# Is the complete Hilbert space of a system, sum of the spans of position, momentum, energy, angular momentum spaces, etc?

What I understand is the following.

1) There is an abstract ket ( vector) which contains all the information about the system and it lives in an abstract vector space, the Hilbert Space which might or might not be infinite dimensional. The position basis/momentum basis/energy basis/ angular momentum basis is the basis of position/momentum/energy/angular momentum operator in this Hilbert space ( please don't go to rigged Hilbert spaces and right now loosely accept that a position eigen ket can be expressed as a linear combination of energy or momentum eigen kets somehow, not going into the details of how). Project the ket onto the basis kets and out of the scalar product construct the position, momentum, energy, angular momentum, etc spaces. The position space and momentum space share the structure of a function space and so accordingly redefine the inner product in them ( #Shankar). The inner product in the original Hilbert Space in defined without reference to any basis. Here inner product is defined w. r. t position space so it's definition changes ( becomes an integral). The energy/ angular momentum basis can have the same definition of inner product as the original Hilbert Space. The operators in the abstract Hilbert space are themselves abstract and need to be given a structure, which they are in these spaces. Wavefunction is in one to one correspondence with the system ket so quantum mechanics proceeds the same way in this function space as in the abstract Hilbert Space. The operators are written in position basis ( their matrix elements are w. r. t position basis, etc).

My questions are ( please address each of the following in your answer) -

1) Momentum operator can be written in position basis. How is energy operator or angular momentum operator written in position basis. Does writing energy operator in position basis means writing it's matrix elements in position basis or something else.

2) Why we don't do quantum mechanics in energy space rather than position space.

3) Can the complete Hilbert Space be considered as the sum ( not direct sum since space of eigenvectors of energy will not be closed under addition so will not be subspace of the original complete space) of spans of position, momentum, energy, angular momentum spaces, etc or not. I ask this because to incorporate Spin we need to taken tensor product of two spaces but here position and momentum operator don't commute so you can't take take a tensor product of their spaces.

4) How is that position space different from our real 3D space.

I couldn't get the answer till now. The already given answer doesn't doesn't answer my doubts.If anybody could help, it would be great.

## 1 Answer

The Weyl relations can be given an irreducible representation via the Schrödinger representation, which von Neumann showed to be unique. Indeed, this is a representation of the position and momentum operators on $$L^2(\mathbb R^n)$$ with the Lebesgue measure in the case of a system with $$n$$ degrees of freedom. In this representation, the position operator acts as a multiplication operator, whereas the momentum operator acts like a differentiation operator.

1) Every other observable quantity is, like in the classical case, an expression that involves momentum and position. For example, the energy of a free particle is proportional to the square of the momentum operator. All the Physics that one needs is then already "encoded" in the Hilbert space of the representation, i.e. $$L^2(\mathbb R^n)$$.

2) The QHO is normally solved in "energy" space. More correctly, one determines a base for the number operator $$N$$, but then again, the energy operator is a function of $$N$$ so modulo some functional calculus we can consider that the energy space.

3) see 1)

4) Strictly speaking, the position space is a Hilbert space, while our space is believed to be Euclidean, with an underlying structure of $$\mathbb R^3$$. However, that $$\mathbb R^3$$ is inferrable from the $$\mathbb R^3$$ of $$L^2(\mathbb R^3)$$. In this sense, then, the two spaces are the same.

• Sorry, I don't have that strong a mathematical background so couldn't connect much. It would be better if you explain a bit without going to L^2 etc.. like a sort of superficial but intuitive explanation at the level of advanced linear algebra. That could help me understand 1) and 3)...I got 2). Just one thing in 2) Does writing Energy operator or for that matter in general any other operator in the the position basis ( or any other basis) means writing it's matrix elements w. r. t to position eigenkets or something else – Shashaank Apr 22 '20 at 12:08
• Could you pls specifically explain the point I made in the above comment. And if you could pls edit your answer without going to Lebsegue measure and L^2, an intuitive explanation rather, it would be quite helpful. As for 4) you say both are same then dimensionality of position space is infinite ( because there are infinite position eigenvectors so the space made by them will be infinite dimensional obviously) but out real space is 3 dimensional and if take just x axis our space is 1 dimensional. So how are both the same.... – Shashaank Apr 22 '20 at 12:13
• I'm afraid that, without a deeper understanding of certain concepts, it is not possible to give a reason as to why the "position Hilbert space" is enough to be able to deal with position, momentum, energy, and any other observable operators. An argument that one can make, a posteriori, is that a system with $n$ degrees of freedom could be described on a Hilbert space that is at least as large as $L^2(\mathbb R^n)$. – Phoenix87 Apr 22 '20 at 12:16
• and how are the postion hilbert space and our space same when position Hilbert space is infinite dimensional whereas ours is 3 dimensional. There dimensional is not the same... Also what subject would I need to refer to for understanding this, functional analysis or anything else... – Shashaank Apr 22 '20 at 12:55