In general,(adsbygoogle = window.adsbygoogle || []).push({});

##\displaystyle{\langle q_{f}|e^{-iHt/\hbar}|q_{i}\rangle=\int\mathcal{D}q(t)\ e^{iS[q]/\hbar}}##

and

##\displaystyle{\langle q_{f}|\hat{Q}(t)|q_{i}\rangle=\int\mathcal{D}q(t)\ e^{iS[q]/\hbar}}\ q(t).##

How can one switch from the above expressions to the following?

##\displaystyle{\langle q_{f}|T\{\hat{Q}(t_{1})\hat{Q}(t_{2})\}|q_{i}\rangle=\int\mathcal{D}q(t)\ e^{iS[q]/\hbar}}\ q(t_{1})q(t_{2})##

In other words, why does the path integral compute time-ordered products?

**Physics Forums | Science Articles, Homework Help, Discussion**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# A Path integral computes time-ordered products

Have something to add?

Draft saved
Draft deleted

Loading...

Similar Threads for Path integral computes |
---|

I A Proof of Kaku (8.18). |

I Path Integrals in Quantum Theory |

I Random measurements in QED? |

A Path Integral of a Spontaneously Broken Theory |

I Feynman path integral |

**Physics Forums | Science Articles, Homework Help, Discussion**