# Is there a simple expression for $[x,e^{ixp}]$?

I'm sure this exists somewhere, but somewhat surprisingly it is not that easy to google.* The commutators $$\left[x,e^{i(ax^2+b(xp+px)+cp^2)}\right]$$ of position and the exponential of a quadratic function of position and momentum are definitely known to death, not least because the theory of gaussian states deals with them all the time in the form of quadratic functions of field quadratures.

I'm particularly interested in functions of the specific type $$\left[x_i,e^{ix_jA_{jk}p_k}\right],$$ where I know that $A_{jj}=0$ so there's no hermiticity issues (Einstein summations on both). Has anyone got a reference handy?

* Hence, for posterity, this question. Any suggestions on making this more search-friendly are welcome.

• Wouldn't brute force Taylor expansion work? After all, the method is always to reduce to some sort of $[A,BC]$ and then move $B$ and $C$ to the left and to the right, respectively, grouping everything back eventually. – gented Oct 8 '15 at 22:30
• More on $[A, \exp(B)]$. – Qmechanic Oct 10 '15 at 21:28

Let's try the Baker–Campbell–Hausdorff formula in the form $$e^X Y e^{-X} = Y + [X, Y] + \frac{1}{2!}[X,[X,Y]] + \frac{1}{3!}[X,[X,[X,Y]]]\; + \;...$$ Take $X = ix_jA_{jk}p_k$ and $Y = x_i$. It is not the simplest case since the first commutator, $$[X, Y] = [ix_jA_{jk}p_k, x_i] = ix_jA_{jk}(-i\delta_{ki}) = x_jA_{ji}$$ does not commute with $X = ix_jA_{jk}p_k$. But, let's look at the higher order commutators: $$[X, [X, Y]] = [ix_jA_{jk}p_k, x_{j'}A_{j'i}] = ix_jA_{jk}[p_k, x_{j'}]A_{j'i} = ix_jA_{jk} (-i \delta_{kj'})A_{j'i} = x_jA_{jk}A_{ki}$$ Similarly, $$[X, [X, [X, Y]]] =x_jA_{jk}A_{kl}A_{li}\;\; \text{etc.}$$
Let's take everything in the BCH formula: $$e^{ix_jA_{jk}p_k} x_i e^{-ix_jA_{jk}p_k} = x_j\delta_{ji} + x_jA_{ji} + \frac{1}{2!} x_jA_{jk}A_{ki} + \frac{1}{3!} x_jA_{jk}A_{kl}A_{li} = x_j \left( e^A \right)_{ji}$$ or, better for our purpose here, $$e^{ix_jA_{jk}p_k} x_i e^{-ix_jA_{jk}p_k} = x_i + x_j (e^A - I)_{ji}$$ Multiply on the left by $e^{ix_jA_{jk}p_k}$, rearrange, and obtain $$\left[x_i, e^{ix_jA_{jk}p_k} \right] = x_j \left( I - e^A \right)_{ji} e^{-ix_l A_{lk} p_k}$$
For the commutator in the question title, $\left[x, e^{ipx}\right]$, the same procedure gives a much simpler formula: $$e^{ixp} x e^{-ixp} = x + [ixp, x] + \frac{1}{2!}[ixp,[ixp,x]] + \frac{1}{3!}[ixp,[ixp,[ixp,x ]]]\; + \;... =\\ = x + x + \frac{1}{2!}x + \frac{1}{3!}x \;+\; ... = x + (e - 1)x$$ and $$\left[x, e^{ixp} \right] = (1-e)xe^{ixp}$$