# Tag Info

0

The step of evaluating your second expression for $Z$ and arriving at the path integral form, which is the first expression you mention, is a nontrivial but standard derivation in the path integral formulation of QFT. It is quite lengthy, but can be found in most books that cover the path integral formulation. The general idea it to split the time interval ...

2

In general the "weighting" of each path $q$ in a path integral is given by $e^{\frac{i}{\hbar}S[q]}$. Then paths for which the action $S$ is stationary with respect to small deviations from the path are the only ones which really contribute because the contributions from those with non stationary $S$ get averaged out as the phase changes very rapidly ...

0

There definitely is a least action principle in quantum mechanics, indeed, the path-integral method is based on it. Feynman's doctoral thesis is titled:" the least action principle in quantum mechanics". Please see, e.g., http://cds.cern.ch/record/101498/files/?ln=en

11

Mathematically, a path integral is a generalization of a multi-dimensional integral. In usual $N$-dimensional integrals, one integrates $$\int dx_1 dx_2 \dots dx_N$$ over a subspace of ${\mathbb R}^N$, an $N$-dimensional integral. A path integral is an infinite-dimensional integral $$\int {\mathcal D}f(y)\, Z[f(y)]$$ over all possible functions $f(y)$ of ...

1

from The Reference Frame: Another wrong expectation that a beginner could have - and usually has - is that if you allow the summation over all trajectories of a particle, the typical particles will move faster than light most of the time and this will automatically result in a violation of the special theory of relativity. So an overzealous ...

0

No, the principle of least action started well before quantum mechanics. It is a variational principle that, when applied to the action of a mechanical system, can be used to obtain the equations of motion for that system. The principle led to the development of the Lagrangian and Hamiltonian formulations of classical mechanics. this new formulation of ...

2

I think, from the way you formulated the question, you lost the context of this trick, and then it indeed doesn't make a lot of sense. The point is that in QFT, you want to compute quantities corresponding to the full interacting Hamiltonian, $H$. In practice, however, we only know the eigenstates of the free Hamiltonian $H_{0}$: the plane waves ...

1

In QFT it is not to get down to the ground state, but to choose a correct propagator (amongst variety of Green's function). In other words, it is applying or taking into account the boundary conditions. However, for "incomplete" systems, decaying may really mean getting down to the ground state due to interaction of some sort like irreversibly absorbing ...

0

Your argument is valid for unitary evolution. However turning the time into the complex plane you make it non-unitary. You may say that you introduce small decay for every state $$e^{-iHt}=e^{-iHt-\eta Ht}$$ with the ground energy state the slowest to decay (stable if you set $E_0=0$)

Top 50 recent answers are included