Mixing
Residue Calculus Integrable Model
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I'm currently working on the derivation of Bethe ansatz equations.
I have an equation like this:
$ z^-1prod_m=1,mneq l^Q-1fracq^2z-z_mz-z_m+q^-2z
prod_m=1,mneq l^Q-1fracq^-2z-z_mz-z_m-(z+z^-1)=E,$
where $zin mathbbC,qinmathbbC$ and $E$ is an energy eigenvalue.
Obviously, the poles exist at $z=0, z=infty , z=z_m$. Since the lhs is meromorphic function and the rhs is a constant, to make them equal, I have to calculate the residues and null them.
For $z=0$ the problem is solved. But I have problem with the two other cases.
I tried to calculate the residue at $z=infty$ and at $z=z_m$ via
$Res(f(z),z=infty)=-Res(-z^-2f(1/z),z=0)$ and $ Res(f(z),z=z_m)=lim_zto z_m(z-z_m)cdots$
The vanishing of the resiude at $z=z_m$ should give equations like this:
$z^2_l=q^Qprod_m=1,mneq l^Q-1 fracq^2 z_l -z_mz_l-q^2z_m$
But in both cases I failed. Please, can anybody help. I am grateful for any ideas.
complex-analysis residue-calculus
asked Jul 31 at 23:08
Herrmann Dirac-Neumann
1
add a comment |Â
up vote
0
down vote
favorite
I'm currently working on the derivation of Bethe ansatz equations.
I have an equation like this:
$ z^-1prod_m=1,mneq l^Q-1fracq^2z-z_mz-z_m+q^-2z
prod_m=1,mneq l^Q-1fracq^-2z-z_mz-z_m-(z+z^-1)=E,$
where $zin mathbbC,qinmathbbC$ and $E$ is an energy eigenvalue.
Obviously, the poles exist at $z=0, z=infty , z=z_m$. Since the lhs is meromorphic function and the rhs is a constant, to make them equal, I have to calculate the residues and null them.
For $z=0$ the problem is solved. But I have problem with the two other cases.
I tried to calculate the residue at $z=infty$ and at $z=z_m$ via
$Res(f(z),z=infty)=-Res(-z^-2f(1/z),z=0)$ and $ Res(f(z),z=z_m)=lim_zto z_m(z-z_m)cdots$
The vanishing of the resiude at $z=z_m$ should give equations like this:
$z^2_l=q^Qprod_m=1,mneq l^Q-1 fracq^2 z_l -z_mz_l-q^2z_m$
But in both cases I failed. Please, can anybody help. I am grateful for any ideas.
complex-analysis residue-calculus
asked Jul 31 at 23:08
Herrmann Dirac-Neumann
1
‘Calculate the residue and then null them’ what do you mean, practically?
– Szeto
Aug 1 at 0:55
I mean that the residues , e.g. at $z=z_m$ vanish if the parameters $lbrace z_l rbrace$ satisfy the above equation $z^2_l=cdots$. Does this help you?
– Herrmann Dirac-Neumann
Aug 1 at 9:49
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm currently working on the derivation of Bethe ansatz equations.
I have an equation like this:
$ z^-1prod_m=1,mneq l^Q-1fracq^2z-z_mz-z_m+q^-2z
prod_m=1,mneq l^Q-1fracq^-2z-z_mz-z_m-(z+z^-1)=E,$
where $zin mathbbC,qinmathbbC$ and $E$ is an energy eigenvalue.
Obviously, the poles exist at $z=0, z=infty , z=z_m$. Since the lhs is meromorphic function and the rhs is a constant, to make them equal, I have to calculate the residues and null them.
For $z=0$ the problem is solved. But I have problem with the two other cases.
I tried to calculate the residue at $z=infty$ and at $z=z_m$ via
$Res(f(z),z=infty)=-Res(-z^-2f(1/z),z=0)$ and $ Res(f(z),z=z_m)=lim_zto z_m(z-z_m)cdots$
The vanishing of the resiude at $z=z_m$ should give equations like this:
$z^2_l=q^Qprod_m=1,mneq l^Q-1 fracq^2 z_l -z_mz_l-q^2z_m$
But in both cases I failed. Please, can anybody help. I am grateful for any ideas.
complex-analysis residue-calculus
asked Jul 31 at 23:08
Herrmann Dirac-Neumann
1
I'm currently working on the derivation of Bethe ansatz equations.
I have an equation like this:
$ z^-1prod_m=1,mneq l^Q-1fracq^2z-z_mz-z_m+q^-2z
prod_m=1,mneq l^Q-1fracq^-2z-z_mz-z_m-(z+z^-1)=E,$
where $zin mathbbC,qinmathbbC$ and $E$ is an energy eigenvalue.
Obviously, the poles exist at $z=0, z=infty , z=z_m$. Since the lhs is meromorphic function and the rhs is a constant, to make them equal, I have to calculate the residues and null them.
For $z=0$ the problem is solved. But I have problem with the two other cases.
I tried to calculate the residue at $z=infty$ and at $z=z_m$ via
$Res(f(z),z=infty)=-Res(-z^-2f(1/z),z=0)$ and $ Res(f(z),z=z_m)=lim_zto z_m(z-z_m)cdots$
The vanishing of the resiude at $z=z_m$ should give equations like this:
$z^2_l=q^Qprod_m=1,mneq l^Q-1 fracq^2 z_l -z_mz_l-q^2z_m$
But in both cases I failed. Please, can anybody help. I am grateful for any ideas.
complex-analysis residue-calculus
asked Jul 31 at 23:08
Herrmann Dirac-Neumann
1
asked Jul 31 at 23:08
Herrmann Dirac-Neumann
1
asked Jul 31 at 23:08
Herrmann Dirac-Neumann
1
asked Jul 31 at 23:08
Herrmann Dirac-Neumann
1
1
‘Calculate the residue and then null them’ what do you mean, practically?
– Szeto
Aug 1 at 0:55
I mean that the residues , e.g. at $z=z_m$ vanish if the parameters $lbrace z_l rbrace$ satisfy the above equation $z^2_l=cdots$. Does this help you?
– Herrmann Dirac-Neumann
Aug 1 at 9:49
add a comment |Â
‘Calculate the residue and then null them’ what do you mean, practically?
– Szeto
Aug 1 at 0:55
I mean that the residues , e.g. at $z=z_m$ vanish if the parameters $lbrace z_l rbrace$ satisfy the above equation $z^2_l=cdots$. Does this help you?
– Herrmann Dirac-Neumann
Aug 1 at 9:49
‘Calculate the residue and then null them’ what do you mean, practically?
– Szeto
Aug 1 at 0:55
‘Calculate the residue and then null them’ what do you mean, practically?
– Szeto
Aug 1 at 0:55
I mean that the residues , e.g. at $z=z_m$ vanish if the parameters $lbrace z_l rbrace$ satisfy the above equation $z^2_l=cdots$. Does this help you?
– Herrmann Dirac-Neumann
Aug 1 at 9:49
I mean that the residues , e.g. at $z=z_m$ vanish if the parameters $lbrace z_l rbrace$ satisfy the above equation $z^2_l=cdots$. Does this help you?
– Herrmann Dirac-Neumann
Aug 1 at 9:49
add a comment |Â
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‘Calculate the residue and then null them’ what do you mean, practically?
– Szeto
Aug 1 at 0:55
I mean that the residues , e.g. at $z=z_m$ vanish if the parameters $lbrace z_l rbrace$ satisfy the above equation $z^2_l=cdots$. Does this help you?
– Herrmann Dirac-Neumann
Aug 1 at 9:49