A Utility of path integral formulation of quantum mechanics

[spaghetti3451]

How does the path integral formulation of quantum mechanics as given by $\langle q_{f}|e^{-iHt/\hbar}|q_{i}\rangle = \int \mathcal{D}q(t)\ e^{iS[q]/\hbar}$ make manifest aspects of quantum mechanics such as symmetries?

 

[Mentz114]

How does the path integral formulation of quantum mechanics as given by $\langle q_{f}|e^{-iHt/\hbar}|q_{i}\rangle = \int \mathcal{D}q(t)\ e^{iS[q]/\hbar}$ make manifest aspects of quantum mechanics such as symmetries?

Insofar as we define a symmetry to be a group of transformations that does not alter the physics - won't these will be inherited from $H$ ?

 

[spaghetti3451]

The symmetries are not inherent in the Hamiltonian formulation i.e. the Hamiltonian.

But they are manifest in the Lagrangian formulation.

The question is why?

 

[bhobba]

The question is why?

Noethers theorem requires the PLA to work. The path integral formulation explains why that is. The interesting thing is quantum theories are expressible in that form . Why is probably trying to tell us something imoportant.

Thanks
bBill

 

[spaghetti3451]

What is PLA?

 

[ddd123]

What is PLA?

Principle of Least Action.

 

[spaghetti3451]

Noethers theorem requires the PLA to work. The path integral formulation explains why that is. The interesting thing is quantum theories are expressible in that form . Why is probably trying to tell us something imoportant.

Thanks
bBill

I understand that Noether's theorem requires the principle of least action to work (since the derivation of Noether's Theorem uses the Euler-Lagrange equations and the Euler-Lagrange equations follow from the principle of least action).

But I don't quite see how the path integral formulation of quantum field theory explains why the principle of least action is required for Noether's theorem to work.

 

[stevendaryl]

I understand that Noether's theorem requires the principle of least action to work (since the derivation of Noether's Theorem uses the Euler-Lagrange equations and the Euler-Lagrange equations follow from the principle of least action).

But I don't quite see how the path integral formulation of quantum field theory explains why the principle of least action is required for Noether's theorem to work.

I think that it's that the path integral formulation explains (in some sense) the principle of least action. When you sum amplitudes over all possible paths, you get destructive interference for paths where the amplitude is very sensitive to the path (a small change in the path makes a big change in the amplitude), so the important paths are those where the amplitude is stationary (a small change in the path makes a negligible change to the amplitude). The paths where the amplitude is stationary are the classical paths--the ones you would get by the principle of least action.

 

[bhobba]

But I don't quite see how the path integral formulation of quantum field theory explains why the principle of least action is required for Noether's theorem to work.

Mathematics is a language. Sometimes it can be translated into English, mostly it cant. This is very likely one of the mostly - its a requirement of the theorem and mathematical reasoning (ie logic) requires conservation laws to follow from symmetry principles. If you try for anything 'English' like you will likely get nowhere.

Thanks
Bill

 

[atyy]

The symmetries are not inherent in the Hamiltonian formulation i.e. the Hamiltonian.

But they are manifest in the Lagrangian formulation.

The question is why?

In classical mechanics, there is a "Hamiltonian Noether's theorem" https://ncatlab.org/nlab/show/Noether's+theorem

In quantum mechanics, the Lagrangian formalism provides an easy way to write Hamiltonians that respect symmetries. However, you can just think of it as a tool to write Hamiltonians. The quantum Lagrangian formalism and the saddle point approximation give you the classical principle of least action.

 

[exmarine]

... When you sum amplitudes over all possible paths, you get destructive interference for paths where the amplitude is very sensitive to the path (a small change in the path makes a big change in the amplitude), so the important paths are those where the amplitude is stationary (a small change in the path makes a negligible change to the amplitude)...

Yes, that is the usual textbook explanation. But it has always seemed inadequate to me. A more convincing argument for me is that the infinite number of non-classical path unit phase vectors must have a uniform phase distribution between 0 and 2π. So they would indeed all sum to zero, leaving only those aligned with the classical path phase.

(Maybe this should be the start of a new thread? Don't know how to do that. Sorry.)

 

[vanhees71]

It's as with the usual method to evaluate integrals with the "method of steepest descent", leading to asymptotic series: You can deform the path in the complex plane such that from a quickly oscillatory behavior you come to a steeply falling function along the deformed path.

 

[radium]

It is also very useful for deriving the Ward identities for currents which are somewhat like the quantum version of Noether's theorem and provide restrictions on your theory. This is important in diagrammatics in QFT. Also, the path integral makes clear that the classical solution is just the saddle point solution and you can easily evaluate fluctuations about the saddle to study phase transitions.