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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: \begin{equation} e^{-H}\partial_j e^{H} = \partial_j + \partial_j H\tag{1} \end{equation}

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

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    $\begingroup$ That's a very strange identity. Where is it from? $\endgroup$
    – ACuriousMind
    Feb 1 at 18:29
  • $\begingroup$ @TobiasFünke it’s the partial derivative with respect to the jth coordinate. So $\partial_j = \partial/\partial x_j$. $\endgroup$ Feb 1 at 18:51
  • $\begingroup$ @ACuriousMind a professor said this was the case in lecture but did not show it. $\endgroup$ Feb 1 at 18:51
  • $\begingroup$ 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. $\endgroup$ Feb 1 at 18:52
  • $\begingroup$ @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. $\endgroup$ Feb 1 at 18:54

2 Answers 2

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

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  • $\begingroup$ I was being very dumb, this is probably what the professor meant. Thank you $\endgroup$ Feb 1 at 18:56
  • $\begingroup$ 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 $\psi(x)$. $\endgroup$ Feb 1 at 18:57
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    $\begingroup$ @Sebastian Riese Yes. I'll edit to make clear. $\endgroup$
    – mike stone
    Feb 1 at 19:08
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    $\begingroup$ Imagine not using align $\endgroup$
    – Kyle Kanos
    Feb 1 at 19:25
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    $\begingroup$ 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. $\endgroup$ Feb 1 at 21:29
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  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.

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