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Commutator proof 
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#1
Mar2710, 10:23 PM

P: 31

1. The problem statement, all variables and given/known data
Show [tex]\left[x,f(p)[/tex][tex]\right)][/tex] = [tex]i\hbar\frac{d}{dp}(f(p))\right.[/tex] 2. Relevant equations I can use [tex]\left[x,p^{n}[/tex][tex]\right)][/tex] = [tex]i\hbar\\n\right.[/tex][tex]p^{n}\right.[/tex] f(p) = [tex]\Sigma[/tex] [tex]f_{n}[/tex][tex]p^{n}[/tex] (power series expansion) 3. The attempt at a solution I started by expanding f(p) to the power series which makes [tex]\left[x,\Sigma\\f_{n}\\p^{n}[/tex][tex]\right)][/tex] and I know I must use the commutator identity [A, BC] = [A,B]C + B[A,C] but the power series cannot be split up into two products(BC) ? So I'm not sure how to go on 


#3
Mar2810, 03:13 AM

P: 31

In a text book it says it can be shown using that equation
Trying a different method: [x, f(p)] = [x,[tex]\sum_{n}\\f_{n}p^{n}[/tex]] = [x,f_{n}p^{n} + [tex]\sum_{n1}\\f_{n}p^{n}[/tex]] using [A, B+C] = [A,B] + [A,C] = [x, f_{n}p^{n}] + [x, [tex]\sum_{n1}\\f_{n}p^{n}[/tex]] using [A, BC] = C[A,B] + B[A,C] = f_{n}[x, p^{n}] + p^{n}[x, f_{n}] + [x, [tex]\sum_{n1}\\f_{n}p^{n}[/tex]] using [x, p^{n}] = i[tex]\hbar[/tex]np^{n1} = f_{n}i[tex]\hbar[/tex]np^{n1} + p^{n}[x, f_{n}] + [x, [tex]\sum_{n1}\\f_{n}p^{n}[/tex]] [x, f_{n}] = 0 as f_{n} is a const. = f_{n}i[tex]\hbar[/tex]np^{n1} + [x, [tex]\sum_{n1}\\f_{n}p^{n}[/tex]] am I on the right track? 


#4
Mar2810, 05:49 AM

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P: 16,099

Commutator proof
I'm curious why you used
[A, rC] = [A,r]C + r[A,C]to pull out a scalar, rather than just using [A, rC] = r [A,C]I'm also curious why you stopped using [A, B + C] = [A,B] + [A,C]after a single addition. But that aside, everything you wrote looks correct. We won't know if you're on the right track until we see where this path leads, though! 


#5
Mar2810, 06:03 AM

P: 31

I see what you mean. [x, f_{n}p^{n}] = f_{n}[x, p^{n}] is fine.
I kept going with the addition and noticed a pattern and managed to solve it. Thanks for the hints :) 


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