If we have a Lagrangian density $\mathcal{L}$ for a scalar field $\phi$ depending on $\phi$, $\partial _{\mu} \phi$, and $\partial _{\mu} \partial _{\nu} \phi$, what is the Hamiltonian? Additionally, what do Hamilton's equations become?
120654
This is a special case of , which, I think, might have drawn little attention because it was too general. In this question, I would like to consider a specific case.  Take a classical system given by the action $$ S[x] = \int dt \left[ \frac{\alpha}{2} \ddot{x}^2 + \frac{\beta}{2} \dot{x}^2 + \frac{\gamma}{2} x^2 \right]. $$  This action is Lagrangian, but it is not what we are usually dealing with in physics, because the Lagrangian contains second time derivatives.  The equations of motion are:  $$ \alpha \ddddot{x} - \beta \ddot{x} + \gamma x = 0. $$  This can be solved as usual by employing Fourier transform:  $$ x(t) = a_1 e^{i \omega_1 t} + a_1^{*} e^{-i \omega_1 t} + a_2 e^{i \omega_2 t} + a_2^{*} e^{-i \omega_2 t}, $$  where $\omega_1^2$ and $\omega_2^2$ are the two roots of $\alpha \omega^4 + \beta \omega^2 + \gamma$, given by  $$ \omega_{1,2}^2 = \frac{- \beta \pm \sqrt{\beta^2 - 4 \alpha \gamma}}{2 \alpha} .$$  Now by definition the phase space is the space of solutions of equations of motion. In this case it is parametrized by two complex co-ordinates $a_{1,2}$, meaning that it is 4-dimensional.  I would like to know if there's a natural way to associate the symplectic structure (Poisson bracket) on this phase space. I.e.,  $$ \{a_1, a_1^{*}\} = ? $$ $$ \{a_1, a_2^{*}\} = ? $$ $$ \{a_2, a_1^{*}\} = ? $$ $$ \{a_2, a_2^{*}\} = ? $$ $$ \{a_1, a_2\} = ? $$ $$ \{a_1^{*}, a_2^{*}\} = ? $$
99563
I'm doing some classical field theory exercises with the Lagrangian $$\mathscr{L} = -\frac{1}{4}F_{\mu \nu}F^{\mu \nu}$$ where $F_{\mu \nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$. To find the conjugate momenta $\pi^\mu_{\ \ \ \nu} = \partial \mathscr{L} / \partial(\partial_\mu A^\nu)$, I can use two methods.  First method: directly apply this to $\mathscr{L}$. We get a a factor of $2$ since there are two $F$'s, and another factor of $2$ since each $F$ contains two $\partial_\mu A_\nu$ terms, giving $$\pi^\mu_{\ \ \ \nu} = -F^\mu_{\ \ \ \nu}.$$  Second method: get $\mathscr{L}$ in terms of $A$ by expanding and integrating by parts, yielding $$\mathscr{L} = \frac{1}{2}(\partial_\mu A^\mu)^2 - \frac{1}{2}(\partial_\mu A^\nu)^2.$$ Differentiating this gets factors of $2$ and gives $$\pi^\mu_{\ \ \ \nu} = \partial_\rho A^\rho \delta^\mu_\nu - \partial^\mu A_\nu.$$  These two answers are different! (They do give the same equations of motion, at least.) I guess that means doing the integration by parts changed the canonical momenta.  Is this something I should be worried about? In particular, I have another exercise that wants me to show that one of the canonical momenta vanishes -- this isn't true for the ones I get from the second method! Plus, my stress-energy tensor is changed too. When a problem asks for "the" canonical momenta, am I forbidden from integrating by parts?
14757
of a particle moving in the presence of a scalar field $\varphi (x)$ is given by $$ S = -m\int e^{\varphi (x)}\sqrt{\eta_{\alpha \beta}\frac{dx^{\alpha}}{d \lambda}\frac{dx^{\beta}}{d \lambda}}d\lambda , $$ where $\lambda$ is the parametrization of the worldline of the particle, ignoring the free field part $\int \eta_{\alpha \beta}\partial^{\alpha}\varphi \partial^{\beta} \varphi d^{4}x$.   How does one derive the equations of motion in terms of the parameter  $$d\tau = \sqrt{\eta_{\alpha \beta}\frac{dx^{\alpha}}{d \lambda}\frac{dx^{\beta}}{d \lambda}} d\lambda.   $$ $u^{\alpha} = \frac{dx^{\alpha}}{d\tau} \Rightarrow u_{\alpha}u_{\beta}\eta^{\alpha \beta} = 1$?  My attempt:  $$ \delta S = 0 \Rightarrow \int \left( \frac{\partial (e^{\varphi } \sqrt{...})}{\partial x^{\alpha}}\delta x^{\alpha} +\frac{\partial (e^{\varphi } \sqrt{...})}{\partial \left( \frac{d x^{\alpha}}{d \lambda } \right)}\frac{d}{d\lambda} \delta x^{\alpha} \right)d\lambda = |d\tau = \sqrt{...}d\lambda | = $$ $$ = \int \left(\sqrt{...}e^{\varphi}\partial_{\alpha}\varphi - \frac{d}{d \lambda} \left(\frac{d x_{\alpha}}{d\tau}e^{\varphi}\right)  \right)\delta x^{\alpha}d \lambda  =  $$ $$ =\int \left( \partial_{\alpha}\varphi - \frac{d^{2}x_{\alpha}}{d \tau^{2}} - \frac{dx_{\alpha}}{d\tau} \frac{d \varphi }{d\tau }\right) \delta x^{\alpha} e^{\varphi} \sqrt{...}d\lambda =  $$ $$ =\int \left( \partial_{\alpha}\varphi - \frac{d u_{\alpha}}{d \tau} - u_{\alpha} u_{\beta} \partial^{\beta} \varphi \right)\delta x^{\alpha} e^{\varphi} \sqrt{...}d\lambda \Rightarrow $$ $$ \partial_{\alpha}\varphi - \frac{d u_{\alpha}}{d \tau} - u_{\alpha} u_{\beta} \partial^{\beta} \varphi = 0 \Rightarrow \partial_{\alpha} \varphi = e^{-\varphi}\frac{d }{d \tau}\left( e^{\varphi } u_{\alpha}\right). $$ Unfortunately, this equation doesn't look like the equation , $$ \frac{d (\varphi u_{\alpha})}{d \tau} = -\partial_{\alpha } \varphi.  $$ I can explain the part of differences by renaming the function, $e^{\varphi } \to \varphi $, in the expression for action (then my equation reduces to the form $ \partial_{\alpha} \varphi = \frac{d }{d \tau}\left( \varphi  u_{\alpha}\right)$), but I can't explain why my equation has the wrong sign.
63899
The usual Schrodinger Lagrangian is $$ \tag 1 i(\psi^{*}\partial_{t}\psi ) + \frac{1}{2m} \psi^{*}(\nabla^2)\psi, $$ which gives the correct equations of motion, with conjugate momentum for $\psi^{*}$ vanishing. This Lagrangian density is not real but differs from a real Lagrangian density  $$ \tag 2 \frac{i}{2}(\psi^{*}\partial_{t}\psi -\psi \partial_{t}\psi^{*} ) + \frac{1}{2m} \psi^{*}(\nabla^2)\psi $$ by a total derivative.  My trouble is that these two lagrangian densities lead to different conjugate momenta and hence when setting equal time commutation relations, I am getting different results (a factor of 2 is causing the problem). I can rescale the fields but then my Hamiltonian also changes. Then applying Heisenberg equation of motion, I don't get the operator Schrodinger equation.   Is it possible to work with the real Lagrangian density and somehow get the correct commutation relations? I would have expected two Lagrangians differing by total derivative terms to give identical commutation relations (since canonical transformations preserve them). But perhaps I am making some very simple error. Unless all conjugate momenta are equivalent for two Lagrangians differing by total derivatives, how does one choose the correct one?   I guess the same thing happens for other first order systems like Dirac Lagrangian also.
49635
I've been trying to find how the energy-momentum tensor changes if we add a total derivative to the lagrangian: $$L\to L+\mathrm d_\mu X^\mu.\tag{1}$$   From the answer key:  $$T^{\mu\nu}\to T^{\mu\nu}+\partial_\lambda F^{\lambda\mu\nu}\tag{2}$$ Where $$F^{\lambda\mu\nu}=\frac{\partial X^\lambda}{\partial(\partial_\mu\phi)}\partial^\nu\phi -\frac{\partial X^\mu}{\partial(\partial_\lambda\phi)}\partial^\nu\phi.\tag{3}$$ Is the above answer right? I'm not able to get this result! any help?
22997
Let us consider Dirac equation $$(i\gamma^\mu\partial_\mu -m)\psi ~=~0$$ as a classical field equation. Is it possible to introduce Poisson bracket on the space of spinors $\psi$ in such a way that Dirac equation becomes Hamiltonian equation   $$\dot{\psi}~=~\{ \psi,H\}_{PB}~?$$  Of course, such Poisson bracket would be graded (super Poisson bracket), but if it exists this would explain on classical level why $\frac{1}{2}$-spinors correspond to fermions.
39837
Why does Quantum Field Theory use usually Lagrangians rather than Hamiltonains?  I heard many reasons, but I'm not sure which is true.   Some say it's just a matter of beauty, so Lagrangians are more beautiful because they don't break/separate the space-time variables (so space-time is a single variable, like in the Klein-Gordon Lagrangian and Hamiltonian).  Some say that Hamiltonians are not always Lorentz invariant.  Could someone explain in a little bit more details?
20148
I try to understand constructing of Hamiltonian mechanics with constraints. I decided to start with the simple case: free relativistic particle. I've constructed hamiltonian with constraint:  $$S=-m\int d\tau \sqrt{\dot x_{\nu}\dot x^{\nu}}.$$  Here $\phi=p_{\mu}p^{\mu}-m^2=0$  $-$ .  Then $$H=H_{0}+\lambda \phi=\lambda \phi.$$   So, I want to show that I can obtain from this Hamiltonian the same equation of motion, as obtained from Lagrangian.  But the problem is that I'm not sure what to do with $\lambda=\lambda(q,p)$. I tried the following thing:  $$\dot x_{\mu}=\{x_{\mu},\lambda \phi\}=\{x_{\mu},\lambda p^2\}-m^2\{x_{\mu},\lambda\}=\lambda\{x_{\mu},p^2\}+p^2\{x_{\mu},\lambda\}-m^2\{x_{\mu},\lambda\}$$$$=2\lambda \eta_{\mu b} p^b+p^2\{x_{\mu},\lambda\}-m^2\{x_{\mu},\lambda\}=2\lambda \eta_{\mu b} p^b+p^2\frac{\partial \lambda}{\partial p^{\mu}}-m^2\frac{\partial \lambda}{\partial p^{\mu}},$$  $$\dot \lambda=\{\lambda, \lambda \phi \}=\{\lambda,\lambda p^2\}-m^2\{\lambda,\lambda\}=\lambda\{\lambda,p^2\}+p^2\{\lambda,p^2\}=2\lambda\eta_{ak}p^{a}\frac{\partial \lambda}{\partial x^{k}},$$  $$\dot p_{\mu}=\{p_{\mu},\lambda p^{2}-m^2\lambda \}=p^{2}\{p_{\mu},\lambda\}-m^2\{p_{\mu},\lambda\}=-p^{2}\frac{\partial \lambda}{\partial x^{\mu}}+m^2\frac{\partial \lambda}{\partial x^{\mu}}.$$  If we recall that $p^2-m^2=0$, then we get from the third equation: $\dot p=0$, and from the first: $$\dot x_{\mu}=2\lambda\eta_{ak}p^{a}.$$  So we have    $\dot x_{\mu}=2\lambda\eta_{\mu b}p^{b}.$ $\dot \lambda=2\lambda\eta_{ak}p^{a}\frac{\partial \lambda}{\partial x^{k}}.$ $\dot p=0.$   But I don't know what to do next. Can you help me?
119901
If two Lagrangian (densities) $\mathcal{L}$ give the same equations of motion, are they equivalent?
15509
Recently, I have been looking at the Lagrange equations that describe particles of different spin. I know lagrange takes the form $L=$kinetic-potential, and it seems these equations do take this form. But I dont see how they correspond to the kinetic or potential. How are these equations derived? The Klein-Gordon equation spin 0 $$L=c^2\partial _\lambda \phi \partial^\lambda \phi ^* -\left(\frac{mc}{\hbar}\right)^2\phi \phi^*$$ Dirac Lagrangian spin 1/2 $$L=i\hbar c \tilde\psi \gamma ^\mu \partial _\mu \psi-mc^2\tilde\psi  \psi $$ The Proca equation spin 1 $$L=-\frac{1}{16 \pi}F^{\mu \nu} F_{\mu \nu} +\frac{1}{8 \pi} \left(\frac{mc}{\hbar}\right)^2A^\mu A_\mu$$ I suppose I mean to say, is there one equation that you can get all three of these equations from?
112980
Is there a way to know if a system described by a known equation of motion admits a Hamiltonian function? Take for example $$ \dot \vartheta_i = \omega_i + J\sum_j \sin(\vartheta_j-\vartheta_i)$$ where $\omega_i$ are constants. How can I know if there exist suitable momenta and a Hamiltonian?
61647
I am reading Peskin and Schroeder, chapter ten, and my Lagrangian is $$ \mathcal{L}=\frac{1}{2}(\partial_\mu\phi_r)^2-\frac{1}{2}m^2\phi_r^2-\frac{\lambda}{4!}z^2\phi^4+\frac{1}{2}\delta_Z(\partial_\mu\phi_r)^2-\frac{1}{2}\delta_m\phi^2_r-\frac{\delta_\lambda}{4!}z^2\phi^4. $$  It was my understanding that terms of power 2 in the fields always give Feynman rules that are propagators. However, it appears that the counterterms with power 2 in the fields give a Feynman rule looking like $$i(p^2\delta_z-\delta_m),$$ instead of something with a denominator that would be more familiar. Like $\frac{i}{p^2-m^2+i\epsilon}$, from the first terms. Why is this the case? Is the idea that any term with power 2 in the fields gives a propagator wrong?
5422
What is the reason for the observation that across the board fields in physics are generally governed by second order (partial) differential equations?    If someone on the street would flat out ask me that question, then I'd probably mumble something about physicists wanting to be able to use the Lagrangian approach. And to allow a positive rotation and translation invariant energy term, which allows for local propagation, you need something like $-\phi\Delta\phi$.  I assume the answer goes in this direction, but I can't really justify why more complex terms in the Lagrangian are not allowed or why higher orders are a physical problem. Even if these require more initial data, I don't see the a priori problem.  Furthermore you could come up with quantities in the spirit of $F\wedge F$ and $F \wedge *F$ and okay yes... maybe any made up scalar just doesn't describe physics or misses valuable symmetries. On there other hand in the whole renormalization business, they seem to be allowed to use lots and lots of terms in their Lagrangians. And if I understand correctly, supersymmetry theory is basically a method of introducing new Lagrangian densities too.   Do we know the limit for making up these objects? What is the fundamental justification for order two?
72049
in the book Quantum Field Theory by Itzykson and Zuber the following derivation for the stress-energy tensor is proposed (p.22):  Assume a Lagrangian density depending on the spacetime coordinates $x$ only through fields and their gradients. Under a translation we have $$\mathcal L (x+a)=\mathcal L[\phi_i(x+a),\partial_\mu\phi_i(x+a)].$$  Consider an infinitesimal $x$-dependent transformation $$\delta\phi_i=\delta a^\mu(x)\partial_\mu\phi_i(x),$$ $$\delta\partial_u\phi_i(x)=\delta a^\nu \partial_\nu\partial_\mu\phi_i(x)+\partial_\mu[\delta a^\nu (x)]\partial_\nu \phi_i(x).$$  The proof then proceeds with variation of the action and integration by parts. But why do we consider an $x$-dependent local transformation instead of a global transformation?
94808
If we have a complex scalar $\phi$ we know that the gauge-invariant interaction with $A$ is given by $A^\mu J_\mu$, where $J$ is the Noether current of the $U(1)$ symmetry of the Lagrangian $$ J_\mu\sim \phi^\dagger\partial_\mu\phi-\phi\partial_\mu \phi^\dagger \tag{1} $$  If we have a real scalar instead, $J=0$ and the field doesn't couple to $A$: real particles aren't charged.  My question is: what if we took $$ J_\mu\sim \phi\partial_\mu\phi \tag{2} $$ to couple $\phi$ to $A$? The interaction is Lorentz invariant and renormalisable, but it is not gauge-invariant, which is probably a bad thing. At what point would this theory break down?  This is probably very naïve, but I think that the Feynman rules for this theory are straightforward. So I guess the theory makes sense at least perturbatively. The theory is probably flawed at a more fundamental level, but I cant't seem to find where (some kind of gauge anomaly perhaps?)
119809
I'm trying to understand infinitesimal canonical transformations and conservation theorems (section 9.6 Goldstein ed3). My specific problem is with understanding eq 9.104,         $\partial H = H(B) - K(A^{'}) $, where $\partial H$ represents the change in the hamiltonian under an infinitesimal canonical transformation (from the active view point ofc). The issue only arises for generating functions that explicitly depend on time.  Here's what I've thought of so far. I understand what $K$ is from the passive viewpoint (or so I think), its the hamiltonian that would furnish Hamilton's equations for the transformed variables $Q=q+ \delta q$ and $P = p + \delta p$. In the specific case of the generating function being a time dependent hamiltonian, $K = H + \delta t \frac{\partial H}{\partial t} = H + \delta t\dot H$, which makes sense because once you transform the variables to what they'll be in time $\delta t$, you'll also need to add on a term that comes from the Hamiltonian being at a later time $t+ \delta t$.   In the book, this is explained as a transformation depending on time changes the hamiltonian, so $H(A) \rightarrow K(A^{'}) $. Which I sorta get. But why not for $H(B)$? And what about the case when the Hamiltonian is time dependent and it is the generating function? Is $H(B)$ at a later time then?   Note: $A$ represents the original variables, $B$ is the transformed variables in the active sense, so in the same phase space as $A$, and $A^{'}$ is the same point as A but in the transformed phase space (passive).
31116
I was reading about the formulation of mechanics in special relativity and found that the action for a massive free point-particle  as $$ S = -mc\int_a^b ds  $$ So, I did a few observations, ie. $$ S = -mc^2 \int_a^b dt $$ and being  $$u_{\mu}u^{\mu} = c^2$$ I wrote the action as $$ S = -\int_a^b m u_{\mu}u^{\mu} dt  $$ which in my opinion resembles more to the classical kinetic energy.  Now, I never saw something like this in a book or text on internet. Everyone seems to work with the classical speed and the old good Euler-Lagrange equations with time as parameter.  So my question is: it's possible to derive the correct equations of relativistic dynamics from this action?  Attempt of solution  The action is $$ S = \int_a^b L(x,v,t) dt $$ and it should be Lorentz invariant (the action or the lagrangian?). Now, changing the parametrization of the path with $d\tau$ instead of $dt$ shouldn't change the integral (with the condition that everything changes accordingly). So I rewrite the action as $$ S = \int_{\alpha}^{\beta} L(x_{\mu}, \frac{dx_{\mu}}{d\tau}, \tau)d\tau $$ and do the variation of this and minimize it: $$ \delta S = \int_{\alpha}^{\beta} \left[ \frac{\partial L}{\partial x_{\mu}}\delta x_{\mu} + \frac{\partial L}{\partial u_{\mu}}\delta u_{\mu}\right] d\tau = 0 $$ Then, using the old manipulations it yields $$ \frac{d}{d\tau}\left( \frac{\partial L}{\partial u_{\mu}} \right) -  \frac{\partial L}{\partial x_{\mu}} = 0 $$  This applied to the lagrangian I wrote $$ L =  m \eta_{\mu\nu}u^{\mu} u^{\nu} $$ seems to yield the correct 4-impulse $$ p_{\nu} = \frac{\partial L}{\partial u_{\mu}} = m \eta_{\mu\nu}u^{\nu} $$  but the hamiltonian (total energy in the text I'm reading) is zero:  $$ H = p_{\mu}u_{\mu} - L = mu_{\mu}u^{\mu} - mu_{\mu}u^{\mu} = 0 $$  I think I did wrong the derivation of the Euler-Lagrange equations, but not sure.