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In a nutshell, Noether's theorem states that for every continuous symmetry a corresponding conserved quantity exists.

Now, the Hamiltonian equations of motion (let's talk about a classical system here) are invarinat under addition of any constant to the Hamiltonian function

$$H\rightarrow H+ \text{const.}$$

Is there a corresponding conserved quantity?

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Yes, it is the $const$ itself: $\frac{d}{dt} const = 0$ ;-) –  Vladimir Kalitvianski May 10 '11 at 20:11

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What you're describing is not a symmetry in the sense of Noether's theorem of the (classical i.e. non relativistic and non quantum) system, but it is a symmetry of our description of the system.

A good example is provided when the Hamiltonian can be written as $H=T+V$ where $T$ is a kinetic energy and $V$ is a potential energy. Modifying the Hamiltonian by adding a constant to the potential energy does not change the behavior of the system in either the classical case or in the non relativistic quantum case.

This sort of symmetry is called a "gauge symmetry" and is discussed, in the quantum mechanics case, at length in Sakurai's quantum textbook "Modern Quantum Mechanics":
http://www.amazon.com/Modern-Quantum-Mechanics-2nd-Sakurai/dp/0805382917

See section (2.6) page 123, "Potentials and Gauge Transformations", which describes the difference between the classical case, where changing the potential has no physical significance, and the quantum case, where changing the potential changes the phase, but doesn't result in physically observable consequences.

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...though changing the phase might bring up the idea of locally changing phases, yielding $U(1)$ and thus electrodynamics –  Tobias Kienzler May 11 '11 at 7:38

Noether's theorem is that for every (differentiable) symmetry of the action (integral of the Lagrangian) there is a conserved quantity. It does not directly say anything about symmetries of the equations of motion, nor about the Hamiltonian.

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Nope, this operation is not a symmetry in the physics sense. A symmetry transformation is a transformation that changes or mixes the values of the basic degrees of freedom such as positions and momenta $x(t)$, $p(t)$ in mechanics or the values of fields such as $\vec E(\vec x,t)$ and $\vec B(\vec x,t)$ in electromagnetism.

The Hamiltonian is not an independent variable or a basic degree of freedom; it is a function of them. You're not changing the values of any quantities that evolve with time; instead, you're changing some formulae for derived and in principle unnecessary auxiliary objects in the theory (in this case the Hamiltonian), claiming that other formulae are preserved. This is not a symmetry so there is no conservation law associated with this operation. You're not "rotating" real objects which is what symmetry transformations should do: you're just redefining auxiliary, derived objects on the paper.

Incidentally, the operation you mention fails to be "harmless" in general relativity because the energy is a source of gravitational field - curvature of space - so if you move it by a constant, you do change physics.

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+1, but it is harmless since it only translates the Hamiltonian which is then no longer equivalent to the energy (if it has been so before) –  Tobias Kienzler May 11 '11 at 7:37
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Well, if one defines a Hamiltonian to differ from energy, so that the gravity may be sourced by yet another formula, then such a Hamiltonian is really a totally bureaucratic concept only and it can't affect anything of physical value. In that case, one may totally change it. Square it or cube it. Define a new $H$ which is the old $H^3$, modify the require formulae appropriately (you modified your Einstein's equations as well, so I am allowed to modify other equations, too). Clearly, if the concepts are not sharply defined to enter equations in a unique way, they're vacuous. –  Luboš Motl May 11 '11 at 9:34
    
you're right of course, there's no sensible reason to modify any formulae into some more complicated at all. If we agree on taking the Hamiltonian as the simplest (in terms of e.g. counting terms) form from which one can obtain the other $H$'es then $H+$const is indeed invalid (and not harmless at all if considered valid) –  Tobias Kienzler May 11 '11 at 10:04

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