Does this definition of operator contradict with gradient as an operator?

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.