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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.

Qmechanic
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JosephSanders
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    That's a very strange identity. Where is it from? – ACuriousMind Feb 01 '24 at 18:29
  • @TobiasFünke it’s the partial derivative with respect to the jth coordinate. So $\partial_j = \partial/\partial x_j$. – JosephSanders Feb 01 '24 at 18:51
  • @ACuriousMind a professor said this was the case in lecture but did not show it. – JosephSanders Feb 01 '24 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. – Sebastian Riese Feb 01 '24 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. – Tobias Fünke Feb 01 '24 at 18:54

2 Answers2

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

mike stone
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  • I was being very dumb, this is probably what the professor meant. Thank you – JosephSanders Feb 01 '24 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 $\psi(x)$. – Sebastian Riese Feb 01 '24 at 18:57
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    @Sebastian Riese Yes. I'll edit to make clear. – mike stone Feb 01 '24 at 19:08
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    Imagine not using align – Kyle Kanos Feb 01 '24 at 19:25
  • @JosephSanders This "identity" is pretty misleading though. For instance, one can set $H=\partial_x$ and easily get misled into a wrong answer of $=(\partial_x^{2} +\partial_x ) |p>$ when it just should be $\partial_x |p>$ where $|p>$ is a momentum eigenstate – Bohan Xu Feb 01 '24 at 19:32
  • @BohanXu Hm that is true. Do you have any insight on when you can use the Leibnitz rule and when it does not work? – JosephSanders Feb 01 '24 at 19:52
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    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. – Tomáš Brauner Feb 01 '24 at 21:29
  • @JosephSanders learning dirac notation and doing quantum mechanics in its natural language that is linear algebra instead of unnatural language that is function, would allow you to answer the question yourself very easily. – Bohan Xu Feb 02 '24 at 01:13
  • @BohanXu This problem came up in classical field theory but yeah I will do that – JosephSanders Feb 02 '24 at 01:25
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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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