I wrote:
> [...]
> Let me know if you need more explanation, or other references.[/color]
David Baker wrote:
> Any references would be much appreciated![/color]
See below.
> I think I do also need some further explanation. Disjointness
> of the representations doesn't seem to be enough, by itself,
> to establish that the two representations don't share a
> bunch of operators in common.[/color]
I said "...disjoint Hilbert spaces...", meaning two Hilbert
spaces which don't have any vectors in common. These are
really Fock spaces, generated cyclically by (different)
creation operators from different vacua.
> For example, consider the usual Fock representation and the
> one we get by using the accelerated "Rindler observer's" choice
> of complex structure. These two representations are disjoint.
> But the (Bogoliubov) transformations between them leave the field
> operators unchanged. That much is assumed in most derivations
> of the Unruh effect.[/color]
I don't understand what you mean. If the Bogoliubov transformation
left the operators "unchanged", it would be the identity - which
is certainly incorrect.
> It just seems straightforwardly true that \phi(f) is a
> well-defined operator on Fock space, so is L(f) I, and so I can
> ask what expectation value their sum takes on in any Fock state
> -- regardless of whether the transformation from \phi to \phi'
> is unitarily implementable. Where does that reasoning go wrong?[/color]
You're trying to apply intuition gained from working in finite
dimensional vector spaces to the infinite dimensional case.
That's unwise. In general, a "+" sign between operators like
that is merely formal. One can't assume that each operator and
their sum are all well-defined on the same Hilbert space. This
is discussed at a pedestrian level in ch6(?) of Klauder's book
"Beyond Conventional Quantization".
Another helpful introductory text is Umezawa's book: "ThermoField
Dynamics and Condensed States" (ch2). He shows how Bogoliubov
transformations and field operator displacements map between
disjoint Fock spaces. For your convenience, I've appended below
an (updated) version of an old spr post where I tried (poorly) to
present the essence of Umezawa's explanation. But do check in your
local library for his book, as it's far more extensive than my
brief summary below.
-strangerep
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The following is mostly summarized from Umezawa's textbook (Ref[1]).
Consider 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 be 0 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, ...> = 0
[a*_i, a*_j] |n1, n2, ...> = 0
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 difference 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 0 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.
The vector space based on {|n1, n2, n3, ...>} is not a Hilbert
space, because we haven't yet equipped it with an inner product.
But as this space is of uncountably infinite dimension, we
can't use familiar Riemann-Lebesgue integration to help us
define an inner product. (There does not exist any translation
invariant sigma-finite measure on such a space.)
Confronted by this impasse, 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: we need only retain those vectors whose total number
of particles is finite. This subset is called the "[0]-set".
It can be shown that the [0]-set *is* countable, and can be
equipped with a well-defined inner product. Then it can be used
for physics, being a separable Hilbert space known as Fock space.
The trouble now is that the choice of a subset of the huge space
{|n1, n2, n3, ...>} to use as our 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 any vector in a given Fock
space cannot be written as a linear combination of vectors from
another Fock space.
To show an example, let's take the usual creation and annihilation
operators in momentum space: a(p), a*(p). Consider what happens
when we try to mix creation and annihilation operators in such
a way that the canonical commutations are preserved..
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 CCRs as a and b
implies the constraint:
A^2 - B^2 = 1.
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 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 the 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):
<vac'| [...any product of a(p), b(p)..] |vac>
After some rather difficult math (see [1]), it can be shown that
ALL such expressions are 0. Therefore, the transformed vacuum
|vac'> of the "alpha,beta" Fock space is not 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 (or "disjoint"), and
what we mean when we say that the representations are "unitarily
inequivalent".
Another example is the so-called "boson field displacement":
alpha(p) = a(p) + c(p)
where c(p) is a (c-number) function of p. Once again, to assess
the effect of this transformation on the vacuum and other states,
we need to find a G such that:
alpha(p) = G^-1 a(p) G
In this case, it turns out that the following G does the trick:
G = exp(-Integral d^3k (c*(k)a(k) - c(k)a*(k)))
Again, a tedious calculation shows that the transformed vacuum
|vac'> = G^-1 |vac>
is orthogonal to the original |vac>, and indeed every vector
generated cyclically by alpha(p) from |vac'> is orthogonal to
every vector generated cyclically by a(p) from |vac>. Thus,
G maps between disjoint Fock spaces.
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Ref [1] Umezawa, Matsumoto & Tachiki.
"Thermo Field Dynamics and Condensed States", North Holland
ISBN 0-444-86361-3
(See ch2 in particular for a better exposition of the above.)
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