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I am supposed to prove an operator G is hermitian but I seem to be proving it isn't. G is defined as

$$G = -i(O-O^t)$$

where $O$ is some linear operator. If I take the transpose of the whole thing I get

$$G^t = -i(O^t - O),$$

which is $-G$. Am I missing something?

Robert Lewis
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boudreaux
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    Hermitian means it is equal to its transpose complex conjugate. If $O$ is real then your G is hermitian, otherwise it is not. – Anders Beta Oct 11 '19 at 22:33

2 Answers2

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The hermitian adjoint is not merely the transpose of an operator; it is the complex conjugate of the transpose; that is, for complex matrices $A$,

$A^\dagger = \overline{A^T} = (\bar A)^T; \tag 1$

so assuming $O$ is a real matrix, and

$G = -i(O - O^T), \tag 2$

we have

$G^\dagger = \overline{(-i(O - O^T))^T} = \overline{-i(O^T - O)}$ $= i(O^T - O) = -i(O - O^T) = G; \tag 3$

$G$ is indeed self-adjoint.

Robert Lewis
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1

Recall that by definition we need to take the conjugate transpose that is

$$G^H=i(O^T-O)=G$$

user
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