11 votes

Hamiltonian of a complex scalar quantum field

Where did I go wrong? Seems like probably there are a number of different places you could have gone wrong. Since this is a homework-like question, I will answer in terms of classical field theory, ...
hft's user avatar
  • 19.8k
5 votes
Accepted

Why do we need a Poisson bracket structure?

The simplest way to motivate the Dirac bracket is to consider a situation where two second-class constraints $f_1,f_2$ have constant non-zero Poisson bracket $\{f_1,f_2\} = c \neq 0$, e.g. when we ...
ACuriousMind's user avatar
  • 125k
4 votes
Accepted

Hamiltonian of a complex scalar quantum field

In field theory texts, e.g. Peskin and Schroeder, one will often see the stress energy tensor written simply as: $$T^{\mu\nu}={\partial\mathcal L\over\partial(\partial_\mu\phi)}\partial^\nu\phi-\delta^...
Albertus Magnus's user avatar
3 votes

Why is the Hamiltonian zero in relativity?

TL;DR: OP's trick makes sense in general. It is not necessarily related to relativity theory. Let us first consider the Lagrangian formalism. It is natural to consider a general situation where the ...
Qmechanic's user avatar
  • 202k
2 votes

Hamiltonian formalism (with symplectic form) for time-dependent Lagrangian

OP's question seems to be partly inspired by the covariant Hamiltonian/phase space formalism, cf. e.g. OP's previous question here. OP's method of generating a worldline (WL) reparametrization ...
Qmechanic's user avatar
  • 202k
2 votes

Changing variables from $\dot{q}$ to $p$ in Lagrangian instead of Legendre Transformation

Obviously, you can write the Lagrangian $L(t,q,\dot{q})$ as a function of $t,q,p$ instead of $t,q,\dot{q}$, by just inverting the relation $$p= \frac{\partial L(t,q,\dot{q})}{\partial \dot{q}}.$$ ...
Valter Moretti's user avatar
1 vote
Accepted

Why $q,p,Q,P$ are Independent Variables when Using Generating Functions?

In Hamiltonian formalism, specifically generating functions, why do the variables $q, p, Q, P$ are treated as independent when finding the equations that arise from the generating function? Let us ...
hft's user avatar
  • 19.8k
1 vote

Why $q,p,Q,P$ are Independent Variables when Using Generating Functions?

As QMechanic described in his answer to the related post, the generating functions are defined on a $4n$ dimensional manifold. By definition you have $4n$ independent coordinates to describe a point ...
Er Jio's user avatar
  • 1,231
1 vote

Why $q,p,Q,P$ are Independent Variables when Using Generating Functions?

The system with $s$ degrees of freedom has a $2s$-dimensional phase space. Hence, as a description of the system state, out of the $4s$ variables - which includes $2s$ old phase space coordinates and $...
xiyao's user avatar
  • 19
1 vote

Regarding Poission structure of Hamiltonian phase space

Well, given an arbitrary $2n$-dimensional symplectic manifold $(M,\{\cdot,\cdot\})$, the Darboux theorem shows that there locally always exist canonical coordinates $(q^1, \ldots, q^n,p_1, \ldots, p_n)...
Qmechanic's user avatar
  • 202k
1 vote
Accepted

Constrained Hamiltonian problems

This is a quite broad question. Briefly speaking, starting from the non-degenerate canonical Poisson bracket structure, according to the Dirac-Bergmann analysis, the constraints can in principle (...
Qmechanic's user avatar
  • 202k
1 vote

Regarding varying the coefficients of constraints in the pre-extended Hamiltonian in Dirac method

The short answer is that at this point in Ref. 1 $u^m$ are new independent variables; not functions of $q$ and $p$. Equivalently, $u^m$ are Lagrange undetermined multipliers. It is possibly helpful to ...
Qmechanic's user avatar
  • 202k
1 vote

What is Phase Space Formulation of QM and does it explain use of complex variables in QM?

I feel like the other two answers are not satisfying from a mathematical point of view, respectively due to their brevity or dogmatic nature. I will describe two ways in which the complex numbers are ...
Cole Comfort's user avatar

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