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Is the derivative with respect to a fermion field Grassmann-odd?

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3

1

$begingroup$

Fermion fields anticommute because they are Grassmann numbers, that is,
beginequation
psi chi = - chi psi.
endequation
I was wondering whether derivatives with respect to Grassmann numbers also anticommute, as in

beginequation
fracpartialpartial psi (bar psi psi) stackrel?= - bar psi fracpartialpartial psi (psi) = - bar psi.
endequation

field-theory differentiation fermions grassmann-numbers superalgebra

share|cite|improve this question

edited May 24 at 17:35

Qmechanic♦

109k122081280

asked May 24 at 17:00

David AlbandeaDavid Albandea

724

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yup, they are defined to anticommute.
  $endgroup$
  – knzhou
  May 24 at 17:03
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  related/possible dup.: Derivative with respect to a spinor of the free Dirac lagrangian.
  $endgroup$
  – AccidentalFourierTransform
  May 24 at 21:14
  

  
  
  
  

add a comment | 

3

1

$begingroup$

Fermion fields anticommute because they are Grassmann numbers, that is,
beginequation
psi chi = - chi psi.
endequation
I was wondering whether derivatives with respect to Grassmann numbers also anticommute, as in

beginequation
fracpartialpartial psi (bar psi psi) stackrel?= - bar psi fracpartialpartial psi (psi) = - bar psi.
endequation

field-theory differentiation fermions grassmann-numbers superalgebra

share|cite|improve this question

edited May 24 at 17:35

Qmechanic♦

109k122081280

asked May 24 at 17:00

David AlbandeaDavid Albandea

724

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yup, they are defined to anticommute.
  $endgroup$
  – knzhou
  May 24 at 17:03
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  related/possible dup.: Derivative with respect to a spinor of the free Dirac lagrangian.
  $endgroup$
  – AccidentalFourierTransform
  May 24 at 21:14
  

  
  
  
  

add a comment | 

$begingroup$

Fermion fields anticommute because they are Grassmann numbers, that is,
beginequation
psi chi = - chi psi.
endequation
I was wondering whether derivatives with respect to Grassmann numbers also anticommute, as in

beginequation
fracpartialpartial psi (bar psi psi) stackrel?= - bar psi fracpartialpartial psi (psi) = - bar psi.
endequation

field-theory differentiation fermions grassmann-numbers superalgebra

share|cite|improve this question

edited May 24 at 17:35

Qmechanic♦

109k122081280

asked May 24 at 17:00

David AlbandeaDavid Albandea

724

$endgroup$

share|cite|improve this question

edited May 24 at 17:35

Qmechanic♦

109k122081280

asked May 24 at 17:00

David AlbandeaDavid Albandea

724

share|cite|improve this question

edited May 24 at 17:35

Qmechanic♦

109k122081280

asked May 24 at 17:00

David AlbandeaDavid Albandea

724

edited May 24 at 17:35

Qmechanic♦

109k122081280

edited May 24 at 17:35

Qmechanic♦

109k122081280

asked May 24 at 17:00

David AlbandeaDavid Albandea

724

asked May 24 at 17:00

David AlbandeaDavid Albandea

724

2 Answers
2

active

oldest

votes

7

$begingroup$

1. Since$^1$ $$(fracpartialpartial zz)~=~1tag1$$
   for any supernumber-valued variable $z$, the Grassmann-parity of the partial derivative $fracpartialpartial z$ should be the same as the Grassmann-parity of $z$ in order for eq. (1) to preserve Grassmann-parity.




2. In superspace $mathbbR^mni(x,theta)$ a functional derivative $fracdeltadelta z(x,theta)$ and its superfield $z(x,theta)$ carry the same (opposite) Grassmann parity if the number $m$ of $theta$'s is even (odd), respectively:
   $$ (fracdeltadelta z(x,theta)z(x^prime,theta^prime))~=~delta^n(x!-!x^prime)delta^m(theta!-!theta^prime) tag2.$$

--

$^1$ The parenthesis on the left-hand side of eq. (1) is supposed to indicate that the derivative $fracpartialpartial z$ does not act past $z$.

share|cite|improve this answer

edited May 24 at 21:09

answered May 24 at 17:33

Qmechanic♦Qmechanic

109k122081280

$endgroup$

add a comment | 

2

$begingroup$

In a Grassmann algebra (or more pedantic, a $mathbbZ_2$-graded algebra), the expression with two equality signs you wrote is ill-defined. One either has a left-derivative, or a right-derivative, which are different operators in terms of results. More precisely,

$$ fracpartial^Lpartial psi left(barpsipsiright) = (-)^epsilon(barpsi)epsilon(psi) barpsi $$

$$ fracpartial^Rpartial psi left(barpsipsiright) = barpsi $$,

where the epsilons are the Grassmann parities of the two variables. if both variables are Grassmann-odd, then the results of the operators are different (one is minus the other).

share|cite|improve this answer

answered May 24 at 21:32

DanielCDanielC

1,7971920

$endgroup$

add a comment | 

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7

$begingroup$

1. Since$^1$ $$(fracpartialpartial zz)~=~1tag1$$
   for any supernumber-valued variable $z$, the Grassmann-parity of the partial derivative $fracpartialpartial z$ should be the same as the Grassmann-parity of $z$ in order for eq. (1) to preserve Grassmann-parity.




2. In superspace $mathbbR^mni(x,theta)$ a functional derivative $fracdeltadelta z(x,theta)$ and its superfield $z(x,theta)$ carry the same (opposite) Grassmann parity if the number $m$ of $theta$'s is even (odd), respectively:
   $$ (fracdeltadelta z(x,theta)z(x^prime,theta^prime))~=~delta^n(x!-!x^prime)delta^m(theta!-!theta^prime) tag2.$$

--

$^1$ The parenthesis on the left-hand side of eq. (1) is supposed to indicate that the derivative $fracpartialpartial z$ does not act past $z$.

share|cite|improve this answer

edited May 24 at 21:09

answered May 24 at 17:33

Qmechanic♦Qmechanic

109k122081280

$endgroup$

add a comment | 

7

$begingroup$

1. Since$^1$ $$(fracpartialpartial zz)~=~1tag1$$
   for any supernumber-valued variable $z$, the Grassmann-parity of the partial derivative $fracpartialpartial z$ should be the same as the Grassmann-parity of $z$ in order for eq. (1) to preserve Grassmann-parity.




2. In superspace $mathbbR^mni(x,theta)$ a functional derivative $fracdeltadelta z(x,theta)$ and its superfield $z(x,theta)$ carry the same (opposite) Grassmann parity if the number $m$ of $theta$'s is even (odd), respectively:
   $$ (fracdeltadelta z(x,theta)z(x^prime,theta^prime))~=~delta^n(x!-!x^prime)delta^m(theta!-!theta^prime) tag2.$$

--

$^1$ The parenthesis on the left-hand side of eq. (1) is supposed to indicate that the derivative $fracpartialpartial z$ does not act past $z$.

share|cite|improve this answer

edited May 24 at 21:09

answered May 24 at 17:33

Qmechanic♦Qmechanic

109k122081280

$endgroup$

add a comment | 

$begingroup$

1. Since$^1$ $$(fracpartialpartial zz)~=~1tag1$$
   for any supernumber-valued variable $z$, the Grassmann-parity of the partial derivative $fracpartialpartial z$ should be the same as the Grassmann-parity of $z$ in order for eq. (1) to preserve Grassmann-parity.




2. In superspace $mathbbR^mni(x,theta)$ a functional derivative $fracdeltadelta z(x,theta)$ and its superfield $z(x,theta)$ carry the same (opposite) Grassmann parity if the number $m$ of $theta$'s is even (odd), respectively:
   $$ (fracdeltadelta z(x,theta)z(x^prime,theta^prime))~=~delta^n(x!-!x^prime)delta^m(theta!-!theta^prime) tag2.$$

--

$^1$ The parenthesis on the left-hand side of eq. (1) is supposed to indicate that the derivative $fracpartialpartial z$ does not act past $z$.

share|cite|improve this answer

edited May 24 at 21:09

answered May 24 at 17:33

Qmechanic♦Qmechanic

109k122081280

$endgroup$

share|cite|improve this answer

edited May 24 at 21:09

answered May 24 at 17:33

Qmechanic♦Qmechanic

109k122081280

answered May 24 at 17:33

Qmechanic♦Qmechanic

109k122081280

answered May 24 at 17:33

Qmechanic♦Qmechanic

109k122081280

2

$begingroup$

In a Grassmann algebra (or more pedantic, a $mathbbZ_2$-graded algebra), the expression with two equality signs you wrote is ill-defined. One either has a left-derivative, or a right-derivative, which are different operators in terms of results. More precisely,

$$ fracpartial^Lpartial psi left(barpsipsiright) = (-)^epsilon(barpsi)epsilon(psi) barpsi $$

$$ fracpartial^Rpartial psi left(barpsipsiright) = barpsi $$,

where the epsilons are the Grassmann parities of the two variables. if both variables are Grassmann-odd, then the results of the operators are different (one is minus the other).

share|cite|improve this answer

answered May 24 at 21:32

DanielCDanielC

1,7971920

$endgroup$

add a comment | 

2

$begingroup$

In a Grassmann algebra (or more pedantic, a $mathbbZ_2$-graded algebra), the expression with two equality signs you wrote is ill-defined. One either has a left-derivative, or a right-derivative, which are different operators in terms of results. More precisely,

$$ fracpartial^Lpartial psi left(barpsipsiright) = (-)^epsilon(barpsi)epsilon(psi) barpsi $$

$$ fracpartial^Rpartial psi left(barpsipsiright) = barpsi $$,

where the epsilons are the Grassmann parities of the two variables. if both variables are Grassmann-odd, then the results of the operators are different (one is minus the other).

share|cite|improve this answer

answered May 24 at 21:32

DanielCDanielC

1,7971920

$endgroup$

add a comment | 

$begingroup$

In a Grassmann algebra (or more pedantic, a $mathbbZ_2$-graded algebra), the expression with two equality signs you wrote is ill-defined. One either has a left-derivative, or a right-derivative, which are different operators in terms of results. More precisely,

$$ fracpartial^Lpartial psi left(barpsipsiright) = (-)^epsilon(barpsi)epsilon(psi) barpsi $$

$$ fracpartial^Rpartial psi left(barpsipsiright) = barpsi $$,

where the epsilons are the Grassmann parities of the two variables. if both variables are Grassmann-odd, then the results of the operators are different (one is minus the other).

share|cite|improve this answer

answered May 24 at 21:32

DanielCDanielC

1,7971920

$endgroup$

share|cite|improve this answer

answered May 24 at 21:32

DanielCDanielC

1,7971920

answered May 24 at 21:32

DanielCDanielC

1,7971920

answered May 24 at 21:32

DanielCDanielC

1,7971920