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Operators are associative as seen here.

But when we try to calculate $[\hat{x}, \hat{p}]$ for example, we use a test function and apply $\hat{p}$ to both $\hat{x}$ and the function, instead of associating $\hat{p}$ with $\hat{x}$ and be done with it (obviously the result would be incorrect).

Edit to elaborate my question:

$[\hat{x}, \hat{p}] f = \hat{x}\hat{p}f - (\hat{p} \hat{x}) f = \hbar/i \bigg(\hat{x} (\partial f/\partial x) - (\partial x/\partial x)f\bigg) = \hbar/i \bigg(\hat{x} \ \partial /\partial x - 1) f\bigg)$

Why can't we do what we did on the second term? Associate the operators before applying them on the test functions?

Qmechanic
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Diaa Eldin Malek
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    As a general trend, mathematicians focus on associative algebras. Non-associative algebras are excruciating and great efforts are taken to avoid them. – Cort Ammon Apr 25 '20 at 08:07
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    Are you joking, right? Lie algebras are all over the place in physics! – MannyC Apr 26 '20 at 07:52

4 Answers4

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Operators act on elements of the Hilbert space, not other operators. $\hat p$ and $\hat x$ are defined by how they act on states, e.g.

$$\hat p: f(x) \mapsto -if'(x)$$ $$\hat x: f(x) \mapsto x\cdot f(x)$$

There is no sense in which $\hat p$ can act on $\hat x$, or vice versa. We can, however, compose operators (also called multiplying them, by overloading the term "multiply") by applying them one after another:

$$(\hat p \circ \hat x): f(x) \mapsto \hat p\big(\hat x(f(x))\big) = \hat p\big(x f(x)\big) = -i\big[x f(x)\big]'$$ $$(\hat x \circ \hat p): f(x) \mapsto \hat x\big(\hat p(f(x))\big) = \hat x\big(-if'(x)\big) = -ixf'(x)$$

where $\hat p \circ \hat x$ is pronounced "$\hat p$ after $\hat x$". It is conventional to drop the $\circ$ notation, but this is what appears in the commutation relations.

J. Murray
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  • I see, but what about operators in matrix form? Do they still belong to Hilbert space if they can act on one another or "multiply" into one operator? – Diaa Eldin Malek Apr 24 '20 at 20:46
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    @DiaaEldinMalek Operators aren't elements of the Hilbert space $\mathcal H$, they live in $End(\mathcal H)$ - the space of endomorphisms on $\mathcal H$ (that is, maps from $\mathcal H$ back to itself). Given any two endomorphisms, you can "multiply" (or compose) them together to get another endomorphism. Note - what I've said requires some slight tweaking if the operators can't act on the whole Hilbert space (like $\hat p$ and $\hat x$, in general), but it is true for finite matrix operators. – J. Murray Apr 24 '20 at 20:53
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Suppose we want to evaluate $$\hat{p} \hat{x} \psi(x).$$ Everything here is perfectly associative, so you can evaluate it two ways, $$\hat{p} \hat{x} \psi(x) = (\hat{p} \hat{x}) \psi(x) = \hat{p} (\hat{x} \psi(x)).$$ The error you made is that you can't compute $\hat{p} \hat{x}$ like this: $$\hat{p} \hat{x} = - i\frac{\partial}{\partial x} \hat{x} \neq - i\frac{\partial}{\partial x} x = - i.$$ The issue is that $\hat{x}$ isn't the same thing as $x$. The operator $\hat{x}$ acts on wavefunctions $\psi(x)$ by multiplying them by $x$, but that isn't the same thing as saying it is the number $x$. The operator $\hat{p} \hat{x}$ needs to be thought of as doing these two operations in order: multiply wavefunctions by $x$, then apply $- i \partial / \partial x$.

knzhou
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  • You could evaluate $(\hat p\hat x)$ through matrix multiplication right? – Charlie Apr 24 '20 at 20:43
  • Not very convenient though as they are infinite dimensional matrices. – MannyC Apr 24 '20 at 20:45
  • @Charlie Yeah, but it would be really annoying. The point of the early-undergrad "acting on functions" notation that OP is using is to avoid having to talk about matrix elements or bases, but to do the multiplication you'll have to. – knzhou Apr 24 '20 at 20:49
  • knzhou & @Manny Ok thank you :) – Charlie Apr 24 '20 at 21:04
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Operators, in general, are associative. Associative means that we can calculate in a different order $$ABC = (AB)C = A(BC)$$ but it does not mean that we can write the operators in a different order. You are confusing it with commutative. We do not, in general have $$AB = BA$$

Consider a simple example. If you reflect an image in the $y$-axis (operator $B$), and then rotate the result by $90^\circ$ (operator $A$) then you will get a different result from rotating the image and then reflecting. In matrices this is $$AB = \begin{pmatrix} 0 & -1 \\ 1 & 0 \\ \end{pmatrix} \begin{pmatrix} -1& 0 \\ 0 &1 \\ \end{pmatrix} \neq \begin{pmatrix} -1& 0 \\ 0 &1 \\ \end{pmatrix} \begin{pmatrix} 0 & -1 \\ 1 & 0 \\ \end{pmatrix} =BA $$ as you can easily check.

penovik
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Charles Francis
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    Oh no, I didn't mean that. I meant when using a test function to evaluate $ \hat{p} (\hat{x} f(x))$, why can't we do $ (\hat{p}\hat{x}) f(x)$?

    edit: That way p is just a derivative with respect to x acting on x, so it's 1. The commutator would evaluate to $x {\partial}/{\partial x} - 1$

     – Diaa Eldin Malek Apr 24 '20 at 20:25
  • I have given an example to show that operators do not in general commute – Charles Francis Apr 24 '20 at 20:41
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No, operators are not all associative. Though in regards to your example, linear operators acting on a separable Hilbert space are. It would be interesting if any new formulation of quantum mechanics can make use of non-associative operators. Some people wrote more ideas about that and other physical applications in the following post.

Non-associative operators in Physics

One of the answers mentioned some work done using deformation quantization as a possible way to work towards that idea (eventually). I wanted to at least mention it since that method is highly related to (a particular) functional calculus, which seems more closely related to the spirit of your question and comes up in commutator computations often.

penovik
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