I don't understand operators

  • Thread starter silimay
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  • #1
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So I was reading from my quantum book (Gasiorowicz) and I ame across this sentence:

[tex] [p^2, x] = p [p, x] + [p, x] p = \frac{2\hbar}{i} p [/tex]

I don't understand this. I know that [tex] p = -i \hbar \frac{\partial}{\partial x} [/tex], but I can't see how to get that expression...I just come up with something like [tex] x {\hbar}^2 \frac{{\partial}^2}{{\partial x}^2} [/tex] when I try multiplying it out.
 

Answers and Replies

  • #2
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try to derive [tex][AB,C] = ?[/tex]
then use [tex][x,p] = i\hbar[/tex]
to find [tex][p^2,x][/tex]
 
  • #3
nrqed
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So I was reading from my quantum book (Gasiorowicz) and I ame across this sentence:

[tex] [p^2, x] = p [p, x] + [p, x] p = \frac{2\hbar}{i} p [/tex]

I don't understand this. I know that [tex] p = -i \hbar \frac{\partial}{\partial x} [/tex], but I can't see how to get that expression...I just come up with something like [tex] x {\hbar}^2 \frac{{\partial}^2}{{\partial x}^2} [/tex] when I try multiplying it out.

All those expressions only make sense if you imagine applying them to some "test function" f(x). For example,

[tex] [x,p_x] f(x) = -i \hbar x \partial_x f(x) + i \hbar \partial_x (x f(x)) [/tex] apply the product rule on the second term and then something will cancel out. At the very end of the calculation (and only then) you may drop the test function f(x).

Using a test function is not the fastest way to prove complicated commutation relations, however. But it's the only way to make sense of these commutation relations. after you have done a few with a test function you will be able to do the more complex cases without the crutch of a test function.
 

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