Matrix commutation

  • Thread starter kreil
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kreil
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Homework Statement


Show, by series expansion, that if A and B are two matrices which do not commute, then [itex]e^{A+B} \ne e^Ae^B[/itex], but if they do commute then the relation holds.


Homework Equations


[tex]e^A=1+A[/tex]
[tex]e^B=1+B[/tex]
[tex]e^{A+B}=1+(A+B)[/tex]


The Attempt at a Solution


Assuming that the first 2 terms in the expansion is sufficient to use here (is it?), I got the following:

[tex]e^Ae^B=(1+A)(1+B)=1+A+B+AB[/tex]

This would be equal if the AB term were not tacked on the end...does this term somehow become zero when the two matrices commute? If I'm on the wrong track please let me know.

Josh
 

Answers and Replies

  • #2
kreil
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I think I figured it out after a little thought...

[tex]e^Ae^B=(1+A)(1+B)=1+A+B+AB-BA=1+A+B+[A,B][/tex]

So if A and B commute, [A,B]=0 and the relation holds, else [A,B] does not equal zero and [itex]e^{A+B} \ne e^Ae^B[/itex]
 

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