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I was studying Quantum Mechanics from the book Quantum Mechanics, Concepts and Applications by Nouredine Zettili. I came across this definition of an operator.

An operator $\hat{A}$ is a mathematical rule that when applied to a ket $\left| \psi \right>$ transforms it into another ket $\left| \psi' \right>$ of the same space ...

(emphasis mine)

Then, in the examples, gradient is stated to be an operator, even though it takes a scalar function $\psi(\vec{r})$ and transforms it into a vector function $\vec{\nabla}\psi(\vec{r})$, which surely belongs to a different space.

What am I missing here? A wikipedia search said that operators is a function from one set of physical states to another.

I am looking for something more authoritative, though, given the previous definition was from a textbook.

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    $\begingroup$ Clearly there is something wrong there (however don't think too much of it). As people have told you at Math SE it is often interchangeable with "linear map" and physicists are incredibly loose when it comes to terminology. $\endgroup$
    – NDewolf
    Aug 25, 2020 at 9:58
  • $\begingroup$ @NDewolf so I see. I was being a bit rigorous with the definition, I guess. Can I leave the question or do I delete it? I'm unfamiliar with the procedure for such quickly resolved questions. $\endgroup$ Aug 25, 2020 at 10:04
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    $\begingroup$ Does this answer your question? Rigorous mathematical definition of vector operator? $-$ the essentials are all there, but maybe the technical level there assumes more fluency than what your question suggests. $\endgroup$ Aug 25, 2020 at 10:10
  • $\begingroup$ On more mundane matters, please do not post images of texts you want to quote, but type it out instead so it is readable for all users and so that it can be indexed by search engines. For formulae, use MathJax instead. $\endgroup$ Aug 25, 2020 at 10:11
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    $\begingroup$ @MyungjinHyun Being rigorous is not a bad thing. Sadly enough, a lot of concepts in ordinary QM requite much more advanced concepts than people want to admit in introductory courses. $\endgroup$
    – NDewolf
    Aug 25, 2020 at 10:20

3 Answers 3

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You are correct $-$ the book is (slightly) abusing notation here in the name of simplicity.

As you point out, the term 'operator' is generally understood to be a linear mapping $\hat O:\mathcal H \to \mathcal H$ from one vector (Hilbert) space to itself.

The gradient, on the other hand, is what's known as a vector operator, which is basically a triplet of operators $\hat O_i:\mathcal H \to \mathcal H$ (with $i=1,2,3$) which "transform as a vector" (in the sense that, if you rotate your reference frame, the $\hat O_i$ in the new frame are a suitable linear combination of the $\hat O_i$ in the old frame). The details of this are explained in depth in the thread Rigorous mathematical definition of vector operator?, but the short answer is that the "vector operator" can be defined as a linear mapping $$ \hat{\vec O} : \mathcal H \to \mathcal H^{\oplus 3} = \mathcal H\oplus \mathcal H \oplus \mathcal H $$ from $\mathcal H$ to the direct sum of $\mathcal H$ with itself, or, equivalently, as a linear mapping into a tensor product with $\mathbb R^3$, $$ \hat{\vec O} : \mathcal H \to \mathcal H \otimes \mathbb R^3 . $$

The components of the vector operator, the $\hat O_i$, can then be obtained from the vector operator $\hat{\vec O}$ in straightforward ways. These $\hat O_i$ are 'proper' linear operators, in the sense that they map $\mathcal H$ into itself.

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  • $\begingroup$ Thanks for the extra mathematical detail, that cleared up my residual confusion. $\endgroup$ Aug 25, 2020 at 12:49
  • $\begingroup$ @MyungjinHyun Glad to help. $\endgroup$ Aug 25, 2020 at 12:52
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I believe the issue here is with a loose use of term operator. Gradient is a mathematical operator, i.e. it is an operator in the same sense as arithmetic operations, divergence operator, curl, integration, etc. - it transforms a mathematical function. It is however not an oprator in the sense of linear algebra (a definition given in the question), which is how operators are understood in quantum mechanics.

Another relevant point (not cancelling the previous one) is that the gradient here is really a shortcut for a collection of three operators: $\partial_x, \partial_y, \partial_z$, which project scalar functions on scalar functions.

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  • $\begingroup$ So the way I understand it: the text does an abuse of language of linear algebra, and uses the term operator a bit too loosely; from the perspective of linear algebra, only a linear map would be an operator. $\endgroup$ Aug 25, 2020 at 10:59
  • $\begingroup$ I don't agree with your first answer. It is an operator on the Hilbert space of L2 functions. But your second point is the right answer. $\endgroup$ Aug 25, 2020 at 11:01
  • $\begingroup$ @EricDavidKramer My first answer takes on a semantic issue: gradient and many other things are called operator well beyond quantum mechanics and physics. I am not familiar with the textbook in question, but I the gradient is called an operator in many different contexts. $\endgroup$
    – Roger V.
    Aug 25, 2020 at 13:22
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    $\begingroup$ I'm just saying that the gradient is a perfectly valid operator (although it's unbounded) in the linear algebra sense, when acting on the hilbert space L2. It may be called an operator in other contexts too but the use here is perfectly valid. $\endgroup$ Aug 25, 2020 at 13:42
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I do not think Zettili has used anything wrong there. The reasons are:

  1. If you go back a few pages you will find it mentioned that Hilbert Space is linear space and further back you would under linear spaces that these consist of two sets vectors and scalars. So $\nabla$ $\Psi$ can be understood as transformation of the scalar elements satisfying the scalar functions to a vector. Both are in same space that is a Hilbert Space which is a linear space.

  2. The inner product also yields a scalar function and as far as I know Hilbert space is complete with respect to inner products so it belongs in it. So here we see the transformed entity is scalar function and it belongs in the same space.

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    $\begingroup$ This is absolutely not correct. Linear (vector) spaces do not contain scalars as elements. $\endgroup$
    – NDewolf
    Aug 25, 2020 at 12:01
  • $\begingroup$ In Zettili it is mentioned: A linear vector space consists of two sets of elements and two algebraic rules: a set of vectors and a set of scalars a, b, c,; a rule for vector addition and a rule for scalar multiplication. Maybe I misinterpreted this but it clearly states that it has a set of scalars $\endgroup$
    – Lost
    Aug 25, 2020 at 12:08
  • $\begingroup$ I don't know how to attach a photo in comments so I did not but the above given is exactly copied from the book. $\endgroup$
    – Lost
    Aug 25, 2020 at 12:08
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    $\begingroup$ The definition of a vector space does include scalars. But these are not elements of the space itself. $\endgroup$
    – NDewolf
    Aug 25, 2020 at 12:29
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    $\begingroup$ I'm getting the feeling that Zettili is not a very big fan of the general mathematical terminology/foundations. $\endgroup$
    – NDewolf
    Aug 25, 2020 at 13:50

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