Dirac distribution δ(x-x)
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When canonically quantising a scalar field without normal ordering, one comes across the following expression:
$$int_mathbb{R}E(x)delta(x-x)dx.$$
Here, $E(x)$ is a smooth unbound function.
I have read that this is infinity because, 'the value of the Dirac function at zero is infinity.'
For obvious reasons, this explanation is very unsatisfying - $delta$ is not a function but a distribution. And the first question therefore is: How does the expression
$$intvarphi(x)delta(x-x)dx$$
mathematically make sense? And for this, let's first assume that $varphi$ is a Schwartz-function, unlike $E(x)$. I know that $intvarphi(x)delta(x-a)dx=varphi(a)$. But here $a$ is a constant!
The second question is then how does the above equation make sense, if you swap $varphi$ for $E$.
mathematical-physics distribution-theory
asked Nov 16 at 10:08
Thomas Wening
12110
add a comment |
up vote
2
down vote
favorite
When canonically quantising a scalar field without normal ordering, one comes across the following expression:
$$int_mathbb{R}E(x)delta(x-x)dx.$$
Here, $E(x)$ is a smooth unbound function.
I have read that this is infinity because, 'the value of the Dirac function at zero is infinity.'
For obvious reasons, this explanation is very unsatisfying - $delta$ is not a function but a distribution. And the first question therefore is: How does the expression
$$intvarphi(x)delta(x-x)dx$$
mathematically make sense? And for this, let's first assume that $varphi$ is a Schwartz-function, unlike $E(x)$. I know that $intvarphi(x)delta(x-a)dx=varphi(a)$. But here $a$ is a constant!
The second question is then how does the above equation make sense, if you swap $varphi$ for $E$.
mathematical-physics distribution-theory
asked Nov 16 at 10:08
Thomas Wening
12110
3
The expression $delta(x-x)=delta(0)$ doesn't make sense. I suspect the QFT argument which led to the first integral above is faulty. Note that $delta(0)$ typically is the volume of spacetime. So there must have been a careless infinite volume limit taken.
– Abdelmalek Abdesselam
Nov 16 at 15:19
Aha! So it is indeed undefined
– Thomas Wening
Nov 16 at 15:29
3
I remember there was a good treatment of this kind of issues in the old book by T.D. Lee amazon.com/Particle-Physics-Introduction-Contemporary-Concepts/… He introduced canonical quantization by working on a lattice and in a box $Omega$. In the course of the explanation you see how $vol(Omega)$ becomes $delta(0)$. This gets even worse when you consider scattering cross sections.
– Abdelmalek Abdesselam
Nov 16 at 15:36
1
I have read one can get rid off this problem by normalising your Hamiltonian by the Volume, i.e. considering an energy density?
– Thomas Wening
Nov 16 at 15:41
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
When canonically quantising a scalar field without normal ordering, one comes across the following expression:
$$int_mathbb{R}E(x)delta(x-x)dx.$$
Here, $E(x)$ is a smooth unbound function.
I have read that this is infinity because, 'the value of the Dirac function at zero is infinity.'
For obvious reasons, this explanation is very unsatisfying - $delta$ is not a function but a distribution. And the first question therefore is: How does the expression
$$intvarphi(x)delta(x-x)dx$$
mathematically make sense? And for this, let's first assume that $varphi$ is a Schwartz-function, unlike $E(x)$. I know that $intvarphi(x)delta(x-a)dx=varphi(a)$. But here $a$ is a constant!
The second question is then how does the above equation make sense, if you swap $varphi$ for $E$.
mathematical-physics distribution-theory
asked Nov 16 at 10:08
Thomas Wening
12110
When canonically quantising a scalar field without normal ordering, one comes across the following expression:
$$int_mathbb{R}E(x)delta(x-x)dx.$$
Here, $E(x)$ is a smooth unbound function.
I have read that this is infinity because, 'the value of the Dirac function at zero is infinity.'
For obvious reasons, this explanation is very unsatisfying - $delta$ is not a function but a distribution. And the first question therefore is: How does the expression
$$intvarphi(x)delta(x-x)dx$$
mathematically make sense? And for this, let's first assume that $varphi$ is a Schwartz-function, unlike $E(x)$. I know that $intvarphi(x)delta(x-a)dx=varphi(a)$. But here $a$ is a constant!
The second question is then how does the above equation make sense, if you swap $varphi$ for $E$.
mathematical-physics distribution-theory
mathematical-physics distribution-theory
asked Nov 16 at 10:08
Thomas Wening
12110
asked Nov 16 at 10:08
Thomas Wening
12110
edited Nov 16 at 11:41
asked Nov 16 at 10:08
Thomas Wening
12110
asked Nov 16 at 10:08
Thomas Wening
12110
asked Nov 16 at 10:08
Thomas Wening
12110
12110
3
The expression $delta(x-x)=delta(0)$ doesn't make sense. I suspect the QFT argument which led to the first integral above is faulty. Note that $delta(0)$ typically is the volume of spacetime. So there must have been a careless infinite volume limit taken.
– Abdelmalek Abdesselam
Nov 16 at 15:19
Aha! So it is indeed undefined
– Thomas Wening
Nov 16 at 15:29
3
I remember there was a good treatment of this kind of issues in the old book by T.D. Lee amazon.com/Particle-Physics-Introduction-Contemporary-Concepts/… He introduced canonical quantization by working on a lattice and in a box $Omega$. In the course of the explanation you see how $vol(Omega)$ becomes $delta(0)$. This gets even worse when you consider scattering cross sections.
– Abdelmalek Abdesselam
Nov 16 at 15:36
1
I have read one can get rid off this problem by normalising your Hamiltonian by the Volume, i.e. considering an energy density?
– Thomas Wening
Nov 16 at 15:41
add a comment |
3
The expression $delta(x-x)=delta(0)$ doesn't make sense. I suspect the QFT argument which led to the first integral above is faulty. Note that $delta(0)$ typically is the volume of spacetime. So there must have been a careless infinite volume limit taken.
– Abdelmalek Abdesselam
Nov 16 at 15:19
Aha! So it is indeed undefined
– Thomas Wening
Nov 16 at 15:29
3
I remember there was a good treatment of this kind of issues in the old book by T.D. Lee amazon.com/Particle-Physics-Introduction-Contemporary-Concepts/… He introduced canonical quantization by working on a lattice and in a box $Omega$. In the course of the explanation you see how $vol(Omega)$ becomes $delta(0)$. This gets even worse when you consider scattering cross sections.
– Abdelmalek Abdesselam
Nov 16 at 15:36
1
I have read one can get rid off this problem by normalising your Hamiltonian by the Volume, i.e. considering an energy density?
– Thomas Wening
Nov 16 at 15:41
3
3
The expression $delta(x-x)=delta(0)$ doesn't make sense. I suspect the QFT argument which led to the first integral above is faulty. Note that $delta(0)$ typically is the volume of spacetime. So there must have been a careless infinite volume limit taken.
– Abdelmalek Abdesselam
Nov 16 at 15:19
The expression $delta(x-x)=delta(0)$ doesn't make sense. I suspect the QFT argument which led to the first integral above is faulty. Note that $delta(0)$ typically is the volume of spacetime. So there must have been a careless infinite volume limit taken.
– Abdelmalek Abdesselam
Nov 16 at 15:19
Aha! So it is indeed undefined
– Thomas Wening
Nov 16 at 15:29
Aha! So it is indeed undefined
– Thomas Wening
Nov 16 at 15:29
3
3
I remember there was a good treatment of this kind of issues in the old book by T.D. Lee amazon.com/Particle-Physics-Introduction-Contemporary-Concepts/… He introduced canonical quantization by working on a lattice and in a box $Omega$. In the course of the explanation you see how $vol(Omega)$ becomes $delta(0)$. This gets even worse when you consider scattering cross sections.
– Abdelmalek Abdesselam
Nov 16 at 15:36
I remember there was a good treatment of this kind of issues in the old book by T.D. Lee amazon.com/Particle-Physics-Introduction-Contemporary-Concepts/… He introduced canonical quantization by working on a lattice and in a box $Omega$. In the course of the explanation you see how $vol(Omega)$ becomes $delta(0)$. This gets even worse when you consider scattering cross sections.
– Abdelmalek Abdesselam
Nov 16 at 15:36
1
1
I have read one can get rid off this problem by normalising your Hamiltonian by the Volume, i.e. considering an energy density?
– Thomas Wening
Nov 16 at 15:41
I have read one can get rid off this problem by normalising your Hamiltonian by the Volume, i.e. considering an energy density?
– Thomas Wening
Nov 16 at 15:41
add a comment |
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3
The expression $delta(x-x)=delta(0)$ doesn't make sense. I suspect the QFT argument which led to the first integral above is faulty. Note that $delta(0)$ typically is the volume of spacetime. So there must have been a careless infinite volume limit taken.
– Abdelmalek Abdesselam
Nov 16 at 15:19
Aha! So it is indeed undefined
– Thomas Wening
Nov 16 at 15:29
3
I remember there was a good treatment of this kind of issues in the old book by T.D. Lee amazon.com/Particle-Physics-Introduction-Contemporary-Concepts/… He introduced canonical quantization by working on a lattice and in a box $Omega$. In the course of the explanation you see how $vol(Omega)$ becomes $delta(0)$. This gets even worse when you consider scattering cross sections.
– Abdelmalek Abdesselam
Nov 16 at 15:36
1
I have read one can get rid off this problem by normalising your Hamiltonian by the Volume, i.e. considering an energy density?
– Thomas Wening
Nov 16 at 15:41