- #1
Tio Barnabe
The generators ##N^{\pm}{}_\mu = \frac{1}{2}(J_\mu \pm iK_\mu)## obey the algebra of ##SU(2)##. On the RHS we see the Lorentz generators of rotations and boosts, respectively.
I considered the case where ##N^{\pm}{}_\mu = (1/2) \sigma_\mu##, i.e. the (1/2, 1/2) representation of the Lorentz group, where ##\sigma_\mu## are the 2x2 Pauli matrices.
Substitution into the equation above for ##N^{\pm}{}_\mu## leads to ##J_\mu = \sigma_\mu; \ K_\mu = 0##. I didn't like this result, since it seems to indicate that boosts can't be done.
Never the less, I procceded to the calculation of ##J_3 = \sigma_3##. I got the result $$\exp(J_3 \theta) = \exp(\sigma_3 \theta) = \begin{pmatrix}\cosh \theta + \sinh \theta&0\\0&\cosh \theta - \sinh \theta\end{pmatrix}$$
When operating with this matrix on an arbritary matrix $$V = v_0 \bf{1} + \sum_i v_i \sigma_i$$ [which (I think) is the most general matrix on the space in question, because it's a 2x2 hermitian matrix, and thus will satisfy the conditions for ##SU(2)##.]
I get for the transformed components ##v_i## exactly the same result I get for a Lorentz boost of a Lorentz four-vector along the 3-direction (z-direction), i.e., $$\begin{pmatrix}\cosh \theta&0&0&\sinh \theta\\0&0&0&0\\0&0&0&0\\\sinh \theta&0&0&\cosh \theta\end{pmatrix} \begin{pmatrix}v_0\\v_1\\v_2\\v_3\end{pmatrix}$$
This is too strange to be correct. What am I missing?
I considered the case where ##N^{\pm}{}_\mu = (1/2) \sigma_\mu##, i.e. the (1/2, 1/2) representation of the Lorentz group, where ##\sigma_\mu## are the 2x2 Pauli matrices.
Substitution into the equation above for ##N^{\pm}{}_\mu## leads to ##J_\mu = \sigma_\mu; \ K_\mu = 0##. I didn't like this result, since it seems to indicate that boosts can't be done.
Never the less, I procceded to the calculation of ##J_3 = \sigma_3##. I got the result $$\exp(J_3 \theta) = \exp(\sigma_3 \theta) = \begin{pmatrix}\cosh \theta + \sinh \theta&0\\0&\cosh \theta - \sinh \theta\end{pmatrix}$$
When operating with this matrix on an arbritary matrix $$V = v_0 \bf{1} + \sum_i v_i \sigma_i$$ [which (I think) is the most general matrix on the space in question, because it's a 2x2 hermitian matrix, and thus will satisfy the conditions for ##SU(2)##.]
I get for the transformed components ##v_i## exactly the same result I get for a Lorentz boost of a Lorentz four-vector along the 3-direction (z-direction), i.e., $$\begin{pmatrix}\cosh \theta&0&0&\sinh \theta\\0&0&0&0\\0&0&0&0\\\sinh \theta&0&0&\cosh \theta\end{pmatrix} \begin{pmatrix}v_0\\v_1\\v_2\\v_3\end{pmatrix}$$
This is too strange to be correct. What am I missing?
