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Commutator proof

  • Thread starter cahill8
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  • #1
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Homework Statement


Show [tex]\left[x,f(p)[/tex][tex]\right)][/tex] = [tex]i\hbar\frac{d}{dp}(f(p))\right.[/tex]


Homework 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)


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
 

Answers and Replies

  • #2
Hurkyl
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and I know I must use the commutator identity [A, BC] = [A,B]C + B[A,C]
How do you know that?
 
  • #3
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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,fnpn + [tex]\sum_{n-1}\\f_{n}p^{n}[/tex]]

using [A, B+C] = [A,B] + [A,C]

= [x, fnpn] + [x, [tex]\sum_{n-1}\\f_{n}p^{n}[/tex]]

using [A, BC] = C[A,B] + B[A,C]

= fn[x, pn] + pn[x, fn] + [x, [tex]\sum_{n-1}\\f_{n}p^{n}[/tex]]

using [x, pn] = i[tex]\hbar[/tex]npn-1

= fni[tex]\hbar[/tex]npn-1 + pn[x, fn] + [x, [tex]\sum_{n-1}\\f_{n}p^{n}[/tex]]

[x, fn] = 0 as fn is a const.

= fni[tex]\hbar[/tex]npn-1 + [x, [tex]\sum_{n-1}\\f_{n}p^{n}[/tex]]

am I on the right track?
 
Last edited:
  • #4
Hurkyl
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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
31
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I see what you mean. [x, fnpn] = fn[x, pn] 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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