Gauge Transformations in Momentum Space?

Eugene Stefanovich

mikem@despammed.com wrote:
> > The
> > wavefunction of the state with definite momentum is "zero"
> everywhere
> > in the position space. However, if you integrate its square over the
> > entire universe you should get 1. You wouldn't say that momentum
> > eigenstates lie in a separate Hilbert space, wouldn't you?
>
> I'm not quite sure what you mean here. My textbooks say that
> <x|p> = exp(ipx), meaning that a position eigenstate |x> is _not_
> orthogonal to a momentum eigenstate |p>. But perhaps you meant
> something else?

If you require that |p> is a normalized vector, then

<x|p> = N exp(ipx)

where the normalization factor N is basically "zero". One can formally
say that N = 1/sqrt(V) where V is the "volume of space", i.e., infinity.
The probability of finding definite momentum particle in each finite
volume W is W/V, which is "zero". However, this does not mean that this
state is "outside" of the Hilbert space. The probability of finding
the particle "somewhere" (i.e. the intergal over V) is V/V = 1.

>
> > 2. There is an infinite number of unitary transformations from
> flavor
> > eigenstates to mass eigenstates. Blasone-Vitiello's transformation
> > changes vacuum, which seems unphysical to me. I would prefer to have
> a
> > unique vacuum without particles of any kind. This is achieved, for
> > example, by the following transformation:
> >
> > U = a_v* a_1 + a_u* a_2
> >
> > where a_1, a_2 are annihilation operators of the mass eigenstates
> > a_v* and a_u* are creation operators of the flavor eigenstates
> > (e.g., a_v* = cos(phi) a_1* + sin(phi) a_2*)
> >
> > It is
> > 1) unitary in the 0-particle and 1-particle sectors
> > 2) transforms a_1, a_2 to a_v and a_u, respectively
> > 3) does not change vacuum.
>
> I stared at this for a while, but I'm still unsure what you mean.
> Did you omit an "exp" and/or an Integral and/or a "phi" in
> your definition of U above?

Sorry, I should have been more specific. The full definition of U is:
1) U = 1 on the vacuum vector |0>
2) U = a_v* a_1 + a_u* a_2 on one-particle subspaces
H_1 (+) H_2 = H_v (+) H_u
3) U = whatever on the rest of the Fock space.

>
> >> [...] The basic idea is that every new vector is orthogonal to
> every
> >> old vector. Therefore, none of the new vectors can be expressed as
> >> superpositions of the old vectors. That's essentially what defines
> a
> >> UIR.
>
> > [...] If I remember well, the idea was to
> >
> > 1) apply a unitary tranformation to the vacuum vector |0'> = U|0>
> > 2) Find components of |0'> in a basis
> > 3) Observe that in some limit all these components tend to zero
> > 4) Conclude that the vector |0'> goes outside the original
> > Hilbert space H in this limit.
> >
> > This seems unfair to me. In these examples, even if all components
> of
> > |0'> tend to zero, their number tends to infinity, and the sum of
> > squares of the components |0'> in any basis in H should remain 1.
>
> I look at it this way: if we have a (continuously-parametrized,
> infinite) basis |b> for H, then any other vector |v> can be expressed
> as an integral superposition, whose coefficients are given by taking
> the inner product between |v> and each respective |b>, i.e:
>
> |v> = Integral db <b|v> |b>
>
> So if <b|v> is 0 for every b , the Integral must be 0, showing that |v>
> cannot be expressed as a superposition of |b>'s. This is quite
> different from the position/momentum case where <x|p> = exp(ipx) which
> is non-zero.

There is full analogy. If we use (as we should) normalized vectors
for |x> and |p>, then we obtain

<x|p> = 1/sqrt(V) exp(ipx)

|p> = Integral dx <x|p> |x>
= 1/sqrt(V) Integral dx exp(ipx) |x>

The wave function (the density of the probability amplitude)
of |p> in the position representation must be

1/sqrt(V) exp(ipx)

because the volume integral of its square is

Integral dx |1/sqrt(V) exp(ipx)|^2 = 1/V Integral dx
= 1

as it should. The function exp(ipx) without the normalization
factor 1/sqrt(V) does not have a probabilistic
interpretation, because the volume integral of its square is infinite.

> > The unitary operator U is explicitly defined within the Hilbert
> space
> > H, so it is beyond me how it can bring any vector outside of H.
>
> If one examines the U carefully, it is not really correct to say
> that it is explicitly defined "within" the Hilbert (Fock) space. A
> crucial step in the construction of Fock space is to restrict it
> to have only state vectors whose total particle number is finite.
> Without this restriction, one cannot define an inner product on the
> space, because the usual Riemann-Lebesgue integral calculus doesn't
> work: we can't approximate an arbitrary vector therein by a countable
> sum arbitrarily closely, as is necessary when defining integrals
> rigorously. Umezawa explains (some of) this. But if you can't get a
> copy, part of it appears in a summary I posted to spr back on
> 15-Dec-2004 in a thread titled "Degenerate vacua in QFT":
>
> http://www.lns.cornell.edu/spr/2004-12/msg0065860.html
>
> modulo some followup corrections by Arnold Neumaier. :-)
>
> The "U" maps vectors in the Fock space into other vectors in the
> larger non-separable space, of which the Fock space is but a
> subspace. The total particle number of those "other vectors" turns
> out to be infinite, proving that they lie outside Fock space, which
> by construction contains only vectors of *finite* total particle
> number.

Thank you for the reference. I've seen similar arguments in other
places, but they do not make much sense to me. I have a strong feeling
that the distinction between separable and non-separable Hilbert spaces
was invented by mathematicians to make life of physicists miserable.
I don't think there is anything wrong with regarding the Hilbert space
of a single particle as non-separable. After all, the number of points
in 3D space is not countable, and one can associate a distinct basis
vector (eigenvector of the position operator) with each such point.

Your arguments could be correct if your DEFINE the Fock space as
having not more than a finite number of particles.
Then, why I am not allowed
to DEFINE the Fock space as having any number of particles from zero
to infinity?
I have thought about these issues and came to a conclusion that there
could be some non-trivial math involved, but it has no significance to
physics. Again, I am not talking about superconductivity and
spontaneously broken vacuum symmetries - the issues I'm not so familiar
with. Maybe there is deep physical truth concerning UIR in these fields,
I just don't know.

For myself, I found another more comfortable attitude to
these issues. This attitude is not frequently discussed in physics
literature, but I found it rather illuminating.
This is based on the so-called
"non-standard analysis" first developed by A. Robinson in 1960.
This is too vast a field to be described in one post, but the basic idea
is to treat finite, "infinitely small" and "infinitely large" quantities
on the same footing. In this approach, the numbers like 1/sqrt(V),
where V is the volume of the entire space, make perfect sense, and there
is no trouble to calculate the integral

Integral dx |1/sqrt(V) exp(ipx)|^2 = 1

even if the integrand is "zero" everywhere.

There are few papers which try to apply this approach to quantum
mechanics. See, for example
A. Friedman, "Non-standard extension of quantum logic and
Dirac's bra-ket formalism of quantum mechanics", Int. J. Theor. Phys.
33 (1994), 307 (he was my student back in old times).

The non-standard analysis is now a well-established branch of
mathematics. I think, its use in QM may clarify some conceptual issues,
but I don't expect any physical discoveries there.

Eugene.

Eugene Stefanovich

mikem@despammed.com wrote:
> > The
> > wavefunction of the state with definite momentum is "zero"
> everywhere
> > in the position space. However, if you integrate its square over the
> > entire universe you should get 1. You wouldn't say that momentum
> > eigenstates lie in a separate Hilbert space, wouldn't you?
>
> I'm not quite sure what you mean here. My textbooks say that
> <x|p> = exp(ipx), meaning that a position eigenstate |x> is _not_
> orthogonal to a momentum eigenstate |p>. But perhaps you meant
> something else?

If you require that |p> is a normalized vector, then

<x|p> = N exp(ipx)

where the normalization factor N is basically "zero". One can formally
say that N = 1/sqrt(V) where V is the "volume of space", i.e., infinity.
The probability of finding definite momentum particle in each finite
volume W is W/V, which is "zero". However, this does not mean that this
state is "outside" of the Hilbert space. The probability of finding
the particle "somewhere" (i.e. the intergal over V) is V/V = 1.

>
> > 2. There is an infinite number of unitary transformations from
> flavor
> > eigenstates to mass eigenstates. Blasone-Vitiello's transformation
> > changes vacuum, which seems unphysical to me. I would prefer to have
> a
> > unique vacuum without particles of any kind. This is achieved, for
> > example, by the following transformation:
> >
> > U = a_v* a_1 + a_u* a_2
> >
> > where a_1, a_2 are annihilation operators of the mass eigenstates
> > a_v* and a_u* are creation operators of the flavor eigenstates
> > (e.g., a_v* = cos(phi) a_1* + sin(phi) a_2*)
> >
> > It is
> > 1) unitary in the 0-particle and 1-particle sectors
> > 2) transforms a_1, a_2 to a_v and a_u, respectively
> > 3) does not change vacuum.
>
> I stared at this for a while, but I'm still unsure what you mean.
> Did you omit an "exp" and/or an Integral and/or a "phi" in
> your definition of U above?

Sorry, I should have been more specific. The full definition of U is:
1) U = 1 on the vacuum vector |0>
2) U = a_v* a_1 + a_u* a_2 on one-particle subspaces
H_1 (+) H_2 = H_v (+) H_u
3) U = whatever on the rest of the Fock space.

>
> >> [...] The basic idea is that every new vector is orthogonal to
> every
> >> old vector. Therefore, none of the new vectors can be expressed as
> >> superpositions of the old vectors. That's essentially what defines
> a
> >> UIR.
>
> > [...] If I remember well, the idea was to
> >
> > 1) apply a unitary tranformation to the vacuum vector |0'> = U|0>
> > 2) Find components of |0'> in a basis
> > 3) Observe that in some limit all these components tend to zero
> > 4) Conclude that the vector |0'> goes outside the original
> > Hilbert space H in this limit.
> >
> > This seems unfair to me. In these examples, even if all components
> of
> > |0'> tend to zero, their number tends to infinity, and the sum of
> > squares of the components |0'> in any basis in H should remain 1.
>
> I look at it this way: if we have a (continuously-parametrized,
> infinite) basis |b> for H, then any other vector |v> can be expressed
> as an integral superposition, whose coefficients are given by taking
> the inner product between |v> and each respective |b>, i.e:
>
> |v> = Integral db <b|v> |b>
>
> So if <b|v> is 0 for every b , the Integral must be 0, showing that |v>
> cannot be expressed as a superposition of |b>'s. This is quite
> different from the position/momentum case where <x|p> = exp(ipx) which
> is non-zero.

There is full analogy. If we use (as we should) normalized vectors
for |x> and |p>, then we obtain

<x|p> = 1/sqrt(V) exp(ipx)

|p> = Integral dx <x|p> |x>
= 1/sqrt(V) Integral dx exp(ipx) |x>

The wave function (the density of the probability amplitude)
of |p> in the position representation must be

1/sqrt(V) exp(ipx)

because the volume integral of its square is

Integral dx |1/sqrt(V) exp(ipx)|^2 = 1/V Integral dx
= 1

as it should. The function exp(ipx) without the normalization
factor 1/sqrt(V) does not have a probabilistic
interpretation, because the volume integral of its square is infinite.

> > The unitary operator U is explicitly defined within the Hilbert
> space
> > H, so it is beyond me how it can bring any vector outside of H.
>
> If one examines the U carefully, it is not really correct to say
> that it is explicitly defined "within" the Hilbert (Fock) space. A
> crucial step in the construction of Fock space is to restrict it
> to have only state vectors whose total particle number is finite.
> Without this restriction, one cannot define an inner product on the
> space, because the usual Riemann-Lebesgue integral calculus doesn't
> work: we can't approximate an arbitrary vector therein by a countable
> sum arbitrarily closely, as is necessary when defining integrals
> rigorously. Umezawa explains (some of) this. But if you can't get a
> copy, part of it appears in a summary I posted to spr back on
> 15-Dec-2004 in a thread titled "Degenerate vacua in QFT":
>
> http://www.lns.cornell.edu/spr/2004-12/msg0065860.html
>
> modulo some followup corrections by Arnold Neumaier. :-)
>
> The "U" maps vectors in the Fock space into other vectors in the
> larger non-separable space, of which the Fock space is but a
> subspace. The total particle number of those "other vectors" turns
> out to be infinite, proving that they lie outside Fock space, which
> by construction contains only vectors of *finite* total particle
> number.

Thank you for the reference. I've seen similar arguments in other
places, but they do not make much sense to me. I have a strong feeling
that the distinction between separable and non-separable Hilbert spaces
was invented by mathematicians to make life of physicists miserable.
I don't think there is anything wrong with regarding the Hilbert space
of a single particle as non-separable. After all, the number of points
in 3D space is not countable, and one can associate a distinct basis
vector (eigenvector of the position operator) with each such point.

Your arguments could be correct if your DEFINE the Fock space as
having not more than a finite number of particles.
Then, why I am not allowed
to DEFINE the Fock space as having any number of particles from zero
to infinity?
I have thought about these issues and came to a conclusion that there
could be some non-trivial math involved, but it has no significance to
physics. Again, I am not talking about superconductivity and
spontaneously broken vacuum symmetries - the issues I'm not so familiar
with. Maybe there is deep physical truth concerning UIR in these fields,
I just don't know.

For myself, I found another more comfortable attitude to
these issues. This attitude is not frequently discussed in physics
literature, but I found it rather illuminating.
This is based on the so-called
"non-standard analysis" first developed by A. Robinson in 1960.
This is too vast a field to be described in one post, but the basic idea
is to treat finite, "infinitely small" and "infinitely large" quantities
on the same footing. In this approach, the numbers like 1/sqrt(V),
where V is the volume of the entire space, make perfect sense, and there
is no trouble to calculate the integral

Integral dx |1/sqrt(V) exp(ipx)|^2 = 1

even if the integrand is "zero" everywhere.

There are few papers which try to apply this approach to quantum
mechanics. See, for example
A. Friedman, "Non-standard extension of quantum logic and
Dirac's bra-ket formalism of quantum mechanics", Int. J. Theor. Phys.
33 (1994), 307 (he was my student back in old times).

The non-standard analysis is now a well-established branch of
mathematics. I think, its use in QM may clarify some conceptual issues,
but I don't expect any physical discoveries there.

Eugene.

Eugene Stefanovich

mikem@despammed.com wrote:
> > The
> > wavefunction of the state with definite momentum is "zero"
> everywhere
> > in the position space. However, if you integrate its square over the
> > entire universe you should get 1. You wouldn't say that momentum
> > eigenstates lie in a separate Hilbert space, wouldn't you?
>
> I'm not quite sure what you mean here. My textbooks say that
> <x|p> = exp(ipx), meaning that a position eigenstate |x> is _not_
> orthogonal to a momentum eigenstate |p>. But perhaps you meant
> something else?

If you require that |p> is a normalized vector, then

<x|p> = N exp(ipx)

where the normalization factor N is basically "zero". One can formally
say that N = 1/sqrt(V) where V is the "volume of space", i.e., infinity.
The probability of finding definite momentum particle in each finite
volume W is W/V, which is "zero". However, this does not mean that this
state is "outside" of the Hilbert space. The probability of finding
the particle "somewhere" (i.e. the intergal over V) is V/V = 1.

>
> > 2. There is an infinite number of unitary transformations from
> flavor
> > eigenstates to mass eigenstates. Blasone-Vitiello's transformation
> > changes vacuum, which seems unphysical to me. I would prefer to have
> a
> > unique vacuum without particles of any kind. This is achieved, for
> > example, by the following transformation:
> >
> > U = a_v* a_1 + a_u* a_2
> >
> > where a_1, a_2 are annihilation operators of the mass eigenstates
> > a_v* and a_u* are creation operators of the flavor eigenstates
> > (e.g., a_v* = cos(phi) a_1* + sin(phi) a_2*)
> >
> > It is
> > 1) unitary in the 0-particle and 1-particle sectors
> > 2) transforms a_1, a_2 to a_v and a_u, respectively
> > 3) does not change vacuum.
>
> I stared at this for a while, but I'm still unsure what you mean.
> Did you omit an "exp" and/or an Integral and/or a "phi" in
> your definition of U above?

Sorry, I should have been more specific. The full definition of U is:
1) U = 1 on the vacuum vector |0>
2) U = a_v* a_1 + a_u* a_2 on one-particle subspaces
H_1 (+) H_2 = H_v (+) H_u
3) U = whatever on the rest of the Fock space.

>
> >> [...] The basic idea is that every new vector is orthogonal to
> every
> >> old vector. Therefore, none of the new vectors can be expressed as
> >> superpositions of the old vectors. That's essentially what defines
> a
> >> UIR.
>
> > [...] If I remember well, the idea was to
> >
> > 1) apply a unitary tranformation to the vacuum vector |0'> = U|0>
> > 2) Find components of |0'> in a basis
> > 3) Observe that in some limit all these components tend to zero
> > 4) Conclude that the vector |0'> goes outside the original
> > Hilbert space H in this limit.
> >
> > This seems unfair to me. In these examples, even if all components
> of
> > |0'> tend to zero, their number tends to infinity, and the sum of
> > squares of the components |0'> in any basis in H should remain 1.
>
> I look at it this way: if we have a (continuously-parametrized,
> infinite) basis |b> for H, then any other vector |v> can be expressed
> as an integral superposition, whose coefficients are given by taking
> the inner product between |v> and each respective |b>, i.e:
>
> |v> = Integral db <b|v> |b>
>
> So if <b|v> is 0 for every b , the Integral must be 0, showing that |v>
> cannot be expressed as a superposition of |b>'s. This is quite
> different from the position/momentum case where <x|p> = exp(ipx) which
> is non-zero.

There is full analogy. If we use (as we should) normalized vectors
for |x> and |p>, then we obtain

<x|p> = 1/sqrt(V) exp(ipx)

|p> = Integral dx <x|p> |x>
= 1/sqrt(V) Integral dx exp(ipx) |x>

The wave function (the density of the probability amplitude)
of |p> in the position representation must be

1/sqrt(V) exp(ipx)

because the volume integral of its square is

Integral dx |1/sqrt(V) exp(ipx)|^2 = 1/V Integral dx
= 1

as it should. The function exp(ipx) without the normalization
factor 1/sqrt(V) does not have a probabilistic
interpretation, because the volume integral of its square is infinite.

> > The unitary operator U is explicitly defined within the Hilbert
> space
> > H, so it is beyond me how it can bring any vector outside of H.
>
> If one examines the U carefully, it is not really correct to say
> that it is explicitly defined "within" the Hilbert (Fock) space. A
> crucial step in the construction of Fock space is to restrict it
> to have only state vectors whose total particle number is finite.
> Without this restriction, one cannot define an inner product on the
> space, because the usual Riemann-Lebesgue integral calculus doesn't
> work: we can't approximate an arbitrary vector therein by a countable
> sum arbitrarily closely, as is necessary when defining integrals
> rigorously. Umezawa explains (some of) this. But if you can't get a
> copy, part of it appears in a summary I posted to spr back on
> 15-Dec-2004 in a thread titled "Degenerate vacua in QFT":
>
> http://www.lns.cornell.edu/spr/2004-12/msg0065860.html
>
> modulo some followup corrections by Arnold Neumaier. :-)
>
> The "U" maps vectors in the Fock space into other vectors in the
> larger non-separable space, of which the Fock space is but a
> subspace. The total particle number of those "other vectors" turns
> out to be infinite, proving that they lie outside Fock space, which
> by construction contains only vectors of *finite* total particle
> number.

Thank you for the reference. I've seen similar arguments in other
places, but they do not make much sense to me. I have a strong feeling
that the distinction between separable and non-separable Hilbert spaces
was invented by mathematicians to make life of physicists miserable.
I don't think there is anything wrong with regarding the Hilbert space
of a single particle as non-separable. After all, the number of points
in 3D space is not countable, and one can associate a distinct basis
vector (eigenvector of the position operator) with each such point.

Your arguments could be correct if your DEFINE the Fock space as
having not more than a finite number of particles.
Then, why I am not allowed
to DEFINE the Fock space as having any number of particles from zero
to infinity?
I have thought about these issues and came to a conclusion that there
could be some non-trivial math involved, but it has no significance to
physics. Again, I am not talking about superconductivity and
spontaneously broken vacuum symmetries - the issues I'm not so familiar
with. Maybe there is deep physical truth concerning UIR in these fields,
I just don't know.

For myself, I found another more comfortable attitude to
these issues. This attitude is not frequently discussed in physics
literature, but I found it rather illuminating.
This is based on the so-called
"non-standard analysis" first developed by A. Robinson in 1960.
This is too vast a field to be described in one post, but the basic idea
is to treat finite, "infinitely small" and "infinitely large" quantities
on the same footing. In this approach, the numbers like 1/sqrt(V),
where V is the volume of the entire space, make perfect sense, and there
is no trouble to calculate the integral

Integral dx |1/sqrt(V) exp(ipx)|^2 = 1

even if the integrand is "zero" everywhere.

There are few papers which try to apply this approach to quantum
mechanics. See, for example
A. Friedman, "Non-standard extension of quantum logic and
Dirac's bra-ket formalism of quantum mechanics", Int. J. Theor. Phys.
33 (1994), 307 (he was my student back in old times).

The non-standard analysis is now a well-established branch of
mathematics. I think, its use in QM may clarify some conceptual issues,
but I don't expect any physical discoveries there.

Eugene.

Eugene Stefanovich

mikem@despammed.com wrote:
> > The
> > wavefunction of the state with definite momentum is "zero"
> everywhere
> > in the position space. However, if you integrate its square over the
> > entire universe you should get 1. You wouldn't say that momentum
> > eigenstates lie in a separate Hilbert space, wouldn't you?
>
> I'm not quite sure what you mean here. My textbooks say that
> <x|p> = exp(ipx), meaning that a position eigenstate |x> is _not_
> orthogonal to a momentum eigenstate |p>. But perhaps you meant
> something else?

If you require that |p> is a normalized vector, then

<x|p> = N exp(ipx)

where the normalization factor N is basically "zero". One can formally
say that N = 1/sqrt(V) where V is the "volume of space", i.e., infinity.
The probability of finding definite momentum particle in each finite
volume W is W/V, which is "zero". However, this does not mean that this
state is "outside" of the Hilbert space. The probability of finding
the particle "somewhere" (i.e. the intergal over V) is V/V = 1.

>
> > 2. There is an infinite number of unitary transformations from
> flavor
> > eigenstates to mass eigenstates. Blasone-Vitiello's transformation
> > changes vacuum, which seems unphysical to me. I would prefer to have
> a
> > unique vacuum without particles of any kind. This is achieved, for
> > example, by the following transformation:
> >
> > U = a_v* a_1 + a_u* a_2
> >
> > where a_1, a_2 are annihilation operators of the mass eigenstates
> > a_v* and a_u* are creation operators of the flavor eigenstates
> > (e.g., a_v* = cos(phi) a_1* + sin(phi) a_2*)
> >
> > It is
> > 1) unitary in the 0-particle and 1-particle sectors
> > 2) transforms a_1, a_2 to a_v and a_u, respectively
> > 3) does not change vacuum.
>
> I stared at this for a while, but I'm still unsure what you mean.
> Did you omit an "exp" and/or an Integral and/or a "phi" in
> your definition of U above?

Sorry, I should have been more specific. The full definition of U is:
1) U = 1 on the vacuum vector |0>
2) U = a_v* a_1 + a_u* a_2 on one-particle subspaces
H_1 (+) H_2 = H_v (+) H_u
3) U = whatever on the rest of the Fock space.

>
> >> [...] The basic idea is that every new vector is orthogonal to
> every
> >> old vector. Therefore, none of the new vectors can be expressed as
> >> superpositions of the old vectors. That's essentially what defines
> a
> >> UIR.
>
> > [...] If I remember well, the idea was to
> >
> > 1) apply a unitary tranformation to the vacuum vector |0'> = U|0>
> > 2) Find components of |0'> in a basis
> > 3) Observe that in some limit all these components tend to zero
> > 4) Conclude that the vector |0'> goes outside the original
> > Hilbert space H in this limit.
> >
> > This seems unfair to me. In these examples, even if all components
> of
> > |0'> tend to zero, their number tends to infinity, and the sum of
> > squares of the components |0'> in any basis in H should remain 1.
>
> I look at it this way: if we have a (continuously-parametrized,
> infinite) basis |b> for H, then any other vector |v> can be expressed
> as an integral superposition, whose coefficients are given by taking
> the inner product between |v> and each respective |b>, i.e:
>
> |v> = Integral db <b|v> |b>
>
> So if <b|v> is 0 for every b , the Integral must be 0, showing that |v>
> cannot be expressed as a superposition of |b>'s. This is quite
> different from the position/momentum case where <x|p> = exp(ipx) which
> is non-zero.

There is full analogy. If we use (as we should) normalized vectors
for |x> and |p>, then we obtain

<x|p> = 1/sqrt(V) exp(ipx)

|p> = Integral dx <x|p> |x>
= 1/sqrt(V) Integral dx exp(ipx) |x>

The wave function (the density of the probability amplitude)
of |p> in the position representation must be

1/sqrt(V) exp(ipx)

because the volume integral of its square is

Integral dx |1/sqrt(V) exp(ipx)|^2 = 1/V Integral dx
= 1

as it should. The function exp(ipx) without the normalization
factor 1/sqrt(V) does not have a probabilistic
interpretation, because the volume integral of its square is infinite.

> > The unitary operator U is explicitly defined within the Hilbert
> space
> > H, so it is beyond me how it can bring any vector outside of H.
>
> If one examines the U carefully, it is not really correct to say
> that it is explicitly defined "within" the Hilbert (Fock) space. A
> crucial step in the construction of Fock space is to restrict it
> to have only state vectors whose total particle number is finite.
> Without this restriction, one cannot define an inner product on the
> space, because the usual Riemann-Lebesgue integral calculus doesn't
> work: we can't approximate an arbitrary vector therein by a countable
> sum arbitrarily closely, as is necessary when defining integrals
> rigorously. Umezawa explains (some of) this. But if you can't get a
> copy, part of it appears in a summary I posted to spr back on
> 15-Dec-2004 in a thread titled "Degenerate vacua in QFT":
>
> http://www.lns.cornell.edu/spr/2004-12/msg0065860.html
>
> modulo some followup corrections by Arnold Neumaier. :-)
>
> The "U" maps vectors in the Fock space into other vectors in the
> larger non-separable space, of which the Fock space is but a
> subspace. The total particle number of those "other vectors" turns
> out to be infinite, proving that they lie outside Fock space, which
> by construction contains only vectors of *finite* total particle
> number.

Thank you for the reference. I've seen similar arguments in other
places, but they do not make much sense to me. I have a strong feeling
that the distinction between separable and non-separable Hilbert spaces
was invented by mathematicians to make life of physicists miserable.
I don't think there is anything wrong with regarding the Hilbert space
of a single particle as non-separable. After all, the number of points
in 3D space is not countable, and one can associate a distinct basis
vector (eigenvector of the position operator) with each such point.

Your arguments could be correct if your DEFINE the Fock space as
having not more than a finite number of particles.
Then, why I am not allowed
to DEFINE the Fock space as having any number of particles from zero
to infinity?
I have thought about these issues and came to a conclusion that there
could be some non-trivial math involved, but it has no significance to
physics. Again, I am not talking about superconductivity and
spontaneously broken vacuum symmetries - the issues I'm not so familiar
with. Maybe there is deep physical truth concerning UIR in these fields,
I just don't know.

For myself, I found another more comfortable attitude to
these issues. This attitude is not frequently discussed in physics
literature, but I found it rather illuminating.
This is based on the so-called
"non-standard analysis" first developed by A. Robinson in 1960.
This is too vast a field to be described in one post, but the basic idea
is to treat finite, "infinitely small" and "infinitely large" quantities
on the same footing. In this approach, the numbers like 1/sqrt(V),
where V is the volume of the entire space, make perfect sense, and there
is no trouble to calculate the integral

Integral dx |1/sqrt(V) exp(ipx)|^2 = 1

even if the integrand is "zero" everywhere.

There are few papers which try to apply this approach to quantum
mechanics. See, for example
A. Friedman, "Non-standard extension of quantum logic and
Dirac's bra-ket formalism of quantum mechanics", Int. J. Theor. Phys.
33 (1994), 307 (he was my student back in old times).

The non-standard analysis is now a well-established branch of
mathematics. I think, its use in QM may clarify some conceptual issues,
but I don't expect any physical discoveries there.

Eugene.

Eugene Stefanovich

mikem@despammed.com wrote:
> > The
> > wavefunction of the state with definite momentum is "zero"
> everywhere
> > in the position space. However, if you integrate its square over the
> > entire universe you should get 1. You wouldn't say that momentum
> > eigenstates lie in a separate Hilbert space, wouldn't you?
>
> I'm not quite sure what you mean here. My textbooks say that
> <x|p> = exp(ipx), meaning that a position eigenstate |x> is _not_
> orthogonal to a momentum eigenstate |p>. But perhaps you meant
> something else?

If you require that |p> is a normalized vector, then

<x|p> = N exp(ipx)

where the normalization factor N is basically "zero". One can formally
say that N = 1/sqrt(V) where V is the "volume of space", i.e., infinity.
The probability of finding definite momentum particle in each finite
volume W is W/V, which is "zero". However, this does not mean that this
state is "outside" of the Hilbert space. The probability of finding
the particle "somewhere" (i.e. the intergal over V) is V/V = 1.

>
> > 2. There is an infinite number of unitary transformations from
> flavor
> > eigenstates to mass eigenstates. Blasone-Vitiello's transformation
> > changes vacuum, which seems unphysical to me. I would prefer to have
> a
> > unique vacuum without particles of any kind. This is achieved, for
> > example, by the following transformation:
> >
> > U = a_v* a_1 + a_u* a_2
> >
> > where a_1, a_2 are annihilation operators of the mass eigenstates
> > a_v* and a_u* are creation operators of the flavor eigenstates
> > (e.g., a_v* = cos(phi) a_1* + sin(phi) a_2*)
> >
> > It is
> > 1) unitary in the 0-particle and 1-particle sectors
> > 2) transforms a_1, a_2 to a_v and a_u, respectively
> > 3) does not change vacuum.
>
> I stared at this for a while, but I'm still unsure what you mean.
> Did you omit an "exp" and/or an Integral and/or a "phi" in
> your definition of U above?

Sorry, I should have been more specific. The full definition of U is:
1) U = 1 on the vacuum vector |0>
2) U = a_v* a_1 + a_u* a_2 on one-particle subspaces
H_1 (+) H_2 = H_v (+) H_u
3) U = whatever on the rest of the Fock space.

>
> >> [...] The basic idea is that every new vector is orthogonal to
> every
> >> old vector. Therefore, none of the new vectors can be expressed as
> >> superpositions of the old vectors. That's essentially what defines
> a
> >> UIR.
>
> > [...] If I remember well, the idea was to
> >
> > 1) apply a unitary tranformation to the vacuum vector |0'> = U|0>
> > 2) Find components of |0'> in a basis
> > 3) Observe that in some limit all these components tend to zero
> > 4) Conclude that the vector |0'> goes outside the original
> > Hilbert space H in this limit.
> >
> > This seems unfair to me. In these examples, even if all components
> of
> > |0'> tend to zero, their number tends to infinity, and the sum of
> > squares of the components |0'> in any basis in H should remain 1.
>
> I look at it this way: if we have a (continuously-parametrized,
> infinite) basis |b> for H, then any other vector |v> can be expressed
> as an integral superposition, whose coefficients are given by taking
> the inner product between |v> and each respective |b>, i.e:
>
> |v> = Integral db <b|v> |b>
>
> So if <b|v> is 0 for every b , the Integral must be 0, showing that |v>
> cannot be expressed as a superposition of |b>'s. This is quite
> different from the position/momentum case where <x|p> = exp(ipx) which
> is non-zero.

There is full analogy. If we use (as we should) normalized vectors
for |x> and |p>, then we obtain

<x|p> = 1/sqrt(V) exp(ipx)

|p> = Integral dx <x|p> |x>
= 1/sqrt(V) Integral dx exp(ipx) |x>

The wave function (the density of the probability amplitude)
of |p> in the position representation must be

1/sqrt(V) exp(ipx)

because the volume integral of its square is

Integral dx |1/sqrt(V) exp(ipx)|^2 = 1/V Integral dx
= 1

as it should. The function exp(ipx) without the normalization
factor 1/sqrt(V) does not have a probabilistic
interpretation, because the volume integral of its square is infinite.

> > The unitary operator U is explicitly defined within the Hilbert
> space
> > H, so it is beyond me how it can bring any vector outside of H.
>
> If one examines the U carefully, it is not really correct to say
> that it is explicitly defined "within" the Hilbert (Fock) space. A
> crucial step in the construction of Fock space is to restrict it
> to have only state vectors whose total particle number is finite.
> Without this restriction, one cannot define an inner product on the
> space, because the usual Riemann-Lebesgue integral calculus doesn't
> work: we can't approximate an arbitrary vector therein by a countable
> sum arbitrarily closely, as is necessary when defining integrals
> rigorously. Umezawa explains (some of) this. But if you can't get a
> copy, part of it appears in a summary I posted to spr back on
> 15-Dec-2004 in a thread titled "Degenerate vacua in QFT":
>
> http://www.lns.cornell.edu/spr/2004-12/msg0065860.html
>
> modulo some followup corrections by Arnold Neumaier. :-)
>
> The "U" maps vectors in the Fock space into other vectors in the
> larger non-separable space, of which the Fock space is but a
> subspace. The total particle number of those "other vectors" turns
> out to be infinite, proving that they lie outside Fock space, which
> by construction contains only vectors of *finite* total particle
> number.

Thank you for the reference. I've seen similar arguments in other
places, but they do not make much sense to me. I have a strong feeling
that the distinction between separable and non-separable Hilbert spaces
was invented by mathematicians to make life of physicists miserable.
I don't think there is anything wrong with regarding the Hilbert space
of a single particle as non-separable. After all, the number of points
in 3D space is not countable, and one can associate a distinct basis
vector (eigenvector of the position operator) with each such point.

Your arguments could be correct if your DEFINE the Fock space as
having not more than a finite number of particles.
Then, why I am not allowed
to DEFINE the Fock space as having any number of particles from zero
to infinity?
I have thought about these issues and came to a conclusion that there
could be some non-trivial math involved, but it has no significance to
physics. Again, I am not talking about superconductivity and
spontaneously broken vacuum symmetries - the issues I'm not so familiar
with. Maybe there is deep physical truth concerning UIR in these fields,
I just don't know.

For myself, I found another more comfortable attitude to
these issues. This attitude is not frequently discussed in physics
literature, but I found it rather illuminating.
This is based on the so-called
"non-standard analysis" first developed by A. Robinson in 1960.
This is too vast a field to be described in one post, but the basic idea
is to treat finite, "infinitely small" and "infinitely large" quantities
on the same footing. In this approach, the numbers like 1/sqrt(V),
where V is the volume of the entire space, make perfect sense, and there
is no trouble to calculate the integral

Integral dx |1/sqrt(V) exp(ipx)|^2 = 1

even if the integrand is "zero" everywhere.

There are few papers which try to apply this approach to quantum
mechanics. See, for example
A. Friedman, "Non-standard extension of quantum logic and
Dirac's bra-ket formalism of quantum mechanics", Int. J. Theor. Phys.
33 (1994), 307 (he was my student back in old times).

The non-standard analysis is now a well-established branch of
mathematics. I think, its use in QM may clarify some conceptual issues,
but I don't expect any physical discoveries there.

Eugene.

Eugene Stefanovich

mikem@despammed.com wrote:
> > The
> > wavefunction of the state with definite momentum is "zero"
> everywhere
> > in the position space. However, if you integrate its square over the
> > entire universe you should get 1. You wouldn't say that momentum
> > eigenstates lie in a separate Hilbert space, wouldn't you?
>
> I'm not quite sure what you mean here. My textbooks say that
> <x|p> = exp(ipx), meaning that a position eigenstate |x> is _not_
> orthogonal to a momentum eigenstate |p>. But perhaps you meant
> something else?

If you require that |p> is a normalized vector, then

<x|p> = N exp(ipx)

where the normalization factor N is basically "zero". One can formally
say that N = 1/sqrt(V) where V is the "volume of space", i.e., infinity.
The probability of finding definite momentum particle in each finite
volume W is W/V, which is "zero". However, this does not mean that this
state is "outside" of the Hilbert space. The probability of finding
the particle "somewhere" (i.e. the intergal over V) is V/V = 1.

>
> > 2. There is an infinite number of unitary transformations from
> flavor
> > eigenstates to mass eigenstates. Blasone-Vitiello's transformation
> > changes vacuum, which seems unphysical to me. I would prefer to have
> a
> > unique vacuum without particles of any kind. This is achieved, for
> > example, by the following transformation:
> >
> > U = a_v* a_1 + a_u* a_2
> >
> > where a_1, a_2 are annihilation operators of the mass eigenstates
> > a_v* and a_u* are creation operators of the flavor eigenstates
> > (e.g., a_v* = cos(phi) a_1* + sin(phi) a_2*)
> >
> > It is
> > 1) unitary in the 0-particle and 1-particle sectors
> > 2) transforms a_1, a_2 to a_v and a_u, respectively
> > 3) does not change vacuum.
>
> I stared at this for a while, but I'm still unsure what you mean.
> Did you omit an "exp" and/or an Integral and/or a "phi" in
> your definition of U above?

Sorry, I should have been more specific. The full definition of U is:
1) U = 1 on the vacuum vector |0>
2) U = a_v* a_1 + a_u* a_2 on one-particle subspaces
H_1 (+) H_2 = H_v (+) H_u
3) U = whatever on the rest of the Fock space.

>
> >> [...] The basic idea is that every new vector is orthogonal to
> every
> >> old vector. Therefore, none of the new vectors can be expressed as
> >> superpositions of the old vectors. That's essentially what defines
> a
> >> UIR.
>
> > [...] If I remember well, the idea was to
> >
> > 1) apply a unitary tranformation to the vacuum vector |0'> = U|0>
> > 2) Find components of |0'> in a basis
> > 3) Observe that in some limit all these components tend to zero
> > 4) Conclude that the vector |0'> goes outside the original
> > Hilbert space H in this limit.
> >
> > This seems unfair to me. In these examples, even if all components
> of
> > |0'> tend to zero, their number tends to infinity, and the sum of
> > squares of the components |0'> in any basis in H should remain 1.
>
> I look at it this way: if we have a (continuously-parametrized,
> infinite) basis |b> for H, then any other vector |v> can be expressed
> as an integral superposition, whose coefficients are given by taking
> the inner product between |v> and each respective |b>, i.e:
>
> |v> = Integral db <b|v> |b>
>
> So if <b|v> is 0 for every b , the Integral must be 0, showing that |v>
> cannot be expressed as a superposition of |b>'s. This is quite
> different from the position/momentum case where <x|p> = exp(ipx) which
> is non-zero.

There is full analogy. If we use (as we should) normalized vectors
for |x> and |p>, then we obtain

<x|p> = 1/sqrt(V) exp(ipx)

|p> = Integral dx <x|p> |x>
= 1/sqrt(V) Integral dx exp(ipx) |x>

The wave function (the density of the probability amplitude)
of |p> in the position representation must be

1/sqrt(V) exp(ipx)

because the volume integral of its square is

Integral dx |1/sqrt(V) exp(ipx)|^2 = 1/V Integral dx
= 1

as it should. The function exp(ipx) without the normalization
factor 1/sqrt(V) does not have a probabilistic
interpretation, because the volume integral of its square is infinite.

> > The unitary operator U is explicitly defined within the Hilbert
> space
> > H, so it is beyond me how it can bring any vector outside of H.
>
> If one examines the U carefully, it is not really correct to say
> that it is explicitly defined "within" the Hilbert (Fock) space. A
> crucial step in the construction of Fock space is to restrict it
> to have only state vectors whose total particle number is finite.
> Without this restriction, one cannot define an inner product on the
> space, because the usual Riemann-Lebesgue integral calculus doesn't
> work: we can't approximate an arbitrary vector therein by a countable
> sum arbitrarily closely, as is necessary when defining integrals
> rigorously. Umezawa explains (some of) this. But if you can't get a
> copy, part of it appears in a summary I posted to spr back on
> 15-Dec-2004 in a thread titled "Degenerate vacua in QFT":
>
> http://www.lns.cornell.edu/spr/2004-12/msg0065860.html
>
> modulo some followup corrections by Arnold Neumaier. :-)
>
> The "U" maps vectors in the Fock space into other vectors in the
> larger non-separable space, of which the Fock space is but a
> subspace. The total particle number of those "other vectors" turns
> out to be infinite, proving that they lie outside Fock space, which
> by construction contains only vectors of *finite* total particle
> number.

Thank you for the reference. I've seen similar arguments in other
places, but they do not make much sense to me. I have a strong feeling
that the distinction between separable and non-separable Hilbert spaces
was invented by mathematicians to make life of physicists miserable.
I don't think there is anything wrong with regarding the Hilbert space
of a single particle as non-separable. After all, the number of points
in 3D space is not countable, and one can associate a distinct basis
vector (eigenvector of the position operator) with each such point.

Your arguments could be correct if your DEFINE the Fock space as
having not more than a finite number of particles.
Then, why I am not allowed
to DEFINE the Fock space as having any number of particles from zero
to infinity?
I have thought about these issues and came to a conclusion that there
could be some non-trivial math involved, but it has no significance to
physics. Again, I am not talking about superconductivity and
spontaneously broken vacuum symmetries - the issues I'm not so familiar
with. Maybe there is deep physical truth concerning UIR in these fields,
I just don't know.

For myself, I found another more comfortable attitude to
these issues. This attitude is not frequently discussed in physics
literature, but I found it rather illuminating.
This is based on the so-called
"non-standard analysis" first developed by A. Robinson in 1960.
This is too vast a field to be described in one post, but the basic idea
is to treat finite, "infinitely small" and "infinitely large" quantities
on the same footing. In this approach, the numbers like 1/sqrt(V),
where V is the volume of the entire space, make perfect sense, and there
is no trouble to calculate the integral

Integral dx |1/sqrt(V) exp(ipx)|^2 = 1

even if the integrand is "zero" everywhere.

There are few papers which try to apply this approach to quantum
mechanics. See, for example
A. Friedman, "Non-standard extension of quantum logic and
Dirac's bra-ket formalism of quantum mechanics", Int. J. Theor. Phys.
33 (1994), 307 (he was my student back in old times).

The non-standard analysis is now a well-established branch of
mathematics. I think, its use in QM may clarify some conceptual issues,
but I don't expect any physical discoveries there.

Eugene.

Eugene Stefanovich

mikem@despammed.com wrote:
> > The
> > wavefunction of the state with definite momentum is "zero"
> everywhere
> > in the position space. However, if you integrate its square over the
> > entire universe you should get 1. You wouldn't say that momentum
> > eigenstates lie in a separate Hilbert space, wouldn't you?
>
> I'm not quite sure what you mean here. My textbooks say that
> <x|p> = exp(ipx), meaning that a position eigenstate |x> is _not_
> orthogonal to a momentum eigenstate |p>. But perhaps you meant
> something else?

If you require that |p> is a normalized vector, then

<x|p> = N exp(ipx)

where the normalization factor N is basically "zero". One can formally
say that N = 1/sqrt(V) where V is the "volume of space", i.e., infinity.
The probability of finding definite momentum particle in each finite
volume W is W/V, which is "zero". However, this does not mean that this
state is "outside" of the Hilbert space. The probability of finding
the particle "somewhere" (i.e. the intergal over V) is V/V = 1.

>
> > 2. There is an infinite number of unitary transformations from
> flavor
> > eigenstates to mass eigenstates. Blasone-Vitiello's transformation
> > changes vacuum, which seems unphysical to me. I would prefer to have
> a
> > unique vacuum without particles of any kind. This is achieved, for
> > example, by the following transformation:
> >
> > U = a_v* a_1 + a_u* a_2
> >
> > where a_1, a_2 are annihilation operators of the mass eigenstates
> > a_v* and a_u* are creation operators of the flavor eigenstates
> > (e.g., a_v* = cos(phi) a_1* + sin(phi) a_2*)
> >
> > It is
> > 1) unitary in the 0-particle and 1-particle sectors
> > 2) transforms a_1, a_2 to a_v and a_u, respectively
> > 3) does not change vacuum.
>
> I stared at this for a while, but I'm still unsure what you mean.
> Did you omit an "exp" and/or an Integral and/or a "phi" in
> your definition of U above?

Sorry, I should have been more specific. The full definition of U is:
1) U = 1 on the vacuum vector |0>
2) U = a_v* a_1 + a_u* a_2 on one-particle subspaces
H_1 (+) H_2 = H_v (+) H_u
3) U = whatever on the rest of the Fock space.

>
> >> [...] The basic idea is that every new vector is orthogonal to
> every
> >> old vector. Therefore, none of the new vectors can be expressed as
> >> superpositions of the old vectors. That's essentially what defines
> a
> >> UIR.
>
> > [...] If I remember well, the idea was to
> >
> > 1) apply a unitary tranformation to the vacuum vector |0'> = U|0>
> > 2) Find components of |0'> in a basis
> > 3) Observe that in some limit all these components tend to zero
> > 4) Conclude that the vector |0'> goes outside the original
> > Hilbert space H in this limit.
> >
> > This seems unfair to me. In these examples, even if all components
> of
> > |0'> tend to zero, their number tends to infinity, and the sum of
> > squares of the components |0'> in any basis in H should remain 1.
>
> I look at it this way: if we have a (continuously-parametrized,
> infinite) basis |b> for H, then any other vector |v> can be expressed
> as an integral superposition, whose coefficients are given by taking
> the inner product between |v> and each respective |b>, i.e:
>
> |v> = Integral db <b|v> |b>
>
> So if <b|v> is 0 for every b , the Integral must be 0, showing that |v>
> cannot be expressed as a superposition of |b>'s. This is quite
> different from the position/momentum case where <x|p> = exp(ipx) which
> is non-zero.

There is full analogy. If we use (as we should) normalized vectors
for |x> and |p>, then we obtain

<x|p> = 1/sqrt(V) exp(ipx)

|p> = Integral dx <x|p> |x>
= 1/sqrt(V) Integral dx exp(ipx) |x>

The wave function (the density of the probability amplitude)
of |p> in the position representation must be

1/sqrt(V) exp(ipx)

because the volume integral of its square is

Integral dx |1/sqrt(V) exp(ipx)|^2 = 1/V Integral dx
= 1

as it should. The function exp(ipx) without the normalization
factor 1/sqrt(V) does not have a probabilistic
interpretation, because the volume integral of its square is infinite.

> > The unitary operator U is explicitly defined within the Hilbert
> space
> > H, so it is beyond me how it can bring any vector outside of H.
>
> If one examines the U carefully, it is not really correct to say
> that it is explicitly defined "within" the Hilbert (Fock) space. A
> crucial step in the construction of Fock space is to restrict it
> to have only state vectors whose total particle number is finite.
> Without this restriction, one cannot define an inner product on the
> space, because the usual Riemann-Lebesgue integral calculus doesn't
> work: we can't approximate an arbitrary vector therein by a countable
> sum arbitrarily closely, as is necessary when defining integrals
> rigorously. Umezawa explains (some of) this. But if you can't get a
> copy, part of it appears in a summary I posted to spr back on
> 15-Dec-2004 in a thread titled "Degenerate vacua in QFT":
>
> http://www.lns.cornell.edu/spr/2004-12/msg0065860.html
>
> modulo some followup corrections by Arnold Neumaier. :-)
>
> The "U" maps vectors in the Fock space into other vectors in the
> larger non-separable space, of which the Fock space is but a
> subspace. The total particle number of those "other vectors" turns
> out to be infinite, proving that they lie outside Fock space, which
> by construction contains only vectors of *finite* total particle
> number.

Thank you for the reference. I've seen similar arguments in other
places, but they do not make much sense to me. I have a strong feeling
that the distinction between separable and non-separable Hilbert spaces
was invented by mathematicians to make life of physicists miserable.
I don't think there is anything wrong with regarding the Hilbert space
of a single particle as non-separable. After all, the number of points
in 3D space is not countable, and one can associate a distinct basis
vector (eigenvector of the position operator) with each such point.

Your arguments could be correct if your DEFINE the Fock space as
having not more than a finite number of particles.
Then, why I am not allowed
to DEFINE the Fock space as having any number of particles from zero
to infinity?
I have thought about these issues and came to a conclusion that there
could be some non-trivial math involved, but it has no significance to
physics. Again, I am not talking about superconductivity and
spontaneously broken vacuum symmetries - the issues I'm not so familiar
with. Maybe there is deep physical truth concerning UIR in these fields,
I just don't know.

For myself, I found another more comfortable attitude to
these issues. This attitude is not frequently discussed in physics
literature, but I found it rather illuminating.
This is based on the so-called
"non-standard analysis" first developed by A. Robinson in 1960.
This is too vast a field to be described in one post, but the basic idea
is to treat finite, "infinitely small" and "infinitely large" quantities
on the same footing. In this approach, the numbers like 1/sqrt(V),
where V is the volume of the entire space, make perfect sense, and there
is no trouble to calculate the integral

Integral dx |1/sqrt(V) exp(ipx)|^2 = 1

even if the integrand is "zero" everywhere.

There are few papers which try to apply this approach to quantum
mechanics. See, for example
A. Friedman, "Non-standard extension of quantum logic and
Dirac's bra-ket formalism of quantum mechanics", Int. J. Theor. Phys.
33 (1994), 307 (he was my student back in old times).

The non-standard analysis is now a well-established branch of
mathematics. I think, its use in QM may clarify some conceptual issues,
but I don't expect any physical discoveries there.

Eugene.

mikem@despammed.com

Eugene Stefanovich wrote in part:

> [...] I don't think there is anything wrong with
> regarding the Hilbert space of a single particle
> as non-separable. After all, the number of points
> in 3D space is not countable, and one can associate
> a distinct basis vector (eigenvector of the position
> operator) with each such point.

That's because we can integrate over a 3D space.
But, (at least with standard integration), we can't
integrate over an infinite dimensional space in the
same way. But about here, my detailed knowledge
starts to dry up so I can't say much more.

> Your arguments could be correct if your DEFINE
> the Fock space as having not more than a finite
> number of particles. Then, why I am not allowed
> to DEFINE the Fock space as having any number
> of particles from zero to infinity?

Only because of the difficulty with performing
standard integration over uncountably-infinite
dimensional spaces.

> [...] See, for example: A. Friedman, "Non-standard
> extension of quantum logic and Dirac's bra-ket
> formalism of quantum mechanics", Int. J. Theor. Phys.
> 33 (1994), 307 [...]

Is this paper on the archive, or somewhere else online?
(It's a pain for me to travel to university libraries these
days.)


Regarding the other items in your post, I need to think
about them for a while before replying.

mikem@despammed.com

Eugene Stefanovich wrote in part:

> [...] I don't think there is anything wrong with
> regarding the Hilbert space of a single particle
> as non-separable. After all, the number of points
> in 3D space is not countable, and one can associate
> a distinct basis vector (eigenvector of the position
> operator) with each such point.

That's because we can integrate over a 3D space.
But, (at least with standard integration), we can't
integrate over an infinite dimensional space in the
same way. But about here, my detailed knowledge
starts to dry up so I can't say much more.

> Your arguments could be correct if your DEFINE
> the Fock space as having not more than a finite
> number of particles. Then, why I am not allowed
> to DEFINE the Fock space as having any number
> of particles from zero to infinity?

Only because of the difficulty with performing
standard integration over uncountably-infinite
dimensional spaces.

> [...] See, for example: A. Friedman, "Non-standard
> extension of quantum logic and Dirac's bra-ket
> formalism of quantum mechanics", Int. J. Theor. Phys.
> 33 (1994), 307 [...]

Is this paper on the archive, or somewhere else online?
(It's a pain for me to travel to university libraries these
days.)


Regarding the other items in your post, I need to think
about them for a while before replying.

mikem@despammed.com

Eugene Stefanovich wrote in part:

> [...] I don't think there is anything wrong with
> regarding the Hilbert space of a single particle
> as non-separable. After all, the number of points
> in 3D space is not countable, and one can associate
> a distinct basis vector (eigenvector of the position
> operator) with each such point.

That's because we can integrate over a 3D space.
But, (at least with standard integration), we can't
integrate over an infinite dimensional space in the
same way. But about here, my detailed knowledge
starts to dry up so I can't say much more.

> Your arguments could be correct if your DEFINE
> the Fock space as having not more than a finite
> number of particles. Then, why I am not allowed
> to DEFINE the Fock space as having any number
> of particles from zero to infinity?

Only because of the difficulty with performing
standard integration over uncountably-infinite
dimensional spaces.

> [...] See, for example: A. Friedman, "Non-standard
> extension of quantum logic and Dirac's bra-ket
> formalism of quantum mechanics", Int. J. Theor. Phys.
> 33 (1994), 307 [...]

Is this paper on the archive, or somewhere else online?
(It's a pain for me to travel to university libraries these
days.)


Regarding the other items in your post, I need to think
about them for a while before replying.

mikem@despammed.com

Eugene Stefanovich wrote in part:

> [...] I don't think there is anything wrong with
> regarding the Hilbert space of a single particle
> as non-separable. After all, the number of points
> in 3D space is not countable, and one can associate
> a distinct basis vector (eigenvector of the position
> operator) with each such point.

That's because we can integrate over a 3D space.
But, (at least with standard integration), we can't
integrate over an infinite dimensional space in the
same way. But about here, my detailed knowledge
starts to dry up so I can't say much more.

> Your arguments could be correct if your DEFINE
> the Fock space as having not more than a finite
> number of particles. Then, why I am not allowed
> to DEFINE the Fock space as having any number
> of particles from zero to infinity?

Only because of the difficulty with performing
standard integration over uncountably-infinite
dimensional spaces.

> [...] See, for example: A. Friedman, "Non-standard
> extension of quantum logic and Dirac's bra-ket
> formalism of quantum mechanics", Int. J. Theor. Phys.
> 33 (1994), 307 [...]

Is this paper on the archive, or somewhere else online?
(It's a pain for me to travel to university libraries these
days.)


Regarding the other items in your post, I need to think
about them for a while before replying.

mikem@despammed.com

Eugene Stefanovich wrote in part:

> [...] I don't think there is anything wrong with
> regarding the Hilbert space of a single particle
> as non-separable. After all, the number of points
> in 3D space is not countable, and one can associate
> a distinct basis vector (eigenvector of the position
> operator) with each such point.

That's because we can integrate over a 3D space.
But, (at least with standard integration), we can't
integrate over an infinite dimensional space in the
same way. But about here, my detailed knowledge
starts to dry up so I can't say much more.

> Your arguments could be correct if your DEFINE
> the Fock space as having not more than a finite
> number of particles. Then, why I am not allowed
> to DEFINE the Fock space as having any number
> of particles from zero to infinity?

Only because of the difficulty with performing
standard integration over uncountably-infinite
dimensional spaces.

> [...] See, for example: A. Friedman, "Non-standard
> extension of quantum logic and Dirac's bra-ket
> formalism of quantum mechanics", Int. J. Theor. Phys.
> 33 (1994), 307 [...]

Is this paper on the archive, or somewhere else online?
(It's a pain for me to travel to university libraries these
days.)


Regarding the other items in your post, I need to think
about them for a while before replying.

mikem@despammed.com

Eugene Stefanovich wrote in part:

> [...] I don't think there is anything wrong with
> regarding the Hilbert space of a single particle
> as non-separable. After all, the number of points
> in 3D space is not countable, and one can associate
> a distinct basis vector (eigenvector of the position
> operator) with each such point.

That's because we can integrate over a 3D space.
But, (at least with standard integration), we can't
integrate over an infinite dimensional space in the
same way. But about here, my detailed knowledge
starts to dry up so I can't say much more.

> Your arguments could be correct if your DEFINE
> the Fock space as having not more than a finite
> number of particles. Then, why I am not allowed
> to DEFINE the Fock space as having any number
> of particles from zero to infinity?

Only because of the difficulty with performing
standard integration over uncountably-infinite
dimensional spaces.

> [...] See, for example: A. Friedman, "Non-standard
> extension of quantum logic and Dirac's bra-ket
> formalism of quantum mechanics", Int. J. Theor. Phys.
> 33 (1994), 307 [...]

Is this paper on the archive, or somewhere else online?
(It's a pain for me to travel to university libraries these
days.)


Regarding the other items in your post, I need to think
about them for a while before replying.

mikem@despammed.com

Eugene Stefanovich wrote in part:

> [...] I don't think there is anything wrong with
> regarding the Hilbert space of a single particle
> as non-separable. After all, the number of points
> in 3D space is not countable, and one can associate
> a distinct basis vector (eigenvector of the position
> operator) with each such point.

That's because we can integrate over a 3D space.
But, (at least with standard integration), we can't
integrate over an infinite dimensional space in the
same way. But about here, my detailed knowledge
starts to dry up so I can't say much more.

> Your arguments could be correct if your DEFINE
> the Fock space as having not more than a finite
> number of particles. Then, why I am not allowed
> to DEFINE the Fock space as having any number
> of particles from zero to infinity?

Only because of the difficulty with performing
standard integration over uncountably-infinite
dimensional spaces.

> [...] See, for example: A. Friedman, "Non-standard
> extension of quantum logic and Dirac's bra-ket
> formalism of quantum mechanics", Int. J. Theor. Phys.
> 33 (1994), 307 [...]

Is this paper on the archive, or somewhere else online?
(It's a pain for me to travel to university libraries these
days.)


Regarding the other items in your post, I need to think
about them for a while before replying.

mikem@despammed.com

Eugene Stefanovich wrote in part:

> [...] I don't think there is anything wrong with
> regarding the Hilbert space of a single particle
> as non-separable. After all, the number of points
> in 3D space is not countable, and one can associate
> a distinct basis vector (eigenvector of the position
> operator) with each such point.

That's because we can integrate over a 3D space.
But, (at least with standard integration), we can't
integrate over an infinite dimensional space in the
same way. But about here, my detailed knowledge
starts to dry up so I can't say much more.

> Your arguments could be correct if your DEFINE
> the Fock space as having not more than a finite
> number of particles. Then, why I am not allowed
> to DEFINE the Fock space as having any number
> of particles from zero to infinity?

Only because of the difficulty with performing
standard integration over uncountably-infinite
dimensional spaces.

> [...] See, for example: A. Friedman, "Non-standard
> extension of quantum logic and Dirac's bra-ket
> formalism of quantum mechanics", Int. J. Theor. Phys.
> 33 (1994), 307 [...]

Is this paper on the archive, or somewhere else online?
(It's a pain for me to travel to university libraries these
days.)


Regarding the other items in your post, I need to think
about them for a while before replying.

mikem@despammed.com

Eugene Stefanovich wrote in part:

> [...] I don't think there is anything wrong with
> regarding the Hilbert space of a single particle
> as non-separable. After all, the number of points
> in 3D space is not countable, and one can associate
> a distinct basis vector (eigenvector of the position
> operator) with each such point.

That's because we can integrate over a 3D space.
But, (at least with standard integration), we can't
integrate over an infinite dimensional space in the
same way. But about here, my detailed knowledge
starts to dry up so I can't say much more.

> Your arguments could be correct if your DEFINE
> the Fock space as having not more than a finite
> number of particles. Then, why I am not allowed
> to DEFINE the Fock space as having any number
> of particles from zero to infinity?

Only because of the difficulty with performing
standard integration over uncountably-infinite
dimensional spaces.

> [...] See, for example: A. Friedman, "Non-standard
> extension of quantum logic and Dirac's bra-ket
> formalism of quantum mechanics", Int. J. Theor. Phys.
> 33 (1994), 307 [...]

Is this paper on the archive, or somewhere else online?
(It's a pain for me to travel to university libraries these
days.)


Regarding the other items in your post, I need to think
about them for a while before replying.

Eugene Stefanovich

<mikem@despammed.com> wrote in message
news:1128407680.618420.49880@g43g2000cwa.googlegroups.com...

> > Your arguments could be correct if your DEFINE
> > the Fock space as having not more than a finite
> > number of particles. Then, why I am not allowed
> > to DEFINE the Fock space as having any number
> > of particles from zero to infinity?
>
> Only because of the difficulty with performing
> standard integration over uncountably-infinite
> dimensional spaces.

I would prefer to generalize the way we integrate things
rather that stick to artificial separable spaces only to discover
that they cannot accomodate the interacting systems we are
most interested in.

> > [...] See, for example: A. Friedman, "Non-standard
> > extension of quantum logic and Dirac's bra-ket
> > formalism of quantum mechanics", Int. J. Theor. Phys.
> > 33 (1994), 307 [...]
>
> Is this paper on the archive, or somewhere else online?
> (It's a pain for me to travel to university libraries these
> days.)

No, it's not on the Internet. I can send you a copy if you
give me your address.

Eugene.