2)$ representation with the Lorentz vector. However, to make the argument complete, let me add something more.

What we have seen so far cannot be a proof because there are other representations with exactly one $j=0$ and one $j=1$ parts: the reducible representations $(0,0) \oplus (1,0) $ and $(0,0) \oplus (0,1)$. But under an arbitrary boost, the $j=0$ and $j=1$ parts of these reducible representations transform independently. (Actually the $j=0$ part, coming from $(0,0)$ is invariant.) This cannot be the case of the Lorentz vector, as space and time components should mix under boost. So we are left with only $(1/2,1/2)$.

I won't bother to explicitly work out how components of $(1/2, 1/2)$ are related to the space and time components of a Lorentz vector, as it is already done in the excellent answer by Andrew McAddams.