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Let me look at the Hamiltonian of a charged particle in a plane in a constant magnetic field ($\vec{B}$) pointing upwards - then in usual notation it is,

$$\hat{H} = \frac{1}{2m}\biggl(\hat{p} + \frac{e}{c}\hat{A}(\hat{r})\biggr)^2$$

To convert this in a Feynman-path-integral language, I pick say a gauge $\vec{A}=(-\frac{B}{2}y,\frac{B}{2}x)$and then in this gauge $\hat{p}$ and $\hat{A}$ commute and that makes rewriting in the path-integral language much easier.

If I put this through the usual process of "deriving" a Feynman path-integral then I would get the expression,

$$\int \bigl[\mathcal{D}\vec{r}\bigr]\bigl[\mathcal{D}\vec{p}\bigr] \exp\biggl[i \int dt \biggl(\vec{p}\cdot\dot{\vec{r}} - \frac{1}{2m}\Bigl(\vec{p}+\frac{e}{c}\vec{A}\Bigr)^2\biggr)\biggr]$$

  • Is it obvious (or true ) that the above expression is independent of the gauge I chose to the calculation? Is there a way to write the system in the path-integral language without explicitly choosing a gauge? (I have worked through calculations in Yang-Mills theory in the Fadeev-Popov gauge which does exactly that in those cases but I can't see a way out here...)

  • Can't I have written down the above path-integral without even going through the usual process of finding infinitesimal transition amplitudes and then collecting them together? I mean how often is it safe to say that for a Hamiltonian $H(p,q)$ the path integral representation of the transition amplitude will be $\int [\mathcal{D}p][\mathcal{D}q]e^{i\int dt (p\dot{q}-H(q,p)) }$?

Now the Heisenberg's equation of motion will tell that,

$$\frac{d\vec{r}}{dt} = \frac{1}{m} \biggl(\vec{p} + \frac{e}{c}\vec{A}\biggr)^2$$

In the Feynman path-integral the position and the momentum vectors are treated to be independent variables and hence it would be wrong to substitute the above expression into the path-integral but outside one can I guess do this substitution and one would get for the action (whatever sits in the exponent),

$$S = \int dt \biggl(\frac{p^2}{2m} - \frac{e^2}{2mc^2}A^2\biggr)$$

  • But the above expression doesn't look right!? The integrand isn't what the Lagrangian should be. right?

Now in the derivation of the Feynman-path-integral if one integrates out the momentum for every infinitesimal transition amplitude and then reconstitutes the path-integral then one would get the expression,

$$\int [\mathcal{D}\vec{r}] \exp\biggl[i\int dt \biggl(\frac{m}{2}\dot{\vec{r}}^2 - \frac{e}{c}\dot{\vec{r}}\cdot\vec{A}\biggr)\biggr]$$

  • Now what seems to sit in the exponent is what is the "correct" Lagrangian - I would think. Why did the answer differ in the two different ways of looking at it?

  • I wonder if given a Hamiltonian its corresponding Lagrangian can be "defined" as whatever pops out in the exponent if that system is put through this Feynman re-writing. In this case keeping to just classical physics I am not sure how to argue that the $\frac{m}{2} \dot{\vec{r}}^2 - \frac{e}{c} \dot{\vec{r}}\cdot\vec{A}$ is the Lagrangian for the system with the Hamiltonian $\frac{1}{2m}\bigl(\vec{p} + \frac{e}{c}\vec{A}(\vec{r})\bigr)^2$

I think I have seen examples on curved space-time where the "classical" Lagrangian differs from what pops out in the exponent when the system is path-integrated.

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