Why is lattice QCD called non-perturbative?
$begingroup$
Like, if you are approximating a smooth structure with a discrete lattice, isn't this like a perturbation from smooth space-time?
If Feynman diagrams are a perturbative method, why are Feynamn diagrams on a lattice/grid called non-perturbative?
quantum-chromodynamics non-perturbative lattice-gauge-theory
edited Dec 25 '18 at 16:44
Qmechanic♦
103k121851173
asked Dec 22 '18 at 21:04
zoobyzooby
1,288514
$endgroup$
add a comment |
$begingroup$
Like, if you are approximating a smooth structure with a discrete lattice, isn't this like a perturbation from smooth space-time?
If Feynman diagrams are a perturbative method, why are Feynamn diagrams on a lattice/grid called non-perturbative?
quantum-chromodynamics non-perturbative lattice-gauge-theory
edited Dec 25 '18 at 16:44
Qmechanic♦
103k121851173
asked Dec 22 '18 at 21:04
zoobyzooby
1,288514
$endgroup$
$begingroup$
My mistake, I assumed lattice QCD was the same as QCD Feynman diagrams on a finite grid.
$endgroup$
– zooby
Dec 22 '18 at 21:35
add a comment |
$begingroup$
Like, if you are approximating a smooth structure with a discrete lattice, isn't this like a perturbation from smooth space-time?
If Feynman diagrams are a perturbative method, why are Feynamn diagrams on a lattice/grid called non-perturbative?
quantum-chromodynamics non-perturbative lattice-gauge-theory
edited Dec 25 '18 at 16:44
Qmechanic♦
103k121851173
asked Dec 22 '18 at 21:04
zoobyzooby
1,288514
$endgroup$
Like, if you are approximating a smooth structure with a discrete lattice, isn't this like a perturbation from smooth space-time?
If Feynman diagrams are a perturbative method, why are Feynamn diagrams on a lattice/grid called non-perturbative?
quantum-chromodynamics non-perturbative lattice-gauge-theory
quantum-chromodynamics non-perturbative lattice-gauge-theory
edited Dec 25 '18 at 16:44
Qmechanic♦
103k121851173
asked Dec 22 '18 at 21:04
zoobyzooby
1,288514
edited Dec 25 '18 at 16:44
Qmechanic♦
103k121851173
asked Dec 22 '18 at 21:04
zoobyzooby
1,288514
edited Dec 25 '18 at 16:44
Qmechanic♦
103k121851173
edited Dec 25 '18 at 16:44
Qmechanic♦
103k121851173
edited Dec 25 '18 at 16:44
Qmechanic♦
103k121851173
103k121851173
asked Dec 22 '18 at 21:04
zoobyzooby
1,288514
asked Dec 22 '18 at 21:04
zoobyzooby
1,288514
asked Dec 22 '18 at 21:04
zoobyzooby
1,288514
1,288514
$begingroup$
My mistake, I assumed lattice QCD was the same as QCD Feynman diagrams on a finite grid.
$endgroup$
– zooby
Dec 22 '18 at 21:35
add a comment |
$begingroup$
My mistake, I assumed lattice QCD was the same as QCD Feynman diagrams on a finite grid.
$endgroup$
– zooby
Dec 22 '18 at 21:35
$begingroup$
My mistake, I assumed lattice QCD was the same as QCD Feynman diagrams on a finite grid.
$endgroup$
– zooby
Dec 22 '18 at 21:35
$begingroup$
My mistake, I assumed lattice QCD was the same as QCD Feynman diagrams on a finite grid.
$endgroup$
– zooby
Dec 22 '18 at 21:35
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
In general, by a perturbative approach, we mean an approximation of the form,
$$f = f_0 + epsilon f_1 + epsilon^2 f_2 + dots$$
where $epsilon$ is the perturbation parameter, for some solution $f$. That is to say, one can approximate the behaviour of the solution by this series.
However, summing all the terms does not mean you will recover the exact solution; in most cases perturbative series are asymptotic series.
On the other hand, lattice QCD is an approach which is not described as perturbation theory because it does not follow this scheme, and in principle one recovers the exact solution in the appropriate limit.
To convince yourself of the distinction, consider the differential equation,
$$frac{mathrm d f}{mathrm dx} = g(x).$$
If we choose to discretize it (very naively), we obtain a linear system,
$$frac{f_{i+1}-f_{i}}{Delta x} = g_i$$
which is a totally different approach to plugging in a series expansion like the one above. That being said, Feynman diagrams are always a perturbative approach, so Feynman diagrams on a lattice are a perturbative approach, but putting the theory on a lattice is not what makes it perturbative.
answered Dec 22 '18 at 21:27
JamalSJamalS
14.2k53184
$endgroup$
$begingroup$
So in general it's just the kind of limit your taking.
$endgroup$
– zooby
Dec 22 '18 at 21:33
1
$begingroup$
@zooby: For a perturbative expansion, the limit does not exist. (This is what "asymptotic series" means.) For lattice QCD, as far as we can tell, the limit exists.
$endgroup$
– Peter Shor
Dec 22 '18 at 22:42
$begingroup$
@Peter do you know if string theory perturbation theory uses asymptotic expansions?
$endgroup$
– zooby
Dec 23 '18 at 0:03
1
$begingroup$
@zooby String perturbation theory replaces Feynman diagrams with a higher dimensional analogue, namely Riemann surfaces. It's still a perturbative expansion.
$endgroup$
– JamalS
Dec 23 '18 at 21:47
1
$begingroup$
zooby didn't ask whether string perturbation theory is perturbative (which is kind of obvious from the name), but whether it's asymptotic.
$endgroup$
– tparker
Dec 25 '18 at 18:07
|
show 1 more comment
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
In general, by a perturbative approach, we mean an approximation of the form,
$$f = f_0 + epsilon f_1 + epsilon^2 f_2 + dots$$
where $epsilon$ is the perturbation parameter, for some solution $f$. That is to say, one can approximate the behaviour of the solution by this series.
However, summing all the terms does not mean you will recover the exact solution; in most cases perturbative series are asymptotic series.
On the other hand, lattice QCD is an approach which is not described as perturbation theory because it does not follow this scheme, and in principle one recovers the exact solution in the appropriate limit.
To convince yourself of the distinction, consider the differential equation,
$$frac{mathrm d f}{mathrm dx} = g(x).$$
If we choose to discretize it (very naively), we obtain a linear system,
$$frac{f_{i+1}-f_{i}}{Delta x} = g_i$$
which is a totally different approach to plugging in a series expansion like the one above. That being said, Feynman diagrams are always a perturbative approach, so Feynman diagrams on a lattice are a perturbative approach, but putting the theory on a lattice is not what makes it perturbative.
answered Dec 22 '18 at 21:27
JamalSJamalS
14.2k53184
$endgroup$
$begingroup$
So in general it's just the kind of limit your taking.
$endgroup$
– zooby
Dec 22 '18 at 21:33
1
$begingroup$
@zooby: For a perturbative expansion, the limit does not exist. (This is what "asymptotic series" means.) For lattice QCD, as far as we can tell, the limit exists.
$endgroup$
– Peter Shor
Dec 22 '18 at 22:42
$begingroup$
@Peter do you know if string theory perturbation theory uses asymptotic expansions?
$endgroup$
– zooby
Dec 23 '18 at 0:03
1
$begingroup$
@zooby String perturbation theory replaces Feynman diagrams with a higher dimensional analogue, namely Riemann surfaces. It's still a perturbative expansion.
$endgroup$
– JamalS
Dec 23 '18 at 21:47
1
$begingroup$
zooby didn't ask whether string perturbation theory is perturbative (which is kind of obvious from the name), but whether it's asymptotic.
$endgroup$
– tparker
Dec 25 '18 at 18:07
|
show 1 more comment
$begingroup$
In general, by a perturbative approach, we mean an approximation of the form,
$$f = f_0 + epsilon f_1 + epsilon^2 f_2 + dots$$
where $epsilon$ is the perturbation parameter, for some solution $f$. That is to say, one can approximate the behaviour of the solution by this series.
However, summing all the terms does not mean you will recover the exact solution; in most cases perturbative series are asymptotic series.
On the other hand, lattice QCD is an approach which is not described as perturbation theory because it does not follow this scheme, and in principle one recovers the exact solution in the appropriate limit.
To convince yourself of the distinction, consider the differential equation,
$$frac{mathrm d f}{mathrm dx} = g(x).$$
If we choose to discretize it (very naively), we obtain a linear system,
$$frac{f_{i+1}-f_{i}}{Delta x} = g_i$$
which is a totally different approach to plugging in a series expansion like the one above. That being said, Feynman diagrams are always a perturbative approach, so Feynman diagrams on a lattice are a perturbative approach, but putting the theory on a lattice is not what makes it perturbative.
answered Dec 22 '18 at 21:27
JamalSJamalS
14.2k53184
$endgroup$
$begingroup$
So in general it's just the kind of limit your taking.
$endgroup$
– zooby
Dec 22 '18 at 21:33
1
$begingroup$
@zooby: For a perturbative expansion, the limit does not exist. (This is what "asymptotic series" means.) For lattice QCD, as far as we can tell, the limit exists.
$endgroup$
– Peter Shor
Dec 22 '18 at 22:42
$begingroup$
@Peter do you know if string theory perturbation theory uses asymptotic expansions?
$endgroup$
– zooby
Dec 23 '18 at 0:03
1
$begingroup$
@zooby String perturbation theory replaces Feynman diagrams with a higher dimensional analogue, namely Riemann surfaces. It's still a perturbative expansion.
$endgroup$
– JamalS
Dec 23 '18 at 21:47
1
$begingroup$
zooby didn't ask whether string perturbation theory is perturbative (which is kind of obvious from the name), but whether it's asymptotic.
$endgroup$
– tparker
Dec 25 '18 at 18:07
|
show 1 more comment
$begingroup$
In general, by a perturbative approach, we mean an approximation of the form,
$$f = f_0 + epsilon f_1 + epsilon^2 f_2 + dots$$
where $epsilon$ is the perturbation parameter, for some solution $f$. That is to say, one can approximate the behaviour of the solution by this series.
However, summing all the terms does not mean you will recover the exact solution; in most cases perturbative series are asymptotic series.
On the other hand, lattice QCD is an approach which is not described as perturbation theory because it does not follow this scheme, and in principle one recovers the exact solution in the appropriate limit.
To convince yourself of the distinction, consider the differential equation,
$$frac{mathrm d f}{mathrm dx} = g(x).$$
If we choose to discretize it (very naively), we obtain a linear system,
$$frac{f_{i+1}-f_{i}}{Delta x} = g_i$$
which is a totally different approach to plugging in a series expansion like the one above. That being said, Feynman diagrams are always a perturbative approach, so Feynman diagrams on a lattice are a perturbative approach, but putting the theory on a lattice is not what makes it perturbative.
answered Dec 22 '18 at 21:27
JamalSJamalS
14.2k53184
$endgroup$
In general, by a perturbative approach, we mean an approximation of the form,
$$f = f_0 + epsilon f_1 + epsilon^2 f_2 + dots$$
where $epsilon$ is the perturbation parameter, for some solution $f$. That is to say, one can approximate the behaviour of the solution by this series.
However, summing all the terms does not mean you will recover the exact solution; in most cases perturbative series are asymptotic series.
On the other hand, lattice QCD is an approach which is not described as perturbation theory because it does not follow this scheme, and in principle one recovers the exact solution in the appropriate limit.
To convince yourself of the distinction, consider the differential equation,
$$frac{mathrm d f}{mathrm dx} = g(x).$$
If we choose to discretize it (very naively), we obtain a linear system,
$$frac{f_{i+1}-f_{i}}{Delta x} = g_i$$
which is a totally different approach to plugging in a series expansion like the one above. That being said, Feynman diagrams are always a perturbative approach, so Feynman diagrams on a lattice are a perturbative approach, but putting the theory on a lattice is not what makes it perturbative.
answered Dec 22 '18 at 21:27
JamalSJamalS
14.2k53184
answered Dec 22 '18 at 21:27
JamalSJamalS
14.2k53184
answered Dec 22 '18 at 21:27
JamalSJamalS
14.2k53184
answered Dec 22 '18 at 21:27
JamalSJamalS
14.2k53184
14.2k53184
$begingroup$
So in general it's just the kind of limit your taking.
$endgroup$
– zooby
Dec 22 '18 at 21:33
1
$begingroup$
@zooby: For a perturbative expansion, the limit does not exist. (This is what "asymptotic series" means.) For lattice QCD, as far as we can tell, the limit exists.
$endgroup$
– Peter Shor
Dec 22 '18 at 22:42
$begingroup$
@Peter do you know if string theory perturbation theory uses asymptotic expansions?
$endgroup$
– zooby
Dec 23 '18 at 0:03
1
$begingroup$
@zooby String perturbation theory replaces Feynman diagrams with a higher dimensional analogue, namely Riemann surfaces. It's still a perturbative expansion.
$endgroup$
– JamalS
Dec 23 '18 at 21:47
1
$begingroup$
zooby didn't ask whether string perturbation theory is perturbative (which is kind of obvious from the name), but whether it's asymptotic.
$endgroup$
– tparker
Dec 25 '18 at 18:07
|
show 1 more comment
$begingroup$
So in general it's just the kind of limit your taking.
$endgroup$
– zooby
Dec 22 '18 at 21:33
1
$begingroup$
@zooby: For a perturbative expansion, the limit does not exist. (This is what "asymptotic series" means.) For lattice QCD, as far as we can tell, the limit exists.
$endgroup$
– Peter Shor
Dec 22 '18 at 22:42
$begingroup$
@Peter do you know if string theory perturbation theory uses asymptotic expansions?
$endgroup$
– zooby
Dec 23 '18 at 0:03
1
$begingroup$
@zooby String perturbation theory replaces Feynman diagrams with a higher dimensional analogue, namely Riemann surfaces. It's still a perturbative expansion.
$endgroup$
– JamalS
Dec 23 '18 at 21:47
1
$begingroup$
zooby didn't ask whether string perturbation theory is perturbative (which is kind of obvious from the name), but whether it's asymptotic.
$endgroup$
– tparker
Dec 25 '18 at 18:07
$begingroup$
So in general it's just the kind of limit your taking.
$endgroup$
– zooby
Dec 22 '18 at 21:33
$begingroup$
So in general it's just the kind of limit your taking.
$endgroup$
– zooby
Dec 22 '18 at 21:33
1
1
$begingroup$
@zooby: For a perturbative expansion, the limit does not exist. (This is what "asymptotic series" means.) For lattice QCD, as far as we can tell, the limit exists.
$endgroup$
– Peter Shor
Dec 22 '18 at 22:42
$begingroup$
@zooby: For a perturbative expansion, the limit does not exist. (This is what "asymptotic series" means.) For lattice QCD, as far as we can tell, the limit exists.
$endgroup$
– Peter Shor
Dec 22 '18 at 22:42
$begingroup$
@Peter do you know if string theory perturbation theory uses asymptotic expansions?
$endgroup$
– zooby
Dec 23 '18 at 0:03
$begingroup$
@Peter do you know if string theory perturbation theory uses asymptotic expansions?
$endgroup$
– zooby
Dec 23 '18 at 0:03
1
1
$begingroup$
@zooby String perturbation theory replaces Feynman diagrams with a higher dimensional analogue, namely Riemann surfaces. It's still a perturbative expansion.
$endgroup$
– JamalS
Dec 23 '18 at 21:47
$begingroup$
@zooby String perturbation theory replaces Feynman diagrams with a higher dimensional analogue, namely Riemann surfaces. It's still a perturbative expansion.
$endgroup$
– JamalS
Dec 23 '18 at 21:47
1
1
$begingroup$
zooby didn't ask whether string perturbation theory is perturbative (which is kind of obvious from the name), but whether it's asymptotic.
$endgroup$
– tparker
Dec 25 '18 at 18:07
$begingroup$
zooby didn't ask whether string perturbation theory is perturbative (which is kind of obvious from the name), but whether it's asymptotic.
$endgroup$
– tparker
Dec 25 '18 at 18:07
|
show 1 more comment
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$begingroup$
My mistake, I assumed lattice QCD was the same as QCD Feynman diagrams on a finite grid.
$endgroup$
– zooby
Dec 22 '18 at 21:35