Physics, asked by skysweetwini954, 1 year ago

Path Integral on Feynman Hibbs: Interaction of EM field and matter, how can we get to equation (9.68) from (9.67)?

Answers

Answered by Anonymous
0
by understanding the question you have to follow the Feynman's Hibbs formula
...........
is

λ(1)MM=iℏ∑k∫tbtaDx(t)ψ∗M(xb)e(i/ℏ)Smatiπkc×∫tbta∫tbtadtds[j¯1,k(t)j¯∗1,k(s)+j¯2,k(t)j¯∗2,k(s)]e−ikc|t−s|ψM(xa)(1)(1)λMM(1)=iℏ∑k∫tatbDx(t)ψM∗(xb)e(i/ℏ)Smatiπkc×∫tatb∫tatbdtds[j¯1,k(t)j¯1,k∗(s)+j¯2,k(t)j¯2,k∗(s)]e−ikc|t−s|ψM(xa)



Answered by sushmita
16
On Feynman Hibbs "Quantum Mechanics and Path Integrals", the equation (9.67) describe the transition amplitude of the matter (for example an atom) to go from the state to the state when it iteracts with the electromagnetic field, considering vacuum to vacuum transitions only. This equation is
Where are the states describing the matter.
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