Grassmann Numbers & Commutation Relations

[RedX]

If you have a Grassman number \eta that anticommutes with the creation and annihilation operators, then is the expression:

<0|\eta|0>

well defined? Because you can write this as:

&lt;1|a^{\dagger} \eta a|1&gt;=-&lt;1| \eta a^{\dagger} a|1&gt;<br /> =-&lt;1|\eta|1&gt;

But if \eta is a constant, then shouldn't:


&lt;0|\eta|0&gt;=&lt;1|\eta|1&gt;=\eta ?

 

[A. Neumaier]

If you have a Grassman number \eta that anticommutes with the creation and annihilation operators, then is the expression:

&lt;0|\eta|0&gt;

well defined? Because you can write this as:

&lt;1|a^{\dagger} \eta a|1&gt;=-&lt;1| \eta a^{\dagger} a|1&gt;<br /> =-&lt;1|\eta|1&gt;

But if \eta is a constant, then shouldn't:


&lt;0|\eta|0&gt;=&lt;1|\eta|1&gt;=\eta ?

Grassmann numbers are operators (though they are called numbers).
&lt;0|\eta|0&gt;=0 is well-defined and vanishes.