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In what sense, the effective action $\Gamma[\phi_c]$ is quantum-corrected classical action $S[\phi]$?

+ 7 like - 0 dislike
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There is a difference between the classical field $\phi(x)$ (that appears in the classical action $S[\phi]$) and the quantity $\phi_c$ defined as $$\phi_c(x)\equiv\langle 0|\hat{\phi}(x)|0\rangle_J$$ which appears in the effective action. Even though $\phi_c(x)$ is referred to as the "classical field", I don't see why $\phi(x)$ and $\phi_c$ should be the same.

Therefore, in what sense, the effective action $\Gamma[\phi_c]$ is quantum-corrected classical action $S[\phi]$? How can we compare the functionals of two different objects (namely, $\phi(x)$ and $\phi_c(x)$) and claim that $\Gamma[\phi_c]$ is a correction over $S[\phi]$.

I apologize if the question and the confusion is unclear.

This post imported from StackExchange Physics at 2017-09-17 12:58 (UTC), posted by SE-user SRS
asked Jun 7 in Theoretical Physics by SRS (65 points) [ no revision ]

2 Answers

+ 6 like - 0 dislike

We want to calculate the path integral $$ Z = \int \mathcal{D}{\phi}\, e^{i \hbar^{-1} S[\phi]} $$ which encodes a transition amplitude between initial and final quantum states.

If we had the effective action $\Gamma[\phi]$ at our disposal, we would have calculated the same result by solving for $$ \phi_c(x):\quad \left. \frac{\delta \Gamma}{\delta \phi} \right|_{\phi=\phi_c} = 0 $$ and plugging it back in the effective action: $$ Z = e^{i \hbar^{-1} \Gamma[\phi_c]}. $$

This is the definition of $\Gamma$.

Note that no path integrals are required at this point. Boundary conditions are implicitly present throughout this answer, encoding the exact states between which the quantum transition occurs. Their existence ensures that there is only one solution $\phi_c$.

Now to why $\phi_c$ is called classical: it solves the e.o.m. given by the action $\Gamma$.

Think of $\Gamma$ as of an object in which all the short-scale properties of the integration measure $\mathcal{D}\phi$ (including renormalization-related issues) are already accounted for. You simply solve the e.o.m. and plug the solution in the exponential and you are done: here is your transition amplitude.

That being said, $\Gamma$ is not classical in the sense that it still describes dynamics of a quantum theory. Only in a different fashion. Simple algebraic manipulations instead of path integrals.

Finally, note how if the path integral is Gaussian, $$\Gamma[\phi] = S[\phi] + \text{const},$$ where $\text{const}$ accounts for the path integral normalization constant. There are no quantum corrections.

In classical theory, however, we solve the e.o.m. w.r.t. $\phi = \phi_c$ for $S[\phi]$, not $\Gamma[\phi]$. Plugging it back into $S[\phi_c]$ gives us the Hamilton function. When the path integral is Gaussian, it doesn't matter if we use $S$ or $\Gamma$, and exponentiating the Hamilton function gives you the transition amplitude. However if we are dealing with an interacting theory, the correct way to do this would be to use $\Gamma$ instead of $S$. In this sense, $\Gamma$ is the quantum-corrected version of $S$.

And yes, it is always true (can be shown using the saddle point approximation formula) that $$ \Gamma[\phi] = S[\phi] + \mathcal{O}(\hbar). $$

Why wouldn't we just use $\Gamma[\phi]$ to define the quantum theory and forget about $S[\phi]$ alltogether? Because $\Gamma$ is non-local and contains infinitely many adjustable parameters. These can be determined from the form of $S[\phi]$ by, well, quantization. That's why it is $S[\phi]$ which defines the theory, not $\Gamma$. $\Gamma$ is to be calculated via path integrals.

UPDATE: It is also important to understand that in naive QFT $\Gamma$ contains divergences, while $S$ doesn't. However, the actual situation is opposite. It is $S$ which contains divergences (divergent bare couplings), which cancel out against the divergences coming from the path integral, rendering a finite (i.e. renormalized) $\Gamma$. That $\Gamma$ should be finite is evident from how we use it to calculate physical properties: we only solve the e.o.m. and plug the result back in $\Gamma$.

Actually, the whole point of renormalization is to make $\Gamma$ finite and well-defined while adjusting only a finite number of diverging couplings in the bare action $S$.

This post imported from StackExchange Physics at 2017-09-17 12:58 (UTC), posted by SE-user Solenodon Paradoxus
answered Jun 7 by Solenodon Paradoxus (60 points) [ no revision ]
+ 5 like - 0 dislike

There is already a good answer. Here we provide a formal proof (via the stationary phase/WKB approximation).

  1. To fix notation, we define the effective/proper action $$ \Gamma[\phi_{\rm cl}]~=~W_c[J]-J_k \phi_{\rm cl}^k, \tag{1}$$ as the Legendre transformation of the generating functional $W_c[J]$ for connected diagrams. We assume that the Legendre transformation is regular, i.e. the formula $$ \phi_{\rm cl}^k~\approx~\frac{\delta W_c[J]}{\delta J_k}\qquad \Leftrightarrow \qquad J_k~\approx~-\frac{\delta \Gamma[\phi_{\rm cl}]}{\delta \phi_{\rm cl}^k} \tag{2}$$ is invertible. Here $J_k$ are the sources and $\phi_{\rm cl}^k$ are the classical fields.

  2. The partition function/path integral is $$ \exp\left\{ \frac{i}{\hbar} W_c[J]\right\}~=~Z[J] ~:=~\int \! {\cal D}\phi~\exp\left\{ \frac{i}{\hbar} \left(S[\phi]+J_k \phi^k\right)\right\} . \tag{3}$$ The first equality in eq. (3) is the linked cluster theorem.

  3. The quantum averaged field is by definition $$ \langle \phi^k \rangle_J ~:=~\frac{1}{Z[J]} \int \! {\cal D}\phi~\phi^k\exp\left\{ \frac{i}{\hbar} \left(S[\phi]+J_{\ell} \phi^{\ell}\right)\right\}$$ $$ ~=~\frac{1}{Z[J]} \frac{\hbar}{i} \frac{\delta }{\delta J_k}\int \! {\cal D}\phi\exp\left\{ \frac{i}{\hbar} \left(S[\phi]+J_{\ell} \phi^{\ell}\right)\right\}~\stackrel{(3)}{=}~\frac{\delta W_c[J]}{\delta J_k}~\stackrel{(2)}{\approx}~\phi_{\rm cl}^k, \tag{4}$$ i.e. it is equal to the classical field.

  4. By formal Fourier transformation of the path integral (3), we get $$ \exp\left\{ \frac{i}{\hbar}S[\phi_{\rm cl}]\right\} ~\stackrel{(3)}{=}~\int \! {\cal D}J~\exp\left\{ \frac{i}{\hbar} \left(W_c[J]-J_k \phi_{\rm cl}^k\right)\right\} $$ $$~\stackrel{(1)}{\sim}~\frac{1}{\sqrt{\det(\ldots)}}\exp\left\{ \frac{i}{\hbar}\Gamma[\phi_{\rm cl}]\right\} \quad\text{for}\quad\hbar~\to~0 \tag{5}$$ in the stationary phase/WKB approximation.

  5. Eq. (5) shows that the effective action $$ \Gamma[\phi_{\rm cl}] ~=~S[\phi_{\rm cl}] +{\cal O}(\hbar) \tag{6}$$ agrees with the action $S$ up to quantum corrections. (The square root factor in eq. (5) only contributes at one-loop and beyond.)

  6. Consider given fixed sources $J_k$. From $$\frac{\delta \Gamma[\phi_{\rm cl}]}{\delta \phi_{\rm cl}^k}~\stackrel{(2)}{\approx}~-J_k~\stackrel{\text{EL eqs.}}{\approx}~\frac{\delta S[\phi]}{\delta \phi^k}, \tag{7} $$ we deduce that the two solutions agree $$ \phi_{\rm cl}^k[J]~\stackrel{(6)+(7)}{\approx}~\phi^k[J] +{\cal O}(\hbar) \tag{8} $$ up to quantum corrections. (We assume that each solution to eq. (7) is unique, due to pertinent boundary conditions.)

  7. From the $\hbar$/loop-expansion, one may deduce that the (amputated) tree diagrams in the effective action $$ \Gamma_{\rm tree}[\phi_{\rm cl}] ~\stackrel{(6)}{=}~S[\phi_{\rm cl}] \tag{9}$$ is the action $S$. (Here we have assumed that the action $S$ has no $\hbar$-dependence.) See also this related Phys.SE post.

This post imported from StackExchange Physics at 2017-09-17 12:58 (UTC), posted by SE-user Qmechanic
answered Jul 25 by Qmechanic (2,790 points) [ no revision ]

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