Schwinger Quantum action

[mhill]

Given the path integral

$$< -\infty | \infty > = N \int D[\phi ]e^{i \int dx (L+J \phi ) }$$

then , it would be true that (Schwinger)

$$\frac{\partial < -\infty | \infty >}{\partial J }= < -\infty |\delta \int L d^{4}x | \infty >$$

If so, could someone provide an exmple with the Kelin-gordon scalar field plus an interaction of the form $$\phi ^{4}$$

 

[lbrits]

This is a consequence of the fact that
$$\int\!\mathcal{D}\phi\, \frac{\delta}{\delta \phi} \left( \dots \right) = 0$$
Essentially, one integrates by parts and hopes that the exponential damps out the boundary in function space, if such a boundary exists. Thus, the equations of motion for $$\phi$$ are obeyed inside the expectation value.

Anyway, using (space) integration by parts, write your action as
$$\int\!d^4x\, \frac{1}{2}\phi ( - \partial^2 - m^2 ) \phi + \frac{\lambda}{4!} \phi^4$$.
You should now be able to find the variation in this action just like taking ordinary derivatives. If you get stuck, tell us what you've done and we can see about help.

I hope someone doesn't just blurt out the answer. This forum seems to be full of keeners like that =)

 

[denian]

I can confirm that the equation presented is known as the Schwinger Quantum action. It is a useful tool for calculating the probability amplitude of a particle moving from one point to another in space-time.

The equation states that the probability amplitude < -\infty | \infty > is equal to the integral over all possible paths of the particle, denoted by D[\phi ], multiplied by the exponential of the action, which is the sum of the Lagrangian (L) and the external source (J) multiplied by the field ( \phi ).

The second equation states that the derivative of the probability amplitude with respect to the external source (J) is equal to the expectation value of the operator \delta \int L d^{4}x between the initial and final states. This means that by varying the external source, we can calculate how the probability amplitude changes.

To provide an example, let's consider the Klein-Gordon scalar field with an interaction term of the form \phi ^{4}. The Lagrangian in this case would be:

L = \frac{1}{2}(\partial_{\mu}\phi)^2 - \frac{1}{2}m^2\phi^2 - \frac{\lambda}{4!}\phi^4

where m is the mass of the particle and \lambda is the coupling constant for the interaction term.

Using the Schwinger Quantum action, we can calculate the probability amplitude for a particle to move from an initial state at x=-\infty to a final state at x=\infty. This would be represented as < -\infty | \infty >.

Taking the derivative with respect to the external source, we get:

\frac{\partial < -\infty | \infty >}{\partial J }= < -\infty |\delta \int L d^{4}x | \infty >

Substituting the Lagrangian for our example, we get:

\frac{\partial < -\infty | \infty >}{\partial J }= < -\infty |\delta \int \left[\frac{1}{2}(\partial_{\mu}\phi)^2 - \frac{1}{2}m^2\phi^2 - \frac{\lambda}{4!}\phi^4 + J\phi\right] d^{4}x | \infty >

Using the properties of the