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r-rashidi

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thank you.

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- Thread starter r-rashidi
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r-rashidi

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thank you.

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"Let [itex] \mathcal{H} [/itex] be the space of all complex valued functions defined on [itex] \mathbb{R} [/itex] which vanich everywhere, except a countable number of points in [itex] \mathbb{R} [/itex] and the values of those points form a square summable sequence. For such functions the inner product

[tex] \langle f, g\rangle =\sum_{x\in\mathbb{R}} f(x)\bar{g}(x) [/tex]

is well defined, because the summation is actually over a countable set. The space [itex] \mathcal{H} [/itex] is not separable, because, for any sequence of functions [itex] f_{n}\in\mathcal{H} [/itex], there are nonzero functions [itex] f [/itex], such that

[tex]\langle f,f_{n}\rangle =0 \ , \ \forall n\in\mathbb{N}[/tex] "

Satisfied ?

As a side note, quantum physics (whether QM or quantum theory of fields) operates ONLY with separable Hilbert spaces.

Daniel.

- #3

Palindrom

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If I may barge in, why is that? Why is separability so important?

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Daniel.

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r-rashidi

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George Jones

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r-rashidi said:

The Fock space constructed from a separable 1-particle state space is itself separable.

Regards,

George

- #7

r-rashidi

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this subject is discucced .)

Regards,

reza.

- #8

George Jones

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r-rashidi said:By Haag's theorem in QFT (Local Q.P by Haag)one can obtain two representations from one algebra that are not unitarity equivalent. but all seperable Hilbert spaces(with infinit daimension) are topologically isomorphe( for example: thorem 2.3.3,Mathematical Quantization,Nik Weaver,).it shows Fock space is nonseprable.(i didn't read all of this book and i don't know this subject is discucced .)reza.

I don't understand what you're saying.

1) an algebra can have representations that are not unitarily equivalent. (OK)

2) all infinite-dimensional separable Hilbert spaces are homeomorphic. (OK)

3) Fock space is not separable. (???)

Are you saying that 3) follows from 1) and 2)? If so, how?

I still stand by the statement I made in my previous post.

A topological space is separable if it contains a countable, dense subset. As dextercioby said, for a Hilbert space [itex]V[/itex], this is equivalent to the existence of a complete orthonormal basis [itex]B[/itex], i.e., a countable set of orthonormal vectors iin terms of which every vector in [itex]V[/itex] can be expressed as a strong limit of linear combinations of elements of [itex]B[/itex].

For an infinite-dimensional Hilbert space [itex]V[/itex] with complete orthonormal basis [itex]B[/itex], let

[tex]

T = \mathbb{C} \oplus V \oplus \left( V \otimes V \right) \oplus \left( V \otimes V \otimes V \right) \oplus ...

[/tex]

A complete orthonormal basis for [itex]V \otimes V[/itex] is [itex]B \times B[/itex], which is countable, since the Cartesian product of countable sets is countable. By induction, every space [itex]V \otimes ... \otimes V \right[/itex] has a countable orthonormal basis. Therefore a basis for [itex]T[/itex] is [itex]\left\{ \psi_{ij} \right\}[/itex], where the index [itex]i[/itex] labels the space of i-particle states and the the index [itex]j[/itex] labels the j-th basis vector for the space of i-particle states.

Again, since the Cartesian product of countable spaces is countable, this basis has a countable number of elements. Hence, [itex]T[/itex] is separable. Fock space, which is the appropriate symmetric or anisymmetric subspace of [tex]T[/itex], is also separable.

Regards,

George

- #9

seratend

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False. If it was true, by induction L_infinity([0,1],dx) is a separable hilbert space because L_n([0,1],dx) for any finite n is separable.George Jones said:A complete orthonormal basis for [itex]V \otimes V[/itex] is [itex]B \times B[/itex], which is countable, since the Cartesian product of countable sets is countable. By induction, every space [itex]V \otimes ... \otimes V \right[/itex] has a countable orthonormal basis.

Seratend.

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- #10

George Jones

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seratend said:False. If it was true, by induction L_infinity([0,1],dx) is a separable hilbert space because L_n([0,1],dx) for any finite n is separable.

I wasn't very explicit, but I meant ... to represent a finite (but arbitrary) number of tensor products. I think that's all I need for my argument.

Regards,

George

- #11

seratend

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George Jones said:I wasn't very explicit, but I meant ... to represent a finite (but arbitrary) number of tensor products. I think that's all I need for my argument.

Regards,

George

{0,1}^n for any finite n is countable however {0,1}^|N is uncountable (same cardinality as |R) what do you conclude?

Seratend.

- #12

r-rashidi

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reza.

- #13

seratend

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r-rashidi said:

reza.

I guess you mean isomorphic. This is only true for separable hilbert spaces with the *same* dimension (dim in |N U {+oO}).

Seratend.

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George Jones

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seratend said:{0,1}^n for any finite n is countable however {0,1}^|N is uncountable (same cardinality as |R) what do you conclude?

I conclude that this example is not relevant for the construct of Fock space.

For [itex]V[/itex] a separable Hilbert space, set

[tex]

T = \sum_{n=0}^\infty \otimes^n V,

[/tex]

with [itex]\otimes^0 V = \mathbb{C}[/itex]. Just as an operational definition has to be given to the infinite sum

[tex]

S = \sum_{n=0}^\infty a_n,

[/tex]

an operational definition has to be given to the expression for [itex]T[/itex]. The infinite sum of real numbers is defined by

[tex]

S = \lim_{n\rightarrow\infty}s_n,

[/tex]

where

[tex]

s_n = \sum_{i=0}^n a_i.

[/tex]

Note that an infinite sum does directly add up an infinite number of terms; infinity only appears in the limit process.

[itex]T[/itex] is the set of sequences [itex]\left\{ \left( w_0 , w_1 , ... , w_i , ... \right) \right\}[/itex] such that:

1) [itex]w_i \in \otimes^i V[/itex];

2)

[tex]

\sum_{i=0}^\infty \|w_i\|^2 <\infty.

[/tex]

Condition 1) gives [itex]T[/itex] as a countably infinite direct sum of vector spaces, while condition 2) allows an inner product to be defined on [itex]T[/itex].

Let [itex]\left\{\psi_{ij} | j \in \mathbb{N} \right\}[/itex] be an orthonormal basis for [itex]\otimes^i V[/itex]. Then [itex]B_T = \left\{\psi_{ij} | \left(i ,j \right) \in \mathbb{N}_0 \times \mathbb{N} \right\}[/itex] is an orthnormal basis for [itex]T[/itex]. Since [itex]\mathbb{N}_0 \times \mathbb{N}[/itex] is countable, [itex]B_T[/itex] is countable, and [itex]T[/itex] is separable.

In order for your example to be relevant, [itex]i[/itex] would have to equal [itex]\mathbb{N}[/itex], which is not possible since [itex]\mathbb{N}[/itex] is not an element of [itex]\mathbb{N}[/itex].

It is well-known that the Fock space based on a separable Hilbert space is itself separable.

Regards,

George

- #15

seratend

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George Jones said:I conclude that this example is not relevant for the construct of Fock space.

...

It is well-known that the Fock space based on a separable Hilbert space is itself separable.

Regards,

George

You are claiming this while the OP thinks the contrary. I am just saying, with some adequate examples, that you have not given any valid proof for the separability property. And I still cannot see any valid demo in your last answer (I do not understand the meaning of your |No symbol nor the following claim).

Countability/uncountability demonstrations need some care in order to be valid (it is easy to make wrong demos such as |R is countable).

So I give another hint: Is [itex]\otimes^\mathbb{N} V[/itex] included in the fock space T?

Seratend.

- #16

r-rashidi

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seratend said:I guess you mean isomorphic. This is only true for separable hilbert spaces with the *same* dimension (dim in |N U {+oO}).

Seratend.

Equivalence means:an unitary map exists between two spaces(isometric topological isomorphism )(unitary equivalence )

reza.

- #17

seratend

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r-rashidi said:Equivalence means:an unitary map exists between two spaces(isometric topological isomorphism )(unitary equivalence )

reza.

: ) An equivalence relation is very general. Therefore it is sometimes usefull to give some precisions.

Seratend.

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vanesch

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George Jones said:In order for your example to be relevant, [itex]i[/itex] would have to equal [itex]\mathbb{N}[/itex], which is not possible since [itex]\mathbb{N}[/itex] is not an element of [itex]\mathbb{N}[/itex].

The problem I see is that although the basis constructed by George (psi_ij) is countable, this constructs a Fock space which, in my opinion, is not closed, and seratend's example of {0,1}^N is relevant, in that {0,1}^N is NOT the same as the union of all {0,1}^n: you need to impose CLOSURE under Cauchy series.

Now, I have to admit that I don't know if this closure requirement is part of the definition of a Fock space. In any case, it doesn't make sense physically to include states with an infinite number of particles.

If the closure requirement is not part of the Fock space, then I think George is right, in that there is a countable basis ; but only, the space is not closed (and is, if I understand correctly, not a true Hilbert space). If the closure requirement is part of the Fock space definition, then I tend to agree with seratend in that this closure makes the space non-separable in the same way as the closure of the union of all {0,1}^n (which are countable) make us have something with same cardinality as the real numbers.

cheers,

Patrick.

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George Jones

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seratend said:You are claiming this while the OP thinks the contrary. I am just saying, with some adequate examples, that you have not given any valid proof for the separability property. And I still cannot see any valid demo in your last answer (I do not understand the meaning of your |No symbol nor the following claim).

Sorry, I should have defined my symbols. [itex]\mathbb{N} := \left\{1, 2, 3, ... \right\}[/itex] and [itex]\mathbb{N}_0 := \left\{0, 1, 2, 3, ... \right\}[/itex]. For each [itex]i[/itex] in [itex]\mathbb{N}_0[/itex], [itex]\psi_{ij}[/itex] is in [itex]\otimes^i V[/itex], and [itex]\left\{ \psi_{ij} \right\}[/itex] is a countable orthonormal basis, labeled by [itex]j[/itex] in [itex]\mathbb{N}[/itex], for [itex]\otimes^i V[/itex]. Therefore, [itex]\psi_{ij}[/itex] = [itex]\psi_{kl}[/itex] iff [itex]\left( i , j \right) = \left( k , l \right)[/itex].

This establishes a bijection between [itex]B_T = \left\{\psi_{ij} | \left(i ,j \right) \in \mathbb{N}_0 \times \mathbb{N} \right\}[/itex] and [itex]\mathbb{N}_0 \times \mathbb{N}[/itex].

seratend said:Countability/uncountability demonstrations need some care in order to be valid (it is easy to make wrong demos such as |R is countable).

I agree.

seratend said:So I give another hint: Is [itex]\otimes^\mathbb{N} V[/itex] included in the fock space T?

No.

Fock space is a formalism for handling an arbitrary, but finite, number of particles, as well as for dealing with a changeable number of particles.

Regards,

George

- #20

vanesch

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George Jones said:Fock space is a formalism for handling an arbitrary, but finite, number of particles, as well as for dealing with a changeable number of particles.

With reference to what I posted, does that mean then, that Fock space is not closed (and hence, not a Hilbert space) ?

cheers,

Patrick.

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George Jones

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vanesch said:The problem I see is that although the basis constructed by George (psi_ij) is countable, this constructs a Fock space which, in my opinion, is not closed, and seratend's example of {0,1}^N is relevant, in that {0,1}^N is NOT the same as the union of all {0,1}^n: you need to impose CLOSURE under Cauchy series..

I don't want it to be the same.

vanesch said:Now, I have to admit that I don't know if this closure requirement is part of the definition of a Fock space. In any case, it doesn't make sense physically to include states with an infinite number of particles.

If the closure requirement is not part of the Fock space, then I think George is right, in that there is a countable basis ; but only, the space is not closed (and is, if I understand correctly, not a true Hilbert space). If the closure requirement is part of the Fock space definition, then I tend to agree with seratend in that this closure makes the space non-separable in the same way as the closure of the union of all {0,1}^n (which are countable) make us have something with same cardinality as the real numbers.

The [itex]T[/itex] that I defined is an inner product space that needs to be completed (made "closed" with respect to Cauchy sequences) in order to obtain a Hilbert space. This has to be done in order for a countable basis to exist, just as the inner product space of square-integrable functions has to be completed in order for a countable orthormal basis to exist.

[itex]T[/itex], when completed, does have a countable orthormal basis. Even when completed, it is only suitable for a finite number of particles.

Regards,

George

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Daniel.

- #23

seratend

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You still have not prooved that the fock space is separable or not (you have just demonstrated a separability on the individual spaces and not the spearability of the fock space).George Jones said:The [itex]T[/itex] that I defined is an inner product space that needs to be completed (made "closed" with respect to Cauchy sequences) in order to obtain a Hilbert space. This has to be done in order for a countable basis to exist, just as the inner product space of square-integrable functions has to be completed in order for a countable orthormal basis to exist.

[itex]T[/itex], when completed, does have a countable orthormal basis. Even when completed, it is only suitable for a finite number of particles.

Regards,

George

With such a deduction a real number which is a cauchy sequence of rational numbers implies |R is a countable set from your deduction. You must understand that the closure modifies the properties of the resulting set at the limit, hence your difficulty to proove the separability.

In addition, you seem to mix the separability with the closure and the existence of a basis ("

Patrick has perfectly underlined the principal problem concerning the proof and you seem not to understand it (T is first an hilbert space, hence closed => think on what really means the notation [itex]T = \sum_{n=0}^\infty \otimes^n V[/itex] to deduce the (no)/separability property).

For example, how can you say that [itex]\otimes^\mathbb{N} V[/itex] is not included in T? If you are able to proove that, prooving the separability is not more complex.

Seratend.

- #24

seratend

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dextercioby said:

Daniel.

Thanks dexter, but the game here is to construct a robust proof without the help of the books! (not so difficult).

Seratend

- #25

vanesch

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dextercioby said:To the ones of you which are not clear with it, try page 53 from the first volume of Reed and Simon.

I'm mathematically not as sophisticated as you are, so maybe you can spell out in a few words what's said there ?

And maybe you can answer my question: is a Fock space simply the sum of all n-particle states, or is closure (under Cauchy sequences) a requirement ?

Is it then correct that in the first case, Fock space has a countable basis, but is not a hilbert space (not closed) and that in the second case, Fock space is closed, but then I'm not sure it is separable (has a countable basis). So what's Fock space in the end ?

- #26

vanesch

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George Jones said:The [itex]T[/itex] that I defined is an inner product space that needs to be completed (made "closed" with respect to Cauchy sequences) in order to obtain a Hilbert space.

Up to here I follow. However, you probably agree that this closure adds (a LOT) of elements to the set, ....

Ah, that's perhaps the key: you have at least a DENSE, countable subset: namely your basis vectors, say, with rational coefficients.

I vaguely remember that there is a theorem that says that a hilbert space that has a countable dense subset is separable, is that the point ?

cheers,

Patrick.

EDIT: I'm beginning to see the point. The basis of the closure is NOT the closure of the basis (which is uncountable)

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- #27

seratend

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vanesch said:EDIT: I'm beginning to see the point. The basis of the closure is NOT the closure of the basis (which is uncountable)

Bravo! (is NOT --> *may be* NOT), hence the additional difficulty.

Seratend.

- #28

George Jones

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No. And you mean an equivalence class of Cauchy sequences of rational numbers.seratend said:With such a deduction a real number which is a cauchy sequence of rational numbers implies |R is a countable set from your deduction.

By snipping the context of the quote, you make me appear to say something that I did not. In particular, I did not say that there exists a countable orthonormal basis for the Hilbert space completion of every inner product space.seratend said:In addition, you seem to mix the separability with the closure and the existence of a basis ("This has to be done in order for a countable basis to exist") of an hilbert space that may be countable or not countable and always exists thanks the axiom of choice.

Regards,

George

- #29

seratend

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Does that mean your are satisfied with your demonstration?

Seratend.

Seratend.

- #30

George Jones

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seratend said:Does that mean your are satisfied with your demonstration?

Exercise: Show that any element of T can be approximated arbitrarily closely by a linear combination of elements of B_T that consists of a finite number of terms.

Regards,

George

- #31

eztum

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knew the (in a sense trivial) example cited here by dextercioby and the non-trivial example of 'almost-periodic functions'. Also everybody knew that the Fock space over a seperable Hilbert space (acting as 'one-particle space') is separable. Today I 'know' that all Hilbert spaces are finite-dimensional! Tempora mutantur!

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