Question

Is there some trick to finding the commutator [x,p^n]

Answer 1

we can apply the mathematical induction principle.
For example, we have the subsequent steps, for \(n=1,\;2,\;3\):
\[\Large \begin{gathered}
\left[ {x,p} \right] = i\hbar \hfill \\
\hfill \\
\left[ {x,{p^2}} \right] = \left[ {x,p} \right]p + p\left[ {x,p} \right] = 2i\hbar p \hfill \\
\hfill \\
\left[ {x,{p^3}} \right] = \left[ {x,p{p^2}} \right] = \left[ {x,{p^2}} \right]p + {p^2}\left[ {x,p} \right] = \hfill \\
\hfill \\
= 2i\hbar pp + i\hbar {p^2} = 3i\hbar {p^2} \hfill \\
\end{gathered} \]
Then I make the subsequent conjecture:
\[\Large \boxed{\left[ {x,{p^n}} \right] = ni\hbar {p^{n - 1}}}\]
then I consider the case \(n+1\), so I get:
\[\Large \begin{gathered}
\left[ {x,{p^{n + 1}}} \right] = \left[ {x,p{p^n}} \right] = p\left[ {x,{p^n}} \right] + \left[ {x,p} \right]{p^n} = \hfill \\
\hfill \\
= pni\hbar {p^{n - 1}} + i\hbar {p^n} = \left( {n + 1} \right)i\hbar {p^n} \hfill \\
\end{gathered} \]
which shows that the above conjecture is true \(\forall n \in \mathbb{N}\)

Answer 2

Awesome, exactly the kind of thing I needed to do stuff with the translation operator's taylor series.

Answer 3

Thanks! :D

Answer 4

:)