Understanding Abstract Kets & Hilbert Space

[mike1000]

Some posts in another thread lead me to a search which ended when I read the following "kets such as $|\psi\rangle$ are elements of abstract Hilbert Space".

That lead me to this paper.

http://www.phy.ohiou.edu/~elster/lectures/qm1_1p2.pdf

"The abstract Hilbert space $l^2$ is given by a set of element $H=\left( |\psi\rangle , |\phi\rangle , |\chi\rangle , \dots \right)$""

I now understand why vanhees71 keeps insisting that it is an "abstract ket".

 

[dextercioby]

But $l^2$ is not an abstract Hilbert space. It's a realization of an abstract Hilbert space by a set of infinite matrices endowed with a certain topology.

 

[vanhees71]

$\ell^2$ is the Hilbert space of square summable complex sequences. It's one realization of the one and only (up to equivalence) separable Hilbert space. You can define the "abstract Hilbert space" as the equivalence classes of all realizations of a separable Hilbert space. In QT usually you have it either as $\ell^2$ or as some $\mathrm{L}^2$ (Hilbert space of square integrable $\mathbb{C}^{2s+1}$ valued functions for a single particle of spin $s$).

 

[mike1000]

It took me forever to figure out what the following notation represented. I have seen it a lot in the forum. $$|A\rangle|B\rangle$$I just discovered what it means and I would like to post it in this thread for future reference.

It is shorthand notation for the tensor product of two vectors, $|A\rangle \otimes |B\rangle$. The $\otimes$ symbol is left out for brevity.

Page 9 and 10 of the following reference
https://ocw.mit.edu/courses/physics...all-2013/lecture-notes/MIT8_05F13_Chap_08.pdf

 

[dextercioby]

I don't think this notation is standard. Show me a textbook which uses it.

 

[Demystifier]

But $l^2$ is not an abstract Hilbert space.

Being abstract is in the eyes of the beholder. A few hundred years ago mathematicians of that time considered negative numbers abstract.

 

[PeterDonis]

Show me a textbook which uses it.

The link @mike1000 gave, while not technically a textbook, I suppose, is lecture notes from MIT's Open Courseware. So it's not unreasonable to expect that its notation is at least somewhat standard.

(I'm also curious what else the notation $\vert A \rangle \vert B \rangle$ might stand for; the implied tensor product is the only usage I'm familiar with, but there are many QM textbooks and much QM literature that I have not read.)

 

[vanhees71]

It's standard notation:
$$|A \rangle \otimes |B \rangle=|A \rangle |B \rangle=|AB \rangle.$$

 

[Demystifier]

It took me forever to figure out what the following notation represented. I have seen it a lot in the forum. $$|A\rangle|B\rangle$$I just discovered what it means and I would like to post it in this thread for future reference.

It is shorthand notation for the tensor product of two vectors, $|A\rangle \otimes |B\rangle$. The $\otimes$ symbol is left out for brevity.

Page 9 and 10 of the following reference
https://ocw.mit.edu/courses/physics...all-2013/lecture-notes/MIT8_05F13_Chap_08.pdf

Sometimes it's useful to use both notations within a single expression. See e.g. my
https://arxiv.org/abs/1112.2034 Eqs. (29)-(30).

 

[mike1000]

How do we know if the eigenvectors of two operators span the same Hilbert Space? What do we mean by "same Hilbert Space"? Obviously spin and momentum are in different HIlbert Spaces because the two spaces have different dimensions. But what about when the dimensions of the two spaces are equal? What if I have two infinite dimensional Hilbert Spaces but one space has a continuous spectrum and the other a discrete spectrum, such as the Hamiltonian which I think has a discrete spectrum in an infinite dimensional space and momentum which has a continuous spectrum in an infinite dimensional space. Does the Hamiltonian operator and the momentum operator span the same Hilbert Space? Or is it appropriate to say that they share the same Hilbert space? And what about the position and momentum operators, do they span the same space(I think they do). And what do we call the common Hilbert space they span?

I guess what I am asking...are all HIlbert spaces which have the same dimension, considered the same Hilbert space and the only thing that distinguishes them is the basis which spans the space?

 

[George Jones]

How do we know if two operators span the same Hilbert Space?

Your use of the term "span" is non-standard. Do you mean "How do we know if two self-adjoint operators operate on the same Hilbert Space?"

 

[PeterDonis]

Obviously spin and momentum are in different HIlbert Spaces because the two spaces have different dimensions.

It's more complicated than that. Consider a free electron. In order to describe completely the state of the electron, we need to specify its momentum (or position) and its spin. So the Hilbert space of the electron, heuristically, is the tensor product of the spin Hilbert space and the momentum (or position) Hilbert space.

Now, suppose I want to measure the spin of this free electron about the $z$ axis. What operator on the electron's Hilbert space corresponds to this measurement? We often write this operator as, say, $S_z$; but that is really incomplete. The full operator is $S_z \otimes I$, where $I$ is the identity operator on the momentum (or position) portion of the Hilbert space, and $S_z$ operates on the spin portion. Similarly, if I want to measure the electron's momentum in the $z$ direction, the full operator would be $I \otimes P_z$, where $I$ here is the identity operator on the spin portion of the Hilbert space.

So it's not that "spin and momentum are in different Hilbert spaces"; it's that if I want to just measure spin or momentum, the operator I use will be a composition of an operator on that portion of the Hilbert space, plus the identity operator on the rest of the Hilbert space.

the Hamiltonian which I think has a discrete spectrum in an infinite dimensional space and momentum which has a continuous spectrum in an infinite dimensional space.

It's more complicated than that. Whether the spectrum of an operator is continuous or discrete depends on the situation. In fact it can be both in the same problem. For example, the spectrum of the Hamiltonian for the electron in the hydrogen atom is discrete in one range (the range of bound states, where the electron's energy is less than the ionization energy of the atom) and continuous in another range (the range of free states, where the electron's energy is greater than the ionization energy of the atom).

Does the Hamiltonian operator and the momentum operator span the same Hilbert Space?

It's more complicated than that. Go back to the case of the electron above. For a free electron, the Hamiltonian operator does not depend on the spin--it only depends on the momentum. So the Hamiltonian, like the momentum operator, only operates on the momentum portion of the Hilbert space--more precisely, the full Hamiltonian operator is a composition of the identity operator on the spin portion of the Hilbert space, with the operator $\hat{p}^2 / 2m$ on the momentum portion.

Now let's put the electron in a magnetic field. Since the electron is charged, its spin couples to the magnetic field, so the electron's spin now affects its energy, i.e., the Hamiltonian operator now does depend on spin. That means the Hamiltonian now does not have the form I just described; it is now a composition of the operator describing how the electron's energy depends on spin (which acts on the spin portion of the Hilbert space) with the operator that describes how the electron's energy depends on momentum (which, btw, no longer has the same form as I gave above, there will now be an additional term that depends on the electromagnetic potential).

what about the position and momentum operators, do they span the same space(I think they do)

That's not the right way to put it. The right way to put it is that both the position and the momentum operators operate on the same portion of the Hilbert space. In the case of the electron above, they both operate on the portion of the Hilbert space that I called the momentum (or position) portion; and the (or position) in parentheses that I kept putting in was because of this.

In systems containing multiple particles, it's more complicated than that, because you have to specify which particle's momentum (or position) you are measuring, and measurements on different particles act on different portions of the overall Hilbert space of the system. And that Hilbert space itself is more complicated than the single particle Hilbert space, both because there are multiple particles and because you have to take quantum statistics into account.

are all HIlbert spaces which have the same dimension, considered the same Hilbert space and the only thing that distinguishes them is the basis which spans the space?

It's more complicated than that. See above for some examples of how. The general answer, I think, is that you need to spend some time working through a good QM textbook.

 

[mike1000]

Your use of the term "span" is non-standard. Do you mean "How do we know if two self-adjoint operators operate on the same Hilbert Space?"

I changed the wording in my post to "How do we know if the eigenvectors of two operators span the same Hilbert Space?" I was thinking that the Hilbert space of an operator was defined as the space spanned by its eigenvectors. But I think I am realizing that space is just a sub-space of the n-dimensional Hilbert space.

For each dimension there is a corresponding Hilbert space. Within each Hilbert space there can be an infinite number of different basis, which can span different sub-spaces of the n-dimensional HIlbert space?

As I think about it, that cannot be correct, because, we know that if a basis is complete, it can describe every vector in the n-dimensional Hilbert space, which means that all complete basis in a n-dimensional Hilbert space can span the entire space, not just a sub space.

Do the eigenvectors of all self-adjoint operators form a complete basis for the Hilbert space that contains them? And wouldn't that imply that I can use the eigenvectors of, say the position operator, to express the eigenvectors of, say the Hamiltonian operator, merely because the eigenvectors of the position operator form a complete basis for infinite dimensional Hilbert space? (I know this is not true, but I do not see why it is not true. If I have a set of vectors that spans the space then I should be able to express any vector in that space using the position basis, even eigenvectors defined by the Hamiltonan operator, because they are both in the same , infinite dimensional Hilbert space and have the same units (probability amplitude))

Obviously, I am confused.

 

[vanhees71]

To specify a basis of your Hilbert space in terms of eigenvectors you need to evaluate the simultaneous eigenvectors of a set of self-adjoint operators representing a complete set of independent compatible observable. Compatible means that all the operators are pairwise commuting and independent means that no operator is the function of all others.

To illustrate this consider a single particle in non-relativistic QT of spin $s \in \{0,1/2,1,\ldots\}$. Then a possible complete set of observables are the three components of position and the $z$ component of spin. The common generalized eigenvectors are denoted with $|\vec{x},\sigma \rangle$, where $\vec{x} \in \mathbb{R}^3$ and $\sigma \in \{-s,-s+1,\ldots,s-1,s \}$.

These eigenvectors are by definition orthonormalized in the generalized sense, i.e.,
$$\langle \vec{x}',\sigma'|\vec{x},\sigma \rangle=\delta^{(3)}(\vec{x}-\vec{x}') \delta_{\sigma,\sigma'}$$
and complete
$$\int_{\mathbb{R}^3} \sum_{\sigma=-s}^s |\vec{x},\sigma \rangle \langle \vec{x},\sigma|=\hat{1}.$$
Any Hilbert space vector is thus uniquely defined by the "components" with respect to this basis, i.e., it's given by a square-integrable $\mathbb{C}^{2s+1}$-valued function, i.e., the components of this $(2s+1)$-dimensional spinor are functions of $\vec{x}$:
$$\psi_{\sigma}(\vec{x})=\langle \vec{x},\sigma|\psi \rangle,$$
and the scalar product between two vectors is
$$\langle \phi|\psi \rangle=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \sum_{\sigma=-s}^s \langle \phi|\vec{x},\sigma \rangle \langle \vec{x},\sigma|\psi \rangle=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \sum_{\sigma=-s}^s \phi_{\sigma}^*(\vec{x}) \psi_{\sigma}(\vec{x}).$$
This shows that the specific realization of the separable Hilbert space (there is, up to equivalence only one such Hilbert space!) depends on the system to be described, i.e., on the operators defining observables. The key to analyze this structure of the Hilbert space is the commutator algebra of the operators. This commutator algebra is specified by the symmetry group of the system. For our one particle it's the Galileo group of Newtonian spacetime. The analysis of the unitary ray representations of this group leads (after some work) to the above basis (usually first one analyzes the group in terms of momentum-spin eigenstates rather than the position-spin eigenstates, but that's a minor detail). Of course, I cannot explain all this within this forum. A very good source on this is Ballentine's book.

 

[mike1000]

To specify a basis of your Hilbert space in terms of eigenvectors you need to evaluate the simultaneous eigenvectors of a set of self-adjoint operators representing a complete set of independent compatible observable. Compatible means that all the operators are pairwise commuting and independent means that no operator is the function of all others.

To illustrate this consider a single particle in non-relativistic QT of spin $s \in \{0,1/2,1,\ldots\}$. Then a possible complete set of observables are the three components of position and the $z$ component of spin. The common generalized eigenvectors are denoted with $|\vec{x},\sigma \rangle$, where $\vec{x} \in \mathbb{R}^3$ and $\sigma \in \{-s,-s+1,\ldots,s-1,s \}$.

These eigenvectors are by definition orthonormalized in the generalized sense, i.e.,
$$\langle \vec{x}',\sigma'|\vec{x},\sigma \rangle=\delta^{(3)}(\vec{x}-\vec{x}') \delta_{\sigma,\sigma'}$$
and complete
$$\int_{\mathbb{R}^3} \sum_{\sigma=-s}^s |\vec{x},\sigma \rangle \langle \vec{x},\sigma|=\hat{1}.$$
Any Hilbert space vector is thus uniquely defined by the "components" with respect to this basis, i.e., it's given by a square-integrable $\mathbb{C}^{2s+1}$-valued function, i.e., the components of this $(2s+1)$-dimensional spinor are functions of $\vec{x}$:
$$\psi_{\sigma}(\vec{x})=\langle \vec{x},\sigma|\psi \rangle,$$
and the scalar product between two vectors is
$$\langle \phi|\psi \rangle=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \sum_{\sigma=-s}^s \langle \phi|\vec{x},\sigma \rangle \langle \vec{x},\sigma|\psi \rangle=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \sum_{\sigma=-s}^s \phi_{\sigma}^*(\vec{x}) \psi_{\sigma}(\vec{x}).$$
This shows that the specific realization of the separable Hilbert space (there is, up to equivalence only one such Hilbert space!) depends on the system to be described, i.e., on the operators defining observables. The key to analyze this structure of the Hilbert space is the commutator algebra of the operators. This commutator algebra is specified by the symmetry group of the system. For our one particle it's the Galileo group of Newtonian spacetime. The analysis of the unitary ray representations of this group leads (after some work) to the above basis (usually first one analyzes the group in terms of momentum-spin eigenstates rather than the position-spin eigenstates, but that's a minor detail). Of course, I cannot explain all this within this forum. A very good source on this is Ballentine's book.

vanhees71, thanks for the time you took to type this response.

Can you tell me, in simple terms, what does it mean when you say "separable Hilbert space"? Why is that an important property? If it did not have the "separable" property how would that cripple it with respect to quantum mechanics? I have looked up the mathematical definition of "separable space" and quite frankly, I cannot figure out why that property is needed for hilbert spaces in quantum mechanics.

 

[mike1000]

Do the eigenvectors of all self-adjoint operators form a complete basis for the Hilbert space that contains them?

The answer is no.

The eigenvectors $|\phi_n\rangle$ of a self-adjoint operator $\hat{A}$\begin{equation}\hat{A}|\phi_n\rangle=a_n|\phi_n\rangle\end{equation} for distinct eigenvalues of $a_n$ do not, in general, form a complete orthonormal system. If they do form a complete orthonormal system {$|\phi_n\rangle$}, then one can use those vectors as a basis for vectors for representing other operators.
see page 8 http://www.phy.ohiou.edu/~elster/lectures/qm1_1p2.pdf

 

[mike1000]

vanhees71, thanks for the time you took to type this response.

Can you tell me, in simple terms, what does it mean when you say "separable Hilbert space"? Why is that an important property? If it did not have the "separable" property how would that cripple it with respect to quantum mechanics? I have looked up the mathematical definition of "separable space" and quite frankly, I cannot figure out why that property is needed for hilbert spaces in quantum mechanics.

According to this, a vector space is "separable" if it contains a dense subset. I do not find this definition very enlightening. Rather it appears to be counter-intuitive. Separate means you can divide into distinct parts. While dense means close together. Together they don't make a whole lot of sense to me. Can someone add a practical aspect to this and not just a purely mathematical aspect?

A Hilbert space being separable means that there exists a set of vectors dense in H and countable. Let {$|\psi_1\rangle,\dots,|\psi_k\rangle,\dots$} be a sequence of vectors in H. If we take out every vector $|\psi_k$ from this sequence which can be represented as a linear combination of the previous vectors, $|\psi_1\rangle,\dots,|\psi_{k-1}\rangle,\dots$, then we obtain a set of linearly independent vectors {$|\phi_1\rangle,\dots,|\phi_n\rangle,\dots$} in which the original sequence is contained. The set {$|\psi_n\rangle$}, is, via construction, dense in H. One can assume that {$|\psi_n\rangle$} is a set of orthogonal vectors.
page 4 http://www.phy.ohiou.edu/~elster/lectures/qm1_1p2.pdf

 

[vanhees71]

A Hilbert space is by definition separable if there exists a countable orthonormal basis. In QM fortunately you have this nice kind of Hilbert space, which makes life pretty easy; in fact so easy that you can treat it in the quite sloppy way physicists do in practice. Only sometimes one gets confused, if one hasn't at least glanced at the more subtle points (like the difference between Hermitean and self-adjoint operators or how to treat the continuous spectrum of unbound self-adjoint operators properly). One way to make the sloppy Dirac formalism rigorous is to introduce the modern concept of the "Rigged Hilbert Space". For a gentle introduction, see Ballentine's book on QM.