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Consider the two hermitian operators $\hat A$ and $\hat H$:

I can prove that the operator $[\hat A,\hat H]$ is non-hermitian as follows:

$$\begin{align} \int\phi^*[\hat A,\hat H]\psi\,dx&=\int\phi^* \hat A\hat H\psi\, dx-\int\phi^* \hat H\hat A\psi\,dx\\ &=\int(\hat H\hat A\phi)^*\psi\,dx-\int(\hat A\hat H\phi)^*\psi\,dx\\ &=\int([\hat H,\hat A]\phi)^*\psi\,dx=-\int([\hat A, \hat H]\phi)^*\psi\,dx \end{align} $$

(The last line would not have the negative sign for an Hermitian operator.)

My question is regarding the operator of an operator $\hat H\hat A$ or $\hat A\hat H$. Do they also correspond to something that is also Hermitian? So if we construct a new operator that is two hermitian operators applied one after the other would that also be Hermitian? In other words, does using two Hermitian operators on the same eigenstate always result in something observable?

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    $\begingroup$ The product of two hermitian operators is hermitian only if they commute. $\endgroup$ – ZeroTheHero Dec 17 '17 at 20:59
  • $\begingroup$ ... if and only if ... $\endgroup$ – Andrew Steane Dec 9 '18 at 17:52
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There is a simple proof that this is not the case: We need

$$ AH \stackrel{!}{=}(AH)^{\dagger} = H^{\dagger}A^{\dagger}=HA $$

in other words, we need $AH=HA$ or else that the two commute.

It is indeed somewhat unfortunate that the product of two Hermitian matrices is not Hermitian - the self-adjoint operators just don't form an algebra with the usual (associative) matrix product.

If you want to define an algebra of Observables, where the product of two Observables can again be seen as an observable, you should try to replace the matrix product by the Jordan product:

$$ A \circ H = \frac{AH + HA}{2} $$

This is clearly a Hermitian operator again. The product is also clearly commutative, though unfortunately not associative. The corresponding algebra is known as Jordan algebra and people tried to do quantum mechanics with it in the beginning. However, not having an associative product turn out to be cumbersome and since it turned out that all but one very special Jordan algebras can be seen as subalgebras of associative matrix algebras, there is no good reason to use them.

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Note that if $A,B$ are Hemitian, i.e. $A=A^{\dagger}$ and $B=B^{\dagger}$, then

$$\left(AB\right)^{\dagger}=B^{\dagger}A^{\dagger}=BA$$

Can you proceed?

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protected by Qmechanic Dec 17 '17 at 21:20

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