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I am looking for the commutator: $$[e^{aq},p]$$ My approach is to Taylor expand the function: $$[\sum_n \frac{1}{n!}(aq)^n,p]$$ I know that $[q^n,p]=ni\hbar q^{n-1}$ So how do I account for $n$ commutators?

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    $\begingroup$ Can you clarify what you mean by "accounting" for $n$ commutators? Are you looking for a proof of the last relation that you mention you already know? $\endgroup$ – joshphysics Feb 8 '14 at 19:45
  • $\begingroup$ Related: physics.stackexchange.com/q/78222/2451 and links therein. $\endgroup$ – Qmechanic Feb 8 '14 at 20:09
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A more general relation than $\left[q^n, p \right] = i\hbar nq^{n-1}$ is $$\left[f(A), B\right] = \left[A, B \right]\frac{\partial f}{\partial A}$$ if $\left[A, \left[A, B \right]\right] =0$. In this case we are okay to use this because $\left[q, p \right] =i\hbar$ is just a number, and that commutes with everything. So you can just apply this formula directly.

Another thing perhaps worth noting is that because the derivative is a linear operator, you can apply it to the individual terms of an infinite sum.

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  • $\begingroup$ How do I calculate the more general case where [A,B] is another operator? Does it go before or after the derivative? $\endgroup$ – psitae Jun 11 '18 at 16:37
  • $\begingroup$ If $[A,B]$ commutes with $A$, then it's the same and does not matter whether it is before or after. If $[A, B]$ does not commute with $A$, the entire expression becomes more complicated in general. Suppose f has a Taylor expansion in powers of $A^n$. Then $[A^n, B] = A^{n-1}[A,B] + [A^{n-1}, B]A $. By recursion this is $[A,B]nA^{n-1}$ if $[A, [A, B]=0$. If not then you can't reorder the terms so the $[A,B]$ ends up sandwiched between $A$s. But it can be done recursively from $[A^n, B] = A^{n-1}[A,B] + [A^{n-1}, B]A $. $\endgroup$ – WhatIAm Jun 12 '18 at 20:18
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It seems to me that the approach formulated by the OP is perfectly sound! The OP demonstrates that he is familiar with the commutator between q^n and p. In order to use this formula, he proposes to apply Taylor series expansion to the target function. That is okay.

The only problem is that the OP got cold feet and stopped his calculation at this point. Actually the end result was almost in sight. Substitution of the special commutator into the Taylor series leads to a formula that can be re-summed easily. It is just another version of the Taylor series of an exponential function!

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If $q$ and $p$ satisfy the canonical commutation relation, $[q,p]=i\hbar$, then you can use the relation between the classical Poisson brackets and commutators: $$ \left[A,B\right]_{classical}\to\frac{1}{i\hbar}\left[A,B\right]\tag{1} $$ I'll assume $A=A(q,p)$ and $B=B(q,p)$ for now. The classical Poisson brackets are give by $$ \left[A,B\right]_{classical}=\frac{\partial A}{\partial q}\frac{\partial B}{\partial p} - \frac{\partial A}{\partial p}\frac{\partial B}{\partial q}\tag{2} $$ Since you have $A=A(q)$ and $B=p$, then Equation (1) via Equation (2) gives $$ \left[A(q),p\right]=i\hbar\frac{\partial A(q)}{\partial q}\tag{3} $$

However, the relationship in (1) isn't always correct (depends on the "niceness" of the function $A(q)$, see this and this). As discussed in another answer of mine (among other places), one would need the Moyal bracket that introduces a correction factor proportional to $\hbar^2$. However, for the current problem of $A(q)\sim\exp(aq)$, (3) would suffice.

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  • $\begingroup$ This is not always true! The evaluation of the bracket in term of the classical Poisson bracket is only valid in general for quadratic functions. See canonical quantization on wikipedia. $\endgroup$ – G. Bergeron Aug 15 '15 at 8:05
  • $\begingroup$ @G.Bergeron: Correct. I discuss this in a later answer to a similar question; I just haven't corrected this one. $\endgroup$ – Kyle Kanos Aug 15 '15 at 12:21

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