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When studying path-integrals one question arose to my mind... Which presentation is more fundamental to calculate the propagator?

The one based on the Hamiltonian (phase space)?

$$K(B|A) = \int \mathcal{D}[p]\mathcal{D}[q] \exp \{ \frac{i}{\hbar} \int dt [ p \dot q - H(p,q) ] \} $$

or the one based on the lagrangian (configurational space)?

$$K(B|A) = \int \mathcal{D}[q] \exp \{ \frac{i}{\hbar} \int dt L \} $$

Reading Feynman thesis we see he affirming that "[...] a method of formulating a quantum analog of systems for which no Hamiltonian, but rather a principle of least action, exists has been worked out. It is a description of this method which constitutes this thesis." He seems to take lagrangian form as more fundamental.

Other authors, like Hatfield or Swanson, seems to take the phase space form as more fundamental. They see the other form as a special case where the $p$ dependence is quadratic.

So, this is my question.
Which one is more trustful? There is any example where one view is privileged?

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    $\begingroup$ The hamiltonian formulation is more fundamental. However, with a simple hamiltonian like $H = \frac{p^2}{2m}+ V(q)$, if you make the integration on $p$, you will easily find the Lagrangian formulation $\endgroup$
    – Trimok
    Commented Oct 19, 2013 at 9:23
  • $\begingroup$ Why hamiltonian is more fundamental? There is some example where following lagragian we find the wrong answer? And what about Feynman thesis? Where he seems to describe an example with no-hamiltonian form? $\endgroup$
    – Erich
    Commented Oct 19, 2013 at 12:13
  • $\begingroup$ Already at a classical level, you will have problem with hamiltonians such as $H =\frac{p^2}{2m} + \lambda pq$. Hamilton equations will give $\dot p/m + \lambda p=0$. This kind of relation cannot be obtained from Euler-Lagrange equations. $\endgroup$
    – Trimok
    Commented Oct 19, 2013 at 17:06
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    $\begingroup$ @Trimok cannot? Sorry if i'm misleading something... but the particular hamiltonian you show seems to be well behaved in a legendre transform... the momenta is $p = m(\dot q - \lambda q)$ (no constraints here) and the lagrangian is simply $L = \frac{m}{2} {\dot q}^2 + \frac{m}{2} \lambda^2 q^2 - \lambda m q \dot q$. Euler-lagrange gives us $\ddot q - \lambda^2 q = 0$ which is exactly the same thing obtained by the canonical equations (if we substitute $p$). $\endgroup$
    – Erich
    Commented Oct 19, 2013 at 20:49
  • $\begingroup$ You are right... : I checked also that, for this particular hamiltonian, the two formulations are equivalent at the classical level. Now, the canonical momentum $p = m (\dot q - \lambda q)$ is, physically, very "special".. $\endgroup$
    – Trimok
    Commented Oct 20, 2013 at 13:22

1 Answer 1

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Comments to the question (v2):

1) The correspondence between Lagrangian (L) and Hamiltonian (H) theories is mired with subtleties. Some general tools for singular Legendre transformations are available, such as Dirac-Bergmann analysis, Faddeev-Jackiw method, etc. But rather than claiming complete understanding and existence of the L-H correspondence, it is probably more fair to say that we have a long list of theories (such as e.g. Yang-mills, Cherns-Simons, GR, etc.), where both sides of the L-H correspondence have been worked out.

2) In general path integrals are poorly understood beyond a perturbative expansion around a Gaussian free theory, so to ponder what happens if the momenta $p$ are not quadratic, is just part of a bigger problem.

3) A fundamental difference between Lagrangian and Hamiltonian theories is that there formally exists a canonical choice of path integral measure in Hamiltonian theories, while the Lagrangian path integral measure traditionally is only fixed modulo gauge-invariant factors. In that sense the Hamiltonian formulation is more fundamental.

In detail, if we assume that the phase space a Hamiltonian theory is equipped with a symplectic two-form

$$\tag{1} \omega~=~\frac{1}{2} dz^I ~\omega_{IJ} \wedge dz^J,$$

there is a canonical measure factor

$$\tag{2} \rho~=~ {\rm Pf}(\omega_{IJ})$$

given by the (super) Pfaffian, at least for finite-dimensional integrals, which under favorable circumstances can be generalized to infinite dimensions. This measure factor $\rho$ is just 1 in Darboux coordinates $(q^1, \ldots, q^n, p_1, \ldots , p_n)$ with $\omega = dp_i \wedge dq^i$.

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  • $\begingroup$ Concerning the Lagrangian-Hamiltonian, what you said about measure made things more clear to me. Anyway, in analysing differences of the approaches it would be nice to compute other path integrals... About non-gaussian path integrals... do you know about any effort in computing cases where the momenta are not quadratic? Or there is some fundamental prohibition? $\endgroup$
    – Erich
    Commented Oct 19, 2013 at 16:31

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