I'm puzzled about a statement from Dirac's book, The principles of quantum mechanics, (§8, p.28):

As a simple examples of this result, it should be noted that, if $\xi$ and $\eta$ are real, in general $\xi\eta$ is not real. This is an important difference from classical mechanics. However, $\xi\eta + \eta\xi$ is real, and so is $i(\xi\eta - \eta\xi)$. Only when $\xi$ and $\eta$ commute is $\xi\eta$ itself also real.

Here $\xi$ and $\eta$ are linear operators, so (I think) these could be represented as matrices. However how can a product of two real matrices be "not real"? What does Dirac mean when he says "is not real"? Is Dirac maybe talking about eigenvalues? So does Dirac mean, that the product of two matrices with real eigenvalues could have imaginary eigenvalues? And does he want to say, that a real symmetric matrix like $\xi\eta + \eta\xi$ and a purely imaginary antisymmetric matrix like $i(\xi\eta - \eta\xi)$ always have real eigenvalues?


1 Answer 1


Flip back a page; Dirac uses real to mean Hermitian ($A^{\dagger} = A$) when talking about linear operators. So you can see that even if $A$ and $B$ are Hermitian, $AB$ won't be Hermitian unless they commute, whereas those linear combinations will be.

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    $\begingroup$ Small comment: This usage of real, while confusing, is partially justified by the fact that Hermitian matrices have strictly real eigenvalues. $\endgroup$
    – Danu
    Mar 1, 2014 at 17:26
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    $\begingroup$ I cannot find the word Hermitian on the page before. But maybe you are referring to this sentence: "A linear operator may equal its adjoint, and it is then called self-adjoint. It corresponds to a real dynamical variable, so it may be called alternatively a real linear operator." So for Dirac a real linear operator is a self-adjoint operator and an imaginary operator is then a non-self-adjoint operator? $\endgroup$
    – asmaier
    Mar 1, 2014 at 17:32
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    $\begingroup$ @asmaier yes, in physics the word "Hermitian" is often used as a synonym for "self-adjoint". $\endgroup$
    – Wildcat
    Mar 1, 2014 at 17:38
  • $\begingroup$ And self-adjoint/Hermitian means the operator has real eigenvalues? $\endgroup$
    – asmaier
    Mar 1, 2014 at 17:48
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    $\begingroup$ Yes, in fact they represent physical observables. $\endgroup$
    – LC7
    Mar 1, 2014 at 18:07

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