Lancaster and Blundell Chapter 24

(24.1) Verify eqn 24.25 and show that eqn 24.29 solves eqn 24.28.

Integrating the term by parts gives as promised, while the term gives . Therefore,

$$-\frac12(\partial_\mu A_\nu\partial^\mu A^\nu-\partial_\mu A_\nu\partial^\nu A^\mu)+\frac12m^2 A^\mu A_\mu=\frac 12A_\mu\left([\partial^2+m^2]\eta_{\mu\nu}-\partial_\mu\partial_\nu\right)A^\nu$$

Evaluating

$$[-(p^2-m^2)g^{\mu\nu}+p^\mu p^\nu]\frac{-i(g_{\nu\lambda}-p_\nu p_\lambda/m^2)}{p^2-m^2}$$

$$=ig^{\mu\nu}(g_{\nu\lambda}-p_\nu p_\lambda/m^2)-i\frac{p^\mu p_\lambda-p^\mu p^2 p_\lambda/m^2}{p^2-m^2}\\=ig^\mu_\lambda – i\frac{p^\mu p_\lambda}{m^2}+i\frac{(p^2-m^2) p^\mu p_\lambda}{m^2(p^2-m^2)}=ig^\mu_\lambda$$

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(24.2) Consider the Lagrangian with a shift

$$\mathcal L=\frac12(\partial_\mu\phi)^2-\frac{m^2}2\phi^2-\frac g8\phi^4+\frac{1}{2g}\left(\sigma-\frac g2\phi^2\right)^2$$

By performing a functional integral over the field , show that doesn’t change the dynamics of the theory.

The Lagrangian density becomes

$$\frac12(\partial_\mu\phi)^2-\frac{m^2}2\phi^2-\frac g8\phi^4+\frac g8\phi^4+\frac 1{4g^2}\sigma^2-\frac12\sigma \phi^2\\=\frac12(\partial_\mu\phi)^2-\frac{m^2}2\phi^2+\frac1{4g^2}\sigma^2-\frac12\sigma\phi^2$$

Identifying and , The path integral over gives a factor of

$$B[\det a]^{-\frac 12}e^{-\frac i2\int d^4x d^4y \left(-\frac{\phi^2}{2}\right)2g\left(-\frac{\phi^2}2\right)}$$

Here, we realize that is diagonal, i.e. a Dirac delta . Thus we recover the term in the Lagrangian, and the overall constant out in front cancels.

The Euler-Lagrange equations for are

$$\frac{\partial \mathcal L}{\partial\sigma}=0$$

since there is no dependence on the derivatives of .

$$\frac1{2g^2}\sigma+\frac{\phi^2}2=0$$

Thus is entirely determined by , and thus there are no dynamical degrees of freedom in . The Feynman diagrams have vertices with 2 particles and 1 particle.

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(24.3) We want to do the integral

$$Z(J)=\int dx\ e^{-\frac12 Ax^2-\frac\lambda{4!}x^4+Jx}$$

$$=\int dx\ e^{-\frac12 Ax^2+Jx}e^{-\frac\lambda{4!}x^4}=\sum_n \frac1{n!}\left(-\frac\lambda{4!}x^4\right)^n\int dx\ e^{-\frac12 Ax^2+Jx}\\=\left[\sum_n\frac1{n!}\left(-\frac\lambda{4!}\frac{\partial^4}{\partial J^4}\right)^n\right]\int dx\ e^{-\frac12 Ax^2+Jx}\\=\left[e^{-\frac\lambda{4!}\frac{\partial^4}{\partial J^4}}\right]\left[\left(\frac{2\pi}A\right)^{\frac12}e^{\frac{J^2}{2A}}\right]$$

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(24.4) By analogy, the generating functional for theory is

$$Z[J]=\left[e^{-\frac\lambda{4!}\int d^4z\ \frac{\delta^4}{\delta J(z)^4}}\right]\mathcal Z_0[J]$$

where

$$\mathcal Z_0[J]=e^{-\frac12\int d^4xd^4y\ J(x)\Delta(x-y)J(y)}$$

Act on with the functional derivative four times.

Applying the definition of the functional derivative,

$$\frac{\delta \mathcal Z_0[J]}{\delta J(z)}=\lim_{\epsilon\rightarrow 0}\frac1\epsilon(Z[J(x)+\epsilon\delta(z-x)]-Z[J(x)])$$

$$=\lim_{\epsilon\rightarrow 0}\frac1\epsilon \left[e^{-\frac12\int d^4xd^4y\left[\left(J(x)+\epsilon\delta(x-z)\right)\Delta(x-y)\left(J(y)+\epsilon\delta(x-y)\right)\right]}-e^{-\frac12\int d^4xd^4y\ J(x)\Delta(x-y)J(y)}\right]$$

Expanding the exponentials and keeping only the terms to first order in , we indeed find

$$=\left[-\int d^4y\ \Delta(z-y)J(x)\right]\mathcal Z_0[J]$$

The rest of this calculation is very tedious, and I skip it. The functional derivative acts as expected with the product rule and chain rules.