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I am currently wrestling with computation of Green's functions in an interacting field theory.

If I use the Yukawa field theory with a scalar $\phi$ field and a complex $\psi$ field as a simple example, I have the following Lagrangian:

$$\mathcal{L}=\frac{1}{2}(\partial_{\mu}\phi)^{2}-\frac{1}{2}M^{2}\phi^{2}+\frac{1}{2}(\partial_{\mu}\psi^{\dagger})(\partial^{\mu}\psi)-\frac{1}{2}m^{2}\psi^{\dagger}\psi-g\psi^{\dagger}\psi\phi$$

If I wanted to find a Green's function like:

$$G_{1}^{(3)}(x_{1},x_{2},x_{3})=\langle\Omega|T\phi(x_{1})\psi(x_{2})\psi^{\dagger}(x_{3})|\Omega\rangle\quad\text{or}\quad G_{2}^{(3)}(x_{1},x_{2},x_{3})=\langle\Omega|T\phi(x_{1})\psi^{\dagger}(x_{2})\psi^{\dagger}(x_{3})|\Omega\rangle$$

I can do this using a generating functional, $Z[J,\eta,\eta^{\dagger}]$:

$$G^{(3)}_{1}(x_{1},x_{2},x_{3})=\frac{1}{Z[0,0,0]}\left.\frac{\delta^{3}Z}{\delta J\delta \eta\delta \eta^{\dagger}}\right|_{J=\eta=\eta^{\dagger}=0},\quad G_{2}^{(3)}(x_{1},x_{2},x_{3})=\frac{1}{Z[0,0,0]}\left.\frac{\delta^{3}Z}{\delta J\delta \eta^{\dagger}\delta \eta^{\dagger}}\right|_{J=\eta=\eta^{\dagger}=0}$$

However, I'm not sure how to derive the generating functional in this case. All of the notes that I can find talk about deriving a generating functional for a free scalar field theory.


I know that we can write a Green's function in terms of the sum of all connected graphs with $n$ external lines. However, in this case I don't have the Feynman rules and so (I think) this doesn't help me

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  • $\begingroup$ Non-linear theories have no Green's function, I believe you mean the n-point correlation function. $\endgroup$ – Slereah Apr 13 '17 at 10:08
  • $\begingroup$ @Slereah A previous examination that I have seen for my course actually asks for a "three-point Green's function that describes the decay $\phi \to \psi\psi^{\dagger}$". However, I think that they do indeed mean the $n$-point correlation function! $\endgroup$ – Thomas Russell Apr 13 '17 at 10:11
  • $\begingroup$ @Slereah It's a quirk that some physicists tend to use "Green's function" and "correlation function" interchangably, sadly. $\endgroup$ – ACuriousMind Apr 13 '17 at 10:59
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The generating functional, as always, is given by

$$ Z[J, \eta, \eta^{\dagger}] = \int \mathcal{D}\phi \mathcal{D} \psi \mathcal{D} \psi^{\dagger} \exp \left[ \frac{i}{\hbar} \int d^4 x \, \left( \mathcal{L}(\phi, \psi, \psi^{\dagger}) +J\phi + \eta^{\dagger}\psi + \psi^{\dagger}\eta\right)\right] = $$ $$ \exp \left[ -\frac{i g}{\hbar} \int d^4 x \, \frac{\delta}{\delta J} \frac{\delta}{\delta \eta} \frac{\delta}{\delta \eta^{\dagger}} \right] \; Z_0[J, \eta, \eta^{\dagger}], $$

where $Z_0$ is the generating functional of the free theory:

$$ Z_0[J, \eta, \eta^{\dagger}] = \hbar \int d^4 x \, \int d^4 y \, \left( \frac{1}{2} J(x) \Delta_M(x, y) J(y) + \eta^{\dagger}(x) \Delta_m(x, y) \eta(y) \right) $$

with $\Delta_m$ the Klein-Gordon propagator with mass $m$.

Your formula for correlations (aka Green's functions of the interacting theory) is correct. You have to Tailor-expand the exponential of the interacting term to the appropriate order in perturbation theory.

Just do the math and you will arrive at the correct expressions for correlation functions. Spoiler alert: the second one vanishes.

Btw you don't need to explicitly write the generating functional to calculate correlations. There's a simpler way: first, note that you can calculate expressions like

$$ \left< F[\phi, \psi, \psi^{\dagger}] \right>_0 = \mathcal{N}_0 \int\mathcal{D}\phi \mathcal{D}\psi \mathcal{D}\psi^{\dagger} \exp\left[ \frac{i}{\hbar} \int d^4 x \, \mathcal{L}_0(\phi, \psi, \psi^{\dagger}) \right] \cdot F[\phi, \psi, \psi^{\dagger}] $$

where $F$ is polynomial in fields with help of Wick's theorem. Also note the diagrammatic interpretation of the terms in the Wick expansion.

Then define

$$ \left< F[\phi, \psi, \psi^{\dagger}] \right> = \mathcal{N} \int\mathcal{D}\phi \mathcal{D}\psi \mathcal{D}\psi^{\dagger} \exp\left[ \frac{i}{\hbar} \int d^4 x \, \mathcal{L}(\phi, \psi, \psi^{\dagger}) \right] \cdot F[\phi, \psi, \psi^{\dagger}] = $$ $$ \frac{\mathcal{N}}{\mathcal{N}_0} \left< F \cdot \exp \left[ - \frac{i g}{\hbar} \int d^4 x \, \phi \psi^{\dagger} \psi \right] \right>_0. $$

and Tailor-expand the exponential of the interacting term to any order (given in advance). You will arrive at the polynomial expression (because both $F$ and the truncated Tailor series are polynomials in $\phi$, $\psi$ and $\psi^{\dagger}$). We already know how to compute these:

The expectation bracket, to each order in the perturbative series, is given by a sum of terms. Each term can be represented graphically as a Feynman diagram.

The expectation brackets are by definition normalized such that $$ \left<1\right>_0 = \left<1\right> = 1, $$

which means that $\mathcal{N}_0$ and $\mathcal{N}$ are related to each other. There's a very generic result, which is: the ratio $\mathcal{N} / \mathcal{N_0}$ corresponds to the product of all bubble graphs (the ones without external legs). These nicely factor out, giving a convenient way to calculate correlations:

The appropriate normalization can be accounted for by simply disregarding graphs with disconnected bubbles.

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  • $\begingroup$ Thank you, this is really helpful; and I'm guessing it also applies to other field theories too? I guessed that the second one was forbidden to all orders! $\endgroup$ – Thomas Russell Apr 14 '17 at 8:36
  • $\begingroup$ @ThomasRussell This indeed applies to all fields. Path integrals can even be defined for anticommuting fields, which gives rise to the generating functional for fermions. $\endgroup$ – Prof. Legolasov Apr 14 '17 at 8:39
  • $\begingroup$ And these path integrals are presumably those using Grassmann numbers? $\endgroup$ – Thomas Russell Apr 18 '17 at 7:19
  • $\begingroup$ @ThomasRussell exactly $\endgroup$ – Prof. Legolasov Apr 18 '17 at 15:59

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