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Grassmann Algebra

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Exploring<strong>Grassmann</strong><strong>Algebra</strong>.nb 43<br />

<strong>Grassmann</strong>Function��Exp�x� Exp�y�, �x, y��, �X, Y��<br />

� Ξ0 �Ψ0 �� Ξ0 �Ψ0 e1 �Ξ1 �Ψ1� �� Ξ0 �Ψ0 e2 �Ξ2 �Ψ2� �<br />

� Ξ0 �Ψ0 �Ξ3 �Ξ2 Ψ1 �Ξ1 Ψ2 �Ψ3� e1 � e2<br />

<strong>Grassmann</strong>Function��Exp�x� Exp�y�, �x, y��, �Y, X��<br />

� Ξ0 �Ψ0 �� Ξ0 �Ψ0 e1 �Ξ1 �Ψ1� �� Ξ0 �Ψ0 e2 �Ξ2 �Ψ2� �<br />

� Ξ0 �Ψ0 �Ξ3 �Ξ2 Ψ1 �Ξ1 Ψ2 �Ψ3� e1 � e2<br />

On the other hand, if we wish to compute the exponential of the sum X + Y we need to use<br />

<strong>Grassmann</strong>Function, because there can be two possibilities for its definition. That is,<br />

although Exp[x+y] appears to be independent of the order of its arguments, an interchange in<br />

the 'ordering' argument from {X,Y} to {Y,X} will change the signs in the result.<br />

<strong>Grassmann</strong>Function��Exp�x � y�, �x, y��, �X, Y��<br />

� Ξ0 �Ψ0 �� Ξ0 �Ψ0 e1 �Ξ1 �Ψ1� �� Ξ0 �Ψ0 e2 �Ξ2 �Ψ2� �<br />

� Ξ0 �Ψ0 �Ξ3 �Ξ2 Ψ1 �Ξ1 Ψ2 �Ψ3� e1 � e2<br />

<strong>Grassmann</strong>Function��Exp�x � y�, �x, y��, �Y, X��<br />

� Ξ0 �Ψ0 �� Ξ0 �Ψ0 e1 �Ξ1 �Ψ1� �� Ξ0 �Ψ0 e2 �Ξ2 �Ψ2� �<br />

� Ξ0 �Ψ0 �Ξ3 �Ξ2 Ψ1 �Ξ1 Ψ2 �Ψ3� e1 � e2<br />

In sum: A function of several <strong>Grassmann</strong> variables may have different results depending on the<br />

ordering of the variables in the function, even if their usual (scalar) form is commutative.<br />

2001 4 5

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