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I was dealing with a problem, which said that,

Supposedly Hamiltonian of a conservative system in classical mechanics is $\omega xp$, where $\omega$ is a constant, and $x$ and $p$ are the position and momentum operator respectively. The corresponding Hamiltonian in quantum mechanics is________.

To solve the problem, an initial step was taken

$$\omega xp= \frac{\omega}{2}(xp+px).$$

stating that the quantum operator should be Hermitian in nature.

What I don't understand is, how the above proceeded step was an important part/property of Hermitian operator, i.e., why the above step was taken?

Qmechanic
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kamer_kane
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2 Answers2

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The hamiltonian of a system has to be an hermitian operator since it's associated to a measurable quantity, the energy. By symply taking the classical hamiltonian$$H=\omega xp$$ and converting it directly to an operatore $$\hat H = \omega\hat x\hat p$$ you can easily see that this operator is not hermitian since, given $\hat x$ and $\hat p$ hermitian, $$\hat H^\dagger = \omega\hat{p}^\dagger\hat{x}^\dagger = \omega \hat p\hat x \neq H$$ since the two operators are now in reverse order. Then you have to make $\hat H$ hermitian.

Classically speaking $$H = \omega xp = \frac{\omega}{2}(xp+px)$$ won't change the hamiltonian in any way since $x$ and $p$ being purely variables, and not operators, they clearly commute. This form of the classical hamiltonian is now more suited for converting it into a quantum hamiltonian by using operators instead of simple functions $$H = \frac{\omega}{2}(\hat x\hat p + \hat p \hat x)$$ and now you can easily convince yourself that this form of the hamiltonian is clearly hermitian.

Such a procedure to take a classical hamiltonian to a suitable quantum hamiltonian is called Weyl ordering.

Davide Morgante
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This is an interesting question because in general there is no unique answer, although there is a "natural" answer for the simplest observables, which include $xp$. The lack of uniqueness is linked to the ordering problem.

The simplest quantization is due to Dirac; he tried to implement the idea that the commutator of two operators representing observables should be $i\hbar$ times the operator corresponding to the Poisson bracket of the two respective classical quantities.

This is based on the observations that \begin{align} [\hat x,\hat p]=i\hbar \widehat{\{x,p\}} \end{align} where $\{\cdot,\cdot\}$ is the classical Poisson bracket. Dirac basically proposed to generalize this to \begin{align} [\hat f,\hat g]=i\hbar\widehat{\{f,g\}} \, . \tag{1} \end{align} subject to the additional assumptions that \begin{align} \widehat{f(x)}=f(\hat x)\, ,\qquad \widehat{g(p)}=g(\hat p) \end{align} The problem is that (1) eventually fails although it does work for $xp$. To get the result you want you need to ask what is $\widehat{xp}$. To get there start with \begin{align} \frac{1}{i\hbar}[x^2,p^2]&=2\left(\hat x\hat p+\hat p\hat x\right) =\widehat{\{x^2,p^2\}}= 4 \widehat{xp} \end{align} from which you get \begin{align} \widehat{xp}=\frac{1}{2}\left(\hat x\hat p+\hat p\hat x\right) \end{align} which is the usual symmetrization rule.

Gronewold in

Groenewold, H.J., 1946. On the principles of elementary quantum mechanics, Physica 12 (1946) 405-460,

showed that (1) eventually fails, i.e. there is no rule consistent with (1) that will allow you to find, for instance, the operator corresponding to $x^2p^2$. This is excellently reviewed in

Chernoff, P.R., 1981. Mathematical obstructions to quantization. Hadronic Journal, 4(3), pp.879-898.

The remarkable thing is that of course the rule (1) works for the ``most common'' observables like angular momentum. However, it is possible to devise classical Hamiltonian for which the quantization is not unique, as per the example of this post.

ZeroTheHero
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