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Let's say I am trying to find the commutator of operators $\mathbf{A}$ and $\mathbf{B}$, and I get $$[\mathbf{A},\mathbf{B}]=\nabla^2 f(x,y,z).\tag{0}$$

There seems to be some ambiguity here.

In operator notation, which usually does not include the test function $g$, what is the notational standard used to distinguish

$$[\mathbf{A},\mathbf{B}] g = \nabla^2 (f(x,y,z)g)\tag{1}$$

and

$$[\mathbf{A},\mathbf{B}] g = (\nabla^2 f(x,y,z))g\tag{2}$$

where $g$ depends on $x$, $y$, and $z$?

Just Some Old Man
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  • The notation that you used in the first equation means the equation in the middle. If you mean it to be the last equation, then you should just write f'(x), or at least the f next to the top differential, to avoid confusion. But only you know which of the two options is correct, because you derived the original expression. –  Nov 04 '21 at 15:58
  • Possible duplicates: Do derivatives of operators act on the operator itself or are they "added to the tail" of operators? and links therein. – Qmechanic Nov 04 '21 at 16:13
  • @Wolphramjonny I am wondering what general conventions distinguish the two. What you proposed only works for a specific case, so I updated the question to inquire about a more troublesome case. – Just Some Old Man Nov 04 '21 at 16:44
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    It is a grammatical problem, not a mathematical one. The notation you used is ambiguous, that is all. –  Nov 04 '21 at 16:45
  • @Wolphramjonny Thank you, that addresses what I was wondering. Can you recommend a less ambiguous standard for notation? – Just Some Old Man Nov 04 '21 at 16:47
  • I am not sure what would be the standard way for an example like the one you used. But I would replace the Laplacian by the momentum operator (divided by $-h^2$). –  Nov 04 '21 at 16:53
  • @Wolphramjonny Even with that change, how would you indicate, for example, a commutator $[\mathbf{A},\mathbf{B}] \psi = (\mathbf{p} f(x,y,z))\psi$? – Just Some Old Man Nov 04 '21 at 17:07
  • sorry, I made a mistake, I would do that only if you mean Eq. (1). In that case you dont need to write $\psi$. Also, I meant the square of the momentum operator –  Nov 04 '21 at 17:14

2 Answers2

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If operators $A$ and $B$ have commutation relation $$[A,B] = g \mathbb I $$ for some function $g:\mathbb R^3 \rightarrow \mathbb C$, then that means $[A,B]\psi = g\psi$, which we might also write $$\bigg([A,B]\psi\bigg)(x,y,z) = g(x,y,z) \psi(x,y,z)$$ My interpretation of OP's expression $(0)$ is precisely this, with $g(x,y,z)\equiv \nabla^2 f(x,y,z)$. However, as OP points out, there is some ambiguity here, so some subsequent clarification on the part of the writer would be warranted. A less ambiguous notation for OP's expressions $(1)$ and $(2)$ might be

$$[A,B] = (\nabla^2 f)\mathbb I + 2(\nabla f) \cdot \nabla + f \nabla^2 \tag{1'}$$ $$[A,B] = (\nabla^2 f) \mathbb I \tag{2'}$$

As is often the case, you need to find the proper balance between notational brevity and clarity which suits your audience and style.

J. Murray
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  • This is what I was looking for. It seems there is no widespread standard notation and clarification from the writer is needed. What is $\mathbb I$? I have not seen that before. – Just Some Old Man Nov 04 '21 at 20:29
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    @JustSomeOldMan $\mathbb I$ is the identity operator on the Hilbert space, to remind us that both sides of $[A,B]=g\mathbb I$ are to be understood as operators. Because the right hand side corresponds to simple multiplication of a wavefunction by the function $g$ it is common to just drop the symbol $\mathbb I$ and write $[A,B]=g$, but for pedagogical clarity I like to keep it in. – J. Murray Nov 05 '21 at 02:20
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If the former solution was what you intended to mean, then you wouldn't write $d/dx f(x)$ -- to my eyes, I would not think from this expression that the derivative is supposed to act on the test function. Instead, you would keep the product rule in mind and write $$ \frac{df}{dx} + f(x) \frac{d}{dx} $$

Zack
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    I don't think this answers the question, frankly. But the question is far from clear. – Gert Nov 04 '21 at 15:33
  • If the derivative was a 4th derivative, this solution would not be appropriate. I am looking for something more general. – Just Some Old Man Nov 04 '21 at 15:49
  • Why not? You can still apply the product rule derivative by derivative and obtain an unambiguous expression in a similar manner. – Zack Nov 04 '21 at 15:56
  • @Gert In what way can I make my question more clear? What notational convention distinguishes equation (1) and (2) in the original question? – Just Some Old Man Nov 04 '21 at 16:46
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    @JustSomeOldMan It would help to use Dirac notation. – Gert Nov 04 '21 at 16:57