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  S-Matrix Elements in Path Integral Formalism

+ 4 like - 0 dislike
1456 views

I have a question related to the connection between the S-Matrix elements and the path integral formalism. In order to formulate the question, I will just work with a scalar field theory for simplicity.

Let us assume that we are given an action $S[\phi]$. In the path integral formalism, we can now define the generating functional \begin{equation} Z[J] \propto \int \mathcal{D}\phi ~ e^{i S[\phi] + \int d^4x~ \phi(x) J(x)} \end{equation} and calculate arbitrary vacuum expectation values \begin{equation} \left<0| \phi(x_1) \ldots \phi(x_n) | 0 \right> \end{equation} using functional derivatives with respect to the source $J$. I also know how to calculate vacuum expectation values in the "canonical quantization formalism" (Wick's theorem etc.). So far so good.

Usually, we are not interested in vevs but rather in S-matrix elements such as $\left<p_1, \ldots, p_n|q_1, \ldots, q_m \right>$ where $p_i$ and $q_j$ are outgoing and ingoing particle momenta. Furthermoe, the transition between $S$-matrix elements and vevs is also clear to me: this is just given by the LSZ reduction formula. So in principle, we are now good to go: we can calculate everything in the path integral formalism and eventually relate this to actual matrix elements using the LSZ formula.

Now come my actual questions:

  1. It seems that there is a more direct relation between the S-matrix elements and the path integral formalism. In fact, on the Wikischolar article about the Slavnov-Taylor identities (written by Dr. Slavnov himself) it is stated that the $S$ matrix can be written as $S = Z[0]$. Where does this come from and how is it to interpret? I am confused because I thought that $S$ was rather a matrix (whose entries, i.e. matrix elements are numbers) and $Z[0]$ is just a number (an evaluated integral). So to me, thsi reads like "matrix = number"... Furthermore, if this equation holds true, how can we obtain the $S$ matrix elements from there?

  2. Even more confusingly, there seems to be another relation to the $S$-matrix element. I have found this in Weinberg Vol. II, chapter 15.7 around equation (15.7.27). There, we have an action that is of the form $I + \delta I$ (the context is here that $I$ is the gauge fixed action of a non-Abelian gauge theory and $\delta I$ is the change due to a small variation in the gauge-fixing condition, but this does not really matter here). It says then: It is a fundamental physical requirement that matrix elements between physical states should be independent of our choice of the gauge-fixing condition, or in other words, of $\delta I$. The change in any matrix element $<\alpha|\beta>$ due to a change $\delta I$ in $I$ is \begin{equation} \delta <\alpha|\beta> ~\propto ~<\alpha|\delta I|\beta>. \end{equation} So now, there seems to be even a relation between the action and the $S$-matrix elements. How does this fit into the entire picture?

My QFT exam is coming up, so thanks a lot for your answers!

This post imported from StackExchange Physics at 2014-08-07 15:19 (UCT), posted by SE-user user56643
asked Aug 4, 2014 in Theoretical Physics by user56643 (20 points) [ no revision ]
retagged Aug 7, 2014

2 Answers

+ 2 like - 0 dislike

IMHO, in the prof Slavnov article, the action path integral formula $(3)$ for $S$ should be understood with constraints about initial and final states. So, in fact, it is a matrix $S_{ij}$.

However, this is not the case for the formula $(4)$ for $Z[J]$. It is a path integral without constraints.

For your second question, just note that $e^{i (S+\delta S)} \approx e^{i S}(1+i\delta S)$, so, with the different definitions of matrix elements and Green functions, you get your result.

This post imported from StackExchange Physics at 2014-08-07 15:19 (UCT), posted by SE-user Trimok
answered Aug 5, 2014 by Trimok (955 points) [ no revision ]
Trimok: Could you be a tad more precise in what you mean by "with the different definitions of matrix elements and Green functions"?

This post imported from StackExchange Physics at 2014-08-07 15:19 (UCT), posted by SE-user PPR
@PPR : As indicated in the question, you get $S$-matrix elements by applying LSZ reduction formulae to Green functions (vacuum expectation values). And vacuum expectation values are just successive derivates of $Z$ relatively to the currents $J(x)$ (up to a global constant)

This post imported from StackExchange Physics at 2014-08-07 15:19 (UCT), posted by SE-user Trimok
+ 0 like - 0 dislike

For your question 1.: note that the integral in $Z[J]$ is performed over only paths connecting the initial state $q_i$ to the final state $q_f$, i.e., $Z[J]$ actually depends on $q_i$ and $q_f$. So, you can view it as a matrix element of $\hat{S}$, namely, $S_{ij}$.

This post imported from StackExchange Physics at 2014-08-07 15:19 (UCT), posted by SE-user Hai-Yao Deng
answered Aug 4, 2014 by Hai-Yao Deng (0 points) [ no revision ]

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