How does canonical quantization work with Grassmann variables?












6














Every quantum field theory textbook I've encountered seems to have the same logical oversight, because of the particular order they cover topics.



First, the books introduce the Dirac Lagrangian,
$$mathcal{L} = bar{psi}(i notpartial - m) psi.$$
To compute the canonical momentum, we note that
$$mathcal{L} supset psi^dagger gamma^0 (i partial_0 gamma^0 psi) = i psi^dagger dot{psi}$$
in mostly negative signature. Therefore, the canonical momentum is
$$frac{partial mathcal{L}}{partial dot{psi}} = i psi^dagger.$$
One then goes on to perform canonical quantization.



Later, the books introduce the Majorana Lagrangian, which in Peskin and Schroeder (problem 3.4) has the form
$$mathcal{L} = chi^dagger i bar{sigma} cdot partial chi + frac{im}{2} (chi^T sigma^2 chi - chi^dagger sigma^2 chi^*).$$
The Majorana mass term vanishes at the classical level because $sigma^2$ is an antisymmetric matrix. The only way out is to postulate that the two-component spinor $chi$ is really a Grassmann variable, so that the two terms in the mass term have the same sign after anticommutation. It is usually stated that, in general, every spinor in a classical Lagrangian has to be a Grassmann number.



However, this contradicts the earlier treatment of the Dirac Lagrangian. If we treat $psi$ as a Grassmann number, then we pick up a sign upon anticommuting the Grassmann derivative, so
$$frac{partial mathcal{L}}{partial dot{psi}} = frac{partial}{partial dot{psi}} (i psi^dagger dot{psi}) = - i psi^dagger frac{partial}{partial dot{psi}} dot{psi} = - i psi^dagger.$$
This extra negative sign completely changes the result of canonical quantization, e.g. it leads to a disastrous negative definite energy. The same problem seems to occur in problem 3.4 of Peskin. If one correctly accounts for the Grassmann sign flip when performing canonical quantization, then one arrives at anticommutation relations that are opposite those given in the problem.



I've searched through a stack of quantum field theory textbooks, and frustratingly, not one of them even mentions this apparent inconsistency, because they all cover the Majorana Lagrangian (and Grassmann numbers) after they've finished the Dirac Lagrangian, so there's no opportunity for this issue to come up. One could avoid this issue by saying that Grassmann numbers only appear in the path integral, but then it becomes impossible to canonically quantize the Majorana theory because the mass term vanishes, which seems even worse. What's going on here?










share|cite|improve this question






















  • I think I have an answer to your question, but just to clarify: your issue is the fact that, if the $psi$ are Grassmann numbers, then for the momentum $Pi$ you may get either $Pi=ibar{psi}$ or $Pi=-ibar{psi}$ on the Dirac lagrangian? (and, consequently, you encounter the same problem on the Majorana one)
    – Luthien
    5 hours ago










  • @Luthien Yes, naively it seems to me that we get the latter, but need the former.
    – knzhou
    5 hours ago












  • Possible duplicates: physics.stackexchange.com/q/186952/2451 , physics.stackexchange.com/q/43502/2451 and links therein.
    – Qmechanic
    4 hours ago


















6














Every quantum field theory textbook I've encountered seems to have the same logical oversight, because of the particular order they cover topics.



First, the books introduce the Dirac Lagrangian,
$$mathcal{L} = bar{psi}(i notpartial - m) psi.$$
To compute the canonical momentum, we note that
$$mathcal{L} supset psi^dagger gamma^0 (i partial_0 gamma^0 psi) = i psi^dagger dot{psi}$$
in mostly negative signature. Therefore, the canonical momentum is
$$frac{partial mathcal{L}}{partial dot{psi}} = i psi^dagger.$$
One then goes on to perform canonical quantization.



Later, the books introduce the Majorana Lagrangian, which in Peskin and Schroeder (problem 3.4) has the form
$$mathcal{L} = chi^dagger i bar{sigma} cdot partial chi + frac{im}{2} (chi^T sigma^2 chi - chi^dagger sigma^2 chi^*).$$
The Majorana mass term vanishes at the classical level because $sigma^2$ is an antisymmetric matrix. The only way out is to postulate that the two-component spinor $chi$ is really a Grassmann variable, so that the two terms in the mass term have the same sign after anticommutation. It is usually stated that, in general, every spinor in a classical Lagrangian has to be a Grassmann number.



However, this contradicts the earlier treatment of the Dirac Lagrangian. If we treat $psi$ as a Grassmann number, then we pick up a sign upon anticommuting the Grassmann derivative, so
$$frac{partial mathcal{L}}{partial dot{psi}} = frac{partial}{partial dot{psi}} (i psi^dagger dot{psi}) = - i psi^dagger frac{partial}{partial dot{psi}} dot{psi} = - i psi^dagger.$$
This extra negative sign completely changes the result of canonical quantization, e.g. it leads to a disastrous negative definite energy. The same problem seems to occur in problem 3.4 of Peskin. If one correctly accounts for the Grassmann sign flip when performing canonical quantization, then one arrives at anticommutation relations that are opposite those given in the problem.



I've searched through a stack of quantum field theory textbooks, and frustratingly, not one of them even mentions this apparent inconsistency, because they all cover the Majorana Lagrangian (and Grassmann numbers) after they've finished the Dirac Lagrangian, so there's no opportunity for this issue to come up. One could avoid this issue by saying that Grassmann numbers only appear in the path integral, but then it becomes impossible to canonically quantize the Majorana theory because the mass term vanishes, which seems even worse. What's going on here?










share|cite|improve this question






















  • I think I have an answer to your question, but just to clarify: your issue is the fact that, if the $psi$ are Grassmann numbers, then for the momentum $Pi$ you may get either $Pi=ibar{psi}$ or $Pi=-ibar{psi}$ on the Dirac lagrangian? (and, consequently, you encounter the same problem on the Majorana one)
    – Luthien
    5 hours ago










  • @Luthien Yes, naively it seems to me that we get the latter, but need the former.
    – knzhou
    5 hours ago












  • Possible duplicates: physics.stackexchange.com/q/186952/2451 , physics.stackexchange.com/q/43502/2451 and links therein.
    – Qmechanic
    4 hours ago
















6












6








6


1





Every quantum field theory textbook I've encountered seems to have the same logical oversight, because of the particular order they cover topics.



First, the books introduce the Dirac Lagrangian,
$$mathcal{L} = bar{psi}(i notpartial - m) psi.$$
To compute the canonical momentum, we note that
$$mathcal{L} supset psi^dagger gamma^0 (i partial_0 gamma^0 psi) = i psi^dagger dot{psi}$$
in mostly negative signature. Therefore, the canonical momentum is
$$frac{partial mathcal{L}}{partial dot{psi}} = i psi^dagger.$$
One then goes on to perform canonical quantization.



Later, the books introduce the Majorana Lagrangian, which in Peskin and Schroeder (problem 3.4) has the form
$$mathcal{L} = chi^dagger i bar{sigma} cdot partial chi + frac{im}{2} (chi^T sigma^2 chi - chi^dagger sigma^2 chi^*).$$
The Majorana mass term vanishes at the classical level because $sigma^2$ is an antisymmetric matrix. The only way out is to postulate that the two-component spinor $chi$ is really a Grassmann variable, so that the two terms in the mass term have the same sign after anticommutation. It is usually stated that, in general, every spinor in a classical Lagrangian has to be a Grassmann number.



However, this contradicts the earlier treatment of the Dirac Lagrangian. If we treat $psi$ as a Grassmann number, then we pick up a sign upon anticommuting the Grassmann derivative, so
$$frac{partial mathcal{L}}{partial dot{psi}} = frac{partial}{partial dot{psi}} (i psi^dagger dot{psi}) = - i psi^dagger frac{partial}{partial dot{psi}} dot{psi} = - i psi^dagger.$$
This extra negative sign completely changes the result of canonical quantization, e.g. it leads to a disastrous negative definite energy. The same problem seems to occur in problem 3.4 of Peskin. If one correctly accounts for the Grassmann sign flip when performing canonical quantization, then one arrives at anticommutation relations that are opposite those given in the problem.



I've searched through a stack of quantum field theory textbooks, and frustratingly, not one of them even mentions this apparent inconsistency, because they all cover the Majorana Lagrangian (and Grassmann numbers) after they've finished the Dirac Lagrangian, so there's no opportunity for this issue to come up. One could avoid this issue by saying that Grassmann numbers only appear in the path integral, but then it becomes impossible to canonically quantize the Majorana theory because the mass term vanishes, which seems even worse. What's going on here?










share|cite|improve this question













Every quantum field theory textbook I've encountered seems to have the same logical oversight, because of the particular order they cover topics.



First, the books introduce the Dirac Lagrangian,
$$mathcal{L} = bar{psi}(i notpartial - m) psi.$$
To compute the canonical momentum, we note that
$$mathcal{L} supset psi^dagger gamma^0 (i partial_0 gamma^0 psi) = i psi^dagger dot{psi}$$
in mostly negative signature. Therefore, the canonical momentum is
$$frac{partial mathcal{L}}{partial dot{psi}} = i psi^dagger.$$
One then goes on to perform canonical quantization.



Later, the books introduce the Majorana Lagrangian, which in Peskin and Schroeder (problem 3.4) has the form
$$mathcal{L} = chi^dagger i bar{sigma} cdot partial chi + frac{im}{2} (chi^T sigma^2 chi - chi^dagger sigma^2 chi^*).$$
The Majorana mass term vanishes at the classical level because $sigma^2$ is an antisymmetric matrix. The only way out is to postulate that the two-component spinor $chi$ is really a Grassmann variable, so that the two terms in the mass term have the same sign after anticommutation. It is usually stated that, in general, every spinor in a classical Lagrangian has to be a Grassmann number.



However, this contradicts the earlier treatment of the Dirac Lagrangian. If we treat $psi$ as a Grassmann number, then we pick up a sign upon anticommuting the Grassmann derivative, so
$$frac{partial mathcal{L}}{partial dot{psi}} = frac{partial}{partial dot{psi}} (i psi^dagger dot{psi}) = - i psi^dagger frac{partial}{partial dot{psi}} dot{psi} = - i psi^dagger.$$
This extra negative sign completely changes the result of canonical quantization, e.g. it leads to a disastrous negative definite energy. The same problem seems to occur in problem 3.4 of Peskin. If one correctly accounts for the Grassmann sign flip when performing canonical quantization, then one arrives at anticommutation relations that are opposite those given in the problem.



I've searched through a stack of quantum field theory textbooks, and frustratingly, not one of them even mentions this apparent inconsistency, because they all cover the Majorana Lagrangian (and Grassmann numbers) after they've finished the Dirac Lagrangian, so there's no opportunity for this issue to come up. One could avoid this issue by saying that Grassmann numbers only appear in the path integral, but then it becomes impossible to canonically quantize the Majorana theory because the mass term vanishes, which seems even worse. What's going on here?







quantum-field-theory dirac-equation spinors classical-field-theory grassmann-numbers






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asked 5 hours ago









knzhouknzhou

42.6k11117205




42.6k11117205












  • I think I have an answer to your question, but just to clarify: your issue is the fact that, if the $psi$ are Grassmann numbers, then for the momentum $Pi$ you may get either $Pi=ibar{psi}$ or $Pi=-ibar{psi}$ on the Dirac lagrangian? (and, consequently, you encounter the same problem on the Majorana one)
    – Luthien
    5 hours ago










  • @Luthien Yes, naively it seems to me that we get the latter, but need the former.
    – knzhou
    5 hours ago












  • Possible duplicates: physics.stackexchange.com/q/186952/2451 , physics.stackexchange.com/q/43502/2451 and links therein.
    – Qmechanic
    4 hours ago




















  • I think I have an answer to your question, but just to clarify: your issue is the fact that, if the $psi$ are Grassmann numbers, then for the momentum $Pi$ you may get either $Pi=ibar{psi}$ or $Pi=-ibar{psi}$ on the Dirac lagrangian? (and, consequently, you encounter the same problem on the Majorana one)
    – Luthien
    5 hours ago










  • @Luthien Yes, naively it seems to me that we get the latter, but need the former.
    – knzhou
    5 hours ago












  • Possible duplicates: physics.stackexchange.com/q/186952/2451 , physics.stackexchange.com/q/43502/2451 and links therein.
    – Qmechanic
    4 hours ago


















I think I have an answer to your question, but just to clarify: your issue is the fact that, if the $psi$ are Grassmann numbers, then for the momentum $Pi$ you may get either $Pi=ibar{psi}$ or $Pi=-ibar{psi}$ on the Dirac lagrangian? (and, consequently, you encounter the same problem on the Majorana one)
– Luthien
5 hours ago




I think I have an answer to your question, but just to clarify: your issue is the fact that, if the $psi$ are Grassmann numbers, then for the momentum $Pi$ you may get either $Pi=ibar{psi}$ or $Pi=-ibar{psi}$ on the Dirac lagrangian? (and, consequently, you encounter the same problem on the Majorana one)
– Luthien
5 hours ago












@Luthien Yes, naively it seems to me that we get the latter, but need the former.
– knzhou
5 hours ago






@Luthien Yes, naively it seems to me that we get the latter, but need the former.
– knzhou
5 hours ago














Possible duplicates: physics.stackexchange.com/q/186952/2451 , physics.stackexchange.com/q/43502/2451 and links therein.
– Qmechanic
4 hours ago






Possible duplicates: physics.stackexchange.com/q/186952/2451 , physics.stackexchange.com/q/43502/2451 and links therein.
– Qmechanic
4 hours ago












1 Answer
1






active

oldest

votes


















5














When you are dealing with Grassmann numbers you have a "left derivative" and a "right" derivative. A left derivative removes the variable from the left, a right derivative removes it from the right.



Let's say we have the function:
begin{equation}
f(theta_1, theta_2)=f_0+f_1theta_1+f_2theta_2+f_3theta_1theta_2
end{equation}

Then the left derivative with respect to $theta_1$ is
begin{equation}
frac{partial_L f}{partialtheta_1}=f_1+f_3theta_2
end{equation}

while the right derivative with respect to $theta_1$ is
begin{equation}
frac{partial_R f}{partialtheta_1}=f_1-f_3theta_2
end{equation}



When you define the conjugate momenta, you may either use left or right derivatives but you have to keep track of your choice when you perform a Legendre transform to get the Hamiltonian. If you define the momentum with left derivatives, i.e. as
begin{equation}
Pi=frac{partial_L L}{partialdot{theta}}
end{equation}

then the canonical Hamiltonian has to be
begin{equation}
H=dot{theta}Pi-L
end{equation}

(you can easily see it's consistent with the definition of momentum with left derivatives) and NOT
begin{equation}
H=Pidot{theta}-L
end{equation}

(which would have worked if we had defined the momentum with right derivatives) as we may be tempted to write.



If I understood correctly, this "sign ambiguity" on the definition of momentum was your problem and this should solve it.






share|cite|improve this answer























  • This is perfect, thanks!
    – knzhou
    4 hours ago










  • Do you have a reference (book/lecture notes) where issues like this are covered in more detail?
    – knzhou
    4 hours ago










  • Glad i could help :) Personally, I studied Grassmann fromalism in "Quantization of Gauge systems" by Henneaux and Teitelboim
    – Luthien
    4 hours ago











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1 Answer
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1 Answer
1






active

oldest

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active

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active

oldest

votes









5














When you are dealing with Grassmann numbers you have a "left derivative" and a "right" derivative. A left derivative removes the variable from the left, a right derivative removes it from the right.



Let's say we have the function:
begin{equation}
f(theta_1, theta_2)=f_0+f_1theta_1+f_2theta_2+f_3theta_1theta_2
end{equation}

Then the left derivative with respect to $theta_1$ is
begin{equation}
frac{partial_L f}{partialtheta_1}=f_1+f_3theta_2
end{equation}

while the right derivative with respect to $theta_1$ is
begin{equation}
frac{partial_R f}{partialtheta_1}=f_1-f_3theta_2
end{equation}



When you define the conjugate momenta, you may either use left or right derivatives but you have to keep track of your choice when you perform a Legendre transform to get the Hamiltonian. If you define the momentum with left derivatives, i.e. as
begin{equation}
Pi=frac{partial_L L}{partialdot{theta}}
end{equation}

then the canonical Hamiltonian has to be
begin{equation}
H=dot{theta}Pi-L
end{equation}

(you can easily see it's consistent with the definition of momentum with left derivatives) and NOT
begin{equation}
H=Pidot{theta}-L
end{equation}

(which would have worked if we had defined the momentum with right derivatives) as we may be tempted to write.



If I understood correctly, this "sign ambiguity" on the definition of momentum was your problem and this should solve it.






share|cite|improve this answer























  • This is perfect, thanks!
    – knzhou
    4 hours ago










  • Do you have a reference (book/lecture notes) where issues like this are covered in more detail?
    – knzhou
    4 hours ago










  • Glad i could help :) Personally, I studied Grassmann fromalism in "Quantization of Gauge systems" by Henneaux and Teitelboim
    – Luthien
    4 hours ago
















5














When you are dealing with Grassmann numbers you have a "left derivative" and a "right" derivative. A left derivative removes the variable from the left, a right derivative removes it from the right.



Let's say we have the function:
begin{equation}
f(theta_1, theta_2)=f_0+f_1theta_1+f_2theta_2+f_3theta_1theta_2
end{equation}

Then the left derivative with respect to $theta_1$ is
begin{equation}
frac{partial_L f}{partialtheta_1}=f_1+f_3theta_2
end{equation}

while the right derivative with respect to $theta_1$ is
begin{equation}
frac{partial_R f}{partialtheta_1}=f_1-f_3theta_2
end{equation}



When you define the conjugate momenta, you may either use left or right derivatives but you have to keep track of your choice when you perform a Legendre transform to get the Hamiltonian. If you define the momentum with left derivatives, i.e. as
begin{equation}
Pi=frac{partial_L L}{partialdot{theta}}
end{equation}

then the canonical Hamiltonian has to be
begin{equation}
H=dot{theta}Pi-L
end{equation}

(you can easily see it's consistent with the definition of momentum with left derivatives) and NOT
begin{equation}
H=Pidot{theta}-L
end{equation}

(which would have worked if we had defined the momentum with right derivatives) as we may be tempted to write.



If I understood correctly, this "sign ambiguity" on the definition of momentum was your problem and this should solve it.






share|cite|improve this answer























  • This is perfect, thanks!
    – knzhou
    4 hours ago










  • Do you have a reference (book/lecture notes) where issues like this are covered in more detail?
    – knzhou
    4 hours ago










  • Glad i could help :) Personally, I studied Grassmann fromalism in "Quantization of Gauge systems" by Henneaux and Teitelboim
    – Luthien
    4 hours ago














5












5








5






When you are dealing with Grassmann numbers you have a "left derivative" and a "right" derivative. A left derivative removes the variable from the left, a right derivative removes it from the right.



Let's say we have the function:
begin{equation}
f(theta_1, theta_2)=f_0+f_1theta_1+f_2theta_2+f_3theta_1theta_2
end{equation}

Then the left derivative with respect to $theta_1$ is
begin{equation}
frac{partial_L f}{partialtheta_1}=f_1+f_3theta_2
end{equation}

while the right derivative with respect to $theta_1$ is
begin{equation}
frac{partial_R f}{partialtheta_1}=f_1-f_3theta_2
end{equation}



When you define the conjugate momenta, you may either use left or right derivatives but you have to keep track of your choice when you perform a Legendre transform to get the Hamiltonian. If you define the momentum with left derivatives, i.e. as
begin{equation}
Pi=frac{partial_L L}{partialdot{theta}}
end{equation}

then the canonical Hamiltonian has to be
begin{equation}
H=dot{theta}Pi-L
end{equation}

(you can easily see it's consistent with the definition of momentum with left derivatives) and NOT
begin{equation}
H=Pidot{theta}-L
end{equation}

(which would have worked if we had defined the momentum with right derivatives) as we may be tempted to write.



If I understood correctly, this "sign ambiguity" on the definition of momentum was your problem and this should solve it.






share|cite|improve this answer














When you are dealing with Grassmann numbers you have a "left derivative" and a "right" derivative. A left derivative removes the variable from the left, a right derivative removes it from the right.



Let's say we have the function:
begin{equation}
f(theta_1, theta_2)=f_0+f_1theta_1+f_2theta_2+f_3theta_1theta_2
end{equation}

Then the left derivative with respect to $theta_1$ is
begin{equation}
frac{partial_L f}{partialtheta_1}=f_1+f_3theta_2
end{equation}

while the right derivative with respect to $theta_1$ is
begin{equation}
frac{partial_R f}{partialtheta_1}=f_1-f_3theta_2
end{equation}



When you define the conjugate momenta, you may either use left or right derivatives but you have to keep track of your choice when you perform a Legendre transform to get the Hamiltonian. If you define the momentum with left derivatives, i.e. as
begin{equation}
Pi=frac{partial_L L}{partialdot{theta}}
end{equation}

then the canonical Hamiltonian has to be
begin{equation}
H=dot{theta}Pi-L
end{equation}

(you can easily see it's consistent with the definition of momentum with left derivatives) and NOT
begin{equation}
H=Pidot{theta}-L
end{equation}

(which would have worked if we had defined the momentum with right derivatives) as we may be tempted to write.



If I understood correctly, this "sign ambiguity" on the definition of momentum was your problem and this should solve it.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 2 hours ago

























answered 4 hours ago









LuthienLuthien

850217




850217












  • This is perfect, thanks!
    – knzhou
    4 hours ago










  • Do you have a reference (book/lecture notes) where issues like this are covered in more detail?
    – knzhou
    4 hours ago










  • Glad i could help :) Personally, I studied Grassmann fromalism in "Quantization of Gauge systems" by Henneaux and Teitelboim
    – Luthien
    4 hours ago


















  • This is perfect, thanks!
    – knzhou
    4 hours ago










  • Do you have a reference (book/lecture notes) where issues like this are covered in more detail?
    – knzhou
    4 hours ago










  • Glad i could help :) Personally, I studied Grassmann fromalism in "Quantization of Gauge systems" by Henneaux and Teitelboim
    – Luthien
    4 hours ago
















This is perfect, thanks!
– knzhou
4 hours ago




This is perfect, thanks!
– knzhou
4 hours ago












Do you have a reference (book/lecture notes) where issues like this are covered in more detail?
– knzhou
4 hours ago




Do you have a reference (book/lecture notes) where issues like this are covered in more detail?
– knzhou
4 hours ago












Glad i could help :) Personally, I studied Grassmann fromalism in "Quantization of Gauge systems" by Henneaux and Teitelboim
– Luthien
4 hours ago




Glad i could help :) Personally, I studied Grassmann fromalism in "Quantization of Gauge systems" by Henneaux and Teitelboim
– Luthien
4 hours ago


















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