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

Derivative with respect to a spinor of the free Dirac lagrangianCompleting the square for Grassmann variablesWhy doesn't this multiplication of Grassmann variables give the expected result?Grassmann variablesGrassmann numbers in the dual spaceClassical Fermion and Grassmann numberDerivative with respect to a spinor of the free Dirac lagrangianHow are supersymmetry transformations even defined?Can you quantize Grassmann-even superfields in the same fashion as Boson fields?Equivalence between Dirac and Majorana action in CFTHow does canonical quantization work with Grassmann variables?

2

$begingroup$

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 7 hours ago

Qmechanic♦

109k122061277

asked 8 hours ago

David AlbandeaDavid Albandea

604

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yup, they are defined to anticommute.
  $endgroup$
  – knzhou
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  related/possible dup.: Derivative with respect to a spinor of the free Dirac lagrangian.
  $endgroup$
  – AccidentalFourierTransform
  4 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

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 7 hours ago

Qmechanic♦

109k122061277

asked 8 hours ago

David AlbandeaDavid Albandea

604

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yup, they are defined to anticommute.
  $endgroup$
  – knzhou
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  related/possible dup.: Derivative with respect to a spinor of the free Dirac lagrangian.
  $endgroup$
  – AccidentalFourierTransform
  4 hours ago
  

  
  
  
  

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 7 hours ago

Qmechanic♦

109k122061277

asked 8 hours ago

David AlbandeaDavid Albandea

604

$endgroup$

share|cite|improve this question

edited 7 hours ago

Qmechanic♦

109k122061277

asked 8 hours ago

David AlbandeaDavid Albandea

604

share|cite|improve this question

edited 7 hours ago

Qmechanic♦

109k122061277

asked 8 hours ago

David AlbandeaDavid Albandea

604

edited 7 hours ago

Qmechanic♦

109k122061277

edited 7 hours ago

Qmechanic♦

109k122061277

asked 8 hours ago

David AlbandeaDavid Albandea

604

asked 8 hours ago

David AlbandeaDavid Albandea

604

2 Answers
2

active

oldest

votes

6

$begingroup$

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 4 hours ago

answered 7 hours ago

Qmechanic♦Qmechanic

109k122061277

$endgroup$

add a comment | 

1

$begingroup$

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 3 hours ago

DanielCDanielC

1,7571920

$endgroup$

add a comment | 

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6

$begingroup$

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 4 hours ago

answered 7 hours ago

Qmechanic♦Qmechanic

109k122061277

$endgroup$

add a comment | 

6

$begingroup$

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 4 hours ago

answered 7 hours ago

Qmechanic♦Qmechanic

109k122061277

$endgroup$

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 4 hours ago

answered 7 hours ago

Qmechanic♦Qmechanic

109k122061277

$endgroup$

share|cite|improve this answer

edited 4 hours ago

answered 7 hours ago

Qmechanic♦Qmechanic

109k122061277

answered 7 hours ago

Qmechanic♦Qmechanic

109k122061277

answered 7 hours ago

Qmechanic♦Qmechanic

109k122061277

1

$begingroup$

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 3 hours ago

DanielCDanielC

1,7571920

$endgroup$

add a comment | 

1

$begingroup$

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 3 hours ago

DanielCDanielC

1,7571920

$endgroup$

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 3 hours ago

DanielCDanielC

1,7571920

$endgroup$

share|cite|improve this answer

answered 3 hours ago

DanielCDanielC

1,7571920

answered 3 hours ago

DanielCDanielC

1,7571920

answered 3 hours ago

DanielCDanielC

1,7571920