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Show that [A,B^{n}]=nB^{n1}[A,B]by dimensionless
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#1
Jun2206, 01:07 PM

P: 464

I'm having trouble figuring out the following commutator relation problem:
Suppose A and B commute with their commutator, i.e., [tex][B,[A,B]]=[A,[A,B]]=0[/tex]. Show that [tex][A,B^{n}]=nB^{n1}[A,B][/tex] I have [tex][A,B^{n}] = AB^{n}  B^{n}A[/tex] and also [tex][A,B^{n}] = AB^{n}  B^{n}A = ABB^{n1}  BB^{n1}A[/tex] I don't know where to go from here. I'm not positive the above relation is correct either. 


#2
Jun2206, 01:31 PM

Sci Advisor
HW Helper
P: 2,884

[A,BC] = B[A,C] + [A,B] C ? It's easy to prove. Just expand out. Now, use with [itex] C= B^{n1} [/itex]. , that is use [itex] [A,B^n] = B[A,B^{n1}] + [A,B] B^{n1} [/itex]. Now, repeat this again on the first term using now [itex] C= B^{n2} [/itex]. You will get a recursion formula that will give you the proof easily. 


#3
Sep2809, 01:08 AM

P: 1

Help! I need to do this same exact problem. But I just don't understand what to do so I was wondering if you wouldn't mind showing the steps that you explained to do in order to get the final solution. Thank you so much!



#4
Jan2711, 06:22 AM

P: 4

Show that [A,B^{n}]=nB^{n1}[A,B]
i also need to answer the same problem for my quantum physics course. thank you.



#5
Jan2711, 01:28 PM

Sci Advisor
P: 1,199

Here's another way:
[tex]AB^n = (AB)B^{n1}[/tex] [tex]=(BA+[A,B])B^{n1}[/tex] [tex]=BAB^{n1} + [A,B]B^{n1}[/tex] Can you understand each step? Now repeat on the first term on the right. Keep going until you end up with [itex]B^n A[/itex] plus some other stuff. According to the statement of the problem, the other stuff should end up being [itex]n[A,B]B^{n1}[/itex]. It's crucial that [itex][A,B][/itex] commutes with [itex]B[/itex] to get this final result. 


#6
Jan3011, 02:50 AM

P: 4

wohow! thanks Avodynefor the reply. got mixed up. my work all ended as a never ending subtraction of powers ofB. thanks for the idea.



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