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Unique wave function for a fixed gauge and a velocity and position PDF

by Garrulo
Tags: fixed, function, gauge, unique, wave
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Garrulo
#1
Apr28-14, 12:25 AM
P: 10
żIs there an unique wavefunction for a system if we know the distribution probability function for variables from the system and first derivatives from these variables and we have the gauge fixed (by external impositions not related with the wavefunction knowledge obviously? (Nothing about hidden variables, of course, I refer)
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tom.stoer
#2
May2-14, 03:04 AM
Sci Advisor
P: 5,364
I am not sure whether I understand correctly. Especially the meaning of
Quote Quote by Garrulo View Post
distribution probability function for variables ... and first derivatives from these variables ...
is not clear to me.

If you have a system which local gauge invariance AND if you have a single physical state ##|\psi\rangle## then the state (or wave function) is unique modulo gauge transformations. That means that instead of a single state ##|\psi\rangle## you have to consider an equivalence class ##[|\psi\rangle]## of states w.r.t. to gauge transformations U

[tex] |\psi\rangle \sim |\psi^\prime\rangle \Leftrightarrow \exists U \in \mathcal{G}:\;|\psi^\prime\rangle = U |\psi\rangle[/tex]

Fixing a gauge means to select one unique representative state from each equivalence class.

So if you start with a unique equivalence class for a physical problem (i.e. an equivalence class of one-dimensional eigenspaces for a not yet gauge-fixed Hamiltonian ##H[A]## with gauge field) then gauge fixing by definition results in a unique eigenstate (up to global transformations ##\exp(i\alpha)##). In that case you don't need any additional conditions.

If you don't have a unique equivalence class then gauge fixing does not provide a unique solution.

Example: if you start with the equivalence class ##[|1s\rangle] = [|100\rangle]## of the Hydrogen atom, then this is still subject to gauge transformations; if you now fix the gauge as usual (resulting in the well-known Hamiltonian with ##\vec{A} = 0##) then this guarantuees a unique state ##|100\rangle## and a unique wave function.

Example2: if you start with the equivalence class of the 4-dim. eigenspace ##[|n=2; l,m\,\text{not fixed}\rangle]##, then gauge fixing still leaves you with a 4-dim. energy eigenspace.

So gauge fixing does not affect the dimension of the subspace "within the equivalence class". You have to reduce this dimension to dim = 1 by some other, physical considerations. What you need is a maximum set of commuting observables, like ##{H,L^2,L_z}## in the case of the Hydrogen atom.

So my summary would be that starting with a Hilbert space ##\mathcal{H}## and a gauge group ##\mathcal{G}## acting on the Hilbert space, the following is required to specify a state (as a one-dim. eigenspace) uniquely:
1) a maximum set of mutually commuting observables ##\{H, \ldots\}##
2) gauge fixing

The probability distribution ##|\psi|^2## does not help b/c you are still free to change the wave function by a local phase. This phase can either be the result of a physical effect or the result of a gauge transformation.

Hope this helps.
Garrulo
#3
May24-14, 06:26 AM
P: 10
In differentiation (not derivation, sorry, in my language differentiate is said "derivar" )of characteristics variables from the system, I refer for exemple for a particle to his position distribution and velocity distribution. But in more general systems, for example, an electromagnetic field without charge, derivative for [itex]\vec{A}[/itex], of course, with uncertainty relation limitations for them. It is said, I know the statistical distribution of the variables x, y ,z.... of the system and the statistical distribution of his differentiation, it is said, from the velocity if it is a particle system


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