0

In quantum mechanics, the Hamiltonian Operator is obtained by looking at the classical Hamiltonian and replacing $(x,p)$ with operators $(\hat x, \hat p)$ and for terms involving the product $xp$, the commutation relation is taken into account before replacing.

But classically the Hamiltonian is not unique, how does quantum mechanics take this into account, how can we be sure that we get the same result and equation in QM no matter which Hamiltonian we choose?

Qmechanic
  • 201,751
  • Possible duplicates: https://physics.stackexchange.com/q/22506/2451, https://physics.stackexchange.com/q/328881/2451, https://physics.stackexchange.com/q/349499/2451 and links therein. – Qmechanic Jun 12 '18 at 07:26
  • What you mean for not unique? In a given coordinate system, the Hamiltonian is unique in classical mechanics up to an additive function of $t$. If the Hamiltonian is assumed to be time independent it is therefore unique up to additive constants. – Valter Moretti Jun 12 '18 at 10:22

0 Answers0