Feynman's path integral formulation

See this video for intuition and also this other one. It explain why it coincides with the principle of least action in the limit.

Quantum mechanics

A quantum particle, according to Feynman ideas, follows all possible paths at the same time. Suppose the particle is detected at point The probability of finding the particle at a given point is the sum of the amplitudes of all the paths that end at that point. The amplitude of a path is given by the exponential of the action of the path divided by $\hbar$.

$$ \mathcal{A} = \exp\left(\frac{i}{\hbar}S\right) $$

where $S$ is the action of the path. The action is defined as

$$ S = \int_{t_1}^{t_2} L(q,\dot{q}) dt $$

where $L$ is the Lagrangian of the system.

If we discretize this process in "finite jumps", all this should be concluded (or, at least, visualized) from the quantum approach to particle motion- an example.

Quantum field theory

How does this connect to Quantum Field Theory?

The path integral formulation extends naturally to fields (quantum fields). Instead of summing over all possible paths of a particle, we sum over all possible field configurations. The action $S$ now involves integrals over space and time, and the Lagrangian $L$ is replaced by a Lagrangian density $\mathcal{L}$ which is a function of the fields and their derivatives.

The amplitude for a field configuration is given by

$$ \mathcal{A} = \exp\left(\frac{i}{\hbar}S[\phi]\right) $$

where $\phi$ represents the fields, and

$$ S[\phi] = \int d^4x \, \mathcal{L}(\phi, \partial_\mu \phi) $$

where $d^4x$ represents integration over spacetime.

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Author of the notes: Antonio J. Pan-Collantes

antonio.pan@uca.es

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