- #1
julian
Gold Member
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- 313
The generators ##\{ L^1, L^2 , L^3 , K^1 , K^2 , K^3 \}## of the Lorentz group satisfy the Lie algebra:
\begin{array}{l}
[L^i , L^j] = \epsilon^{ij}_{\;\; k} L^k \\
[L^i , K^j] = \epsilon^{ij}_{\;\; k} K^k \\
[K^i , K^j] = \epsilon^{ij}_{\;\; k} L^k
\end{array}
It has the Casimirs
[tex]
C_1 = \sum_i (K^i K^i - L^i L^i) , \qquad C_2 = \sum_i K^i L^i
[/tex]
I wish to prove that the Casimirs commute with all of the generators of the Lie algebra. It is easy to prove that ##[C_1 , L^j] = 0##, ##[C_2 , L^j]## and ##[C_2 , K^j] = 0##. However, I'm having more trouble proving ##[C_1 , K^j] = 0##. What I obtain, for example for ##j=1##, is:
[tex]
[C_1 , K^1] = 2 [L^2 , K^3]_+ - 2 [L^3 , K^2]_+
[/tex]
where ##[\cdot , \cdot]_+## is the anti-commutator. I'm not sure how to prove directly that this vanishes. However, there may be an indirect way of proving ##[C_1 , K^1] = 0##. First note:
\begin{array}{l}
[C_1 , C_2] = \sum_i [K^i K^i - L^i L^i , C_2] \\
\sum_i (K^i [K^i , C_2] + [K^i , C_2] K^i - L^i [L^i , C_2] - [L^i , C_2] L^i) \\
= 0
\end{array}
where we have used ##[C_2 , L^j]## and ##[C_2 , K^j] = 0##. Next write
\begin{array}{l}
0 = [C_1 , C_2] \\
= \sum_i [C_1 , K^i L^i] \\
= \sum_i ([C_1 , K^i] L^i + K^i [C_1 , L^i]) \\
= \sum_i [C_1 , K^i] L^i .
\end{array}
where we have used ##[C_1 , L^j] = 0##.
Is it possible to use this to prove ##[C_1 , K^1] = 0##?
I would prefer to prove first that the Casimirs commutate with all the generators first and then conclude the two Casimirs commute, but if this is what I have to resort to...
\begin{array}{l}
[L^i , L^j] = \epsilon^{ij}_{\;\; k} L^k \\
[L^i , K^j] = \epsilon^{ij}_{\;\; k} K^k \\
[K^i , K^j] = \epsilon^{ij}_{\;\; k} L^k
\end{array}
It has the Casimirs
[tex]
C_1 = \sum_i (K^i K^i - L^i L^i) , \qquad C_2 = \sum_i K^i L^i
[/tex]
I wish to prove that the Casimirs commute with all of the generators of the Lie algebra. It is easy to prove that ##[C_1 , L^j] = 0##, ##[C_2 , L^j]## and ##[C_2 , K^j] = 0##. However, I'm having more trouble proving ##[C_1 , K^j] = 0##. What I obtain, for example for ##j=1##, is:
[tex]
[C_1 , K^1] = 2 [L^2 , K^3]_+ - 2 [L^3 , K^2]_+
[/tex]
where ##[\cdot , \cdot]_+## is the anti-commutator. I'm not sure how to prove directly that this vanishes. However, there may be an indirect way of proving ##[C_1 , K^1] = 0##. First note:
\begin{array}{l}
[C_1 , C_2] = \sum_i [K^i K^i - L^i L^i , C_2] \\
\sum_i (K^i [K^i , C_2] + [K^i , C_2] K^i - L^i [L^i , C_2] - [L^i , C_2] L^i) \\
= 0
\end{array}
where we have used ##[C_2 , L^j]## and ##[C_2 , K^j] = 0##. Next write
\begin{array}{l}
0 = [C_1 , C_2] \\
= \sum_i [C_1 , K^i L^i] \\
= \sum_i ([C_1 , K^i] L^i + K^i [C_1 , L^i]) \\
= \sum_i [C_1 , K^i] L^i .
\end{array}
where we have used ##[C_1 , L^j] = 0##.
Is it possible to use this to prove ##[C_1 , K^1] = 0##?
I would prefer to prove first that the Casimirs commutate with all the generators first and then conclude the two Casimirs commute, but if this is what I have to resort to...
Last edited: