Why is lattice QCD called non-perturbative?












2












$begingroup$


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







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$endgroup$












  • $begingroup$
    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
















2












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














2












2








2


1



$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 25 '18 at 16:44









Qmechanic♦

103k121851173




103k121851173










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


















  • $begingroup$
    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










1 Answer
1






active

oldest

votes


















6












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









6












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















6












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














6












6








6





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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


















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