Correlation functions

  • Thread starter RedX
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Should one view correlation functions as:

[tex](1) \; \langle T{\phi(x)\phi(y)}\rangle [/tex]

or

[tex](2) \; \langle T{\phi(x)\phi(y)}\rangle - \langle \phi(x)\rangle \langle\phi(y)}\rangle [/tex]

with the second term being zero?

(2) makes more sense as it really measures whether the two fields are independent (the correlation), while (1) just measures the product and that's it.

Yet in field theory, 2-point correlation functions have the interpretation as Green's functions (i.e., Huygen wavelets). Therefore correlation functions also have the interpretation as the amplitude to go from x to y. That implies that (1) makes better sense.

Also, in Ising's model of a ferromagnet, you can calculate spin correlation functions. The partition function is:

[tex]Z=\exp[-\beta(\Sigma_{ij} \epsilon s_is_j)] [/tex]

Can the spin correlation functions [tex]\langle s_l s_m\rangle [/tex] be interpreted as the Green's function of a spin wave? But to be interpreted as such, doesn't there need to be a kinetic term for the spins (or spin field), similar to [tex]\partial_\mu \phi \partial^\mu \phi[/tex] for scalar fields?
 

A. Neumaier

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Should one view correlation functions as:

[tex](1) \; \langle T{\phi(x)\phi(y)}\rangle [/tex]

or

[tex](2) \; \langle T{\phi(x)\phi(y)}\rangle - \langle \phi(x)\rangle \langle\phi(y)}\rangle [/tex]

with the second term being zero?

(2) makes more sense as it really measures whether the two fields are independent (the correlation), while (1) just measures the product and that's it.
Usually, one takes the expectations in the vacuum state; then <phi(x)>=0. So both are the same. In case of broken symmetry, one introduces a shifted field H(x)=phi(x)-<phi(x)>, and then looks at the time-ordered correlations for that, and gets your second formula when undoing the shift transformation (except that you missed a second T in your formula).
 

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