## Degenerate vacua in QFT

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no, scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\nIn the thread titled\n\n"The vacuum involving spontaneously broken Higgs fields"\n\nDr Baez says:\n\n"It goes back to the fact that in quantum field\ntheory, the canonical commutation relations have many inequivalent\nrepresentations. It turns out we can\'t think of all the different\nnoninvariant vacua as states in the same Hilbert space"\n\nI\'m wondering what does he mean by "the different vacua are not\nstates in the same hilbert space" ?\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">&nbsp;&nbsp;View this Usenet post in original ASCII form </a></div><P></jabberwocky>In the thread titled

"The vacuum involving spontaneously broken Higgs fields"

Dr Baez says:

"It goes back to the fact that in quantum field
theory, the canonical commutation relations have many inequivalent
representations. It turns out we can't think of all the different
noninvariant vacua as states in the same Hilbert space"

I'm wondering what does he mean by "the different vacua are not
states in the same hilbert space" ?

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pirillo wrote: > In the thread titled > > "The vacuum involving spontaneously broken Higgs fields" > > Dr Baez says: >> It goes back to the fact that in quantum field theory, the >> canonical commutation relations have many inequivalent >> representations. It turns out we can't think of all the different >> noninvariant vacua as states in the same Hilbert space" > I'm wondering what does he mean by "the different vacua are not > states in the same hilbert space" ? I'm sure Dr Baez can give a more complete answer than I can. But as I've been trying to learn about this stuff during the past 12 months, I'll offer a contribution below. This is mostly taken from Umezawa's textbook (Ref[1])... Considering the state of a many-body system, where each particle can be in any state labelled by "i", and we write "$n_i$" for the number of particles in state "i". The state of the many-body system is identified when we specify $n_i$ for all i, and is denoted by a ket like this: $|n1, n2, n3, .$..>. For bosons, each $n_i$ can take any non-negative integer value. For fermions, they can only by or 1. In what follows below, I'll assume we're talking about bosons. We then construct the set of all possible such kets and denote the set ${|n1, n2, n3, ...>}.$ Then we can introduce creation and annihilation operators, $a*_i$ and $a_i$ respectively, to take us between different elements of this set: $$a_i |n1, .[/itex]...$, n_i, .$..> $= \sqrt(n_i) |n1, .$...$, n_i - 1, .$..> $a*_i |n1, .$...$, n_i, .$..> $= \sqrt(n_i + 1) |n1, .$...$, n_i + 1, .$..> These definitions imply the usual commutation relations: $[a_i, a*_j] |n1, n2, .$..> $= \delta_ij |n1, n2, .$..> $[a_i, a_j] |n1, n2, .$..> = $[a*_i, a*_j] |n1, n2, .$.$.> =$$ You could be forgiven for thinking that this just looks like the usual construction of QFT Fock space that one finds in many textbooks. But the crucial thing here is that the set [itex]{|n1, n2, n3, ...>}$ is non-countable. I.e: there does NOT exist a 1:1 mapping between elements of this set and the integers. This is easiest to see for the case of fermions where each $n_i$ in any $|n1, n2, n3, .$..> is a or a 1. Since the set is infinite, this is just a binary-number representation of the interval (0,1) of the real line, and we know there are infinitely more real numbers in this interval than all the integers. A similar argument applies for the boson case. A Hilbert space based on ${|n1, n2, n3, ...>}$ is called "non-separable". This means that it does *not* have a countable basis $e_k$ such that any vector in the space can be approximated arbitrarily closely by a linear combination of the $e_k$ like: $Sum{k=0,inf} c_k e_k$. This causes lots of problems when applying the usual math associated with (separable) Hilbert space, such as integration, etc. Certain important theorems no longer hold, since their proofs rely on being able to approach any element of the space arbitrarily closely by such linear combinations. This also creates problems with the definition of an inner product, which we must have to get a useful Hilbert space. It is the source of many of the profound differences between QM and QFT. Confronted by this, one then invokes physical ideas, arguing that the set ${|n1, n2, n3, ...>}$ is unnecessarily large, and that we really only need a subset such that $Sum{i=0,inf} n_i =$ finite. I.e: the subset whose total number of particles is finite. It is called the "$[0]-set$"$. It$ can be shown that the $[0]-set *is*$ countable, and (after being equipped with an inner product) can therefore be used to construct a separable Hilbert space that we all know and love, namely Fock space. The trouble now is that the choice of a subset of the whole space ${|n1, n2, n3, ...>}$ to use as out Fock space is not unique. There is an infinite number of such subsets, all orthogonal to each other. I.e: there is an infinity of different separable Fock spaces lying in the larger non-separable ${|n1, n2, n3, ...>}$ space. Each such Fock space furnishes a representation of the canonical (anti-)commutation relations, but they're orthogonal to each other. We say that two such Fock spaces are "unitarily inequivalent", in the sense that no vectors in one Fock space cannot be written as a linear combinations involving vector(s) from the other Fock space. To show an example, let's assume the usual passage to momentum space so that we get creation and annihilation operators a(p), $a*(p)$ as usual. We can now imagine mixing creation and annihilation operators. Suppose we have two sets of them, called $a, a*, b, b*$. Let's mix them as shown below to get new operators $\alpha, \alpha*, \beta,\beta*:\alpha(p) =$ A a(p) $- B b*(-p)\beta(p) =$ A b(p) $- B a*(-p)$ where in general, A and B may depend on p. Demanding that the new operators $\alpha$ and $\beta$ satisfy the same commutation relations as a and b imposes the constraint: $$A^2 - B^2 = 1[/itex]. Such transformations are called Bogoliubov transformations because they preserve the canonical commutation relations (CCRs). In other words, the $\alpha$ and $\beta$ operators form another representation of the CCRs. To see what transformation is induced on the *states* by this Bogoliubov transformation, we must find an operator G such that: $\alpha(p) = G^-1 a(p) G\beta(p) = G^-1 b(p) G$$ Writing [itex]A = cosh(\theta)$ and $B = sinh(\theta),$ and remembering that $\theta$ may depend on p), it turns out that G is given by: G $= \exp$ Integral $d^{3k} \theta(k) [a(k)b(-k) - b*(-k)a*(k)]$ Now we consider that transformed vacuum state, i.e: |vac'> $= G^-1$ |vac> and compute the inner product between it and all the other vectors of the original Fock space based on a(p) and b(p): After some fairly difficult math (see [1]), it can be shown that ALL such expressions are . Therefore, the transformed vacuum |vac'> of the "$\alpha,\beta$" Fock space cannot be a linear combination of *any* vectors in the "a,b" Fock space. This is what we mean when we say that the two Fock spaces are orthogonal, and what we mean when we say that they're "unitarily inequivalent". Hopefully, it's now clear what is meant by "the different vacua are not states in the same hilbert space". ------------------------------------------------------------ Ref [1] Umezawa, Matsumoto & Tachiki. "Thermo Field Dynamics and Condensed States", North Holland ISBN $0-444-86361-3$



On Wed, 15 Dec 2004 18:37:55 $+0000,$ mikem wrote: [...] > To see what transformation is induced on the *states* by this Bogoliubov > transformation, we must find an operator G such that: > > $\alpha(p) = G^-1 a(p) G$ > $\beta(p) = G^-1 b(p) G$ > > Writing $A = cosh(\theta)$ and $B = sinh(\theta),$ and remembering that $\theta$ > may depend on p), it turns out that G is given by: > > G $= \exp$ Integral $d^{3k} \theta(k) [a(k)b(-k) - b*(-k)a*(k)]$ > > Now we consider that transformed vacuum state, i.e: > > |vac'> $= G^-1$ |vac> > > and compute the inner product between it and all the other vectors of the > original Fock space based on a(p) and b(p): > > > > After some fairly difficult math (see [1]), it can be shown that ALL such > expressions are . Therefore, the transformed vacuum > |vac'> of the "$\alpha,\beta$" Fock space cannot be a linear combination > of *any* vectors in the "a,b" Fock space. This is what we mean when we say > that the two Fock spaces are orthogonal, and what we mean when we say that > they're "unitarily inequivalent". > > Hopefully, it's now clear what is meant by "the different vacua are not > states in the same hilbert space". > > ------------------------------------------------------------ > > Ref [1] Umezawa, Matsumoto & Tachiki. "Thermo Field Dynamics and Condensed > States", North Holland ISBN $0-444-86361-3$ I can see that by applying this transformation G to the Fock space you are working with, you get another Fock space orthogonal to it in the larger non-separable Hilbert space. However, I don't see why they are not unitarily equivalent, unless I'm missing something simple. After all, you do have the unitary transformation G between the two Fock spaces that preserves the CCRs. Is this not unitary equivalence? BTW, if $a,b,a*,b*$ are operators on the first Fock space, how can operator built out of them not preserve it? Are the $a,b,a*,b*$ operators defined in somewhat more generality on the larger non-separable Hilbert space? Perhaps I'm misunderstanding something here. Thanks. Igor

## Degenerate vacua in QFT

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no, scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>mikem@despammed.com wrote:\n\n&gt; This is mostly taken from Umezawa\'s textbook (Ref[1])...\n&gt;\n&gt; Considering the state of a many-body system, where each particle\n&gt; can be in any state labelled by "i", and we write "n_i" for the\n&gt; number of particles in state "i". The state of the many-body system\n&gt; is identified when we specify n_i for all i, and is denoted by a\n&gt; ket like this: |n1, n2, n3, ...&gt;. For bosons, each n_i can take any\n&gt; non-negative integer value. For fermions, they can only by 0 or 1.\n&gt; In what follows below, I\'ll assume we\'re talking about bosons.\n&gt;\n&gt; We then construct the set of all possible such kets and denote the\n&gt; set {|n1, n2, n3, ...&gt;}. Then we can introduce creation and\n&gt; annihilation operators, a*_i and a_i respectively, to take us\n&gt; between different elements of this set:\n&gt;\n&gt; a_i |n1, ...., n_i, ...&gt; = sqrt(n_i) |n1, ...., n_i - 1, ...&gt;\n&gt;\n&gt; a*_i |n1, ...., n_i, ...&gt; = sqrt(n_i + 1) |n1, ...., n_i + 1, ...&gt;\n&gt;\n&gt; These definitions imply the usual commutation relations:\n&gt;\n&gt; [a_i, a*_j] |n1, n2, ...&gt; = delta_ij |n1, n2, ...&gt;\n&gt;\n&gt; [a_i, a_j] |n1, n2, ...&gt; = 0\n&gt;\n&gt; [a*_i, a*_j] |n1, n2, ...&gt; = 0\n&gt;\n&gt; You could be forgiven for thinking that this just looks like the\n&gt; usual construction of QFT Fock space that one finds in many\n&gt; textbooks. But the crucial thing here is that the set\n&gt; {|n1, n2, n3, ...&gt;} is non-countable. I.e: there does NOT exist a\n&gt; 1:1 mapping between elements of this set and the integers. This is\n&gt; easiest to see for the case of fermions where each n_i in any\n&gt; |n1, n2, n3, ...&gt; is a 0 or a 1. Since the set is infinite, this is\n&gt; just a binary-number representation of the interval (0,1) of the\n&gt; real line, and we know there are infinitely more real numbers in\n&gt; this interval than all the integers. A similar argument applies for\n&gt; the boson case.\n&gt;\n&gt; A Hilbert space based on {|n1, n2, n3, ...&gt;} is called\n&gt; "non-separable". This means that it does *not* have a countable\n&gt; basis e_k such that any vector in the space can be approximated\n&gt; arbitrarily closely by a linear combination of the e_k like:\n&gt; Sum{k=0,inf} c_k e_k.\n\nWhat you have so far is not a Hilbert space (and hence the question of\nseparability does not really matter), but just a representation\nof the CCR on a vector space. To define a Hilbert space one must\ndefine an inner product, and there are many inequivalent ways of doing so.\n\nThe most interesting ones are those which have a vacuum state from which\nall others are generated by applying to the vacuum a sequence of modified\ncreation operators (alpha, beta in the reaminder of your mail) which are\n(in the simplest case) linear combinations of the original creation and\nannihilation operators.\n\nThe resulting states form a subspace of the big space you mentioned,\nand on this subspace there is a good inner product. The point is now that\nfor inequivalent choices of the modified creation operators, one gets\ninequivalent Hilbert spaces, all subspaces of the same non-Hilbert space,\nintersecting only in 0. Thus each Hilbert space contains only its native\nvacuum, but none of the other vacua.\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">&nbsp;&nbsp;View this Usenet post in original ASCII form </a></div><P></jabberwocky>mikem@despammed.com wrote:

> This is mostly taken from Umezawa's textbook (Ref[1])...
>
> Considering the state of a many-body system, where each particle
> can be in any state labelled by "i", and we write "$n_i$" for the
> number of particles in state "i". The state of the many-body system
> is identified when we specify $n_i$ for all i, and is denoted by a
> ket like this: $|n1, n2, n3, .$..>. For bosons, each $n_i$ can take any
> non-negative integer value. For fermions, they can only by or 1.
> In what follows below, I'll assume we're talking about bosons.
>
> We then construct the set of all possible such kets and denote the
> set ${|n1, n2, n3, ...>}.$ Then we can introduce creation and
> annihilation operators, $a*_i$ and $a_i$ respectively, to take us
> between different elements of this set:
>
> $a_i |n1, .$...$, n_i, .$..> $= \sqrt(n_i) |n1, .$...$, n_i - 1, .$..>
>
> $a*_i |n1, .$...$, n_i, .$..> $= \sqrt(n_i + 1) |n1, .$...$, n_i + 1, .$..>
>
> These definitions imply the usual commutation relations:
>
> $[a_i, a*_j] |n1, n2, .$..> $= \delta_ij |n1, n2, .$..>
>
> $[a_i, a_j] |n1, n2, .$..> =
>
> $[a*_i, a*_j] |n1, n2, .$.$.> =$
>
> You could be forgiven for thinking that this just looks like the
> usual construction of QFT Fock space that one finds in many
> textbooks. But the crucial thing here is that the set
> ${|n1, n2, n3, ...>}$ is non-countable. I.e: there does NOT exist a
> 1:1 mapping between elements of this set and the integers. This is
> easiest to see for the case of fermions where each $n_i$ in any
> $|n1, n2, n3, .$..> is a or a 1. Since the set is infinite, this is
> just a binary-number representation of the interval (0,1) of the
> real line, and we know there are infinitely more real numbers in
> this interval than all the integers. A similar argument applies for
> the boson case.
>
> A Hilbert space based on ${|n1, n2, n3, ...>}$ is called
> "non-separable". This means that it does *not* have a countable
> basis $e_k$ such that any vector in the space can be approximated
> arbitrarily closely by a linear combination of the $e_k$ like:
> $Sum{k=0,inf} c_k e_k$.

What you have so far is not a Hilbert space (and hence the question of
separability does not really matter), but just a representation
of the CCR on a vector space. To define a Hilbert space one must
define an inner product, and there are many inequivalent ways of doing so.

The most interesting ones are those which have a vacuum state from which
all others are generated by applying to the vacuum a sequence of modified
creation operators $(\alpha, \beta$ in the reaminder of your mail) which are
(in the simplest case) linear combinations of the original creation and
annihilation operators.

The resulting states form a subspace of the big space you mentioned,
and on this subspace there is a good inner product. The point is now that
for inequivalent choices of the modified creation operators, one gets
inequivalent Hilbert spaces, all subspaces of the same non-Hilbert space,
intersecting only in . Thus each Hilbert space contains only its native
vacuum, but none of the other vacua.

Arnold Neumaier



Igor Khavkine wrote: > On Wed, 15 Dec 2004 18:37:55 $+0000,$ mikem wrote: > > [...] > > >>To see what transformation is induced on the *states* by this Bogoliubov >>transformation, we must find an operator G such that: >> $>>\alpha(p) = G^-1 a(p) G>>\beta(p) = G^-1 b(p) G$ >> >>Writing $A = cosh(\theta)$ and $B = sinh(\theta),$ and remembering that $\theta$ >>may depend on p), it turns out that G is given by: >> >>G $= \exp$ Integral $d^{3k} \theta(k) [a(k)b(-k) - b*(-k)a*(k)]$ >> >>Now we consider that transformed vacuum state, i.e: >> >>|vac'> $= G^-1$ |vac> >> >>and compute the inner product between it and all the other vectors of the >>original Fock space based on a(p) and b(p): >> >> >> >>After some fairly difficult math (see [1]), it can be shown that ALL such >>expressions are . Therefore, the transformed vacuum >>|vac'> of the "$\alpha,\beta$" Fock space cannot be a linear combination >>of *any* vectors in the "a,b" Fock space. This is what we mean when we say >>that the two Fock spaces are orthogonal, and what we mean when we say that >>they're "unitarily inequivalent". >> > I can see that by applying this transformation G to the Fock space you are > working with, you get another Fock space orthogonal to it in the larger > non-separable Hilbert space. However, I don't see why they are not > unitarily equivalent, unless I'm missing something simple. After all, you > do have the unitary transformation G between the two Fock spaces that > preserves the CCRs. Is this not unitary equivalence? These operators are formally unitary, which means they define *-automorphisms of the algebra. But they are not really unitary, since there is no inner product on the big space they live on, and 'unitary' means 'norm-preserving'. But what is not defined cannot be preserved. > BTW, if $a,b,a*,b*$ are operators on the first Fock space, how can operator > built out of them not preserve it? Are the $a,b,a*,b*$ operators defined in > somewhat more generality on the larger non-separable Hilbert space? They are defined on the large vector space spanned by all formal linear combinations of arbitrary basis states with definite particle numbers. This large space contains all the Hilbert spaces but is itself not a Hilbert space. If one discretizes (i.e., restricts to a finite number of modes) then the Bogoliubov transforms become true unitary transforms in the finite-mode Fock space, and all representations become equivalent. Arnold Neumaier



Igor Khavkine wrote: > I can see that by applying this transformation G to the Fock space > you are working with, you get another Fock space orthogonal to it > in the larger non-separable Hilbert space. However, I don't see why > they are not unitarily equivalent, unless I'm missing something > simple. After all, you do have the unitary transformation G between > the two Fock spaces that preserves the CCRs. Is this not unitary > equivalence? I was (and still am) a bit perplexed about this terminology. I think it depends on the rigorous definition of "unitary transformation", which involves domains, and so forth. The definition also needs a well-defined inner product to preserve, i.e: the usual inner product on Hilbert spaces. But (IIUC) such an inner product is not well-defined on the larger non-separable space because of the difficulties in doing the usual calculus on that space. So I think of "unitary inequivalence" as signifying that although the transformation G appears superficially unitary, it is not really so. > BTW, if $a,b,a*,b*$ are operators on the first Fock space, how can > operator built out of them not preserve it? Because Fock space is defined in such a way that the total number of particles in any given vector is finite. But the G transformation involves creating an infinite number of particles in the original Fock space and therefore cannot preserve it. > Are the $a,b,a*,b*$ operators defined in somewhat more generality > on the larger non-separable Hilbert space? [...] That's my understanding. One must be super-careful with this stuff because plenty of familiar results don't hold, so intuition gained while working in separable Hilbert spaces spaces is unreliable. Example: naively, one might think that the G transformation with infinitesimal $\theta$ is "infinitesimally close" to the identity, and so one might (erroneously) think that the transformation group is identity-connected. But the math shows that even if $\theta$ is only infinitesimally different from 0, you still get taken to a totally orthogonal Fock space. The source and target vectors have overlap.



On Fri, 17 Dec 2004 13:48:53 $+0000,$ Arnold Neumaier wrote: > mikem@despammed.com wrote: > >> This is mostly taken from Umezawa's textbook (Ref[1])... >> >> Considering the state of a many-body system, where each particle can be >> in any state labelled by "i", and we write "$n_i$" for the number of >> particles in state "i". The state of the many-body system is identified >> when we specify $n_i$ for all i, and is denoted by a ket like this: $|n1,$ >> n2, n3, ...>. For bosons, each $n_i$ can take any non-negative integer >> value. For fermions, they can only by or 1. In what follows below, >> I'll assume we're talking about bosons. >> >> We then construct the set of all possible such kets and denote the set $>> {|n1, n2, n3, ...>}.$ Then we can introduce creation and annihilation >> operators, $a*_i$ and $a_i$ respectively, to take us between different >> elements of this set: >> $>> a_i |n1, ...., n_i, ...> = \sqrt(n_i) |n1, .$...$, n_i - 1, .$..> >> $>> a*_i |n1, ...., n_i, ...> = \sqrt(n_i + 1) |n1, .$...$, n_i + 1, .$..> >> >> These definitions imply the usual commutation relations: >> $>> [a_i, a*_j] |n1, n2, ...> = \delta_ij |n1, n2, .$..> >> $>> [a_i, a_j] |n1, n2, ...> =$ >> $>> [a*_i, a*_j] |n1, n2, .$.$.> =$ >> >> You could be forgiven for thinking that this just looks like the usual >> construction of QFT Fock space that one finds in many textbooks. But the >> crucial thing here is that the set ${|n1, n2, n3, ...>}$ is non-countable. $>> I$.e: there does NOT exist a 1:1 mapping between elements of this set and >> the integers. This is easiest to see for the case of fermions where each $>> n_i$ in any $>> |n1, n2, n3, .$..> is a or a 1. Since the set is infinite, this is >> just a binary-number representation of the interval (0,1) of the real >> line, and we know there are infinitely more real numbers in this >> interval than all the integers. A similar argument applies for the boson >> case. >> >> A Hilbert space based on ${|n1, n2, n3, ...>}$ is called "non-separable". >> This means that it does *not* have a countable basis $e_k$ such that any >> vector in the space can be approximated arbitrarily closely by a linear >> combination of the $e_k$ like: $Sum{k=0,inf} c_k e_k$. > > What you have so far is not a Hilbert space (and hence the question of > separability does not really matter), but just a representation of the CCR > on a vector space. To define a Hilbert space one must define an inner > product, and there are many inequivalent ways of doing so. There are always different choices for an inner product on a vector space to make it into a (pre-)Hilbert space. What's wrong with just making each $|n1,n2,n3,$...> a unit vector orthogonal to all the other ones? Is the dimension of the space "too big" for this to work? Thanks. Igor



On Fri, 17 Dec 2004 13:49:00 $+0000,$ Arnold Neumaier wrote: > If one discretizes (i.e., restricts to a finite number of modes) then the > Bogoliubov transforms become true unitary transforms in the finite-mode > Fock space, and all representations become equivalent. > I'm curious as to how the use of a Bogoliubov transformation here to change between different Fock subspaces of the large space of all states of definite particle number relates to its use in superconductivity theory where it is used to transform the particle spectrum from electron to quasiparticle states. In a normal metal, the ground state is given by the Fermi sea: |vac> $= prod_{E(k) < \mu} c*_k |0>,$ where |0> is the zero particle state, E(k) is the energy dispersion relation, \mu is the chemical potential and $c*_k$ is the electron creation operator for the state with momentum k. After a mean field approximation the effective Hamiltonian contains terms of the form c*c as well as $c*c*$ and cc. Hence, to diagonalize it, we introduce new creation-annihilation operators $$b_k = u_k c_k + v_k c*_(-k)b*_(-k) = v*_k c_(-k) + u*_k c*_k ,$$ where $|u_k|^2 + |v_k|^2 = 1$ in order to preserve the canonical anti-commutation relations. In terms of these new operators, the ground state now has the form |vac> $= prod_{E(k) < \mu} (u_k + v_k b*_k b*_(-k))|0>$. Are there two in equivalent Fock spaces lurking somewhere in the background here? Thanks. Igor



Igor Khavkine wrote: > On Fri, 17 Dec 2004 13:48:53 $+0000,$ Arnold Neumaier wrote: > >>>A Hilbert space based on ${|n1, n2, n3, ...>}$ is called "non-separable". >>>This means that it does *not* have a countable basis $e_k$ such that any >>>vector in the space can be approximated arbitrarily closely by a linear >>>combination of the $e_k$ like: $Sum{k=0,inf} c_k e_k$. >> >>What you have so far is not a Hilbert space (and hence the question of >>separability does not really matter), but just a representation of the CCR >>on a vector space. To define a Hilbert space one must define an inner >>product, and there are many inequivalent ways of doing so. > > There are always different choices for an inner product on a vector > space to make it into a (pre-)Hilbert space. What's wrong with just > making each $|n1,n2,n3,$...> a unit vector orthogonal to all the other > ones? Is the dimension of the space "too big" for this to work? It probably gives a useless topology. In infinite dimensions, choosing the right topology is essential for doing anything useful. Arnold Neumaier



Igor Khavkine wrote: > On Fri, 17 Dec 2004 13:49:00 $+0000,$ Arnold Neumaier wrote: > > >>If one discretizes (i.e., restricts to a finite number of modes) then the >>Bogoliubov transforms become true unitary transforms in the finite-mode >>Fock space, and all representations become equivalent. >> > > > I'm curious as to how the use of a Bogoliubov transformation here to > change between different Fock subspaces of the large space of all states > of definite particle number relates to its use in superconductivity theory > where it is used to transform the particle spectrum from electron to > quasiparticle states. > > In a normal metal, the ground state is given by the Fermi sea: > > |vac> $= prod_{E(k) < \mu} c*_k |0>,$ > > where |0> is the zero particle state, E(k) is the energy dispersion > relation, \mu is the chemical potential and $c*_k$ is the electron creation > operator for the state with momentum k. After a mean field approximation > the effective Hamiltonian contains terms of the form c*c as well as $c*c*$ > and cc. Hence, to diagonalize it, we introduce new creation-annihilation > operators > > $b_k = u_k c_k + v_k c*_(-k)$ > $b*_(-k) = v*_k c_(-k) + u*_k c*_k ,$ > > where $|u_k|^2 + |v_k|^2 = 1$ in order to preserve the canonical > anti-commutation relations. In terms of these new operators, the ground > state now has the form > > |vac> $= prod_{E(k) < \mu} (u_k + v_k b*_k b*_(-k))|0>$. > > Are there two in equivalent Fock spaces lurking somewhere in the > background here? Of course. This is explained in Detail in Umezawa's book. The change of c/a operators is always nonunitary in the above situation. Arnold Neumaier



Arnold Neumaier wrote: > Igor Khavkine wrote: > >> On Fri, 17 Dec 2004 13:48:53 $+0000,$ Arnold Neumaier wrote: >> >>>> A Hilbert space based on ${|n1, n2, n3, ...>}$ is called "non-separable". >>>> This means that it does *not* have a countable basis $e_k$ such that any >>>> vector in the space can be approximated arbitrarily closely by a linear >>>> combination of the $e_k$ like: $Sum{k=0,inf} c_k e_k$. >>> >>> What you have so far is not a Hilbert space (and hence the question of >>> separability does not really matter), but just a representation of >>> the CCR >>> on a vector space. To define a Hilbert space one must define an inner >>> product, and there are many inequivalent ways of doing so. >> >> There are always different choices for an inner product on a vector >> space to make it into a (pre-)Hilbert space. What's wrong with just >> making each $|n1,n2,n3,$...> a unit vector orthogonal to all the other >> ones? Is the dimension of the space "too big" for this to work? > > It probably gives a useless topology. In infinite dimensions, choosing > the right topology is essential for doing anything useful. Even worse, it seems to me that your Hilbert space does not even contain the Bogoliubov transformed vacuum |vac'> $= G^-1$ |vac>, (1) where G $= \exp$ Integral $d^{3k} \theta(k) [a(k)b(-k) - b*(-k)a*(k)]$ (formulas as in mikem's mail), since there is no reason to expect that the formal infinite sum obtained by expanding the right hand side of (1) converges. One needs the vector space of all formal linear combinations sum $\psi(n1,n2,n3,...) |n1,n2,n3,...>$ with _arbitrary_ complex $\psi(n1,n2,n3,...)$ to make G and $G^-1$ well-defined, and this vector space has no natural Hilbert space structure. Your recipe no longer works. See also the new FAQ entry 'Inequivalent representations $of CCR/CAR'$ in my theoretical physics FAQ at http://www.mat.univie.ac.at/~neum/physics-faq.txt where I discuss this in a little more detail (but G has the inverse meaning). Arnold Neumaier



Igor Khavkine wrote: >> What's wrong with just making each $|n1,n2,n3,$...> a unit >> vector orthogonal to all the other ones? [...] Arnold Neumaier wrote: > One needs the vector space of all formal linear combinations > > sum $\psi(n1,n2,n3,...) |n1,n2,n3,...>$ > > with _arbitrary_ complex $\psi(n1,n2,n3,...)$ to make G and $G^-1$ > well-defined, and this vector space has no natural Hilbert > space structure. Your recipe no longer works. [...] One aspect of all this that I'm still hazy about is precisely why we can't just integrate over all the n1, n2, n3, etc. I read that it stems from the fact that a translationally invariant measure does not exist on an infinite dimensional space. But I don't understand the proof. I've looked at the words in Glimme & Jaffe p136 sect A.4, but it's too brief and opaque for me. Could you explain it to me in more pedagogical terms? Also, G&J then go on to discuss Gaussian measures on infinite dimensional spaces. So I don't understand why we can't integrate sensibly over $|n1,n2,n3,$...> space using a Gaussian measure.



mikem@despammed.com wrote: > Arnold Neumaier wrote: > >>One needs the vector space of all formal linear combinations >> sum $\psi(n1,n2,n3,...) |n1,n2,n3,...>$ >>with _arbitrary_ complex $\psi(n1,n2,n3,...)$ to make G and $G^-1$ >>well-defined, and this vector space has no natural Hilbert >>space structure. Your recipe no longer works. [...] > > One aspect of all this that I'm still hazy about is precisely > why we can't just integrate over all the n1, n2, n3, etc. The nk were occupation numbers; the index k in nk would be the momenta, and nk is the number of particles with momentum k. In the continuum limit, one would have an uncountable number of entries in the ket... Thus your question does not make immediate sense. > I read that it stems from the fact that a translationally > invariant measure does not exist on an infinite dimensional > space. But I don't understand the proof. I've looked at the > words in Glimme & Jaffe p136 sect A.4, but it's too brief and > opaque for me. Could you explain it to me in more pedagogical > terms? It is only asserted there, without proof. Instead an intuitive illustration is given why one should expect it to be true: contributions of variables integrated over later in an infinity of repeated 1D integrations must necesarily be highly suppressed in order that the integral converges. But translation invariance would require that all dimensions are treated on equal footing. > Also, G&J then go on to discuss Gaussian measures on > infinite dimensional spaces. So I don't understand why > we can't integrate sensibly over $|n1,n2,n3,$...> space > using a Gaussian measure. One could, and does indeed so in free theories. But it means a restriction on the $\psi(n1,n2,n3,...)$ to have the integral converge. The standard covariance functions correspond exactly to a free theory in Fock space, thus $\psi(n1,n2,n3,...)$ must vanish if $n1+n2+n3+$... diverges, and the sum of absolute squares of $\psi$ evaluated at the remaining, countably many occupation sequences must also converge. This defines a fairly small subspace of the large vector space... More generally, Gaussian measures correspond to so-called generalized free theories. Interacting theories need non-Gaussian measures, and their construction occupies the second third of the book by Glimm and Yaffe. Arnold Neumaier



This thread has become so long and complicated, I had to go back to the top to remember what the original question was about. Now, isn't there a short coherent answer to the original question. I.e., I thought that even when a theory has many groundstates, to begin with, for that theory, there is just one Hilbert space H, which all these different groundstates are elements of. Are you saying that there are some kind of anihilation, creation operators $A_i$ on H such that when applied to two different grounstates repeatedly they generate two independent subspaces--one for each Hilbert space?



pirillo wrote: > This thread has become so long and complicated, > I had to go back to the top to remember what the > original question was about. Now, isn't there a short > coherent answer to the original question. I.e., I > thought that even when a theory has many groundstates, > to begin with, for that theory, there is just one > Hilbert space H, which all these different groundstates > are elements of. Are you saying that there are some > kind of anihilation, creation operators $A_i$ on H such that > when applied to two different grounstates repeatedly > they generate two independent subspaces--one for each > Hilbert space? > It is not clear to me how to make sense of your question. If the ground state is degenerate (i.e., at a critical point of the theory), the relevant Hilbert space is a direct integral of the Hilbert spaces formed by the cyclic representations of the algebra of observables acting on the various vacua. Arnold Neumaier



pirillo wrote: > This thread has become so long and complicated, I had to go back to > the top to remember what the original question was about. Now, isn't > there a short coherent answer to the original question. I.e., I > thought that even when a theory has many groundstates, to begin > with, for that theory, there is just one Hilbert space H, which all > these different groundstates are elements of. To be precise, your original posting contained: > Dr Baez says: > >> "It goes back to the fact that in quantum field theory, the >> canonical commutation relations have many inequivalent >> representations. It turns out we can't think of all the >> different noninvariant vacua as states in the same Hilbert >> space" > > I'm wondering what does he mean by "the different vacua are not > states in the same hilbert space" ? I then tried to answer this by explaining what "inequivalent representations" are. In short, they correspond to different (orthogonal) Hilbert spaces within a larger non-Hilbert space which is difficult to work with mathematically. Each "different noninvariant vacua" mentioned by Dr Baez lives in one these different Hilbert spaces. So no, these different vacua do not live in the same Hilbert (Fock) space. > Are you saying that there are some kind of anihilation, creation > operators $A_i$ on H such that when applied to two different > grounstates repeatedly they generate two independent subspaces > --one for each Hilbert space? Hmmm. It's not entirely sensible to speak of a "ground state" until we have defined the Hilbert space in which it lives. So I'll try to adjust your paragraph above accordingly... Instead of "some kind of annihilation, creation operators", I would have said "different sets of annihilation, creation operators". And one should also specify them as acting on the larger non-Hilbert space, of which the various Hilbert spaces mentioned above are subspaces. But otherwise your statement is sort-og on the right track: we can find different sets of annihilation, creation operators, and a vacuum state for each set, and thereby generate different orthogonal Hilbert (Fock) spaces corresponding to each set.



I wrote: >> One aspect of all this that I'm still hazy about is precisely >> why we can't just integrate over all the n1, n2, n3, etc. Arnold Neumaier replied: > [...] Thus your question does not make immediate sense. That's why I said I was "hazy" about this. If I could compose a fully sensible question, I probably wouldn't need to ask it.. :-) But anyway, your answer about Gaussian measures made some things click in my mind. Thanks. > More generally, Gaussian measures correspond to > so-called generalized free theories. Interacting > theories need non-Gaussian measures, and their > construction occupies the second third of the > book by Glimm and Yaffe.

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