Gauge Transformations in Momentum Space?

Eugene Stefanovich

mikem@despammed.com wrote:

> A different, more recent, area is neutrino oscillations.
> Blasone et al have shown that the Fock space of
> definite flavour states is unitarily inequivalent to
> that definite mass states. See, for example,
> hep-ph/9501263, and also the review article by
> Capolupo: hep-th/0408228. This means that to
> understand the QFT of neutrino oscillations fully,
> we need to understand UIRs and disjoint Fock
> spaces.

OK, let's skip superconductivity and talk about neutrinos.
I looked at Blasone-Vitiello paper. This is a good example of
what seems confusing about UIR for me.

They find a unitary transformation which makes flavor
eigenstates (or creation-annihilation operators) from
mass eigenstates (or creation-annihilation operators).
This transformation also changes the vacuum vector.
In particular, it makes the new vacuum |0'> orthogonal to the old vacuum
|0>.

I have two questions:

1. In my opinion this construction does not mean that the
new vacuum lies in a different Fock state. This wouldn't be the
case even if all components of |0'> in the old basis were "zero"
in the limit of infinite volume.
Each of the components may tend to zero, but the number of
components tends to infinity. So that if you correctly sum up
the infinite number of "zeros" you should still get a vector
of unit norm.

In my view, this is not dissimilar to the normalized plane wave.
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 think that in order to evaluate correctly the expressions like
"zero probability density" x "infinite volume" one should be
careful with limits.
The "nonstandard analysis" may be useful there.

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 am sure I am missing some important point regarding UIR.
Could you please let me know what this point is?

Eugene.

Eugene Stefanovich

mikem@despammed.com wrote:

> A different, more recent, area is neutrino oscillations.
> Blasone et al have shown that the Fock space of
> definite flavour states is unitarily inequivalent to
> that definite mass states. See, for example,
> hep-ph/9501263, and also the review article by
> Capolupo: hep-th/0408228. This means that to
> understand the QFT of neutrino oscillations fully,
> we need to understand UIRs and disjoint Fock
> spaces.

OK, let's skip superconductivity and talk about neutrinos.
I looked at Blasone-Vitiello paper. This is a good example of
what seems confusing about UIR for me.

They find a unitary transformation which makes flavor
eigenstates (or creation-annihilation operators) from
mass eigenstates (or creation-annihilation operators).
This transformation also changes the vacuum vector.
In particular, it makes the new vacuum |0'> orthogonal to the old vacuum
|0>.

I have two questions:

1. In my opinion this construction does not mean that the
new vacuum lies in a different Fock state. This wouldn't be the
case even if all components of |0'> in the old basis were "zero"
in the limit of infinite volume.
Each of the components may tend to zero, but the number of
components tends to infinity. So that if you correctly sum up
the infinite number of "zeros" you should still get a vector
of unit norm.

In my view, this is not dissimilar to the normalized plane wave.
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 think that in order to evaluate correctly the expressions like
"zero probability density" x "infinite volume" one should be
careful with limits.
The "nonstandard analysis" may be useful there.

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 am sure I am missing some important point regarding UIR.
Could you please let me know what this point is?

Eugene.

Eugene Stefanovich

mikem@despammed.com wrote:

> A different, more recent, area is neutrino oscillations.
> Blasone et al have shown that the Fock space of
> definite flavour states is unitarily inequivalent to
> that definite mass states. See, for example,
> hep-ph/9501263, and also the review article by
> Capolupo: hep-th/0408228. This means that to
> understand the QFT of neutrino oscillations fully,
> we need to understand UIRs and disjoint Fock
> spaces.

OK, let's skip superconductivity and talk about neutrinos.
I looked at Blasone-Vitiello paper. This is a good example of
what seems confusing about UIR for me.

They find a unitary transformation which makes flavor
eigenstates (or creation-annihilation operators) from
mass eigenstates (or creation-annihilation operators).
This transformation also changes the vacuum vector.
In particular, it makes the new vacuum |0'> orthogonal to the old vacuum
|0>.

I have two questions:

1. In my opinion this construction does not mean that the
new vacuum lies in a different Fock state. This wouldn't be the
case even if all components of |0'> in the old basis were "zero"
in the limit of infinite volume.
Each of the components may tend to zero, but the number of
components tends to infinity. So that if you correctly sum up
the infinite number of "zeros" you should still get a vector
of unit norm.

In my view, this is not dissimilar to the normalized plane wave.
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 think that in order to evaluate correctly the expressions like
"zero probability density" x "infinite volume" one should be
careful with limits.
The "nonstandard analysis" may be useful there.

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 am sure I am missing some important point regarding UIR.
Could you please let me know what this point is?

Eugene.

Eugene Stefanovich

mikem@despammed.com wrote:

> A different, more recent, area is neutrino oscillations.
> Blasone et al have shown that the Fock space of
> definite flavour states is unitarily inequivalent to
> that definite mass states. See, for example,
> hep-ph/9501263, and also the review article by
> Capolupo: hep-th/0408228. This means that to
> understand the QFT of neutrino oscillations fully,
> we need to understand UIRs and disjoint Fock
> spaces.

OK, let's skip superconductivity and talk about neutrinos.
I looked at Blasone-Vitiello paper. This is a good example of
what seems confusing about UIR for me.

They find a unitary transformation which makes flavor
eigenstates (or creation-annihilation operators) from
mass eigenstates (or creation-annihilation operators).
This transformation also changes the vacuum vector.
In particular, it makes the new vacuum |0'> orthogonal to the old vacuum
|0>.

I have two questions:

1. In my opinion this construction does not mean that the
new vacuum lies in a different Fock state. This wouldn't be the
case even if all components of |0'> in the old basis were "zero"
in the limit of infinite volume.
Each of the components may tend to zero, but the number of
components tends to infinity. So that if you correctly sum up
the infinite number of "zeros" you should still get a vector
of unit norm.

In my view, this is not dissimilar to the normalized plane wave.
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 think that in order to evaluate correctly the expressions like
"zero probability density" x "infinite volume" one should be
careful with limits.
The "nonstandard analysis" may be useful there.

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 am sure I am missing some important point regarding UIR.
Could you please let me know what this point is?

Eugene.

Eugene Stefanovich

mikem@despammed.com wrote:

> A different, more recent, area is neutrino oscillations.
> Blasone et al have shown that the Fock space of
> definite flavour states is unitarily inequivalent to
> that definite mass states. See, for example,
> hep-ph/9501263, and also the review article by
> Capolupo: hep-th/0408228. This means that to
> understand the QFT of neutrino oscillations fully,
> we need to understand UIRs and disjoint Fock
> spaces.

OK, let's skip superconductivity and talk about neutrinos.
I looked at Blasone-Vitiello paper. This is a good example of
what seems confusing about UIR for me.

They find a unitary transformation which makes flavor
eigenstates (or creation-annihilation operators) from
mass eigenstates (or creation-annihilation operators).
This transformation also changes the vacuum vector.
In particular, it makes the new vacuum |0'> orthogonal to the old vacuum
|0>.

I have two questions:

1. In my opinion this construction does not mean that the
new vacuum lies in a different Fock state. This wouldn't be the
case even if all components of |0'> in the old basis were "zero"
in the limit of infinite volume.
Each of the components may tend to zero, but the number of
components tends to infinity. So that if you correctly sum up
the infinite number of "zeros" you should still get a vector
of unit norm.

In my view, this is not dissimilar to the normalized plane wave.
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 think that in order to evaluate correctly the expressions like
"zero probability density" x "infinite volume" one should be
careful with limits.
The "nonstandard analysis" may be useful there.

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 am sure I am missing some important point regarding UIR.
Could you please let me know what this point is?

Eugene.

Eugene Stefanovich

mikem@despammed.com wrote:

> A different, more recent, area is neutrino oscillations.
> Blasone et al have shown that the Fock space of
> definite flavour states is unitarily inequivalent to
> that definite mass states. See, for example,
> hep-ph/9501263, and also the review article by
> Capolupo: hep-th/0408228. This means that to
> understand the QFT of neutrino oscillations fully,
> we need to understand UIRs and disjoint Fock
> spaces.

OK, let's skip superconductivity and talk about neutrinos.
I looked at Blasone-Vitiello paper. This is a good example of
what seems confusing about UIR for me.

They find a unitary transformation which makes flavor
eigenstates (or creation-annihilation operators) from
mass eigenstates (or creation-annihilation operators).
This transformation also changes the vacuum vector.
In particular, it makes the new vacuum |0'> orthogonal to the old vacuum
|0>.

I have two questions:

1. In my opinion this construction does not mean that the
new vacuum lies in a different Fock state. This wouldn't be the
case even if all components of |0'> in the old basis were "zero"
in the limit of infinite volume.
Each of the components may tend to zero, but the number of
components tends to infinity. So that if you correctly sum up
the infinite number of "zeros" you should still get a vector
of unit norm.

In my view, this is not dissimilar to the normalized plane wave.
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 think that in order to evaluate correctly the expressions like
"zero probability density" x "infinite volume" one should be
careful with limits.
The "nonstandard analysis" may be useful there.

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 am sure I am missing some important point regarding UIR.
Could you please let me know what this point is?

Eugene.

Eugene Stefanovich

mikem@despammed.com wrote:

> A different, more recent, area is neutrino oscillations.
> Blasone et al have shown that the Fock space of
> definite flavour states is unitarily inequivalent to
> that definite mass states. See, for example,
> hep-ph/9501263, and also the review article by
> Capolupo: hep-th/0408228. This means that to
> understand the QFT of neutrino oscillations fully,
> we need to understand UIRs and disjoint Fock
> spaces.

OK, let's skip superconductivity and talk about neutrinos.
I looked at Blasone-Vitiello paper. This is a good example of
what seems confusing about UIR for me.

They find a unitary transformation which makes flavor
eigenstates (or creation-annihilation operators) from
mass eigenstates (or creation-annihilation operators).
This transformation also changes the vacuum vector.
In particular, it makes the new vacuum |0'> orthogonal to the old vacuum
|0>.

I have two questions:

1. In my opinion this construction does not mean that the
new vacuum lies in a different Fock state. This wouldn't be the
case even if all components of |0'> in the old basis were "zero"
in the limit of infinite volume.
Each of the components may tend to zero, but the number of
components tends to infinity. So that if you correctly sum up
the infinite number of "zeros" you should still get a vector
of unit norm.

In my view, this is not dissimilar to the normalized plane wave.
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 think that in order to evaluate correctly the expressions like
"zero probability density" x "infinite volume" one should be
careful with limits.
The "nonstandard analysis" may be useful there.

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 am sure I am missing some important point regarding UIR.
Could you please let me know what this point is?

Eugene.

Eugene Stefanovich

mikem@despammed.com wrote:

> A different, more recent, area is neutrino oscillations.
> Blasone et al have shown that the Fock space of
> definite flavour states is unitarily inequivalent to
> that definite mass states. See, for example,
> hep-ph/9501263, and also the review article by
> Capolupo: hep-th/0408228. This means that to
> understand the QFT of neutrino oscillations fully,
> we need to understand UIRs and disjoint Fock
> spaces.

OK, let's skip superconductivity and talk about neutrinos.
I looked at Blasone-Vitiello paper. This is a good example of
what seems confusing about UIR for me.

They find a unitary transformation which makes flavor
eigenstates (or creation-annihilation operators) from
mass eigenstates (or creation-annihilation operators).
This transformation also changes the vacuum vector.
In particular, it makes the new vacuum |0'> orthogonal to the old vacuum
|0>.

I have two questions:

1. In my opinion this construction does not mean that the
new vacuum lies in a different Fock state. This wouldn't be the
case even if all components of |0'> in the old basis were "zero"
in the limit of infinite volume.
Each of the components may tend to zero, but the number of
components tends to infinity. So that if you correctly sum up
the infinite number of "zeros" you should still get a vector
of unit norm.

In my view, this is not dissimilar to the normalized plane wave.
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 think that in order to evaluate correctly the expressions like
"zero probability density" x "infinite volume" one should be
careful with limits.
The "nonstandard analysis" may be useful there.

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 am sure I am missing some important point regarding UIR.
Could you please let me know what this point is?

Eugene.

Eugene Stefanovich

mikem@despammed.com wrote:

> A different, more recent, area is neutrino oscillations.
> Blasone et al have shown that the Fock space of
> definite flavour states is unitarily inequivalent to
> that definite mass states. See, for example,
> hep-ph/9501263, and also the review article by
> Capolupo: hep-th/0408228. This means that to
> understand the QFT of neutrino oscillations fully,
> we need to understand UIRs and disjoint Fock
> spaces.

OK, let's skip superconductivity and talk about neutrinos.
I looked at Blasone-Vitiello paper. This is a good example of
what seems confusing about UIR for me.

They find a unitary transformation which makes flavor
eigenstates (or creation-annihilation operators) from
mass eigenstates (or creation-annihilation operators).
This transformation also changes the vacuum vector.
In particular, it makes the new vacuum |0'> orthogonal to the old vacuum
|0>.

I have two questions:

1. In my opinion this construction does not mean that the
new vacuum lies in a different Fock state. This wouldn't be the
case even if all components of |0'> in the old basis were "zero"
in the limit of infinite volume.
Each of the components may tend to zero, but the number of
components tends to infinity. So that if you correctly sum up
the infinite number of "zeros" you should still get a vector
of unit norm.

In my view, this is not dissimilar to the normalized plane wave.
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 think that in order to evaluate correctly the expressions like
"zero probability density" x "infinite volume" one should be
careful with limits.
The "nonstandard analysis" may be useful there.

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 am sure I am missing some important point regarding UIR.
Could you please let me know what this point is?

Eugene.

Eugene Stefanovich

mikem@despammed.com wrote:

> A different, more recent, area is neutrino oscillations.
> Blasone et al have shown that the Fock space of
> definite flavour states is unitarily inequivalent to
> that definite mass states. See, for example,
> hep-ph/9501263, and also the review article by
> Capolupo: hep-th/0408228. This means that to
> understand the QFT of neutrino oscillations fully,
> we need to understand UIRs and disjoint Fock
> spaces.

OK, let's skip superconductivity and talk about neutrinos.
I looked at Blasone-Vitiello paper. This is a good example of
what seems confusing about UIR for me.

They find a unitary transformation which makes flavor
eigenstates (or creation-annihilation operators) from
mass eigenstates (or creation-annihilation operators).
This transformation also changes the vacuum vector.
In particular, it makes the new vacuum |0'> orthogonal to the old vacuum
|0>.

I have two questions:

1. In my opinion this construction does not mean that the
new vacuum lies in a different Fock state. This wouldn't be the
case even if all components of |0'> in the old basis were "zero"
in the limit of infinite volume.
Each of the components may tend to zero, but the number of
components tends to infinity. So that if you correctly sum up
the infinite number of "zeros" you should still get a vector
of unit norm.

In my view, this is not dissimilar to the normalized plane wave.
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 think that in order to evaluate correctly the expressions like
"zero probability density" x "infinite volume" one should be
careful with limits.
The "nonstandard analysis" may be useful there.

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 am sure I am missing some important point regarding UIR.
Could you please let me know what this point is?

Eugene.

Eugene Stefanovich

mikem@despammed.com wrote:

> A different, more recent, area is neutrino oscillations.
> Blasone et al have shown that the Fock space of
> definite flavour states is unitarily inequivalent to
> that definite mass states. See, for example,
> hep-ph/9501263, and also the review article by
> Capolupo: hep-th/0408228. This means that to
> understand the QFT of neutrino oscillations fully,
> we need to understand UIRs and disjoint Fock
> spaces.

OK, let's skip superconductivity and talk about neutrinos.
I looked at Blasone-Vitiello paper. This is a good example of
what seems confusing about UIR for me.

They find a unitary transformation which makes flavor
eigenstates (or creation-annihilation operators) from
mass eigenstates (or creation-annihilation operators).
This transformation also changes the vacuum vector.
In particular, it makes the new vacuum |0'> orthogonal to the old vacuum
|0>.

I have two questions:

1. In my opinion this construction does not mean that the
new vacuum lies in a different Fock state. This wouldn't be the
case even if all components of |0'> in the old basis were "zero"
in the limit of infinite volume.
Each of the components may tend to zero, but the number of
components tends to infinity. So that if you correctly sum up
the infinite number of "zeros" you should still get a vector
of unit norm.

In my view, this is not dissimilar to the normalized plane wave.
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 think that in order to evaluate correctly the expressions like
"zero probability density" x "infinite volume" one should be
careful with limits.
The "nonstandard analysis" may be useful there.

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 am sure I am missing some important point regarding UIR.
Could you please let me know what this point is?

Eugene.

Eugene Stefanovich

mikem@despammed.com wrote:

> A different, more recent, area is neutrino oscillations.
> Blasone et al have shown that the Fock space of
> definite flavour states is unitarily inequivalent to
> that definite mass states. See, for example,
> hep-ph/9501263, and also the review article by
> Capolupo: hep-th/0408228. This means that to
> understand the QFT of neutrino oscillations fully,
> we need to understand UIRs and disjoint Fock
> spaces.

OK, let's skip superconductivity and talk about neutrinos.
I looked at Blasone-Vitiello paper. This is a good example of
what seems confusing about UIR for me.

They find a unitary transformation which makes flavor
eigenstates (or creation-annihilation operators) from
mass eigenstates (or creation-annihilation operators).
This transformation also changes the vacuum vector.
In particular, it makes the new vacuum |0'> orthogonal to the old vacuum
|0>.

I have two questions:

1. In my opinion this construction does not mean that the
new vacuum lies in a different Fock state. This wouldn't be the
case even if all components of |0'> in the old basis were "zero"
in the limit of infinite volume.
Each of the components may tend to zero, but the number of
components tends to infinity. So that if you correctly sum up
the infinite number of "zeros" you should still get a vector
of unit norm.

In my view, this is not dissimilar to the normalized plane wave.
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 think that in order to evaluate correctly the expressions like
"zero probability density" x "infinite volume" one should be
careful with limits.
The "nonstandard analysis" may be useful there.

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 am sure I am missing some important point regarding UIR.
Could you please let me know what this point is?

Eugene.

Eugene Stefanovich

mikem@despammed.com wrote:

> A different, more recent, area is neutrino oscillations.
> Blasone et al have shown that the Fock space of
> definite flavour states is unitarily inequivalent to
> that definite mass states. See, for example,
> hep-ph/9501263, and also the review article by
> Capolupo: hep-th/0408228. This means that to
> understand the QFT of neutrino oscillations fully,
> we need to understand UIRs and disjoint Fock
> spaces.

OK, let's skip superconductivity and talk about neutrinos.
I looked at Blasone-Vitiello paper. This is a good example of
what seems confusing about UIR for me.

They find a unitary transformation which makes flavor
eigenstates (or creation-annihilation operators) from
mass eigenstates (or creation-annihilation operators).
This transformation also changes the vacuum vector.
In particular, it makes the new vacuum |0'> orthogonal to the old vacuum
|0>.

I have two questions:

1. In my opinion this construction does not mean that the
new vacuum lies in a different Fock state. This wouldn't be the
case even if all components of |0'> in the old basis were "zero"
in the limit of infinite volume.
Each of the components may tend to zero, but the number of
components tends to infinity. So that if you correctly sum up
the infinite number of "zeros" you should still get a vector
of unit norm.

In my view, this is not dissimilar to the normalized plane wave.
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 think that in order to evaluate correctly the expressions like
"zero probability density" x "infinite volume" one should be
careful with limits.
The "nonstandard analysis" may be useful there.

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 am sure I am missing some important point regarding UIR.
Could you please let me know what this point is?

Eugene.

mikem@despammed.com

Eugene Stefanovich wrote:

> 1. In my opinion [the Blasone-Vitiello] construction does not mean
> that the new vacuum lies in a different Fock state. This wouldn't be
> the case even if all components of |0'> in the old basis were "zero"
> in the limit of infinite volume. Each of the components may tend to
> zero, but the number of components tends to infinity. So that if you
> correctly sum up the infinite number of "zeros" you should still get
> a vector of unit norm.
>
> In my view, this is not dissimilar to the normalized plane wave. 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?

> 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?

>> [...] 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.

> 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.

mikem@despammed.com

Eugene Stefanovich wrote:

> 1. In my opinion [the Blasone-Vitiello] construction does not mean
> that the new vacuum lies in a different Fock state. This wouldn't be
> the case even if all components of |0'> in the old basis were "zero"
> in the limit of infinite volume. Each of the components may tend to
> zero, but the number of components tends to infinity. So that if you
> correctly sum up the infinite number of "zeros" you should still get
> a vector of unit norm.
>
> In my view, this is not dissimilar to the normalized plane wave. 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?

> 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?

>> [...] 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.

> 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.

mikem@despammed.com

Eugene Stefanovich wrote:

> 1. In my opinion [the Blasone-Vitiello] construction does not mean
> that the new vacuum lies in a different Fock state. This wouldn't be
> the case even if all components of |0'> in the old basis were "zero"
> in the limit of infinite volume. Each of the components may tend to
> zero, but the number of components tends to infinity. So that if you
> correctly sum up the infinite number of "zeros" you should still get
> a vector of unit norm.
>
> In my view, this is not dissimilar to the normalized plane wave. 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?

> 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?

>> [...] 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.

> 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.

mikem@despammed.com

Eugene Stefanovich wrote:

> 1. In my opinion [the Blasone-Vitiello] construction does not mean
> that the new vacuum lies in a different Fock state. This wouldn't be
> the case even if all components of |0'> in the old basis were "zero"
> in the limit of infinite volume. Each of the components may tend to
> zero, but the number of components tends to infinity. So that if you
> correctly sum up the infinite number of "zeros" you should still get
> a vector of unit norm.
>
> In my view, this is not dissimilar to the normalized plane wave. 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?

> 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?

>> [...] 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.

> 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.