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Unformatted text preview: Physics 6210/Spring 2007/Lecture 20 Lecture 20 Relevant sections in text: § 2.6, 3.1 Gauge transformations (cont.) Our proof that the spectrum of the Hamiltonian does not change when the potentials are redefined by a gauge transformation also indicates how we are to use our model so that all probabilities are unaffected by gauge transformations. We decree that if | ψ i is the state vector of a particle in an EM field described by the potentials ( φ, ~ A ), then | ψ i = e iq ¯ hc f ( ~ X ) | ψ i is the state vector of the particle when using the gauge transformed potentials ( φ , ~ A ). Note that this is a unitary transformation. Let us now see why this prescription works. For a particle, all observables are functions of the position and momentum operators. Here “momentum” means either canonical or mechanical. The position observable is represented (in the Schr¨ odinger picture) by the usual operator ~ X , no matter the gauge. Any observable function G of the position has an expectation value which does not change under a gauge transformation: h ψ | G ( ~ X ) | ψ i = h ψ | e- iq ¯ hc f ( ~ X ) G ( ~ X ) e iq ¯ hc f ( ~ X ) | ψ i = h ψ | G ( ~ X ) | ψ i . The momentum operator is where things get more interesting. The mechanical momentum is a gauge-invariant observable. But it is represented by an operator which changes under a gauge transformation! Indeed, we have ~ Π = ~ p- q c ~ A, ~ Π = ~ p- q c ( ~ A + ∇ f ) . However, it is straightforward to check that (exercise) ~ Π e iq ¯ hc f ( ~ X ) | ψ i = e iq ¯ hc f ( ~ X ) ~ Π | ψ i ....
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