“Hidden” theta-term in Hamiltonian formulation of Yang-Mills theoryKugo and Ojima's Canonical Formulation...

What is the optimal strategy for the Dictionary Game?

Function pointer with named arguments?

'It addicted me, with one taste.' Can 'addict' be used transitively?

How can I get this effect? Please see the attached image

Rivers without rain

How can I print the prosodic symbols in LaTeX?

On The Origin of Dissonant Chords

"You've called the wrong number" or "You called the wrong number"

Why did some of my point & shoot film photos come back with one third light white or orange?

Check if a string is entirely made of the same substring

Are there physical dangers to preparing a prepared piano?

Why boldmath fails in a tikz node?

How do I deal with a coworker that keeps asking to make small superficial changes to a report, and it is seriously triggering my anxiety?

Elements that can bond to themselves?

How much cash can I safely carry into the USA and avoid civil forfeiture?

Coordinate my way to the name of the (video) game

Initiative: Do I lose my attack/action if my target moves or dies before my turn in combat?

Why must Chinese maps be obfuscated?

Can I criticise the more senior developers around me for not writing clean code?

Which big number is bigger?

How to fry ground beef so it is well-browned

Does a large simulator bay have standard public address announcements?

Don’t seats that recline flat defeat the purpose of having seatbelts?

Mistake in years of experience in resume?



“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 2 hours ago









      LCFLCF

      68749




      68749






















          1 Answer
          1






          active

          oldest

          votes


















          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









          $endgroup$














            Your Answer








            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "151"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: false,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: null,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f476230%2fhidden-theta-term-in-hamiltonian-formulation-of-yang-mills-theory%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            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












              $begingroup$

              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






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Physics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid …



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f476230%2fhidden-theta-term-in-hamiltonian-formulation-of-yang-mills-theory%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    -

                    Popular posts from this blog

                    Gersau Kjelder | Navigasjonsmeny46°59′0″N 8°31′0″E46°59′0″N...

                    Image
                    Tidlegare republikkarSkipingar i 1433Kommunar i SchwyzDistrikt i SchwyzNedleggingar i 1798Stader ved Vierwaldstättersee kommunekantonenSchwyzSveitskm²VierwaldstätterseeLauerz Gersau Kommunevåpenet til Gersau Plassering Styresmakter Land   Sveits Kanton kantonen Schwyz Distrikt Gersau Språk tysk Geografi Flatevidd  - kommune 23,70 km² Innbyggjarar  - kommune (desember 2010)    - folketettleik 2 094   88,4 /km² Koordinatar 46°59′0″N 8°31′0″E  Koordinatar: 46°59′0″N 8°31′0″E  Høgd over havet 435 moh Diverse annan informasjon Postnummer 6442 Nettstad: www.gersau.ch Freistaat und Republik Gersau Fristaten og republikken Gersau By-republikk og alliert protektorat i Det gamle sveitsiske eidsforundet ←   ← 1433–1798 1814–1818 →   → Hovudstad Gersau Styreform Republikk Historisk periode Tidleg moderne tid  - Kjøpte fridom frå
                    Read more

                    Hestehale Innhaldsliste Hestehale på kvinner | Hestehale på menn | Galleri | Sjå òg |...

                    Image
                    Frisyrar frisyrehårhårstrikkhårklemmerøyrethesthovudverkeuropeisk motejobbtreningfletterparykkarungdomsopprørmotkulturmusikararmotorsykkelentusiastarpolitietpolitiskforretningverksemdfiskeflette Hestehale sett bakfrå. Hestehale på ein mannleg indianarleiar. Hestehale er ein frisyre der ein samlar det meste eller alt av langt hår [1] i bakhovudet med ein hårstrikk eller liknande. Avhengig av mote kan det òg vere vanleg å sette opp hestehalen med hårklemmer eller å ha hestehalen hengande på den eine sida av hovudet ved å samle håret ved øyret. Hestehale er ein vanleg frisyre for jenter og for gutar med langt hår. Han har fått namnet sitt etter likskapen med halen på ein hest. For stramme hestehaler kan føre til hårtap, [2] i nokre høve også hovudverk. [3] Innhaldsliste 1 Hestehale på kvinner 2 Hestehale på menn 3 Galleri 4 Sjå òg 5 Bakgrunnsstoff 6 Kjelder Hestehale på kvinner | Ungjente med hestehale som held håret unna ansiktet
                    Read more

                    What is the “three and three hundred thousand syndrome”?Who wrote the book Arena?What five creatures were...

                    Image
                    What happens when the centripetal force is equal and opposite to the centrifugal force? Why is "la Gestapo" feminine? Nested Dynamic SOQL Query How are passwords stolen from companies if they only store hashes? Exposing a company lying about themselves in a tightly knit industry: Is my career at risk on the long run? Can "few" be used as a subject? If so, what is the rule? Why are there no stars visible in cislunar space? categorizing a variable turns it from insignificant to significant Do I need an EFI partition for each 18.04 ubuntu I have on my HD? What are the consequences of changing the number of hours in
                    Read more