ps05soln - PHYS651: Problem Set 5 Solutions 1 . L =- 1 4 F...

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Unformatted text preview: PHYS651: Problem Set 5 Solutions 1 . L =- 1 4 F μν F μν =- 1 2 ( ∂ μ A ν )( ∂ μ A ν- ∂ ν A μ ) Applying the Euler-Lagrange equation yields ∂ L ∂A ν = 0 , ∂ L ∂ ( ∂ μ A ν ) =- ( ∂ μ A ν- ∂ ν A μ ) =- F μν ⇒ ∂ μ F μν = 0 (Maxwell equations) Remember that ∂ μ = ( ∂ t , ~ ∇ ) and ∂ μ = ( ∂ t ,- ~ ∇ ). For ν = 0 we get 0 = ∂ i F i = ∇ i E i = ~ ∇ · ~ E and for ν = j 0 = ∂ t F j + ∇ i F ij =- ∂ t E j + jik ∇ i B k = h- ˙ ~ E + ( ~ ∇ × ~ B ) i j 2 . Explicit calculation: Define h A i := h | A | i . ( ∂ μ ∂ μ + m 2 ) θ ( x- y ) h φ ( x ) φ ( y ) i = ∂ μ ∂ μ θ ( x- y ) h φ ( x ) φ ( y ) i (1) + 2 ∂ μ θ ( x- y ) [ ∂ μ h φ ( x ) φ ( y ) i ] + θ ( x- y ) ( ∂ μ ∂ μ + m 2 ) h φ ( x ) φ ( y ) i | {z } vanishes by EoM The statement ( ∂ 2 + m 2 ) D F ( x- y ) =- iδ (4) ( x- y ) is implying an integration in the sense that for a test function f ( x ) Z d 4 x ( ∂ 2 + m 2 ) D F ( x- y ) f ( x ) =- if ( y ) and in this sense we can integrate by parts in equation (1), so that ( ∂ μ ∂ μ + m 2 ) θ ( x- y ) h φ ( x ) φ ( y ) i =- δ ( x- y ) h π ( x ) φ ( y ) i + 2 δ ( x- y ) h π ( x ) φ ( y ) i = δ ( x- y ) h π ( x ) φ ( y ) i where we have used the canonical momentum π ( x ) = ˙ φ ( x ). For the time- ordered product we hence obtain ( ∂ μ ∂ μ + m 2 ) D F ( x- y ) = ( ∂ μ ∂ μ + m 2 ) θ ( x- y ) h φ ( x ) φ ( y ) i + θ ( y- x ) h φ ( y ) φ ( x ) i = δ ( x- y ) h | [ π ( x ) , φ ( y )] | i = δ ( x- y )(- i ) δ (3) ( x- y ) =- iδ (4) ( x- y ) Now let us take the Fourier transform of this equation. Z d 4 p e- ipx ( ∂ μ ∂ μ + m 2 ) D F ( x ) = Z d 4 p e- ipx (- i ) δ (4) ( x ) =- i On the other hand we can integrate by parts twice and let the ∂ 2 operator act on exp(- ipx ). Hence- i =...
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This note was uploaded on 09/28/2008 for the course PHYS 651 taught by Professor Tye,henry during the Fall '03 term at Cornell University (Engineering School).

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ps05soln - PHYS651: Problem Set 5 Solutions 1 . L =- 1 4 F...

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