The lie algebra of $\mathrm{SU}(n)$ is composed by the $n \times n$ antihermitian matrix with null trace over the real field, but physicists prefer to use hermitian matrix. Does this mean taking the algebra over the complex field? If so, why do you do it this way?
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2We want operators to be Hermitian because they have real eigenvalues. So instead of writing an element of $SU(2)$ as $e^X$ with $X$ antihermitian, we write it as $e^{iX}$ with $X$ Hermitian. This is different than complexifying the algebra – John Donne Sep 20 '18 at 14:05
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1FWIW, $u(N)=u(N,\mathbb{C})$ is a real Lie algebra. In particular, its underlying vector space is a real vector space. Related: https://physics.stackexchange.com/q/321230/2451 – Qmechanic Sep 20 '18 at 14:10
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Thank you all, I was immagining something like what you had say, but I was a little bit confused by an assertion on Wikipedia, that I now understood well. – ACA Sep 20 '18 at 15:28
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Does this mean taking the algebra over the complex field?
No.
- The antihermitian matrices are a vector space over the real numbers, with an algebra structure given by the commutator $(A,B)\mapsto [A,B]$.
- The hermitian matrices are also a vector space over the real numbers, and they also have an algebra structure which is now given by $i$ times the commutator $(A,B)\mapsto i[A,B]$.
In both cases, the matrices act on complex-valued vectors, but the vector space (and therefore the algebra) that they form is only over the real numbers.
The reason physicists prefer this convention is that it keeps the eigenvalues real.
Emilio Pisanty
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