“Hidden” theta-term in Hamiltonian formulation of Yang-Mills theoryTheta-dependence of massive Schwinger modelKugo 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
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“Hidden” theta-term in Hamiltonian formulation of Yang-Mills theory
Theta-dependence of massive Schwinger modelKugo 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
$begingroup$
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?
quantum-field-theory hamiltonian-formalism topology yang-mills
edited Apr 26 at 18:46
Qmechanic♦
108k122041255
asked Apr 26 at 18:30
LCFLCF
71749
$endgroup$
add a comment |
$begingroup$
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?
quantum-field-theory hamiltonian-formalism topology yang-mills
edited Apr 26 at 18:46
Qmechanic♦
108k122041255
asked Apr 26 at 18:30
LCFLCF
71749
$endgroup$
add a comment |
$begingroup$
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?
quantum-field-theory hamiltonian-formalism topology yang-mills
edited Apr 26 at 18:46
Qmechanic♦
108k122041255
asked Apr 26 at 18:30
LCFLCF
71749
$endgroup$
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
quantum-field-theory hamiltonian-formalism topology yang-mills
edited Apr 26 at 18:46
Qmechanic♦
108k122041255
asked Apr 26 at 18:30
LCFLCF
71749
edited Apr 26 at 18:46
Qmechanic♦
108k122041255
asked Apr 26 at 18:30
LCFLCF
71749
edited Apr 26 at 18:46
Qmechanic♦
108k122041255
edited Apr 26 at 18:46
Qmechanic♦
108k122041255
edited Apr 26 at 18:46
Qmechanic♦
108k122041255
108k122041255
asked Apr 26 at 18:30
LCFLCF
71749
asked Apr 26 at 18:30
LCFLCF
71749
asked Apr 26 at 18:30
LCFLCF
71749
71749
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Apr 26 at 18:45
knzhouknzhou
48.6k12133235
$endgroup$
$begingroup$
Thanks for great answer! So the "physical" effect we see in the Hamiltonian formalism is the altered dispersion relation of the particle moving in the external field?
$endgroup$
– LCF
Apr 26 at 22:11
$begingroup$
@LCF Sure, I suppose you can put it that way. In any case, the Hamiltonian is just a different function of $x$ and $p$.
$endgroup$
– knzhou
Apr 26 at 22:31
add a comment |
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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
$$H = fracp^22m = frac12 m v^2, quad p = mv.$$
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.
answered Apr 26 at 18:45
knzhouknzhou
48.6k12133235
$endgroup$
$begingroup$
Thanks for great answer! So the "physical" effect we see in the Hamiltonian formalism is the altered dispersion relation of the particle moving in the external field?
$endgroup$
– LCF
Apr 26 at 22:11
$begingroup$
@LCF Sure, I suppose you can put it that way. In any case, the Hamiltonian is just a different function of $x$ and $p$.
$endgroup$
– knzhou
Apr 26 at 22:31
add a comment |
$begingroup$
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.
answered Apr 26 at 18:45
knzhouknzhou
48.6k12133235
$endgroup$
$begingroup$
Thanks for great answer! So the "physical" effect we see in the Hamiltonian formalism is the altered dispersion relation of the particle moving in the external field?
$endgroup$
– LCF
Apr 26 at 22:11
$begingroup$
@LCF Sure, I suppose you can put it that way. In any case, the Hamiltonian is just a different function of $x$ and $p$.
$endgroup$
– knzhou
Apr 26 at 22:31
add a comment |
$begingroup$
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.
answered Apr 26 at 18:45
knzhouknzhou
48.6k12133235
$endgroup$
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.
answered Apr 26 at 18:45
knzhouknzhou
48.6k12133235
answered Apr 26 at 18:45
knzhouknzhou
48.6k12133235
answered Apr 26 at 18:45
knzhouknzhou
48.6k12133235
answered Apr 26 at 18:45
knzhouknzhou
48.6k12133235
48.6k12133235
$begingroup$
Thanks for great answer! So the "physical" effect we see in the Hamiltonian formalism is the altered dispersion relation of the particle moving in the external field?
$endgroup$
– LCF
Apr 26 at 22:11
$begingroup$
@LCF Sure, I suppose you can put it that way. In any case, the Hamiltonian is just a different function of $x$ and $p$.
$endgroup$
– knzhou
Apr 26 at 22:31
add a comment |
$begingroup$
Thanks for great answer! So the "physical" effect we see in the Hamiltonian formalism is the altered dispersion relation of the particle moving in the external field?
$endgroup$
– LCF
Apr 26 at 22:11
$begingroup$
@LCF Sure, I suppose you can put it that way. In any case, the Hamiltonian is just a different function of $x$ and $p$.
$endgroup$
– knzhou
Apr 26 at 22:31
$begingroup$
Thanks for great answer! So the "physical" effect we see in the Hamiltonian formalism is the altered dispersion relation of the particle moving in the external field?
$endgroup$
– LCF
Apr 26 at 22:11
$begingroup$
Thanks for great answer! So the "physical" effect we see in the Hamiltonian formalism is the altered dispersion relation of the particle moving in the external field?
$endgroup$
– LCF
Apr 26 at 22:11
$begingroup$
@LCF Sure, I suppose you can put it that way. In any case, the Hamiltonian is just a different function of $x$ and $p$.
$endgroup$
– knzhou
Apr 26 at 22:31
$begingroup$
@LCF Sure, I suppose you can put it that way. In any case, the Hamiltonian is just a different function of $x$ and $p$.
$endgroup$
– knzhou
Apr 26 at 22:31
add a comment |
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