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Recently I found myself in a state similar to that which @senator found himself here. I too have been reading Dirac's Lectures on Physics and am particularly confused by the notion of Hamiltonians without classical analogues.

The way I understand it, at Second Quantisation one always starts with the classical Lagrangian to end up with the Quantum Mechanical Lagrangian and so within this capacity I do not see how each Hamiltonian cannot have a classical analogue.

Is this still the route taken to obtaining Hamiltonians which do not have classical analogues and if so what are some examples of this?

If it is the case that at each instance of finding a Hamiltonian one starts from a classical Lagrangian, is that not wrong given that Classical Mechanics is a subset of Quantum Mechanics.

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marked as duplicate by Qmechanic quantum-mechanics Mar 8 '18 at 9:17

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Any quantum system with a finite-dimensional space of states - such as the space of states of a particle with non-zero spin in a magnetic field that is fixed in place - has a Hamiltonian without classical analogue, since there are no position or momentum operators on a finite-dimensional space of states due to $[x,p] = \mathrm{i}\hbar$ being impossible there. Such a system has no notion of canonical positions and momenta, hence no classical Hamiltonian or Lagrangian analogue, but is a perfectly valid quantum system nevertheless.

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  • $\begingroup$ what about a classical spin lattice (à la Ising)? $\endgroup$ – AccidentalFourierTransform Feb 16 '18 at 20:01
  • $\begingroup$ @AccidentalFourierTransform What about that? That's not a Hamiltonian or Lagrangian system in the standard sense of classical mechanics, it too has no canonical positions or momenta, which is the sense which the quote by Dirac that OP refers to uses. $\endgroup$ – ACuriousMind Feb 16 '18 at 20:05
  • $\begingroup$ Fair enough. One usually does speak of the Ising Hamiltonian, although it's true it's not a Hamiltonian in the standard sense. $\endgroup$ – AccidentalFourierTransform Feb 16 '18 at 20:08
  • $\begingroup$ @ACuriousMind in such a case wouldn't you still start from a classical Lagrangian though ? $\endgroup$ – Jake Xuereb Feb 16 '18 at 20:28
  • $\begingroup$ @JakeXuereb No, why would you think so? $\endgroup$ – ACuriousMind Feb 16 '18 at 20:35

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