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According to my textbook, it says that $i( LK-KL )$ represents a real quantity when $K$ and $L$ represent a real quantity. $K$ and $L$ are matrices. It says that this is because of basic rules. However, I was not able to recall my memories. Can anyone show the proof of this?

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2 Answers 2

up vote 6 down vote accepted

The statement is that if $K$ and $L$ are Hermitian operators – which means $$ K = K^\dagger, \quad L = L^\dagger$$ and it implies that the eigenvalues of $K,L$ are real and the eigenvectors with different eigenvalues are orthogonal to each other, then $i(KL-LK)$ (the same as yours) is also Hermitian. This is easily proved by computing the Hermitian conjugate of this $i(KL-LK)$ because the result is the same as this operator itself: $$ [i(KL-LK)]^\dagger = i^\dagger (KL-LK)^\dagger = (-i) (L^\dagger K^\dagger - K^\dagger L^\dagger) = (-i)(LK-KL) = i(KL-LK).$$ I used $(AB)^\dagger = B^\dagger A^\dagger$ and $i^\dagger = i^* = -i$.

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If $K$ and $L$ represent real quantities, it means that they do not change under Hermitian conjugation. Use this and the properties of Hermitian conjugation to prove the same for $-i(LK-KL)$.

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