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We could also perform a Taylor series expansion about φ=0 or some other value which is not the VEV. But in that case, we have to include ] in all our analyses, complicating our calculations. This is done, for example, if we don't know what the true VEV is. We could also perform a Taylor series expansion about φ=0 or some other value which is not the VEV. But in that case, we have to include ] in all our analyses, complicating our calculations. This is done, for example, if we don't know what the true VEV is.




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Revision as of 00:46, 10 April 2005

Everything in the following article also applies to statistical mechanics. However, the signs and factors of i are different in that case.

In quantum field theory, given the partition function in terms of the source field J, Z, the energy functional is its logarithm.

E [ J ] i ln Z [ J ] {\displaystyle E\equiv i\ln Z}

Some physicists use W instead where W=-E. See sign conventions

Just as Z is interpreted as the generating functional of the time ordered VEVs (see path integral formulation), E is the generator of "connected" time ordered VEVs where connected here is interpreted in the sense of the cluster decomposition theorem which means that these functions approach zero at large spacelike separations, or in approximations using feynman diagrams, connected components of the graph.

< ϕ ( x 1 ) ϕ ( x n ) > c o n = ( i ) n + 1 δ n E δ J ( x 1 ) δ J ( x n ) | J = 0 {\displaystyle <\phi (x_{1})\cdots \phi (x_{n})>_{con}=(-i)^{n+1}\left.{\frac {\delta ^{n}E}{\delta J(x_{1})\cdots \delta J(x_{n})}}\right|_{J=0}}

or

< ϕ i 1 ϕ i n > c o n = ( i ) n + 1 E , i 1 i n | J = 0 {\displaystyle <\phi ^{i_{1}}\cdots \phi ^{i_{n}}>_{con}=(-i)^{n+1}E^{,i_{1}\dots i_{n}}|_{J=0}}

in the deWitt notation

Then the n-point correlation function is the sum over all the possible partitions of the fields involved in the product into products of connected correlation functions. To clarify with an example,

< ϕ ( x 1 ) ϕ ( x 2 ) ϕ ( x 3 ) >=< ϕ ( x 1 ) ϕ ( x 2 ) ϕ ( x 3 ) > c o n + < ϕ ( x 1 ) ϕ ( x 2 ) > c o n < ϕ ( x 3 ) > c o n + < ϕ ( x 1 ) ϕ ( x 3 ) > c o n < ϕ ( x 2 ) > c o n + < ϕ ( x 1 ) > c o n < ϕ ( x 2 ) ϕ ( x 3 ) > c o n + < ϕ ( x 1 ) > c o n < ϕ ( x 2 ) > c o n < ϕ ( x 3 ) > c o n {\displaystyle <\phi (x_{1})\phi (x_{2})\phi (x_{3})>=<\phi (x_{1})\phi (x_{2})\phi (x_{3})>_{con}+<\phi (x_{1})\phi (x_{2})>_{con}<\phi (x_{3})>_{con}+<\phi (x_{1})\phi (x_{3})>_{con}<\phi (x_{2})>_{con}+<\phi (x_{1})>_{con}<\phi (x_{2})\phi (x_{3})>_{con}+<\phi (x_{1})>_{con}<\phi (x_{2})>_{con}<\phi (x_{3})>_{con}}

Assuming E is a convex functional (which is debatable), the Legendre transformation gives a one-to-one correspondence between the configuration space of all source fields and its dual vector space, the configuration space of all φ fields. φ here is a classical field and not a quantum field operator.

Slightly out of the usual sign conventions for Legendre transforms, the value

ϕ = δ δ J E [ J ] {\displaystyle \phi =-{\delta \over \delta J}E}

or

ϕ i = E , i {\displaystyle \phi ^{i}=-E^{,i}}

is associated to J. This agrees with the time ordered VEV <φ>J. The Legrendre transform of E is the effective action

Γ [ ϕ ] = < J , ϕ > E [ J ] {\displaystyle \Gamma =-<J,\phi >-E}

or

Γ [ ϕ ] = J i ϕ i E [ J ] {\displaystyle \Gamma =-J_{i}\phi ^{i}-E}

where

ϕ = δ δ J E [ J ] {\displaystyle \phi =-{\delta \over \delta J}E}

and

J = δ δ ϕ Γ [ ϕ ] {\displaystyle J=-{\delta \over \delta \phi }\Gamma }

or

J i = Γ , i {\displaystyle J_{i}=-\Gamma _{,i}} .

There are some caveats, though, the major one being we don't have a true one-to-one correspondence between the dual configuration spaces.

If we perform a Taylor series expansion of Γ about the VEV, φ=<φ>, the coefficients give us the amputated correlation functions, also called one particle irreducible correlation functions or 1PI correlation functions.

We could also perform a Taylor series expansion about φ=0 or some other value which is not the VEV. But in that case, we have to include tadpoles in all our analyses, complicating our calculations. This is done, for example, if we don't know what the true VEV is.

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