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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.

Qmechanic
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Jake Xuereb
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  • This post (v4) seems like a list question. Related: https://physics.stackexchange.com/q/276028/2451 and links therein. – Qmechanic Feb 16 '18 at 20:10

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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.

ACuriousMind
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  • what about a classical spin lattice (à la Ising)? – AccidentalFourierTransform Feb 16 '18 at 20:01
  • @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. – ACuriousMind Feb 16 '18 at 20:05
  • Fair enough. One usually does speak of the Ising Hamiltonian, although it's true it's not a Hamiltonian in the standard sense. – AccidentalFourierTransform Feb 16 '18 at 20:08
  • @ACuriousMind in such a case wouldn't you still start from a classical Lagrangian though ? – Jake Xuereb Feb 16 '18 at 20:28
  • @JakeXuereb No, why would you think so? – ACuriousMind Feb 16 '18 at 20:35
  • Even for the case you mentioned I believe, and please correct me if I am wrong, that the spinor symplectic group can be mapped to the symplectic group from Hamiltonian mechanics. Wouldn't that show an analogue @ACuriousMind I thought so as that is the approach I am familiar with for Second Quantisation. That being said I am very very new to this stuff and I would very much appreciate way pointing – Jake Xuereb Feb 16 '18 at 20:37
  • @JakeXuereb What spinor symplectic group? I'm not talking about spinors, and there is no quantization here. Just take the spin-1/2 representation of SU(2), and the Hamiltonian $H = \sigma_i B_i$ where $\sigma_i$ are the Pauli matrices and $\vec B$ is a vector. This is the actual Hamiltonian of a stationary spin-1/2 particle in an external magnetic field, and there is no classical Hamiltonian or Lagrangian mechanical system whose quantization it could possibly be. If you disagree, please specify the classical system you think it corresponds to. – ACuriousMind Feb 16 '18 at 21:15
  • @ACuriousMind Thank you very much for specifying. I will get to reading as it is clear I have a hole somewhere in my understanding ! Thanks for the invaluable way pointing. – Jake Xuereb Feb 16 '18 at 21:39
  • Not sure I follow. What's wrong with angular momentum Hamiltonians - say very naively $H= a L^2- b L_x$? (v.g. http://iopscience.iop.org/article/10.1088/0031-8949/22/5/007/meta) Why would these not have a classical analogue? They get around your arguments because the Hamiltonian contains products of $x_ip_j$. – ZeroTheHero Feb 16 '18 at 22:14
  • @ZeroTheHero There is nothing wrong with angular momentum Hamiltonians that are functions of $x$ and $p$ on an infinite-dimensional space of states. It's the finite-dimensionality, not the spin, that's the issue, spin is just the most straightforward example of where such finite-dimensional spaces occur – ACuriousMind Feb 16 '18 at 22:16
  • along the lines of @ZeroTheHero, isn't $H \sim B\cdot L$ the classical analogue of $H \sim B\cdot\sigma$ ? and if not why ? Apologies if you already explained this in your response to ZeroTheHero – Jake Xuereb Feb 16 '18 at 22:52
  • @JakeXuereb It is an analogue, but not in the sense that the quantization of the former as a classical Hamiltonian would yield the latter. – ACuriousMind Feb 16 '18 at 22:57
  • Ok so yeah, what I was initially getting at was do Hamiltonians exist which do not have classical analogue of any form ,which imo should be the case given that CM is a subset in a sense of QM, and if so could you mention some examples ? What would the starting point of derriving such a Hamiltonian ? – Jake Xuereb Feb 16 '18 at 23:01
  • @ACuriousMind so even if the dynamics might be restricted to a finite dimensional subspace you still want to think of Hilbert space as infinite-dimensional? – ZeroTheHero Feb 16 '18 at 23:16
  • @user929304 I explicitly said the particle is "fixed in place". The spin $s$ of the particle is also fixed, so the space of states we're looking at here is merely $2s+1$-dimensional. You may well argue that "in reality" the particle still has a position and could potentially be elsewhere but I think my point is that you can at least have subsystems of quantum systems that are fine as quantum systems but have no classical analogue. – ACuriousMind Mar 19 '18 at 19:28