Math Is Fun Forum

Talvon

Member

Registered: 2006-11-15

Posts: 16

Proof of [AB,C]=A[B,C]+[A,C]B?

I have to prove the theory above, the problem is I am a physics student and I am only aware that this is an application of ring theory and is Leibniz algebra by chance from another forum researching this problem, and so far I haven't been able to find a proof.

I tried it using twists on the commutator rule from quantum mechanics ([A,B]=AB-BA - I don't know how to handle commutators with powers in them ), and I got:

Q²R-RQ² => QR[Q]-[Q]

Which isn't working <_<

Any pointers would be greatly appreciated, as I haven't studied ring theory and it isn't part of my syllabus, so I don't know where to start

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Proof of [AB,C]=A[B,C]+[A,C]B?

Just simply expand each one:

[AB, C] = ABC - CAB
A[B,C] = A(BC - CB) = ABC - ACB
[A,C]B = (AC - CA)B = ACB - CAB

So we get:

[AB, C] = ABC - CAB = ABC + (ACB - ACB) - CAB = (ABC - ACB) + (ACB - CAB) = A(BC - CB) + (AC - CA)B = A[B,C] + [A,C]B

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

Talvon

Member

Registered: 2006-11-15

Posts: 16

Re: Proof of [AB,C]=A[B,C]+[A,C]B?

Thanks for that, I can follow the proof but I'm unsure where A[B,C] and [A,C]B initially come from. Is it from the initial expansion, ie:

Does [AB, C] = ABC - CAB equal the same as [AB, C] = A[B,C] - [C,A]B?

Also, is a substitution made for [A,C]B in that it initially starts as [C,A]B but [C,A]= -[A,C]? It would explain the addition of the 2 rather than the subtraction

Thanks again

Edit:

Never mind I read it wrong, I missed the point of adding in the (ACB-ACB) bracket, so doing it again with that in mind and re-arranging it it worked out :

So its ok to just put the factor of (ACB-ACB) in there, as it would just cancel itself without the re-arrangement anyway?

Cheers

Last edited by Talvon (2008-10-28 04:11:03)

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Proof of [AB,C]=A[B,C]+[A,C]B?

So its ok to just put the factor of (ACB-ACB) in there, as it would just cancel itself without the re-arrangement anyway?

Yes.

You are right that [A, C] = -[C, A]:

[A,C] = AC - CA = -(-AC + CA) = -(CA - AC) = -[C, A]

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."