1
$\begingroup$

So, I'm considering Lagrangian: $$L=\frac{1} {2}e^{q_1}\dot{q}_2^2. $$ I obtain the primary constraint $\phi=p_1=0$. The canonical Hamiltonian is $H_c=\frac{1} {2}p_2^2e^{-q_1} $, and the total Hamiltonian is $H_T=\frac{1} {2}p_2^2e^{-q_1} +vp_1$, where $v$ is some arbitrary multiplier. When checking the consistency condition we obtain: $$\dot{\phi} =[\phi, H_T] =\frac{1} {2} p_2^2e^{-q_1}.$$ So, I obtain a secondary constrait $\chi=p_2e^{-q_1} =0$. But the Jacobian $J=\frac{\partial\phi_m} {\partial(q, p) } $ doesn't have the constant rank, so I suppose I should write this constraint as $\chi=p_2=0$? But then when trying to find the generator of symmetry by the algorithm $G_1=\phi$, $G_0+\{\phi,H_T\}=a\phi$ and $ \{G_0,H_T\} =b\phi$, I get that $a=-v$, that is the arbitrary constant $v$ (and $ \epsilon $ and $\dot{\epsilon}$ as it should be) is part of the generator.

Can somebody please tell me what am I doing wrong?

$\endgroup$
  • $\begingroup$ Rank jump where? $\endgroup$ – Qmechanic May 30 '18 at 17:08
1
$\begingroup$

Lagrangian $L=\frac{1}{2}e^{q_{1}}(\dot{q}_{2})^{2}$. Canonical momentum $p_{1}=\frac{\partial L}{\partial \dot{q}_{1}}=0$. This is the primary constraint $\phi_{1}(q,p)=p_{1}\approx 0$. The other canonical momentum is $p_{2}=\frac{\partial L}{\partial \dot{q}_{2}}=e^{q_{1}}\dot{q}_{2}$. The Hamiltonian is, $$ H=p_{1}\dot{q}_{1}+p_{2}\dot{q}_{2}-L=e^{q_{1}}(\dot{q}_{2})^{2}-\frac{1}{2}e^{q_{1}}(\dot{q}_{2})^{2}=\frac{1}{2}e^{q_{1}}(\dot{q}_{2})^{2} =\frac{1}{2}(p_{2})^{2}e^{-q_{1}} $$ The total Hamiltonian is $H_{T}=H+\nu(q,p)\phi_{1}(q,p)=H+\nu(q,p) p_{1}$. Here, $\nu(q,p)$ is an arbitrary function. Now check the consistency condition, $$ 0\approx\dot{\phi}_{1}= [\phi_{1},H_{T}]_{PB}\approx[\phi_{1},H]_{PB}+[\phi_{1},\phi_{1}]_{PB}\nu=[p_{1},H]_{PB}=-\frac{\partial p_{1}}{\partial p_{1}}\frac{\partial H}{\partial q_{1}}=H $$ This means there is a secondary constraint $\phi_{2}(q,p)=H$. The system is rather strange because the Hamiltonian is a secondary constraint. The constraints are both first class ( a first class variable has weakly zero PBs with all the constraints). The only PB which needs checking is $[\phi_{1},\phi_{2}]_{PB}=[p_{1},H]_{PB}=H\approx 0$. This problem appears, with a slight change of notation, in section 1.2.2 of "Quantization of Gauge Systems" by Henneaux and Teitelboim entitled "A Counterexample to the Dirac Conjecture". The primary first class constraints are generators of canonical transformations which change the q's and p's but do not change the physical state. Dirac's conjecture is that the secondary first class constraints are also generators of canonical transformations that change the q's and p's but do not change the physical state. Let's act on the q's and p's with the primary first class constraint $\phi_{1}$ to see what variables are part of the physical state. $$ \frac{dq_{1}}{d\epsilon}=[q_{1},\phi_{1}]_{PB}=[q_{1},p_{1}]_{PB}=1 $$ So $q_{1}$ changes so it cannot be part of the physical state. This is confirmed by the EoM, $$ \dot{q}_{1}=[q_{1},H_{T}]_{PB}\approx [q_{1},H]_{PB}+[q_{1},p_{1}]_{PB}\nu=\nu $$ So $q_{1}$ is an arbitrary function and so it cannot be part of the physical state. Henneaux and Teitelboim say that $q_{1}$ is "pure gauge". Returning to the transformations, $$ \frac{dq_{2}}{d\epsilon}=[q_{2},\phi_{1}]_{PB}=[q_{2},p_{1}]_{PB}=0 $$ $q_{2}$ is not changed so it is part of the physical state. Similarly, the momenta are unchanged so they are also physical. Now let's see if the secondary first class constraint $\phi_{2}=H$ changes the q's and p's but does not change the physical state. The interesting case is what happens to $q_{2}$, because we've already seen that it's physical. $$ \frac{dq_{2}}{d\epsilon}=[q_{2},\phi_{2}]_{PB}=[q_{2},H]_{PB}=p_{2}e^{-q_{1}} $$ Now, the secondary constraint is $\phi_{2}=H=\frac{1}{2}(p_{2})^{2}e^{-q_{1}}\approx 0$ which implies $p_{2}\approx 0$. If we put this in the transformation of $q_{2}$, the result is weakly zero $\frac{dq_{2}}{d\epsilon}\approx 0$. So, nothing bad happens, $q_{2}$ is physical and the transformation generated by the secondary first class constraint does not change it. However, Henneaux and Teitelboim get a contradiction of Dirac's conjecture by taking the secondary constraint as $\phi_{2}(q,p)=p_{2}$ instead of $\phi_{2}=H$. Henneaux and Teitelboim's secondary first class constraint changes the physical variable $q_{2}$ according to, $$ \frac{dq_{2}}{d\epsilon}=[q_{2},\phi_{2}]_{PB}=[q_{2},p_{2}]_{PB}=1 $$ This contradict's Dirac's conjecture because the secondary first class constraint is only supposed to change the q's and p's which are not part of the physical state. However, this counter example seems rather contrived because, if the secondary constraint is $H$ then nothing goes wrong.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.