0
$\begingroup$

I've just read how in quantum mechanics, the commutator $[x, p] = i \hbar$, which is supposedly in contrast to classical physics. But if classical physics doesn't even use wavefunctions and operators (as far as I have learned), then what are the position and momentum operators? Is $x$ the identity operator, $p = m \frac{d}{dx}$, and they operate on the position $x$?

$\endgroup$
  • $\begingroup$ Read where? Which page? $\endgroup$ – Qmechanic Aug 17 '18 at 4:12
  • $\begingroup$ Griffiths Intro to Quantum, like the first or second chapter. $\endgroup$ – Faraz Masroor Aug 18 '18 at 23:42
1
$\begingroup$

In classical physics, specifically in Hamilton's formulation of classical mechanics, the Poisson bracket plays a crucial role.

Hamilton's equations of motion as defined here, equations 1, can be expressed in terms of the poisson bracket (which is nice because it is a canonical invariant), equations 3. The Poisson bracket appears in the time evolution equation, equation 5, which governs the evolution of a function of generalized coordinates.

As you said, in classical mechanics we don't have a wavefunction, so we don't have to fuss about which representation we wish to express the wavefunction in (position, momentum, spin, etc...). Rather, in classical mechanics the quantities of interest are not operators, they are functions of a set of (well-behaved) generalized coordinates. Since they are not operators there are no "operations."

So, in classical physics, the commutator of the classical observables of position and momentum you mentioned is zero, because multiplication of scalar functions is commutative. However in quantum mechanics, since observables are operators the commutator is not zero, and hence we have an uncertainty relation for position and momentum.

$\endgroup$

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.