I'd like to ask whether or not quantum operators are associative. Take for example (in one dimension) the momentum operator $\hat{p} = -i \hbar (\mathrm{d}/\mathrm{d}x) $ and the position operator $\hat{x} = x$ acting on some test function $f$ as $$\hat{p} \hat{x} f(x) \, .$$ Does it make a difference whether the order in which the operators act is $(\hat{p}\hat{x}) f(x)$ or $\hat{p} (\hat{x} f(x))$? For example, in the first case, the derivative from the momentum operator acts on the position operator producing $1$, clearly a different result from the latter case in which the derivatve operator acts on the product $xf(x)$. Does this make the operators non-associative, or am I missing something?
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$\begingroup$ Your mistake occurs precisely at the point where you say "producing 1". $\endgroup$– WillOCommented Oct 16, 2022 at 13:31
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2 Answers
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Operators do not act on operators; operators act on states (or wavefunctions, in your example). As such, the order of writing down the operators before a state determines their order of operation. The operator $\hat{x}$ acts on the wavefunction $f(x)$ to produce the wavefunction $xf(x)$. Then the operator $\hat{p}$ acts on the wavefunction $xf(x)$, to produce the wavefunction $-i\hbar\frac{d}{dx}(xf(x))$.
answered Oct 21, 2018 at 22:30
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$\begingroup$ You seem to have a very special context in mind. In particular, in the Heisenberg picture this is certainly not the case, but it is not even entirely correct in the Schrödinger picture. $\endgroup$– user178876Commented Oct 21, 2018 at 22:32
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1$\begingroup$ @marmot Care to elaborate? I don't see anything wrong with this answer. $\endgroup$ Commented Oct 21, 2018 at 22:33
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$\begingroup$ @AccidentalFourierTransform Just consider, say. a translation operator and some creation operator that creates something at the position $x$. $\hat{T}_y \hat{a}(x) |-\rangle$, where $\hat{T}_y$ is the operator that generates as translation by $y$. Does it act on the operator or not? $\endgroup$– user178876Commented Oct 21, 2018 at 22:36
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1$\begingroup$ @marmot The concept of associativity is really sort of moot here, which is what I was trying to get across in this answer. This is because $(\hat{p}\hat{x})f(x)=\hat{p}(\hat{x}f(x))$ by definition, because that's the only sensible definition consistent with the fact that operators act on states. You can say this situation demonstrates associativity if and only if you also say that the statement $(f\circ g)(x)=f(g(x))$ demonstrates associativity. Most people, I think, consider the latter to be the definition of composition of functions, and so the same should apply here. $\endgroup$ Commented Oct 21, 2018 at 22:47
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1$\begingroup$ In more mathematical terms (but in agreement with the answer), associativity is a property of (binary) relations, not of maps. And an operator is a map. One may ask if product or sum of (bounded) operators are associative, but not if operators thenselves are. $\endgroup$– yuggibCommented Oct 22, 2018 at 7:04
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Composition of differential operators$^1$ $$ (\hat{A} \circ \hat{B}) \circ \hat{C} ~=~\hat{A} \circ (\hat{B} \circ \hat{C}) \tag{1}$$ on the class$^2$ of smooth functions is associative, since both the LHS and RHS of eq. (1) is defined by acting on a smooth function $f$ in the same order, namely as $$ \hat{A}(\hat{B}(\hat{C}(f))). \tag{2} $$
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$^1$ More generally, composition of composable functions is associative.
$^2$ For unbounded operators, the issue of domains may hamper composition and associativity.
answered Oct 16, 2022 at 12:44