2
$\begingroup$

As we all learn in Basic Quantum Mechanics (the first quantization), we promote the classical variables to operators. Say, the classical hamiltonian $\frac{p^2}{2m}+V(x)$ to $-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+V(x)$, where we have replaced $p$ by $-i\hbar \frac{\partial}{\partial x}$ in coordinate representation. But if the classical Hamiltonian of a conservative system $H = \omega x p$, where $\omega$ is a constant, this method fails as the constructed Hamiltonian of the form $-i\hbar \omega x\frac{\partial}{\partial x}$ is not hermitian.

What is wrong here?

$\endgroup$
8
$\begingroup$

This problem is rooted in the way we derive quantum mechanics from the classical theory. Basically a quantum theory is a more complex object than a classical theory, so you will need to specify more information about it in order to it to be well defined. Thus after you have defined the hilbert space and it's operators you will need to choose an ordering prescription that will generalize your operators in the case of non commuting objects and that will reduce to the classical case when $\hat{x}$ and $\hat{p}$ (and so on) are not operators. So a possible choice for $\hat{H}$ is $$ \hat{H}=\frac{\omega}{2}(\hat{x}\hat{p}+\hat{p}\hat{x})$$ is the symmetrized ordering (also called weyl quantization) but really it's kinda aribtrary as long as $\hat{H}$ is hermitian (since it's observable) and reduces to it's classical analogue when: $$\hat{p}\rightarrow p\\\hat{x} \rightarrow x$$ A method to deal with those cases was developed by Weyl, Wigner and others and has generated various methods of quantization wich are more elegant than the "brute force" canonical quantization.

$\endgroup$
5
$\begingroup$

When you have a product of operators in the classical Hamiltonian, the ordering of the operators in the quantum prescription is ambiguous. In the general case, suppose we have some classical operator $o=px$. As you noted, simply writing $O=PX$ is not Hermitian. The standard procedure is to symmetrize the sum, so that $$O=\frac{PX+XP}{2}$$ As noted by Shankar on page 120 of Principles of Quantum Mechanics, there is no general procedure for three or more operators.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.