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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:



$$
mathcal{H}=frac{1}{g^2}text{tr}(mathbf{E}^2+mathbf{B}^2)=g^2text{tr}(mathbf{pi}-frac{theta}{8pi^2}mathbf{B})^2+frac{1}{g^2}text{tr}(mathbf{B}^2).
$$



Here, $g$ is the gauge coupling, $E_i=dot{A}_i$ is the non-Abelian electric field, $B_i=-frac{1}{2}epsilon_{ijk}F^{jk}$ the non-Abelian magnetic field, ${F}_{munu}$ is the gluon field strength, and
$$
mathbf{pi}=frac{partial mathcal{L}}{partial mathbf{dot{A}}}= frac{1}{g^2}mathbf{E}+frac{theta}{8pi^2}mathbf{B}
$$
is the momentun conjugate to $mathbf{A}$ (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:
$$
mathcal{L}= -frac{1}{2g^2}text{tr}(F^{munu}F_{munu})+frac{theta}{16pi^2}text{tr}(F^{munu}tilde{F}_{munu})=frac{1}{g^2}text{tr}(mathbf{dot{A}}^2-mathbf{B}^2)-frac{theta}{4pi^2}text{tr}(mathbf{dot{A}} mathbf{B}),
$$
where $tilde{F}_{munu}$ 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











$endgroup$

















    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:



    $$
    mathcal{H}=frac{1}{g^2}text{tr}(mathbf{E}^2+mathbf{B}^2)=g^2text{tr}(mathbf{pi}-frac{theta}{8pi^2}mathbf{B})^2+frac{1}{g^2}text{tr}(mathbf{B}^2).
    $$



    Here, $g$ is the gauge coupling, $E_i=dot{A}_i$ is the non-Abelian electric field, $B_i=-frac{1}{2}epsilon_{ijk}F^{jk}$ the non-Abelian magnetic field, ${F}_{munu}$ is the gluon field strength, and
    $$
    mathbf{pi}=frac{partial mathcal{L}}{partial mathbf{dot{A}}}= frac{1}{g^2}mathbf{E}+frac{theta}{8pi^2}mathbf{B}
    $$
    is the momentun conjugate to $mathbf{A}$ (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:
    $$
    mathcal{L}= -frac{1}{2g^2}text{tr}(F^{munu}F_{munu})+frac{theta}{16pi^2}text{tr}(F^{munu}tilde{F}_{munu})=frac{1}{g^2}text{tr}(mathbf{dot{A}}^2-mathbf{B}^2)-frac{theta}{4pi^2}text{tr}(mathbf{dot{A}} mathbf{B}),
    $$
    where $tilde{F}_{munu}$ 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











    $endgroup$















      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:



      $$
      mathcal{H}=frac{1}{g^2}text{tr}(mathbf{E}^2+mathbf{B}^2)=g^2text{tr}(mathbf{pi}-frac{theta}{8pi^2}mathbf{B})^2+frac{1}{g^2}text{tr}(mathbf{B}^2).
      $$



      Here, $g$ is the gauge coupling, $E_i=dot{A}_i$ is the non-Abelian electric field, $B_i=-frac{1}{2}epsilon_{ijk}F^{jk}$ the non-Abelian magnetic field, ${F}_{munu}$ is the gluon field strength, and
      $$
      mathbf{pi}=frac{partial mathcal{L}}{partial mathbf{dot{A}}}= frac{1}{g^2}mathbf{E}+frac{theta}{8pi^2}mathbf{B}
      $$
      is the momentun conjugate to $mathbf{A}$ (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:
      $$
      mathcal{L}= -frac{1}{2g^2}text{tr}(F^{munu}F_{munu})+frac{theta}{16pi^2}text{tr}(F^{munu}tilde{F}_{munu})=frac{1}{g^2}text{tr}(mathbf{dot{A}}^2-mathbf{B}^2)-frac{theta}{4pi^2}text{tr}(mathbf{dot{A}} mathbf{B}),
      $$
      where $tilde{F}_{munu}$ 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











      $endgroup$




      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:



      $$
      mathcal{H}=frac{1}{g^2}text{tr}(mathbf{E}^2+mathbf{B}^2)=g^2text{tr}(mathbf{pi}-frac{theta}{8pi^2}mathbf{B})^2+frac{1}{g^2}text{tr}(mathbf{B}^2).
      $$



      Here, $g$ is the gauge coupling, $E_i=dot{A}_i$ is the non-Abelian electric field, $B_i=-frac{1}{2}epsilon_{ijk}F^{jk}$ the non-Abelian magnetic field, ${F}_{munu}$ is the gluon field strength, and
      $$
      mathbf{pi}=frac{partial mathcal{L}}{partial mathbf{dot{A}}}= frac{1}{g^2}mathbf{E}+frac{theta}{8pi^2}mathbf{B}
      $$
      is the momentun conjugate to $mathbf{A}$ (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:
      $$
      mathcal{L}= -frac{1}{2g^2}text{tr}(F^{munu}F_{munu})+frac{theta}{16pi^2}text{tr}(F^{munu}tilde{F}_{munu})=frac{1}{g^2}text{tr}(mathbf{dot{A}}^2-mathbf{B}^2)-frac{theta}{4pi^2}text{tr}(mathbf{dot{A}} mathbf{B}),
      $$
      where $tilde{F}_{munu}$ 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




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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 2 hours ago









      Qmechanic♦

      108k122011255




      108k122011255










      asked 3 hours ago









      LCFLCF

      68749




      68749






















          1 Answer
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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
          $$H = frac{p^2}{2m} = frac12 m v^2, quad p = mv.$$
          On the other hand, the Hamiltonian for a particle in a magnetic field is
          $$H = frac{(p-eA)^2}{2m} = frac12 m v^2, quad p = m v + e A.$$
          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
            $$H = frac{p^2}{2m} = frac12 m v^2, quad p = mv.$$
            On the other hand, the Hamiltonian for a particle in a magnetic field is
            $$H = frac{(p-eA)^2}{2m} = frac12 m v^2, quad p = m v + e A.$$
            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
              $$H = frac{p^2}{2m} = frac12 m v^2, quad p = mv.$$
              On the other hand, the Hamiltonian for a particle in a magnetic field is
              $$H = frac{(p-eA)^2}{2m} = frac12 m v^2, quad p = m v + e A.$$
              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
                $$H = frac{p^2}{2m} = frac12 m v^2, quad p = mv.$$
                On the other hand, the Hamiltonian for a particle in a magnetic field is
                $$H = frac{(p-eA)^2}{2m} = frac12 m v^2, quad p = m v + e A.$$
                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
                $$H = frac{p^2}{2m} = frac12 m v^2, quad p = mv.$$
                On the other hand, the Hamiltonian for a particle in a magnetic field is
                $$H = frac{(p-eA)^2}{2m} = frac12 m v^2, quad p = m v + e A.$$
                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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