I know what are Hermitian operators and anti-hermitian operators.but as i have studied quantum mechanics most of the operators we had dealt with them are hermitian and this question popped up in my mind that does any anti-hermitian operator exists in quantum mechanics?
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1$\begingroup$ Sure, all exponents in unitary operators are firmly antihermitean. $\endgroup$ – Cosmas Zachos Oct 2 '20 at 14:53
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4$\begingroup$ If $A$ is anti-hermitian then $H = i A$ will be hermitian, so when anti-hermitian operators do arise we will often choose to consider equivalent hermitian ones instead $\endgroup$ – By Symmetry Oct 2 '20 at 14:57
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$\begingroup$ Exactly, in standard QM you put an extra "i" in front of an Hermitean operator to have the exponent of the unitary one. Operator are regarded as "observables", so you want real spectra (hence Hermitean stuff). $\endgroup$ – Quillo Oct 2 '20 at 15:01
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$\begingroup$ Possible duplicate: physics.stackexchange.com/q/82613/2451 $\endgroup$ – Qmechanic♦ Oct 2 '20 at 16:28
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