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If an expectation value is purely imaginary, then the real component is obviously 0. Because expectation values are real quantities, does this mean that the expected value must be 0? I feel like this pretty self-explanatory but I want to be able to confidently put this on my assignment.

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  • $\begingroup$ What is an expectation value? Is there a context? $\endgroup$
    – Steeven
    Jan 23 at 23:46
  • $\begingroup$ Your question is contradictory. You assume the possibility of an imaginary EV but then acknowledge that such a thing cannot exist. You can have an imaginary EV but it will not correspond to any observable/Hermitian operator. $\endgroup$
    – joseph h
    Jan 24 at 0:23
  • $\begingroup$ @Drjh That's an answer. $\endgroup$
    – Bill N
    Jan 24 at 0:31
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If an expectation value is imaginary, it means the expectation value is imaginary. An expectation value is just a statistical attribute of the distribution of observables.

Hermitian operators have real observables, in the form of real eigenvalues; and correspondingly the expected value of those operators must be real as well. Since every actual observable is real, all Quantum Operators which correspond to an observable are Hermitian, like Spin, Position, Momentum, ... , all the good stuff.

Now in your case, the fact that you get an imaginary E.V. means that there are imaginary eigenvalues to the operator. So the operator is not Hermitian. That's is fine and we can still calculate its E.V., mean, uncertainty...but the Operator does not correspond to a real life observable.

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  • $\begingroup$ "If an expectation value is imaginary, it means the expectation value is imaginary." -- What?? $\endgroup$ Jan 24 at 1:04
  • $\begingroup$ And of course you can measure expectation values of non-hermitian operators. It just takes a bit more effort. $\endgroup$ Jan 24 at 1:06
  • $\begingroup$ Yes Norbz, I agree, we can complex valued observables with Normal Operators; the pretext of the question however suggested that there is something we must "fix" about non real EV, which my answer is that we don't have to fix anything, we just know from this that the operator is not Hermitian $\endgroup$ Jan 24 at 1:11
  • $\begingroup$ My initial sentence was a literary device to show that infact there is no adjustment we have to make to the EV by "only taking the real part" or "because real part is 0 the whole thing is zero", the sentence is a tautology meant to show there is nothing further to be drawn from the antecedent. $\endgroup$ Jan 24 at 1:14
  • $\begingroup$ This is not restricted to normal operators. You can measure the expectation value of any operator, it just takes twice as many measurements. $\endgroup$ Jan 24 at 12:13

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