It is simple to show that under the gauge transformation {→A→→A+∇χϕ→ϕ−∂χ∂tψ→ψexp(iqχℏ)
The Schrodinger equation [−ℏ22m(∇−iq→Aℏ)2+qϕ]ψ=iℏ∂∂tψ
gives back the same equation.
How does it follow that the probability current is gauge invariant?
It is simple to show that under the gauge transformation {→A→→A+∇χϕ→ϕ−∂χ∂tψ→ψexp(iqχℏ)
How does it follow that the probability current is gauge invariant?
What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...
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