# Are these two square root & non-square root worldline actions equivalent at quantum level? (Kleinert's method)

The relativistic particle action is (in its more natural form)

$$A=McS=Mc\int_{\lambda_a}^{\lambda_b}d\lambda\sqrt{x'^2(\lambda)}.\tag{19.12}$$

That action doesn't lend itself easily to a calculation of the corresponding path integral, so (as done in many papers and textbooks) we can use the alternative form that is more suitable since it is quadratic:

$$\bar{A}=\int_{\lambda_a}^{\lambda_b}d\lambda \left[\frac{Mc}{2h(\lambda)}x'^2(\lambda)+h(\lambda)\frac{Mc}{2}\right].\tag{19.10}$$

Although the $\bar{A}$ describes the same classical physics as the original action (19.12), I wonder that it may lead to a different quantum physics. The textbooks give some arguments for equivalence but they don't actually "calculate" the integral, which seems imposible due to the square root. Kleinert's book1 "Path Integrals in Quantum Mechanics, Statistics, Polymer Physics, and ..." is the only one I have found (if you know another reference dealing with the equivalence issue please tell me), page 1424, Appendix 19A. It seems that he got

$$(x_b|x_a)=N''\int\frac{d^D k}{(2\pi)^D}\frac{1}{k^2+M_Rc^2/\hbar^2}e^{ik}(x_b-x_a)\tag{19A.15b}$$

with

$$z \equiv \nu\log\nu , \qquad \nu\equiv \frac{D+1}{\bar{\epsilon}}\lambda_C, \qquad \bar{\epsilon} = \frac{S}{N+1},\tag{19A.13}$$ $$M_R=M(1+z)^{1/2} .\tag{19A.16}$$ So, in principle, they are different, similar but different.

So, my questions are:

1. What is happening here? I mean with the mass, it looks like regularization or renormalization.

2. Is it necessary to calculate both path integrals in order to show they are equivalent?

3. Has the path integral been calculated correctly (by Kleinert) without cheating nor using any hocus-pocus process? Or what has he done?

1 If you don't have he book, please read this draft http://users.physik.fu-berlin.de/~kleinert/b5/psfiles/pthic19.pdf, page 1436.

• Non-Gaussian path integrals are hocus-pocus by definition, or more precisely, by absence of definition :) You might find interesting Polyakov's treatment of this in his "Gauge fields and strings", chapter on random surfaces. He shows that there's a hocus-pocus which kinda justifies saying that any path integral with a field which enters the Lagrangian without derivatives is equivalent to the one for the reduced action. – Prof. Legolasov Aug 19 '17 at 5:03
• @SolenodonParadoxus what do you mean by the reduced action? and how can a lagrangia have no derivatives? – Pathy Aug 19 '17 at 5:41
• in your question the Lagrangian has no derivatives of $h$ in it. Reduced action is what you end up with when you solve the algebraic (classical) equation of motion for the field $h$ (or generally, the field which enters without derivatives) and plug this back in the action. For the relativistic particle this gives the square-root action. – Prof. Legolasov Aug 19 '17 at 8:21
• @SolenodonParadoxus yes off course, h is just like a lagrange multiplier, it is not a true degree of freedom. – Pathy Aug 19 '17 at 14:51
• Which is exactly my point. So Polyakov's handwaving is a more general claim, a special case of which is that your two actions give the same quantum dynamics. – Prof. Legolasov Aug 20 '17 at 8:42

1. Let us start by summarizing Kleinert's results for a relativistic massive point$^1$ particle:

• On one hand, Kleinert finds no mass-renormalization (19.30) for the Hamiltonian phase space path integral (19.16) with a non-square root action (19.10).
• On the other hand, Kleinert finds mass-renormalization (19A.15b) with for the Lagrangian path integral (19A.2) with a square root action (19.12).
2. Note that there doesn't seem to be a truly independent definition of the path integral (19A.2) with a square root action (19.12). Even Kleinert starts his derivation by immediately going to a non-square root integral representation (19A.3).

3. The reason why the results (19.30) & (19A15b) are different is not an issue of square root vs. non-square root action per se. Rather it is an issue of the Lagrangian vs. the Hamiltonian path integral formulation. There is no canonical choice of path integral measure within the Lagrangian formulation. (Recall e.g. the famous Feynman fudge factor.) The Hamiltonian formulation is more fundamental in this respect, cf. e.g. this Phys.SE post.

In the Lagrangian path integral (19A.2), Kleinert (from the onset) omits the pertinent path integral measure factor (which can only be deduced in the Hamiltonian formulation). Hence there is no reason the results should agree.

In turn, omitting the correct path integral measure factor introduces renormalization, even in point mechanics.

4. If one performs the Dirac-Bergmann analysis of both the Lagrangian square root action (19.12) and non-square root action (19.10) (which both possesses a world-line (WL) reparametrization gauge symmetry), and accounts for the pertinent first-class constraints, etc, one eventually ends up with one and the same Hamiltonian action formulation, cf. e.g. this, this & this Phys.SE posts and links therein.

In conclusion, since the Lagrangian square root action (19.12) and non-square root action (19.10) have identical Hamiltonian action, then any path integral quantization scheme (that is consistent with the Hamiltonian formulation) must be identical as well.

References:

1. H. Kleinert, Path Integrals in Quantum Mechanics, Statistics, Polymer Physics, and Financial Markets, 4th edition; Chapter 19 (pdf).

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$^1$ For the analogous Polyakov vs. Nambu-Goto question in string theory, see e.g. this Phys.SE post.