By Stetz A.W.

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Ni + 1, . . = ni + 1 Now replace ni everywhere by ni − 1. . , ni − 1, . . |ai | . . , ni , . . = √ √ ni . . , ni , . . | . . , ni , . . 13) The effect of ai on the state | . . , ni , . . has been to produce a state in which the number of particles in the i’th state has been reduced by one. Eqn. 13) also tells us what the normalization must be. In short ai | . . , ni , . . = √ ni | . . , ni − 1, . . 2. BOSON STATES 65 Of course if ni = 0, the result is identically zero. ai | .

62) with J = λ = 0. 4. 70) in the case of two fields. 62) with λ = 0. 70) reveals that what we have just calculated is ϕ(x1 )ϕ(x2 ) . Let’s pause for a moment to see how all the pieces fit together. 70) equals G(x1 , x2 ). That proof required nothing more than functional differentiation. 68) led us to hope that ϕ(x1 )ϕ(x2 ) would be equal to −D(x1 − x2 )/i. 3. Put it another way: we have proved that the analogy works correctly in the case of (71). 72) Wick So here’s the procedure: make a list of all possible ways of pairing up the indices i, j, · · · , k, l.

FUNCTIONAL TAYLOR SERIES 31 We can use the shortcut to evaluate a more difficult and important case from a later chapter. The following expression occurs naturally when we calculate correlation functions using the path integral formalism. 10) All you need to know about this is that x and y are 4-vectors and D(x) = D(−x). We use functional differentiation to pull D(x − y) out of the functional. 11) The first derivative is found as follows: δF = lim →0 δJ(x ) 1 − i 2 d4 xd4 y[J(x) + δ (4) (x − x )]D(x − y) ×[J(y) + δ (4) (y − x )] + = −i i 2 d4 xd4 yJ(x)D(x − y)J(y) d4 xD(x − x )J(x) Notice that both J’s are incremented by the term.

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