Grassmann Numbers & Commutation Relations

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 2K views
RedX
Messages
963
Reaction score
3
If you have a Grassman number [tex]\eta[/tex] that anticommutes with the creation and annihilation operators, then is the expression:

[tex]<0|\eta|0>[/tex]

well defined? Because you can write this as:

[tex]<1|a^{\dagger} \eta a|1>=-<1| \eta a^{\dagger} a|1><br /> =-<1|\eta|1>[/tex]

But if [tex]\eta[/tex] is a constant, then shouldn't:


[tex]<0|\eta|0>=<1|\eta|1>=\eta[/tex] ?
 
Physics news on Phys.org
RedX said:
If you have a Grassman number [tex]\eta[/tex] that anticommutes with the creation and annihilation operators, then is the expression:

[tex]<0|\eta|0>[/tex]

well defined? Because you can write this as:

[tex]<1|a^{\dagger} \eta a|1>=-<1| \eta a^{\dagger} a|1><br /> =-<1|\eta|1>[/tex]

But if [tex]\eta[/tex] is a constant, then shouldn't:


[tex]<0|\eta|0>=<1|\eta|1>=\eta[/tex] ?

Grassmann numbers are operators (though they are called numbers).
[tex]<0|\eta|0>=0[/tex] is well-defined and vanishes.