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In general, in quantum mechanics we can prove probability current or the Schrodinger equation and other quantities are gauge invariant. However, the Hamiltonian isn't gauge invariant. Under a gauge transformation, the Hamiltonian operator changes(or have i understood wrong?) Does this mean that the Hamiltonian doesn't describes a true physical quantity like in classical mechanics?Closing, if the above are correct, do they have any affect on the principle of least action?

Thank you.

Note: The Hamiltonian is: $$H_f = {1 \over 2m} [P- qA(R,t)]^2 +qU(R,t) $$After a gauge transformation: $$H_g = {1 \over 2m} [P- qA'(R,t)]^2 +qU'(R,t) $$. Thus, we have $$H_f \neq H_g $$

Constantine Black
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2 Answers2

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Does this mean that the Hamiltonian doesn't describe a true physical quantity like in classical mechanics?

Even in classical mechanics, Hamiltonian for one particle in external field EM is function

$$ H(\mathbf r,\mathbf p) = \frac{(\mathbf p - \frac{q}{c}\mathbf A(\mathbf r, t))^2}{2m} + q\phi(\mathbf r,t) $$

where $\mathbf A,\phi$ are any of the valid functions that describe the same external EM field.

The form of the Hamiltonian (dependence on the potentials and $\mathbf r,\mathbf p) is unique, but its value is not; it depends on the choice of the above two functions ("choice of gauge").

This non-uniqueness is no big deal, as Hamiltonian function is mostly a theoretical concept that is useful to formulate the laws and derive other laws; it non-uniqueness is not necessary for that use.

In classical physics, the laws and their consequences can be formulated also in a gauge-independent way with EM fields $\mathbf E,\mathbf B$ only. The potentials and the Hamiltonian can be avoided.

The situation with gauge-dependence is similar to the one with kinetic energy having value that depends on the inertial system it is evaluated for. The value is frame-dependent, but it poses no problem for its use.

Ján Lalinský
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  • @ JanLalinsky: Thank you. a question: if I understand correctly, what you say is that the Hamiltonian, even in classical mechanics is a tool function containing scalar functions we call potentials? But. isn't through the Hamiltonian or Lagrangian formulation we derive the principle of least action? Isn't that, in our theory, a physical principal concerning the laws, or an interpretation of them, of the universe? – Constantine Black Apr 20 '15 at 09:14
  • A correction: Scalar or vector functions we call potentials. – Constantine Black Apr 20 '15 at 09:26
  • Yes, it is a tool (similar to, say, a computer program that allows one to set up representation of an electric circuit in a computer and simulate its behaviour - an imperfect representation of what actually happens, in principle dispensable thing). – Ján Lalinský Apr 20 '15 at 19:17
  • Is it correct to connect the above with the principle of least action? – Constantine Black Apr 21 '15 at 08:41
  • What do you mean? Hamilton's principle and likes can be formulated in several ways, one of which uses the Hamiltonian function. It is more common to formulate Hamilton's principle and likes with Lagrangian function, though. – Ján Lalinský Apr 21 '15 at 21:12
  • Using Hamilton's function or the Lagrangian function, my concern is how the above statement and explanation of your answer effect the interpretation of Hamilton's action principle. Is Hamilton's principle a principle regarding the way nature works? If yes, how, if the above functions(L,H) are just tools? – Constantine Black Apr 22 '15 at 11:11
  • The Hamilton principle is an abstract way to formulate equations of motion. These are tools as well. We have no direct handle on the true ways of nature. – Ján Lalinský Apr 22 '15 at 21:44
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The issue is that the electromagnetic field and its gauge transformations are treated classically here - they are not operators of the quantum theory, but "tacked on" because we want to describe how a quantum object interacts with the electromagnetic field without treating the EM field itself as a quantum object.

"Gauge-invariance" in this half-quantized theory is manifested by the Schrödinger equation being gauge invariant, i.e. by the dynamics being gauge-independent. The Hamiltonian itself is indeed not "gauge-invariant" because the gauge field has not been quantized, and we have not passed to a space of states where the physical states are gauge invariant.

The proper way to obtain manifest gauge invariance of the theory is to quantize the electromagnetic field, i.e. let $A^\mu$ become an operator-valued quantum field. This, however, belongs to the realm of full-blown QFT, and is hence not done in the ad-hoc approach of classical quantum mechanics to interactions with the electromagnetic field. (It is also not necessary to get many nice results)

ACuriousMind
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  • Thanks for the answer. What do you mean with "It is also not necessary to get many nice results". – Constantine Black Apr 19 '15 at 14:16
  • @ConstantineBlack: Uh...precisely what I say. It is often not necessary to quantize the EM field to obtain the proper behaviour of a quantum object interacting with the field. – ACuriousMind Apr 19 '15 at 14:21
  • Do we need or wish the Hamiltonian(in classical or quantum mechanics) to be gauge invariant? If so, can you suggest a place that makes an analysis on the subject? You wrote in your comment above that generally the Hamiltonian is gauge invariant- is the discussion here http://physics.stackexchange.com/questions/94699/gauge-invariance-of-the-hamiltonian-of-the-electromagnetic-field?lq=1 different? This post states that we care for the observables to be gauge invariant and that the hamiltonian is merely a mathematical theoretical tool. Suggestion for reading?(classical + quantum). – Constantine Black Apr 20 '15 at 15:25
  • Maybe I should post the above as a different question? – Constantine Black Apr 20 '15 at 15:27
  • @ConstantineBlack: The definite (and quite hard) treatment of gauge theories in the classical and quantum context is Quantization of Gauge Theories by Henneaux and Teitelboim. Indeed, the Hamiltonian need not be "manifestly" gauge invariant, it may be that there are second-class constraints whose Poisson bracket/commutator with the Hamiltonian is non-zero. These brackets/commutators are themselves only sums of constraints, though, and hence vanish on the constraint surface/physical space of states. – ACuriousMind Apr 20 '15 at 15:38
  • ....."Quantization of Gauge Systems" by Henneaux and Teitelboim. – Frobenius Jul 24 '16 at 14:16