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Chapter 11.4 of Peskin & Schroeder's book discussed the computation of effective action, but I don't understand some details of derivation. The book first split the Lagrangian into normal ones and counterterms. $L=L_1+\delta L$. The source term has also been split as $J=J_1+\delta J$. And $\frac{\delta L_1}{\delta \phi}|_{\phi=\phi_{cl}} + J_1=0$, $\phi_{cl}=\langle \Omega|\phi(x)|\Omega\rangle$.

Then the generating function is \begin{eqnarray*} Z[J]=\mathrm{e}^{-i E[J]}=\int D\phi \mathrm{e}^{i\int d^4x(L_1+J_1\phi)} \mathrm{e}^{i\int d^4x(\delta L+ \delta J\phi)} \end{eqnarray*}

So,expand $\phi(x)=\phi_{cl}(x) + \eta(x)$, \begin{eqnarray*} \int d^4x(L_1+J_1\phi) &=& \int d^4x(L_1[\phi_{cl}]+J_1\phi_{cl}) + \int d^4x \eta(x)(\frac{\delta L_1}{\delta \phi} + J_1) \\ &+&\frac{1}{2} \int d^4x d^4y \eta(x) \eta(y) \frac{\delta^2 L_1}{\delta\phi(x) \delta\phi(y)} \\ &+& \frac{1}{3!} \int d^4x d^4y d^4z \eta(x) \eta(y) \eta(z) \frac{\delta^3 L_1}{\delta\phi(x) \delta\phi(y) \delta\phi(z)} + \cdots \end{eqnarray*}

and, \begin{eqnarray*} \delta L+ \delta J\phi= (\delta L[\phi_{cl}]+ \delta J\phi_{cl}) + (\delta L[\phi_{cl}+\eta]-\delta L[\phi_{cl}] + \delta J \eta) \end{eqnarray*} After integrating over the quadratic term of $\eta$ and collecting constant terms,we get \begin{eqnarray*} -iE[J]&=&i \int d^4x (L_1[\phi_{cl}] + J_1\phi_{cl}) -\frac{1}{2} \mathrm{log} \ \mathrm{det}[-\frac{\delta^2 L_1}{\delta \phi \delta \phi}] \\ &+& \{\mathrm{connected \ diagrams}\} + i \int d^4x (\delta L[\phi_{cl}]+ \delta J\phi_{cl}) \end{eqnarray*} My queation is:

how to convert terms like $\frac{1}{3!} \int d^4x d^4y d^4z \eta(x) \eta(y) \eta(z) \frac{\delta^3 L_1}{\delta\phi(x) \delta\phi(y) \delta\phi(z)}$ and $ \delta L[\phi_{cl}+\eta]-\delta L[\phi_{cl}] + \delta J \eta$ into connected diagrams. I don not know the derivation of this in details.

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The question has been posted nearly one year and received no answer. Now, I think I could try to give an answer based on my study in this year. Since I think the discussion on effective action of chapter 11 of An Introduction to Quantum Field Theory (Peskin & Schroeder) is not very clear, I will reorganize all the material in my way. Another reference is Quantum Field Theory (Mark Srednicki).

If the proof of an equation can be found in Peskin's book, I will not repeat the proof here. So please read the answer with Peskin's book.

My definition of metric is $\mathrm{diag}(-1,1,1,1)$

Effective action

The path integral of a quantum field with external source is $$Z[J] = e^{-iE[J]} = \int \mathcal{D} \phi \exp\left[ i\int d^4x (\mathcal{L}[\phi] + J \phi) \right]$$

Define $\phi_{\mathrm{cl}} (x) \equiv \langle \Omega | \phi(x) | \Omega \rangle_{J}$, we can derive that $$\frac{\delta}{\delta J(x)} E[J] = - \phi_{\mathrm{cl}}(x)$$ We now define effective action as $$\Gamma[\phi_{\mathrm{cl}}] \equiv -E[J] - \int d^4y J(y) \phi_{\mathrm{cl}}(y)$$ Suppose $\mathcal{L}$ is invariant under transformation $U$, i.e. $\mathcal{L}(U\phi) = \mathcal{L}(\phi) $. We have $$U\phi_{\mathrm{cl}}(x) = \langle \Omega | U\phi(x) | \Omega\rangle_{J} = \frac{\int \mathcal{D}\phi e^{i \int \mathcal{L}(\phi)+J\phi}U\phi(x)}{\int \mathcal{D}\phi e^{i \int \mathcal{L}(\phi)+J\phi}}$$ Define $J' = \frac{J\phi}{U\phi}$, and we suppose the measure of path integral is invariant under transformation $U$, then we have $$U\phi_{\mathrm{cl}}(x) = \frac{\int \mathcal{D}U\phi e^{i \int \mathcal{L}(U\phi)+J'U\phi}U\phi(x)}{\int \mathcal{D}U\phi e^{i \int \mathcal{L}(U\phi)+J'U\phi}} = \frac{\int \mathcal{D}\phi e^{i \int \mathcal{L}(\phi)+J'\phi}\phi(x)}{\int \mathcal{D}\phi e^{i \int \mathcal{L}(\phi)+J'\phi}} = \langle \Omega | \phi(x) | \Omega\rangle_{J'}$$ On the one hand, we have $$\Gamma[U\phi_{\mathrm{cl}}] = E[J'] - \int d^4y J'(y)U\phi_{\mathrm{cl}}(y) = E[J'] - \int d^4y J(y)\phi_{\mathrm{cl}}(y)$$ On the other hand, we have \begin{eqnarray} Z[J'] &=& \int \mathcal{D}\phi \exp \left[ i \int d^4x \mathcal{L}(\phi)+J'\phi \right] = \int \mathcal{D}U\phi \exp \left[ i \int d^4x \mathcal{L}(U\phi)+J'U\phi \right] \nonumber \\ &=& \int \mathcal{D}\phi \exp \left[ i \int d^4x \mathcal{L}(\phi)+J\phi \right] = Z[J] \nonumber \end{eqnarray} So, $E[J] = E[J']$ and evidently $$\Gamma(U\phi_{\mathrm{cl}}) = E[J] - \int d^4y J(y)\phi_{\mathrm{cl}}(y) = \Gamma(\phi_{\mathrm{cl}})$$ We have proven that effective action is invariant under transformation $U$.

We can further verify that $$\frac{\delta}{\delta \phi_{\mathrm{cl}}(x)} \Gamma[\phi_{\mathrm{cl}}] = -J(x)$$ If the external source is set to zero, the effective action satisfy the equation $$\frac{\delta}{\delta \phi_{\mathrm{cl}}(x)} \Gamma[\phi_{\mathrm{cl}}] = 0$$ The solution to this equation are the values of $\langle \phi(x) \rangle$ in the stable quantum states of the theory. For a translational-invariant vacuum state, we will find a solution in which $\phi_{\mathrm{cl}}$ is independent of $x$. For simplicity, will assume the vacuum state in our following discussion are all translational-invariant. So, if $T$ is the time extent of the region and $V$ is its three dimensional volume, we can define the effective potential of the field by $$\Gamma[\phi_{\mathrm{cl}}] = -(VT) \cdot V_{\mathrm{eff}}(\phi_{\mathrm{cl}})$$ The condition that $\Gamma[\phi_{\mathrm{cl}}]$ has an extreme then reduces to the simple equation $$\frac{\partial}{\partial \phi_{\mathrm{cl}}} V_{\mathrm{eff}}(\phi_{\mathrm{cl}}) = 0$$ A system with spontaneously broken symmetry will have several minimum of $V_{\mathrm{eff}}$, all with the same energy by virtue of the symmetry. The choice of one among these vacuum is the spontaneous symmetry breaking. Note that $V_{\mathrm{eff}}(\phi_{\mathrm{cl}})$ share the same symetry with origianl Lagrangian even if the state of vacuum is spontaneous symmetry breaking.

Computation of the effective action

Decompose the Lagrangian into a piece depending on renormalized parameters and one containing the counter-terms $$\mathcal{L} = \mathcal{L}_1 + \delta \mathcal{L}$$ Define $J_1$ by $$\left. \frac{\delta \mathcal{L}_1}{\delta \phi} \right|_{\phi = \phi_{\mathrm{cl}}} + J_1(x) = 0$$ Define $\delta J$ by $$J(x) = J_1(x) + \delta J(x)$$ So, we have $$e^{-iE[J]} = \int \mathcal{D}\phi e^{i\int d^4x (\mathcal{L}_1 + J_1\phi)} e^{i\int d^4x (\delta \mathcal{L} + \delta J \phi)}$$ Replace $\phi$ by $\phi_{\mathrm{cl}}+\eta$, \begin{eqnarray} \int d^4x \, (\mathcal{L}_1 + J_1\phi) &=& \int d^4x \, (\mathcal{L}_1[\phi_{\mathrm{cl}}] + J_1\phi_{\mathrm{cl}}) + \int d^4x \, \eta(x) \left( \frac{\delta \mathcal{L}_1}{\delta \phi} + J_1 \right) \nonumber \\ &+& \frac{1}{2} \int d^4x \, d^4y \, \eta(x) \eta(y) \frac{\delta^2 \mathcal{L}_1}{\delta \phi(x) \delta \phi(y)} \nonumber \\ &+& \frac{1}{3!} \int d^4x \, d^4y \, d^4z \, \eta(x) \eta(y) \eta(z) \frac{\delta^3 \mathcal{L}_1}{\delta \phi(x) \delta \phi(y) \delta \phi(z)} + \cdots \nonumber \end{eqnarray} The term linear in $\eta$ vanishes by definition of $J_1$. Then, put back the effects of the counter-term Lagrangian,writing it as $$(\delta \mathcal{L}[\phi_{\mathrm{cl}}] + \delta J \phi_{\mathrm{cl}} ) + ( \delta \mathcal{L}[\phi_{\mathrm{cl}} + \eta] - \delta\mathcal{L} [\phi_{\mathrm{cl}}] + \delta J \eta)$$ Define $$\mathcal{L}_2 = \left(\frac{1}{3!} \int d^4x \, d^4y \, d^4z \, \eta(x) \eta(y) \eta(z) \frac{\delta^3 \mathcal{L}_1}{\delta \phi(x) \delta \phi(y) \delta \phi(z)} + \cdots \right) + ( \delta \mathcal{L}[\phi_{\mathrm{cl}} + \eta] - \delta\mathcal{L} [\phi_{\mathrm{cl}}] + \delta J \eta)$$ So $$e^{-iE[J]} = C_1 e^{i\int \mathcal{L}_2(\frac{1}{i} \frac{\delta}{\delta I})} \left. \int \mathcal{D}\eta e^{i\int \left(\frac{1}{2} \eta \frac{\delta^2 \mathcal{L}_1}{\delta \phi \delta \phi} \eta + I\eta \right)} \right|_{I=0}$$ where $$C_1 \equiv \exp \left[ i \int ( \mathcal{L}_1 [\phi_{\mathrm{cl}}] + J_1\phi_{\mathrm{cl}} + \delta \mathcal{L}[\phi_{\mathrm{cl}}] + \delta J \phi_{\mathrm{cl}} )\right]$$ If we define propagator $D_F$ as $$D_F \equiv i \left( \frac{\delta^2 \mathcal{L}_1}{\delta \phi \delta \phi}\right)^{-1}$$ We have $$Z[J] = e^{-iE[J]} =C_1 Z_0[0] e^{i\int \mathcal{L}_2(\frac{1}{i} \frac{\delta}{\delta I})} \left. \int \mathcal{D}\eta e^{i\int \left(-\frac{1}{2} I D_F I \right)} \right|_{I=0}$$ where $$Z_0[0] \equiv \int \mathcal{D}\eta e^{ \frac{i}{2}\int \eta \left( \frac{\delta^2 \mathcal{L}_1}{\delta \phi \delta \phi}\right) \eta }$$ Peskin represented $Z_0[0]$ by the method of functional determinant in this step. But I will introduce this method when working with specific example and try to make this strange "determinant" more natural.

Recall in the perturbation theory for path integral, we can get a perturbation expansion for $iE[J]$ using connected Feynman diagram. The proof can be found in section 9 of Quantum Field Theory (Mark Srednicki). So we have $$-iE[J] = i \int ( \mathcal{L}_1 [\phi_{\mathrm{cl}}] + J_1\phi_{\mathrm{cl}} + \delta \mathcal{L}[\phi_{\mathrm{cl}}] + \delta J \phi_{\mathrm{cl}} ) + \log(Z_0[0]) + \mbox{ connected diagrams }$$ Note that the propagator of the diagram is given by $D_F$, the vertex of the diagram is given by $\mathcal{L}_2$. This is the procedure that converts terms like $\frac{1}{3!} \int d^4x d^4y d^4z \eta(x) \eta(y) \eta(z) \frac{\delta^3 L_1}{\delta\phi(x) \delta\phi(y) \delta\phi(z)}$ and $ \delta L[\phi_{cl}+\eta]-\delta L[\phi_{cl}] + \delta J \eta$ into connected diagrams.

From this equation, $\Gamma$ follows directly: $$\Gamma[\phi_{\mathrm{cl}}] = \int d^4x \mathcal{L}_1[\phi_{\mathrm{cl}}] -i\log(Z_0[0]) -i \mbox{ connected diagrams} + \int d^4x \delta\mathcal{L}[\phi_{\mathrm{cl}}]$$

Notice that there are no terms remaining that depend explicitly on $J$; thus, $\Gamma$ is expressed as a function of $\phi_{\mathrm{cl}}$ , as it should be. The Feynman diagrams contributing to $\Gamma[\phi_{\mathrm{cl}}]$ have no external lines, and the simplest ones turn out to have two loops. The lowest-order quantum correction to $\Gamma$ is given by the functional determinant.

The last term provides a set of counter-terms that can be used to satisfy the renormalization conditions on $\Gamma$ and, in the process, to cancel divergences that appear in the evaluation of the functional determinant and the diagrams. The renormalization conditions will determine all of the counter-terms in $\delta \mathcal{L}$. However, the formalism we have constructed contains a new counter-term $\delta J$. That coefficient is determined by $\langle \eta \rangle = 0$. In practice, we will satisfy this condition by simply ignoring any one-particle-irreducible one-point diagram, since any such diagram will be cancelled by adjustment of $\delta J$.

Linear sigma model

We begin again with the Lagrangian $$\mathcal{L}_1 = -\frac{1}{2} \partial_{\mu} \phi^i \partial^{\mu}\phi^i + \frac{1}{2} \mu^2 (\phi^i)^2 - \frac{\lambda}{4} [(\phi^i)^2]^2$$ Expand about the classical field $\phi^i = \phi_{\mathrm{cl}}^i + \eta^i$, and we assume the vacuum is translational invariant. Then we have $$\mathcal{L}_1 = -\frac{1}{2}(\partial_{\mu}\eta)^2 + \frac{1}{2}\mu^2(\eta^i)^2 - \frac{\lambda}{2}[(\phi_{\mathrm{cl}}^2)(\eta^i)^2+ 2(\phi_{\mathrm{cl}}^i\eta^i)^2] + \cdots$$ From the terms quadratic in $\eta$, we can read off $$\frac{\delta^2 \mathcal{L}_1}{\delta\phi^i\delta\phi^j} = \partial^2\delta_{ij} + \mu^2\delta_{ij} - \lambda[(\phi_{\mathrm{cl}}^k)^2\delta_{ij} + 2\phi_{\mathrm{cl}}^i \phi_{\mathrm{cl}}^j]$$ We choose the vacuum state by demanding $\phi_{\mathrm{cl}}^i$ points in the $N$th direction $$\phi_{\mathrm{cl}}^i = (0,\cdots,\phi_{\mathrm{cl}})$$ Then the operator is just equal to the Klein-Gordon operator $(\partial^2-m_i^2)$, where $$m_i^2 = \begin{cases} \lambda\phi_{\mathrm{cl}}^2-\mu^2 \quad i=1,\cdots,N-1 \\ 3\lambda\phi_{\mathrm{cl}}^2-\mu^2 \quad i=N \end{cases}$$ $$Z_0[0] \equiv \prod_{i=1}^{N} Z_i = \prod_{i=1}^{N} \int \mathcal{D}\eta e^{ \frac{i}{2}\int \eta \left( \partial^2-m_i^2\right) \eta }$$ Here, $m_i$ is a function of $\phi_{cl}$. We want to get $\log Z_0[0]$ as a function of $\phi_{cl}$ and the constant infinite shift of $\log Z_0[0]$ will be dropped in our calculation. We treat $-\frac{1}{2}m_i^2\eta^2$ as a perturbation, so we have $$Z_i \propto e^{-\frac{im_i^2}{2}(\frac{1}{i} \frac{\delta}{\delta I})^2} \left. \int \mathcal{D}\eta e^{i\int \left(-\frac{1}{2} I S_F I \right)} \right|_{I=0}$$ where $$S_F(x-y) = \int \frac{d^4p}{(2\pi)^4} \frac{-i}{p^2}e^{ip(x-y)}$$ Now we can have the following Feynamn rules.

  1. A line from $x$ to $y$ is associated with $S_F(x-y)$
  2. A vertex joining two lines at $x$ is associated with $-im_i^2\int d^4x$

So we have $$\log Z_i = \sum_{I}C_I$$ where $C_I$ represents connected diagram without external source. A connected diagram without external source must have the following form Connected Feymann diagram without external source So, we have $$C_n = \frac{1}{2n}\int \prod_{k=1}^{n} \frac{d^4p_k d^4x_k}{(2\pi)^4} \frac{-m_i^2}{p_k^2} \exp(ip_k(x_k-x_{k+1})) = \int d^4p \delta(0) \left(-\frac{m_i^2}{p^2} \right)^n$$ $$\log Z_i = -\frac{1}{2}VT \int \frac{d^4p}{(2\pi)^4} \sum -\frac{1}{n} \left(-\frac{m_i^2}{p^2} \right)^n = -\frac{1}{2}VT\int \frac{d^4p}{(2\pi)^4} \log(1+\frac{m_i^2}{p^2})$$ Recall the Gaussian integral $$\int _{-\infty }^{\infty }e^{\left(-{\frac {i}{2}}\sum \limits _{i,j=1}^{n}A_{ij}x_{i}x_{j}\right)}\,d^{n}x = {\sqrt {\frac {(-2\pi i)^{n}}{\det A}}} $$ So formally, we have $$\log Z_i = -\frac{1}{2}\log \det (-\partial_x^2 + m_i^2)\delta(x-y)$$ Define $$M(x-y) \equiv (-\partial_x^2 + m_i^2)\delta(x-y) \quad M_0(x,y) \equiv -\partial_x^2\delta(x-y) \quad M_1(y,z) \equiv \delta(y-z) + im_i^2D_F(y-z)$$ We can verify that $$M(x,z) = \int d^4y M_0(x-y)M_1(y-z)$$ So, $$\log \det M = \log \det M_0 + \log \det M_1 \approx \log \det M_1$$ We drop out $\log \det M_0$ because it do not contain $m(\phi_{\mathrm{cl}})$. Furthermore, we have $M_1 = I - G$, where $I = \delta(x-y)$ is the identity matrix and $G = -im_i^2D_F$. So $$\log \det M_1 = \mathrm{Tr} \log M_1 = \mathrm{Tr} \log(I-G) = -\frac{1}{n}\sum_{n=1}^{\infty} \mathrm{Tr} G_n$$ So we have $$\log Z_i = -\frac{1}{2}\log \det M = \frac{1}{2n} \mathrm{Tr} G_n = \sum_n C_n$$ We then reproduce the result above using diagram perturbation method. Though the definition of functional determinant is not very rigorous, we may trust it as an effective way to calculate functional Gaussian integral.

The following calculation needs tricks of wick rotation and dimension regularization and you can refer to the equation 11.72 of Peskin's book for details. Here, I just list the final result: $$\log Z_0[0] = \frac{i}{2}\frac{\Gamma(-\frac{d}{2})}{(4\pi)^{d/2}}(m^2)^{\frac{d}{2}}VT$$ So, up to one loop corrections, we can get $$V_{\mathrm{eff}} = -\frac{1}{2} \mu^2 \phi_{\mathrm{cl}}^2 + \frac{\lambda}{4} \phi_{\mathrm{cl}}^4 - \frac{1}{2}\frac{\Gamma(-\frac{d}{2})}{(4\pi)^{d/2}}[(N-1)(\lambda\phi_{\mathrm{cl}}^2-\mu^2)^{\frac{d}{2}} + (3\lambda\phi_{\mathrm{cl}}^2-\mu^2)^{\frac{d}{2}}] + \frac{1}{2}\delta_{\mu}\phi_{\mathrm{cl}}^2 + \frac{1}{4}\delta_{\lambda}\phi_{\mathrm{cl}}^4$$ And if we want $V_{\mathrm{eff}}$ is finite for terms involving $\phi_{\mathrm{cl}}$, we can get $$\delta_{\lambda} = \frac{2\lambda^2(N+8)}{(4\pi)^2} \times \frac{1}{4-d} + \mbox{ finite terms }$$ $$\delta_{\mu} = -\frac{2\lambda\mu^2(N+2)}{(4\pi)^2} \times \frac{1}{4-d} + \mbox{ finite terms }$$

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  • $\begingroup$ Hey, there's something I don't get. When we call the vacuum expectation value of the field "classical field", what exactly do we mean by that? does it solve the classical equations of motion? Because it seems like you have to define J_1 to make $\phi_{cl}$ follow the equations of motion. $\endgroup$ – P. C. Spaniel Feb 15 at 5:52
  • $\begingroup$ Here is my personal understandings: A quantum field is a field of operators. It satisfies the same equation of motion as that of a classical field. So the expectation value of the quantum field must solve the classical equations of motion. Here, the quantum field has a source $J(x)$. $\phi_{\rm cl}(x)$ must solve the equation of motion with current $J(x)$. We should note that the dynamics of $\phi_{\rm cl}(x)$ is determined by $\mathcal{L}$ and $J$, not by $\mathcal{L}_1$. So we define the current $J_1$ to make $\phi_{\rm cl}(x)$ ollow the equations of motion generated by $\mathcal{L}_1$. $\endgroup$ – Eric Yang Apr 17 at 3:50

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