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“Hidden” theta-term in Hamiltonian formulation of Yang-Mills theory


Kugo and Ojima's Canonical Formulation of Yang-Mills using BRSTAre the Yang-Mills equation and its generalization gauge invariant?Infinitesimal gauge invariance of Yang--Mills LagrangianDoubts about the theta angle and the ground state energy density in Euclidean Yang-Mills theoryIs there an argument for using the $theta$-vacuum for a Yang-Mills theory that works regardless of the presence of fermions?Uniqueness of Yang-Mills theoryInterpretation of the field strength tensor in Yang-Mills TheoryYang-Mills vs Einstein-Hilbert ActionWhy are two different gauge transformations of $A_mu=0$ in $U(1)$ gauge thoery equivalent?Compactification of space in Hamiltonian formulation of Yang-Mills theory













3












$begingroup$


I've read in David Tong's lecture notes on gauge theory that the Hamiltonian of Yang-Mills theory does not depend on the angular parameter $theta$, because it can be absorbed in the electric field:



$$
mathcalH=frac1g^2texttr(mathbfE^2+mathbfB^2)=g^2texttr(mathbfpi-fractheta8pi^2mathbfB)^2+frac1g^2texttr(mathbfB^2).
$$



Here, $g$ is the gauge coupling, $E_i=dotA_i$ is the non-Abelian electric field, $B_i=-frac12epsilon_ijkF^jk$ the non-Abelian magnetic field, $F_munu$ is the gluon field strength, and
$$
mathbfpi=fracpartial mathcalLpartial mathbfdotA= frac1g^2mathbfE+fractheta8pi^2mathbfB
$$

is the momentun conjugate to $mathbfA$ (see pp. 39 and 40 of the lecture notes).



In contrast, it's well known that the Yang-Mills Lagrangian contains a topological $theta$-term:
$$
mathcalL= -frac12g^2texttr(F^munuF_munu)+fractheta16pi^2texttr(F^munutildeF_munu)=frac1g^2texttr(mathbfdotA^2-mathbfB^2)-fractheta4pi^2texttr(mathbfdotA mathbfB),
$$

where $tildeF_munu$ is the Hodge dual of $F_munu$, and the last equality holds for $A_0=0$ and $D_iE_i=0$.



How is this possible? Tong mentions that the $theta$-dependence in the Hamiltonian formalism is somehow hidden in the structure of the Poisson bracket, but he gives no detailed explanation. Why doesn't it appear in the Hamiltonian itself? The $theta$-term gives rise to the infamous strong CP problem, so how can we explicitly compute this $theta$-dependence of Yang-Mills in this formalism?










share|cite|improve this question











$endgroup$
















    3












    $begingroup$


    I've read in David Tong's lecture notes on gauge theory that the Hamiltonian of Yang-Mills theory does not depend on the angular parameter $theta$, because it can be absorbed in the electric field:



    $$
    mathcalH=frac1g^2texttr(mathbfE^2+mathbfB^2)=g^2texttr(mathbfpi-fractheta8pi^2mathbfB)^2+frac1g^2texttr(mathbfB^2).
    $$



    Here, $g$ is the gauge coupling, $E_i=dotA_i$ is the non-Abelian electric field, $B_i=-frac12epsilon_ijkF^jk$ the non-Abelian magnetic field, $F_munu$ is the gluon field strength, and
    $$
    mathbfpi=fracpartial mathcalLpartial mathbfdotA= frac1g^2mathbfE+fractheta8pi^2mathbfB
    $$

    is the momentun conjugate to $mathbfA$ (see pp. 39 and 40 of the lecture notes).



    In contrast, it's well known that the Yang-Mills Lagrangian contains a topological $theta$-term:
    $$
    mathcalL= -frac12g^2texttr(F^munuF_munu)+fractheta16pi^2texttr(F^munutildeF_munu)=frac1g^2texttr(mathbfdotA^2-mathbfB^2)-fractheta4pi^2texttr(mathbfdotA mathbfB),
    $$

    where $tildeF_munu$ is the Hodge dual of $F_munu$, and the last equality holds for $A_0=0$ and $D_iE_i=0$.



    How is this possible? Tong mentions that the $theta$-dependence in the Hamiltonian formalism is somehow hidden in the structure of the Poisson bracket, but he gives no detailed explanation. Why doesn't it appear in the Hamiltonian itself? The $theta$-term gives rise to the infamous strong CP problem, so how can we explicitly compute this $theta$-dependence of Yang-Mills in this formalism?










    share|cite|improve this question











    $endgroup$














      3












      3








      3





      $begingroup$


      I've read in David Tong's lecture notes on gauge theory that the Hamiltonian of Yang-Mills theory does not depend on the angular parameter $theta$, because it can be absorbed in the electric field:



      $$
      mathcalH=frac1g^2texttr(mathbfE^2+mathbfB^2)=g^2texttr(mathbfpi-fractheta8pi^2mathbfB)^2+frac1g^2texttr(mathbfB^2).
      $$



      Here, $g$ is the gauge coupling, $E_i=dotA_i$ is the non-Abelian electric field, $B_i=-frac12epsilon_ijkF^jk$ the non-Abelian magnetic field, $F_munu$ is the gluon field strength, and
      $$
      mathbfpi=fracpartial mathcalLpartial mathbfdotA= frac1g^2mathbfE+fractheta8pi^2mathbfB
      $$

      is the momentun conjugate to $mathbfA$ (see pp. 39 and 40 of the lecture notes).



      In contrast, it's well known that the Yang-Mills Lagrangian contains a topological $theta$-term:
      $$
      mathcalL= -frac12g^2texttr(F^munuF_munu)+fractheta16pi^2texttr(F^munutildeF_munu)=frac1g^2texttr(mathbfdotA^2-mathbfB^2)-fractheta4pi^2texttr(mathbfdotA mathbfB),
      $$

      where $tildeF_munu$ is the Hodge dual of $F_munu$, and the last equality holds for $A_0=0$ and $D_iE_i=0$.



      How is this possible? Tong mentions that the $theta$-dependence in the Hamiltonian formalism is somehow hidden in the structure of the Poisson bracket, but he gives no detailed explanation. Why doesn't it appear in the Hamiltonian itself? The $theta$-term gives rise to the infamous strong CP problem, so how can we explicitly compute this $theta$-dependence of Yang-Mills in this formalism?










      share|cite|improve this question











      $endgroup$




      I've read in David Tong's lecture notes on gauge theory that the Hamiltonian of Yang-Mills theory does not depend on the angular parameter $theta$, because it can be absorbed in the electric field:



      $$
      mathcalH=frac1g^2texttr(mathbfE^2+mathbfB^2)=g^2texttr(mathbfpi-fractheta8pi^2mathbfB)^2+frac1g^2texttr(mathbfB^2).
      $$



      Here, $g$ is the gauge coupling, $E_i=dotA_i$ is the non-Abelian electric field, $B_i=-frac12epsilon_ijkF^jk$ the non-Abelian magnetic field, $F_munu$ is the gluon field strength, and
      $$
      mathbfpi=fracpartial mathcalLpartial mathbfdotA= frac1g^2mathbfE+fractheta8pi^2mathbfB
      $$

      is the momentun conjugate to $mathbfA$ (see pp. 39 and 40 of the lecture notes).



      In contrast, it's well known that the Yang-Mills Lagrangian contains a topological $theta$-term:
      $$
      mathcalL= -frac12g^2texttr(F^munuF_munu)+fractheta16pi^2texttr(F^munutildeF_munu)=frac1g^2texttr(mathbfdotA^2-mathbfB^2)-fractheta4pi^2texttr(mathbfdotA mathbfB),
      $$

      where $tildeF_munu$ is the Hodge dual of $F_munu$, and the last equality holds for $A_0=0$ and $D_iE_i=0$.



      How is this possible? Tong mentions that the $theta$-dependence in the Hamiltonian formalism is somehow hidden in the structure of the Poisson bracket, but he gives no detailed explanation. Why doesn't it appear in the Hamiltonian itself? The $theta$-term gives rise to the infamous strong CP problem, so how can we explicitly compute this $theta$-dependence of Yang-Mills in this formalism?







      quantum-field-theory hamiltonian-formalism topology yang-mills






      share|cite|improve this question















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      share|cite|improve this question




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      edited 2 hours ago









      Qmechanic♦

      108k122011255




      108k122011255










      asked 2 hours ago









      LCFLCF

      68749




      68749




















          1 Answer
          1






          active

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          $begingroup$

          It's not particularly strange or unusual for a term to seemingly not appear in the Hamiltonian, but still have a physical effect. For example, the Hamiltonian for a free particle is
          $$H = fracp^22m = frac12 m v^2, quad p = mv.$$
          On the other hand, the Hamiltonian for a particle in a magnetic field is
          $$H = frac(p-eA)^22m = frac12 m v^2, quad p = m v + e A.$$
          They are of course different functions of $(x, p)$, reflecting the fact that the dynamics are different, but if you naively write the Hamiltonian in terms of the "physical" variables $(x, v)$ then you get the same result in both cases, since magnetic fields do no work. There is of course still an effect, because $v$ is related to $p$ in a different way. Your example with the $theta$-term is just a more complicated version of this, so there's nothing really weird to resolve.






          share|cite|improve this answer









          $endgroup$













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            $begingroup$

            It's not particularly strange or unusual for a term to seemingly not appear in the Hamiltonian, but still have a physical effect. For example, the Hamiltonian for a free particle is
            $$H = fracp^22m = frac12 m v^2, quad p = mv.$$
            On the other hand, the Hamiltonian for a particle in a magnetic field is
            $$H = frac(p-eA)^22m = frac12 m v^2, quad p = m v + e A.$$
            They are of course different functions of $(x, p)$, reflecting the fact that the dynamics are different, but if you naively write the Hamiltonian in terms of the "physical" variables $(x, v)$ then you get the same result in both cases, since magnetic fields do no work. There is of course still an effect, because $v$ is related to $p$ in a different way. Your example with the $theta$-term is just a more complicated version of this, so there's nothing really weird to resolve.






            share|cite|improve this answer









            $endgroup$

















              4












              $begingroup$

              It's not particularly strange or unusual for a term to seemingly not appear in the Hamiltonian, but still have a physical effect. For example, the Hamiltonian for a free particle is
              $$H = fracp^22m = frac12 m v^2, quad p = mv.$$
              On the other hand, the Hamiltonian for a particle in a magnetic field is
              $$H = frac(p-eA)^22m = frac12 m v^2, quad p = m v + e A.$$
              They are of course different functions of $(x, p)$, reflecting the fact that the dynamics are different, but if you naively write the Hamiltonian in terms of the "physical" variables $(x, v)$ then you get the same result in both cases, since magnetic fields do no work. There is of course still an effect, because $v$ is related to $p$ in a different way. Your example with the $theta$-term is just a more complicated version of this, so there's nothing really weird to resolve.






              share|cite|improve this answer









              $endgroup$















                4












                4








                4





                $begingroup$

                It's not particularly strange or unusual for a term to seemingly not appear in the Hamiltonian, but still have a physical effect. For example, the Hamiltonian for a free particle is
                $$H = fracp^22m = frac12 m v^2, quad p = mv.$$
                On the other hand, the Hamiltonian for a particle in a magnetic field is
                $$H = frac(p-eA)^22m = frac12 m v^2, quad p = m v + e A.$$
                They are of course different functions of $(x, p)$, reflecting the fact that the dynamics are different, but if you naively write the Hamiltonian in terms of the "physical" variables $(x, v)$ then you get the same result in both cases, since magnetic fields do no work. There is of course still an effect, because $v$ is related to $p$ in a different way. Your example with the $theta$-term is just a more complicated version of this, so there's nothing really weird to resolve.






                share|cite|improve this answer









                $endgroup$



                It's not particularly strange or unusual for a term to seemingly not appear in the Hamiltonian, but still have a physical effect. For example, the Hamiltonian for a free particle is
                $$H = fracp^22m = frac12 m v^2, quad p = mv.$$
                On the other hand, the Hamiltonian for a particle in a magnetic field is
                $$H = frac(p-eA)^22m = frac12 m v^2, quad p = m v + e A.$$
                They are of course different functions of $(x, p)$, reflecting the fact that the dynamics are different, but if you naively write the Hamiltonian in terms of the "physical" variables $(x, v)$ then you get the same result in both cases, since magnetic fields do no work. There is of course still an effect, because $v$ is related to $p$ in a different way. Your example with the $theta$-term is just a more complicated version of this, so there's nothing really weird to resolve.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 2 hours ago









                knzhouknzhou

                47.4k11131230




                47.4k11131230



























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