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A relativistic free particle has the Hamiltonian in general:

$$ \mathcal{H} = \sqrt{{\bf p}^2c^2+m^2c^4}.$$

I read somewhere that says, it is possible to go further and say that the EoM are Hamilton's equations. But it is not done as there is "less interest" in such a discussion.

Is there something deeper to this? Like another formalism is ''better''.

(My guess is a more trivial one though. That is, it is not useful because the equations just get very cluttered and ugly)

Qmechanic
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Candy Man
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    I'm...not sure what your question is. The e.o.m exist, but they get quite ugly due to the square root. We usually introduce the einbein to get a more convenient Hamiltonian formalism. What is the question about that? – ACuriousMind Apr 20 '16 at 17:54
  • Related: https://physics.stackexchange.com/q/194877/2451 and links therein. – Qmechanic Sep 03 '16 at 21:07
  • You can construct Hamiltonian which produces correct EOMs. For example consider $H(p,x)=g^{\mu \nu}(x) p_{\mu} p_{\nu}$. There are no square roots, but you lose reparametrization invariance. – Blazej Sep 17 '16 at 23:37
  • See this: https://physics.stackexchange.com/a/694087/226902 – Quillo Mar 20 '23 at 18:31

2 Answers2

6

I) Here we will assume that OP is talking about a relativistic point particle with zero spin in a $d$-dimensional Minkowski spacetime with metric $\eta_{\mu\nu}$ of sign convention $(−,+,\ldots,+)$. Also we put $c=1$ for simplicity.

Note that the relativistic point particle has world-line reparametrization invariance, which is a gauge symmetry/redundancy in the formulation. We are (to a large extent) free to parametrize the world-line of the point particle in any way we wish. Let us call the world-line parameter for $\tau$ (which does not have to be the proper time). This gauge freedom can be encoded in an einbein field $e=e(\tau)>0$. The resulting Hamiltonian Lagrangian is$^1$

$$ L_H~:=~ p_{\mu} \dot{x}^{\mu} - \underbrace{\frac{e}{2}(p^2+m^2)}_{\text{Hamiltonian}}, \tag{1} $$

cf. e.g. this Phys.SE post. Here dot means differentiation wrt. $\tau$. The square of the momentum vector is $$\begin{align} p^2~:=~& \eta^{\mu\nu} p_{\mu} p_{\nu}\cr ~=~&-(p^0)^2+{\bf p}^2\cr ~=~&-2p^+p^- + {\bf p}_{\perp}^2, \end{align}\tag{2}$$ where we have used light-cone coordinates in the last expression.

II) Static gauge $x^0=\tau$. If we integrate out $p^0$ and $e$, we get OP's square root model $$\begin{align} \left. L_H\right|_{x^0=\tau} \quad\stackrel{p^0}{\longrightarrow}&\quad {\bf p}\cdot \dot{\bf x}- \underbrace{\left(\frac{1}{2e} + \frac{e}{2}({\bf p}^2+m^2)\right)}_{\text{Hamiltonian}}\cr\cr \quad\stackrel{e}{\longrightarrow}&\quad {\bf p}\cdot \dot{\bf x} - \underbrace{\sqrt{{\bf p}^2+m^2}}_{\text{Hamiltonian}} .\end{align}\tag{3} $$

For sufficiently short$^2$ times $\Delta \tau=\tau_f-\tau_i$, the path integral becomes$^3$

$$\begin{align}& \langle {\bf x}_f,\tau_f \mid {\bf x}_i,\tau_i\rangle\cr ~=~&i\hbar\Delta\tau\int_{\mathbb{R_+}} \!\frac{\mathrm{d}e}{2} \int_{\mathbb{R}^d} \!\frac{\mathrm{d}^dp}{(2\pi\hbar)^d} \exp\left[\frac{i}{\hbar}\left( p_{\mu} \Delta x^{\mu} -\underbrace{\frac{e}{2}(p^2+m^2)}_{\text{Hamiltonian}}\Delta\tau\right)\right]\cr ~=~& \int_{\mathbb{R}^{d-1}} \!\frac{\mathrm{d}^{d-1}{\bf p}}{(2\pi\hbar)^{d-1}} i\hbar\Delta\tau\int_{\mathbb{R_+}} \!\frac{\mathrm{d}e}{2} ~\underbrace{\frac{1}{\sqrt{2\pi\hbar ie\Delta\tau}}}_{\text{Gauss. } p^0\text{-int.}} \cr &\exp\left[\frac{i}{\hbar}\left( {\bf p}\cdot \Delta {\bf x} -\underbrace{\left( \frac{1}{2e} + \frac{e}{2}({\bf p}^2+m^2)\right)}_{\text{Hamiltonian}}\Delta\tau\right) \right]\cr ~\stackrel{(6)}{=}~& \int_{\mathbb{R}^{d-1}} \!\frac{\mathrm{d}^{d-1}{\bf p}}{(2\pi\hbar)^{d-1}} \frac{\hbar}{2\sqrt{{\bf p}^2+m^2}} \exp\left[\frac{i}{\hbar}\left( {\bf p}\cdot \Delta {\bf x} - \Delta \tau \underbrace{\sqrt{{\bf p}^2+m^2}}_{\text{Hamiltonian}}\right)\right]\cr ~=~&i\hbar\Delta\tau\int_{\mathbb{R_+}} \!\frac{\mathrm{d}e}{2} ~\underbrace{\frac{1}{(2\pi\hbar ie\Delta\tau)^{d/2}}}_{\text{Gauss. } p\text{-int.}} \exp\left[\frac{i}{2\hbar}\left( \frac{(\Delta x)^2}{e\Delta\tau} - m^2e\Delta\tau\right) \right]\cr ~\stackrel{(6)}{=}~&\frac{1}{(2\pi)^{d/2}}\Big(\frac{m/\hbar}{ \sqrt{(\Delta x)^2}}\Big)^{\frac{d}{2}-1}K_{\frac{d}{2}-1}\Big(\frac{m}{\hbar}\sqrt{(\Delta x)^2}\Big) ,\end{align} \tag{4} $$ which also happens to be the standard scalar propagator $\langle\Omega|T[\phi (x_f)\phi (x_i)]|\Omega\rangle $ in QFT/2nd quantization, cf. e.g. Refs. 1-3. From a 2nd quantized perspective, the $e$-integration in eq. (4) is a Schwinger parametrization of the Fourier transformed propagator $$\frac{i}{\hbar}\langle\Omega|T[\widetilde{\phi} (p_f)\widetilde{\phi} (p_i)]|\Omega\rangle~=~\frac{\hbar^2}{p_f^2+m^2-i\epsilon}(2\pi\hbar)^d\delta^d(p_f\!+\!p_i). \tag{5} $$ As is well-known, eq. (4) is Lorentz covariant and falls off exponentially outside the light-cone. In eq. (4) we have used the integrals $$\begin{align} \int_{\mathbb{R}_+} \!\frac{\mathrm{d}e}{e^{1+\nu}}\exp\left[-ae-\frac{b}{e}\right] ~=~&2\left(\frac{a}{b}\right)^{\nu/2} K_{\nu}\left(2\sqrt{ab}\right),\cr \int_{\mathbb{R}_+} \!\frac{\mathrm{d}e}{e^{1-\nu}}\exp\left[-ae-\frac{b}{e}\right] ~=~&2\left(\frac{b}{a}\right)^{\nu/2} K_{\nu}\left(2\sqrt{ab}\right),\cr \int_{\mathbb{R}_+} \!\frac{\mathrm{d}e}{\sqrt{e}}\exp\left[-ae-\frac{b}{e}\right] ~=~&\sqrt{\frac{\pi}{a}} \exp\left[-2\sqrt{ab}\right],\cr {\rm Re}(a), {\rm Re}(b)~>~&0.\end{align}\tag{6} $$

III) Light-cone gauge $x^+=\tau$. If we integrate out $p^-$ and $e$, we get
$$\begin{align} \left. L_H\right|_{x^+=\tau} \quad\stackrel{p^-,~e}{\longrightarrow}\quad & -p^+\cdot \dot{x}^- +{\bf p}_{\perp}\cdot \dot{\bf x}_{\perp}\cr & - \underbrace{\frac{{\bf p}_{\perp}^2+m^2}{2p^+}}_{\text{Hamiltonian}} .\end{align}\tag{7} $$

IV) We stress that the Euler-Lagrange (EL) equations for either of the Hamiltonian Lagrangians (1), (3), and (7) lead to Hamilton's equations. The point is now that physical quantities should not depend on the choice of gauge-fixing. We are free to use the most convenient gauge choice. Each formulation (1), (3), and (7) are valid, and have their pros and cons. The static gauge choice (3) is disfavored because of the square root.

References:

  1. M.E. Peskin & D.V. Schroeder, An Intro to QFT; eq. (2.50).

  2. M.D. Schwartz, QFT and the Standard Model; eq. (6.25).

  3. O. Corradini & C. Schubert, Spinning Particles in QM & QFT, arXiv:1512.08694; subsection 1.5.1, eqs. (1.160-162)

  4. T. Padmanabhan, QFT: The Why, What and How, 2016; subsections 1.3.1 + 1.4.4.

--

$^1$ Strictly speaking, there are also Faddeev-Popov ghost terms and gauge-fixing terms, which we have ignored for simplicity. These action terms are consistently generated in the BFV formulation, cf. e.g. my Phys.SE post here. The normalization factor in eq. (4) can be derived via Gaussian integration in the BFV formulation over the 2 bosonic variables $x^0$, $B$; and the 4 fermionic variables $\bar{C}$, $P$, $C$, $\bar{P}$.

$^2$ Here we just consider a single time slice for simplicity. The full path integral is the continuum limit of multiple time slice discretizations with insertion of corresponding completeness relations. It turns out that the result (4) for the free theory does not depend on the number of time slice discretizations.

$^3$ Here we use the Feynman $i\epsilon$-prescription ${\rm Re}(i\Delta\tau)>0$. The Gaussian integration over $p^0_E=i p^0_M$ becomes damped after a Wick-rotation $\tau_E=i\tau_M$, $x^0_E=ix^0_M$ to Euclidean signature.

Qmechanic
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  • Notes for later: Observation: We Wick rotate in target space $p^0_E=i p^0_M$, while ${\bf p}$ is not Wick rotated, cf. e.g. https://physics.stackexchange.com/q/275918/2451 , https://physics.stackexchange.com/q/313921/2451 – Qmechanic Sep 17 '16 at 18:10
  • Notes for later: $\quad{\rm Re}(i\Delta \tau)>0$; $\quad \int_{\mathbb{R}+} !\frac{\mathrm{d}e}{e^{1-\nu}}\exp\left[-ae-\frac{b}{e}\right] ~=~2\left(\frac{b}{a}\right)^{\nu/2} K{\nu}\left(2\sqrt{ab}\right) $; $\quad \int_{\mathbb{R}+} !\frac{\mathrm{d}e}{e^{1+\nu}}\exp\left[-ae-\frac{b}{e}\right] ~=~2\left(\frac{a}{b}\right)^{\nu/2} K{\nu}\left(2\sqrt{ab}\right) $; $\qquad {\rm Re}(a), {\rm Re}(b)~>~0 $; Integral (6) is a modified Bessel function $K_{1/2}$. – Qmechanic Jul 09 '19 at 16:25
  • Notes for later: Here in point mechanics the momentum is in target space. In FT it is of interest to Fourier transform in the world volume. E.g. the Klein-Gordon (KG) Hamiltonian density ${\cal H} =\frac{1}{2}\left( \pi^2+(\nabla\phi)^2 +m^2\phi^2\right)$ gets its square root from a Lorentz covariant square root normalization of creation & annihilation operators. – Qmechanic Dec 14 '20 at 14:24
  • Notes for later: The KG Lagrangian density ${\cal L}=\frac{1}{2}\left(\pm(\partial\phi)^2 -m^2\phi^2\right)$ has Fourier transform $\widetilde{\cal L}=\frac{1}{2}\left(\pm k^2 -m^2\right)\widetilde{\phi}(k)\widetilde{\phi}(-k)$. From the on-shell EL eq. $\left(\pm k^2 -m^2\right)\widetilde{\phi}(k)\approx 0$ we get the dispersion relation $\pm k^2 \approx m^2$. – Qmechanic Dec 14 '20 at 14:39
  • Notes for later: Greens function for Helmholtz operator $(-\nabla^2+m^2) G({\bf r}) = \delta^d({\bf r})$ in $d$ dimensions (Euclidean signature). – Qmechanic Oct 07 '21 at 07:35
  • $G({\bf r}) =\int_{\mathbb{R}^d}!\frac{d^dk}{(2\pi)^d} \frac{e^{i{\bf k}\cdot {\bf r}}}{k^2+m^2}$ $=\int_{\mathbb{R}^d}! \frac{d^dk}{(2\pi)^d} e^{i {\bf k}\cdot {\bf r}} \int_{\mathbb{R}+}!\frac{de}{2}~\exp\left[-\frac{e}{2}(k^2+m^2)\right]$ $=\int{\mathbb{R}+}!\frac{de}{2}~\frac{1}{(2\pi e)^{d/2}}~e^{-S(e)}$ $=\frac{1}{(2\pi)^{d/2}}\Big(\frac{m}{r}\Big)^{\frac{d}{2}-1}K{\frac{d}{2}-1}(mr)$ $\longrightarrow \frac{r^{2-d}}{(d-2){\rm Vol}(\mathbb{S}^{d-1})}$ $=\frac{\Gamma(d/2-1)}{4\pi^{d/2}}r^{2-d}$ for $m\to 0^+$. Here ${\rm Vol}(\mathbb{S}^{d-1}) =2\frac{\pi^{d/2}}{\Gamma(d/2)}$. – Qmechanic Oct 07 '21 at 10:23
  • Here the action $S(e)=\frac{em^2}{2}+\frac{r^2}{2e}\approx mr$. Stationary point: $0\approx S^{\prime}(e)=\frac{m^2}{2}-\frac{r^2}{2e^2}$, so that $e\approx r/m$. And $S^{\prime\prime}(e)=\frac{r^2}{e^3}\approx \frac{m^3}{r}$. Steepest descent method yields $G({\bf r}) \sim \frac{1}{2}~\frac{1}{(2\pi e)^{d/2}}~\sqrt{\frac{2\pi}{S^{\prime\prime}(e)}}e^{-S(e)} = \frac{1}{(2\pi)^{d/2}}\Big(\frac{m}{r} \Big)^{\frac{d}{2}-1} \sqrt{\frac{\pi}{2mr}}e^{-mr}$ for $mr\gg 1$. Idea: Substitution $e=\exp(u)$ changes integration region to whole $\mathbb{R}$. – Qmechanic Oct 07 '21 at 21:46
  • Wilsonian effective action is the generator of connected diagrams: $\exp\left{-\frac{1}{\hbar}W_c[J^H,\phi_L] \right} := \int ! {\cal D}\frac{\phi_H}{\sqrt{\hbar}}~\exp\left{ \frac{1}{\hbar} \left(-S[\phi_L+\phi_H]+J^H_k \phi_H^k\right)\right}$, where $\sqrt{k_L^2+m^2} < \Lambda \equiv \Lambda_L < \sqrt{k_H^2+m^2}< \Lambda_H$. Note that the cut-off is species dependent $m\ll\Lambda$. Non-$x$-local action terms are exponentially suppressed. Hm. The mass $m$ might run with $\Lambda$. https://physics.stackexchange.com/q/602474/2451 https://physics.stackexchange.com/q/254260/2451 – Qmechanic Oct 08 '21 at 20:59
  • Low/light action: $S_L[\phi_L]=S_{L,2}[\phi_L]+\ldots$, where $S_{L,2}[\phi_L] =\frac{1}{2}\int_{\mathbb{R}^d}!d^dr~\phi_L({\bf r})(-\nabla^2+m^2)\exp\left[\frac{-\nabla^2+m^2}{2\Lambda^2}\right]\phi_L({\bf r})$ $=\frac{1}{2}\int_{\mathbb{R}^d}!\frac{d^dk}{(2\pi)^d}~\widetilde{\phi}L(-{\bf k})(k^2+m^2)\underbrace{\exp\left[\frac{k^2+m^2}{2\Lambda^2}\right]}{=1/K}\widetilde{\phi}L({\bf k})$. High/heavy action: $S_H[\phi_H]=S{H,2}[\phi_H]+\ldots$, where – Qmechanic Oct 08 '21 at 21:50
  • $S_{H,2}[\phi_H]=\frac{1}{2}\int_{\mathbb{R}^d}!\frac{d^dk}{(2\pi)^d}~\widetilde{\phi}H(-{\bf k})(k^2+m^2) \frac{1}{1-K}\widetilde{\phi}_H({\bf k})$. Low/light modes: $G_L({\bf r}) =\int{\mathbb{R}^d}!\frac{d^dk}{(2\pi)^d}\frac{e^{i{\bf k}\cdot{\bf r}}}{k^2+m^2}\underbrace{\exp\left[-\frac{k^2+m^2}{2\Lambda^2}\right]}{=K}$ $=\int{\mathbb{R}^d}!\frac{d^dk}{(2\pi)^d}e^{i{\bf k}\cdot{\bf r}}\int_{\mathbb{R}+}!\frac{de}{2}~ \exp\left[-\frac{1}{2}(e+\Lambda^{-2})(k^2+m^2)\right]$ $=\int{\Lambda^{-2}}^{\infty}!\frac{de}{2}~\frac{1}{(2\pi e)^{d/2}}~e^{-S(e)}$. – Qmechanic Oct 09 '21 at 12:05
  • High/heavy modes: $G_H({\bf r})=G({\bf r})-G_L({\bf r})$ $=\int_0^{\Lambda^{-2}}!\frac{de}{2}~\frac{1}{(2\pi e)^{d/2}}~e^{-S(e)}$ $=\Lambda^{d-2}\int_0^1!\frac{de}{2}~\frac{1}{(2\pi e)^{d/2}}~e^{-S(\Lambda^{-2}e)}$ $=\frac{r^{2-d}}{(2\pi )^{d/2}}\int_{(r\Lambda)^2}^{\infty}!\frac{du}{2}~u^{d/2-2}~e^{-S(r^2/u)}$ $=\frac{r^{2-d}}{(2\pi )^{d/2}}\int_{(r\Lambda)^2}^{\infty}! \frac{du}{2}~\underbrace{\frac{u^{d/2-2}}{1-(mr/u)^2}}{\text{diff}}~\underbrace{(1-(mr/u)^2)e^{-S(r^2/u)}}{\text{int}}$. High/heavy action is not natural. – Qmechanic Oct 10 '21 at 19:59
  • Case $\frac{m}{\Lambda}>r\Lambda\gg 1 \Rightarrow \Lambda^{-2}> r/m$, i.e. stationary point is inside integration interval. Then $m\gg \Lambda \gg r^{-1}$ dominates $\Lambda$: $\quad G_H({\bf r}) \approx G({\bf r})$ for $mr\gg 1$. 2. Case $ r\Lambda > \frac{m}{\Lambda} \gg 1 \Rightarrow \Lambda^{-2}< r/m$, i.e. stationary point is outside integration interval. $\quad G_H({\bf r}) \stackrel{\text{IBP}}{\sim}\frac{\Lambda^{d-2}}{(2\pi)^{d/2}((r\Lambda)^2-(m/\Lambda)^2)}\exp\left[-\frac{m^2}{2\Lambda^2} -\frac{(r\Lambda)^2}{2}\right]$ for $r\Lambda\gg 1$.
    • – Qmechanic Oct 10 '21 at 20:39
  • Case $\frac{m}{\Lambda}\ll r\Lambda$: $\quad G_H({\bf r})\approx\int_0^{\Lambda^{-2}}!\frac{de}{2}~\frac{1}{(2\pi e)^{d/2}}~\exp\left[-\frac{r^2}{2e}\right]$ $=\int_0^{2\Lambda^{-2}}!\frac{de}{4}~\frac{1}{(\pi e)^{d/2}}~\exp\left[-\frac{r^2}{e}\right]$ $=\int_{\Lambda^2/2}^{\infty} !\frac{du}{4\pi^{d/2}}~u^{d/2-2}~e^{-ur^2}$ $=\frac{r^{2-d}}{4\pi^{d/2}}\int_{(r\Lambda)^2/2}^{\infty}!du~\underbrace{u^{d/2-2}}{\text{diff}}~\underbrace{e^{-u}}{\text{int}}$
    • – Qmechanic Oct 13 '21 at 18:48
  • $=\frac{r^{2-d}}{4\pi^{d/2}}\Gamma\Big(\frac{d}{2}-1,\frac{(r\Lambda)^2}{2}\Big)$ $\stackrel{\text{IBP}}{\sim}\frac{\Lambda^{d-4}}{(2\pi)^{d/2}r^2}\exp\left[-\frac{(r\Lambda)^2}{2}\right]$ for $r\Lambda\gg 1$. – Qmechanic Oct 22 '21 at 19:33
  • Hi @Qmechanic, there is a thing I would like to ask you. Are these subjects part of your teaching activity? Research activity? Both? I am not able to focus on your research field. You seem to me a theoretical physicists but sometimes you come out with issues which are, at least in my country, more appropriate for mathematical physics. – Valter Moretti Aug 27 '22 at 14:25
  • Hi @Valter Moretti. Some but not all subjects are related to research, teaching & supervision. Formally, I have a Ph.D. in theoretical physics, but with some courses in mathematical physics. – Qmechanic Aug 27 '22 at 14:57
  • Thanks @Qmechanics,... you are a hybrid like me. – Valter Moretti Aug 27 '22 at 15:26