# Why does $e^{-H}\partial_j e^{H} = \partial_j + \partial_jH$?

I apologize if this is a dumb question but I have really thought about this a while and I can’t understand it. I have tried to prove this using the power series of the exponential function but I did not get anywhere. I really don't understand why the first term is there. So can someone help me understand why the following is true: $$$$e^{-H}\partial_j e^{H} = \partial_j + \partial_j H\tag{1}$$$$

Where $$H$$ is an arbitrary time independent Hamiltonian that can be taken to be a scalar field for our purposes.

• That's a very strange identity. Where is it from? Feb 1 at 18:29
• @TobiasFünke it’s the partial derivative with respect to the jth coordinate. So $\partial_j = \partial/\partial x_j$. Feb 1 at 18:51
• @ACuriousMind a professor said this was the case in lecture but did not show it. Feb 1 at 18:51
• The formula $e^{A}Be^{-A} = e^{[A, \cdot]} B$ comes to mind (for operators $A$ and $B$ on a Hilbert space. With this you can try to work out, how the commutators have to look for the right hand side to be correct. Feb 1 at 18:52
• @SebastianRiese But the time derivative is not an operator on the Hilbert space. That being said, I don't know what $\partial_j$ means, even after the comment of OP, i.e. whether or not $j$ could also mean the time coordinate. Feb 1 at 18:54

It's just Leibnitz rule. For any $$\psi(x)$$ we have $$e^{-H(x)}\partial_x e^{H(x)} \psi(x) = e^{-H(x)} (\partial_x e^{H(x)})\psi(x) + e^{-H(x)} e^{H(x)}\partial_x \psi\\ = e^{-H(x)}e^{H(x)} \partial_x H\psi(x)+ \partial_x \psi\\ = (\partial_x H )\psi(x) + \partial_x \psi\\ =\left[(\partial_x H) +\partial_x\right] \psi\\ \equiv (H'+\partial_x)\psi$$

• I was being very dumb, this is probably what the professor meant. Thank you Feb 1 at 18:56
• With the $\partial_x$ in the last line not acting on the $\psi(x)$ to the right? If I read this expression I would expect that it acts on the $H(x)$ and the $\psi(x)$. Feb 1 at 18:57
• @Sebastian Riese Yes. I'll edit to make clear. Feb 1 at 19:08
• Imagine not using align Feb 1 at 19:25
• This derivation is only correct under the assumption that $H(x)$ commutes with $\partial_xH(x)$, for instance when $H(x)$ only depends on the coordinate operator. Feb 1 at 21:29
1. OP's identity$$^1$$ \begin{align} e^{-H}\partial e^H ~\equiv~&e^{-[H,\cdot]}\partial\cr ~\equiv~& \partial+[\partial, H] +\frac{1}{2}[[\partial, H],H] +\frac{1}{6}[[[\partial, H],H],H] +\ldots\cr ~\stackrel{?}{=}~&\partial+[\partial, H]\end{align}\tag{1} is not true in general.

2. Example: If $$H=ax\partial$$, then the left-hand side $$e^{-H}\partial e^H=e^{a}\partial$$ is different from the right-hand side $$\partial +[\partial,H]=(1+a)\partial$$.

3. OP's identity (1) becomes true if we additionally assume that $$H$$ commutes with $$[\partial,H]$$.

$$^1$$ Note that the $$\partial\equiv\frac{\partial}{\partial x}$$ operator notation is ambiguous, cf. e.g. this Phys.SE post. In particular, we interpret OP's notation $$\partial H$$ on the right-hand side to effectively mean $$[\partial,H]$$, cf. the Leibniz rule. OP's identity (1) is in general wrong with any interpretation.