Field operators in canonically transformed representations of the CCRs

[David Baker]

Here's a question about inequivalent representations of the CCRs...

For a given Hilbert space representation, what is it that determines
which set of field operators \phi(x), or \phi(f) if we want to get
rigorous a la Wightman, gives us THE field operators for that
representation. For example, say I use a canonical transformation to
go from the Fock representation to one of the "coherent
representations," that is I transform

\phi(f) -> \phi'(f) = \phi(f) + L(f) I

for L some linear functional and I the identity operator. If L(f) is
not bounded for normalized f, the resulting coherent representation is
unitarily inequivalent to the Fock representation. But the field
operators \phi'(f) are linear combinations of the Fock field operators
and scalar multiples of the identity operator, so of course these are
also operators in the Fock representation. And they satisfy the same
commutation relations as the Fock field operators do.

So what makes it the case that the operators \phi(f) are the Fock
field operators, while the \phi'(f) are the field operators in the
coherent representaiton, when both sets of operators are well-defined
on both representations?

 

[sr]

David Baker wrote:

> Here's a question about inequivalent representations of the CCRs...

Ah, now there's a subject dear to my heart. :-)


> For a given Hilbert space representation, what is it that
> determines which set of field operators \phi(x), or \phi(f)
> if we want to get rigorous a la Wightman, gives us THE field
> operators for that representation. For example, say I use a
> canonical transformation to go from the Fock representation to
> one of the "coherent representations," that is I transform
>
> \phi(f) -> \phi'(f) = \phi(f) + L(f) I
>
> for L some linear functional and I the identity operator.
> If L(f) is not bounded for normalized f, the resulting
> coherent representation is unitarily inequivalent to the Fock
> representation. But the field operators \phi'(f) are linear
> combinations of the Fock field operators and scalar multiples of
> the identity operator, so of course these are also operators in
> the Fock representation.

No, that's incorrect. See below.


> And they satisfy the same commutation relations as the Fock
> field operators do.
>
> So what makes it the case that the operators \phi(f) are the
> Fock field operators, while the \phi'(f) are the field operators
> in the coherent representation, when both sets of operators
> are well-defined on both representations?

They do satisfy the same CCRs, but they're not proper operators
in the original Fock space. I.e: the "coherent representation" fields
you mentioned are not well-defined on the original Fock space.

Fock space is made up of vectors corresponding to finite numbers
of particles (a restriction which enables an inner product to
exist on the Fock space). But the transformation you described
above (also known as a "field displacement") maps such vectors
into other vectors which correspond to an infinite number of
particles, and are thus orthogonal to the entire original Fock
space. I.e: the field displacement transformation maps between
disjoint Hilbert spaces, and is therefore not a well-defined
"operator" in the strict mathematical meaning of that word.
(Some authors call them "improper operators" to emphasize this.)

Let me know if you need more explanation, or other references.

- strangerep.

 

[David Baker]

> I.e: the field displacement transformation maps between
> disjoint Hilbert spaces, and is therefore not a well-defined
> "operator" in the strict mathematical meaning of that word.
> (Some authors call them "improper operators" to emphasize this.)
>
> Let me know if you need more explanation, or other references.

Any references would be much appreciated!

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. 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.

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?

 

[Arnold Neumaier]

David Baker schrieb:
> Here's a question about inequivalent representations of the CCRs...
>
> For a given Hilbert space representation, what is it that determines
> which set of field operators \phi(x), or \phi(f) if we want to get
> rigorous a la Wightman, gives us THE field operators for that
> representation. For example, say I use a canonical transformation to
> go from the Fock representation to one of the "coherent
> representations," that is I transform
>
> \phi(f) -> \phi'(f) = \phi(f) + L(f) I
>
> for L some linear functional and I the identity operator. If L(f) is
> not bounded for normalized f, the resulting coherent representation is
> unitarily inequivalent to the Fock representation. But the field
> operators \phi'(f) are linear combinations of the Fock field operators
> and scalar multiples of the identity operator, so of course these are
> also operators in the Fock representation. And they satisfy the same
> commutation relations as the Fock field operators do.
>
> So what makes it the case that the operators \phi(f) are the Fock
> field operators, while the \phi'(f) are the field operators in the
> coherent representaiton, when both sets of operators are well-defined
> on both representations?

The right field operators have finite expectations in the states of
the representation.Arnold Neumaier

 

[sr]

I wrote:

> [...]
> Let me know if you need more explanation, or other references.

David Baker wrote:

> Any references would be much appreciated!

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.

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.

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?

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

--------------------------------------------------------------
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.

------------------------------------------------------
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.)
------------------------------------------------------------

 

[cstar]

David Baker says:

> 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?

Strangerep says:

> 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.

Here Baker is correct: if L(f) were a finite real number, and if \phi(f) were a self-adjoint (possibly unbounded) operator, then \phi (f)+L(f)I would be a self-adjoint (possibly unbounded) operator. It is true that in general "+" is merely formal for unbounded operators -- but this is not the general case, because L(f)I is a bounded operator.

I believe what is happening here is more complicated:

Let S be the space of testfunctions (e.g. the classical phase space). Then a quantization includes a map S \ni f -> \phi (f), where the \phi (f) satisfy the CCRs. However, a quantization also includes embedding k of S as a dense subset of a Hilbert space H. (Here H is the "one-particle space".) If two quantizations are disjoint, then we have two one-particle spaces:

k : S -> H
k': S -> H'

and the embeddings k and k' are not compatible -- so there are elements (i.e. testfunctions) in H that have no "counterpart" in H', and vice versa.

The scalar function L is originally defined on S. When we embed S into H or H', it may, or may not be extendible to all of H (or H'). My (educated) guess about this situation is that L is not extendible to the coherent state one-particle space H', and so L(f) is not well-defined for all f in H'. In this case, \phi '(f) = \phi(f)+L(f) doesn't make sense because L(f) is not really a scalar.