I have a question regarding the effect of quantum mechanical operators. The definition that I'm familiar with says that an operator $A$ acts on a vector from a Hilbert space, $|\psi\rangle$, and the result is another vector, $|\psi'\rangle$:
$$A |\psi\rangle = |\psi'\rangle.$$
However, in class I've also seen operators applied to scalar-valued functions, such as the momentum operator in position space:
$$P = – i \hbar \frac{\partial}{\partial x},\quad P \psi(x) = – i \hbar \frac{\partial}{\partial x} \psi(x)$$
However, mathematically, that doesn't make sense to me, since $P$ should operate on a vector, not a (scalar) function! I know that wave functions can be viewed as the coefficients of a state vector when the vector is written in a particular basis, such as
$$|\psi\rangle
= \int_{-\infty}^\infty |x\rangle \langle x| \,\mathrm{d}x\; |\psi\rangle
= \int_{-\infty}^\infty |x\rangle \langle x|\psi\rangle \,\mathrm{d}x
= \int_{-\infty}^\infty \psi(x) |x\rangle \,\mathrm{d}x$$
with
$$\psi(x) = \langle x|\psi\rangle$$
but, as far as I can tell, the operator should still act on a vector, not on a function alone.
On the Wikipedia article "Operator (physics)", under "Linear operators in wave mechanics", I found the following:
$$A \psi(x) = A \langle x | \psi \rangle = \langle x | A | \psi \rangle$$
However, that last step seems dubious to me. Is it valid to just swap the operator into the inner product like that? In general, I don't see the mathematical meaning of expressions like $A\psi(x)$ or $A \langle x|$, since $A$ can neither act on a scalar value nor on a bra. Does it only work for self-adjoint operators, since we'd then have $A^\dagger = A$, which might help with $A$ acting on bras?
Best Answer
I really appreciate this question. You are perfectly right and your confusion is understandable. Sadly, the physics world is somewhat sloppy in their use of notation at times.
Of course, when writing $P\psi(x)$ one does not intent to apply the operator to a scalar, but the $P$ is applied to the ket vector $\psi$.
But now comes the major source of confusion, IMO. While to you the wave function really is just
and the state vector is element of some abstract Hilbert space, many people will argue that the Hilbert space is actually a function space. Which one precisely depends on the system under consideration but specifically for a free particle it'll be $L^2(\mathbb R,\mathbb C)$, i.e. the space of square-integrable function from $\mathbb R$ to $\mathbb C$. So in this space, the actual function $\psi$ is the vector. That space is linear (a.k.a. a vector space) and it also has a scalar product, defined via the integral. It is also complete which means that one can insert identities as you did above. That's actually what a Hilbert space is (to a mathematician): a complete vector space with a scalar product.
When taking on this view, it is legitimate to apply an operator to a function. Indeed, only now does the identification $P=-i\hbar\partial_x$ make any sense. It is a differential operator which can be applied to functions.
The advantage of the Dirac bra-ket-notation is that it allows you to step back from any concrete realizations of the underlying Hilbert space. This has some advantages:
So why should we bother working in function spaces?
$^1$ For an overview of what can go terribly wrong in quantum mechanics upon getting to comfortable with Dirac's notation, I recommend this excellent article: F. Gieres, Mathematical surprises and Dirac's formalism in quantum mechanics, arXiv:quant-ph/9907069
Edit in response to @Jyothi's comment:
The action of the momentum operator as a derivative may be written as $\langle x|\hat P|\psi\rangle = -i\hbar\frac{\partial}{\partial x}\langle x|\psi\rangle$.
Now, consider the following calculation of the action of the operator product $\hat P\hat X$:
$$ \langle x|\hat P\hat X|\psi\rangle = \int\mathrm dx'\,\langle x|\hat P\hat X|x'\rangle\langle x'|\psi\rangle = \int\mathrm dx'\,x'\langle x|\hat P|x'\rangle\langle x'|\psi\rangle\\ = -i\hbar\int\mathrm dx'\,x'\psi(x')\frac{\partial}{\partial x}\delta(x-x')\\ = +i\hbar\int\mathrm dx'\,x'\psi(x')\frac{\partial}{\partial x'}\delta(x-x')\\ =-i\hbar\int\mathrm dx'\,\delta(x-x')\frac{\partial}{\partial x'}(x'\psi(x'))\\ =-i\hbar\frac{\partial}{\partial x}(x\psi(x))\\ = -i\hbar\Bigl(\psi(x)+x\frac{\partial}{\partial x}\psi(x)\Bigr)\\ =-i\hbar\langle x|\psi\rangle + x\langle x|\hat P|\psi\rangle\\ =-i\hbar\langle x|\psi\rangle + \langle x|\hat X\hat P|\psi\rangle\\ =\langle x|(-i\hbar+\hat X\hat P)|\psi\rangle. $$ In the third line, I rewrote a derivative for $x$ as an derivative for $x'$, yielding a minus sign. In the step after, integration by parts was used, arguing that the boundary terms vanish at $\pm\infty$. By assuming this holds for general $|\psi\rangle$, one can equate the operators on the LHS and RHS, giving $\hat P\hat X=-i\hbar+\hat X\hat P$ or $[\hat X,\hat P]=i\hbar$.
This "calculation" is severely handwavy, none of it is mathematically rigorous. For example, I'm taking derivatives of $\delta$ which is not defined. But I believe the gist of this calculation could be made rigorous in a distributional sense. Also, in my mind the commutator is actually more fundamental and this, if anything, shows that the wave function representation is compatible with that.