The Principle of Least Action in Quantum Physics

[lugita15]

In Volume II Chapter 19 of his Lectures on Physics, Feynman discusses the principle of least action and its role in quantum mechanics. He says the following on page 19-9:
"The complete quantum mechanics (for the nonrelativistic case and neglecting electron spin) works as follows: The probabilty that a particle starting at point 1 at the time t_{1} will arrive at point 2 at the time t_{1} is the square of a probability amplitude. For every x(t) we could have-for every possible imaginary trajectory-we have to calculate an amplitude. Then we add them all together. What do we take for the amplitude for each path? Our action integral tells us what the amplitude for a single path ought to be. The amplitude is proportional to some constant times e^{iS/\hbar}, where S is the action for that path."
Why does Feynman make the qualification "for the nonrelativistic case and neglecting electron spin"? Doesn't the same principle apply even relativistically in calculating the probability amplitude for a path?

Any help would be greatly appreciated.
Thank You in Advance.

 

[lbrits]

Relativistic quantum mechanics demands a quantized field approach, due to properties of the Lorentz group. So, in QFT, the path integral approach is still used, but it is a sum over field configurations and not particle configurations.

Incidentally, you can incorporate spin in the path integral approach for ordinary quantum mechanics. Feynman apparently tried this but couldn't make it work. You end up with a point particle that has N=2 worldline supersymmetry. It's not nearly as elegant as the spinless case.

 

[lugita15]

Relativistic quantum mechanics demands a quantized field approach, due to properties of the Lorentz group. So, in QFT, the path integral approach is still used, but it is a sum over field configurations and not particle configurations.

Could you please elaborate on your explanation?

 

[Hans de Vries]

"The complete quantum mechanics (for the nonrelativistic case and neglecting electron spin) works as follows: The probabilty that a particle starting at point 1 at the time t_{1} will arrive at point 2 at the time t_{1} is the square of a probability amplitude. For every x(t) we could have-for every possible imaginary trajectory-we have to calculate an amplitude. Then we add them all together. What do we take for the amplitude for each path? Our action integral tells us what the amplitude for a single path ought to be. The amplitude is proportional to some constant times e^{iS/\hbar}, where S is the action for that path."
Why does Feynman make the qualification "for the nonrelativistic case and neglecting electron spin"? Doesn't the same principle apply even relativistically in calculating the probability amplitude for a path?

Any help would be greatly appreciated.
Thank You in Advance.

The principle of least Action holds perfectly well in the relativistic case. The classical
mechanical approach is just the limit of the relativistic case.

In the relativistic case the principle of Least action can be interpreted as the
principle of least proper time: The particle follows the path which minimizes its aging.

Most elementary, in relativistic quantum mechanics, we have the de Broglie waves
which are the eigen functions of the Klein Gordon equation and which transform
relativistically correct.

Here applies what Feynman mentions above, the principle of least action becomes
the principle of least phase. The particle chooses the path which minimizes the phase
changes.

It also applies for a particle with spin, as long as the spin points in the same direction
along all the paths. The latter is not necessarily the case, which then requires additional
mathematics.


See this part of a chapter of my book on the Lagrangian:
http://physics-quest.org/Book_Chapter_Lagrangian.pdf


Regards, Hans.

 

[lightarrow]

The principle of least
See this part of a chapter of my book on the Lagrangian:
http://physics-quest.org/Book_Chapter_Lagrangian.pdf

Interesting.
(I think you have to correct the expression for kinetic energy in eq. (17.13), pag9).

 

[Hans de Vries]

Interesting.

Thanks for your interest lightarrow.

(I think you have to correct the expression for kinetic energy in eq. (17.13), pag9).

The point is that the relativistic Hamiltonian and lagrangian also include the restmass
energy. The term of the relativistic Lagrangian you are looking at is a combination.
It can be considered as the kinetic term minus the restmass energy in the non-
relativistic limit.

http://physics-quest.org/Book_Chapter_Lagrangian.pdf


Regards, Hans

 

[lightarrow]

The point is that the relativistic Hamiltonian and lagrangian also include the restmass
energy. The term of the relativistic Lagrangian you are looking at is a combination.
It can be considered as the kinetic term minus the restmass energy in the non-
relativistic limit.

Sorry, can't grasp it, can you explain better? The relativistic kinetic energy is

\frac{mc^2}{\sqrt{1 - v^2/c^2}}\ -\ mc^2

How does it become

-mc^2\sqrt{1 - v^2/c^2} ?

Thank you.

 

[Hans de Vries]

Sorry, can't grasp it, can you explain better? The relativistic kinetic energy is

\frac{mc^2}{\sqrt{1 - v^2/c^2}}\ -\ mc^2

How does it become

-mc^2\sqrt{1 - v^2/c^2} ?

Thank you.

-mc^2\sqrt{1 - v^2/c^2}~~~~~~~~~~~~~~~~~ : is the relativistic Lagrangian of a free particle

-mc^2\sqrt{1 - v^2/c^2} ~-~ V(q)~~~~: is the Lagrangian in a potential field V


The kinetic term is this one:

T ~=~ \frac{1}{2}pv ~=~ \frac{1}{2}\frac{mv^2}{\sqrt{1-\frac{v^2}{c^2}}} ~\approx ~ \frac{1}{2}mv^2<br />


See also equation 17.7 of http://physics-quest.org/Book_Chapter_Lagrangian.pdf

I made some improvements to the text.


Regards, Hans

 

[Hans de Vries]

I added a new section which derives the Lagrangian density of the scalar Klein Gordon field
in a rather unique way: Directly from the Lagrangian of the classical relativistic particle.http://physics-quest.org/Book_Chapter_Lagrangian.pdf (see the last section)That, is. The Euler Lagrange equation is not needed. We use it the other way around
to show that the Klein Gordon equation is the scalar quantum field representation
corresponding with the classical relativistic particle (The only assumption is that the
field is a scalar field)
Regards, Hans.

 

[nrqed]

Relativistic quantum mechanics demands a quantized field approach, due to properties of the Lorentz group. So, in QFT, the path integral approach is still used, but it is a sum over field configurations and not particle configurations.

Actually, it is possible to do relativistic quantum mechanics following a "first quantized" approach in which one starts with an actual path integral (over x(tau) where tau can be seen as Schwinger's proper time). Feynman even mentioned that in one of his classic papers and gave the formula for the example of the propagation of a scalar field in an electromagnetic background field (plus on-loop EM graphs included).


Of course, the price to pay is that if one starts writing multiloop diagrams, the rules for the vertices are arbitrary as opposed to the second quantized approach which gives the rules for any number of loops. But using unitarity, one can in principle reconstruct how the rules for different number of loops are related.

This is the basis of the whole Bern-Kosower string inspired approach to n-gluon amplitude calculations.



My main point is that one could do QED, phi^4, etc calculations using integrals over paths (not over fields).

 

[lightarrow]

-mc^2\sqrt{1 - v^2/c^2}~~~~~~~~~~~~~~~~~ : is the relativistic Lagrangian of a free particle

-mc^2\sqrt{1 - v^2/c^2} ~-~ V(q)~~~~: is the Lagrangian in a potential field V


The kinetic term is this one:

T ~=~ \frac{1}{2}pv ~=~ \frac{1}{2}\frac{mv^2}{\sqrt{1-\frac{v^2}{c^2}}} ~\approx ~ \frac{1}{2}mv^2<br />

Thanks, it's clear now.