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I hope this fits this section. This doesn't all fit into the title, but this comes from a homework on conformal field theory, and I am slightly stumped on it. I just can't seem to get anything sensible out of it at the end, but it may be because I've just done something wrong (even though I've calculated it several times, or that I just dont understand the question correctly.

We have a 2D field theory defined by the action:

[tex] S = \frac{1}{2\pi\alpha'} \int d^2z \partial\bar{\Phi}\partial\Phi [/tex]

where [itex]\partial = \frac{d}{dz}[/itex] is the holomorhpic derivative. We also have the identification:

[tex] \Phi \equiv \Phi + 2\pi R [/tex] (1)

Note that [itex] \Phi(z,\bar{z}) [/itex], where [itex]z[/itex] and [itex]\bar{z}[/itex] are treated as independent variables.

a) Find a general classical solution for [itex]\Phi[/itex] using the complex coordi-

nates [itex]z[/itex] and [itex]\bar{z}[/itex] that is consistent with (1).

The variation of the action should be 0:

[tex] \delta S = 0 [/tex]

If [itex]\Phi(\bar{z})[/itex] is real-valued on its restriction to the real numbers, then

[tex]\Phi(\bar{z}) = \bar{\Phi}(z) [/tex]

My attempt at a solution first of all finds the equations of motion for [itex]\Phi(z,\bar{z})[/itex] using that the variation of the action should be 0:

[tex] 0 = \delta S = \frac{1}{2\pi\alpha'} \int d^2z \delta(\partial\bar{\Phi}\partial\Phi) = \frac{1}{2\pi\alpha'} \int d^2z (\partial(\delta\bar{\Phi})\partial\Phi + \partial\bar{\Phi}\partial(\delta\Phi)) [/tex]

Then using partial integration, and realizing that the boundary terms are zero due to the definition of the variation, one gets:

[tex] \frac{1}{2\pi\alpha'} \int d^2z (\delta\bar{\Phi}\partial^2\Phi + \partial^2\bar{\Phi}\delta\Phi) = 0 [/tex]

From which follows that we get the equations of motion [itex]\partial^2\Phi = 0[/itex] (2) and [itex] \partial^2 \bar{\Phi} = 0 [/itex] (3). Solving (2) by integrating twice, we get (note that integration constants depend on [itex]\bar{z}[/itex]:

[tex] \Phi(z,\bar{z}) = z\alpha(\bar{z}) + \beta(\bar{z}) [/tex] (4)

and then we must have:

[tex] \bar{\Phi} = \bar{z}\bar{\alpha}(z) + \bar{\beta}(z) [/tex]

Now, here comes my problems. Using (3), we get:

[tex] \partial^2 \bar{\Phi} = \partial^2(\bar{z}\bar{\alpha}(z) + \bar{\beta}(z)) =0 [/tex]

Which once again after integrating twice gives a similar expression (the constants are just complex constants in ℂ this time):

[tex] \bar{z}\bar{\alpha}(z) + \bar{\beta}(z) =az+b [/tex]

Isolating \bar{\beta}(z) and taking the complex conjugate, we have:

[tex] \beta(\bar{z}) = \bar{a}\bar{z}+\bar{b} - z\alpha(\bar{z}) [/tex]

Which if we put this into our expression (4) for [itex] \Phi[/itex] gives:

[tex] \Phi(z,\bar{z}) = z\alpha(\bar{z}) + \beta(\bar{z}) = z\alpha(\bar{z}) + \bar{a}\bar{z}+\bar{b} - z\alpha(\bar{z}) = \bar{a}\bar{z}+\bar{b} [/tex]

Now, if I have done this correctly, I don't see how I am supposed to fit this with criterion (1) in the statement of the problem. As a matter of fact, I'm not sure I understand the criterion since it seems to just cancel the Phis on both sides to give [itex] 0 \equiv 2\pi R[/itex] which I am pretty sure is not the case. I am also later supposed to split [itex]\Phi[/itex] into a holomorphic and anti-holomorphic part, which I should be able to derive left- and right-momentum from. My final solution here seems like it only has an antiholomorphic part to begin with. This is where I am kind of stuck, and not sure if I've done this correctly or not.

--- Thanks to anyone that takes time to read this through!

## Homework Statement

We have a 2D field theory defined by the action:

[tex] S = \frac{1}{2\pi\alpha'} \int d^2z \partial\bar{\Phi}\partial\Phi [/tex]

where [itex]\partial = \frac{d}{dz}[/itex] is the holomorhpic derivative. We also have the identification:

[tex] \Phi \equiv \Phi + 2\pi R [/tex] (1)

Note that [itex] \Phi(z,\bar{z}) [/itex], where [itex]z[/itex] and [itex]\bar{z}[/itex] are treated as independent variables.

a) Find a general classical solution for [itex]\Phi[/itex] using the complex coordi-

nates [itex]z[/itex] and [itex]\bar{z}[/itex] that is consistent with (1).

## Homework Equations

The variation of the action should be 0:

[tex] \delta S = 0 [/tex]

If [itex]\Phi(\bar{z})[/itex] is real-valued on its restriction to the real numbers, then

[tex]\Phi(\bar{z}) = \bar{\Phi}(z) [/tex]

## The Attempt at a Solution

My attempt at a solution first of all finds the equations of motion for [itex]\Phi(z,\bar{z})[/itex] using that the variation of the action should be 0:

[tex] 0 = \delta S = \frac{1}{2\pi\alpha'} \int d^2z \delta(\partial\bar{\Phi}\partial\Phi) = \frac{1}{2\pi\alpha'} \int d^2z (\partial(\delta\bar{\Phi})\partial\Phi + \partial\bar{\Phi}\partial(\delta\Phi)) [/tex]

Then using partial integration, and realizing that the boundary terms are zero due to the definition of the variation, one gets:

[tex] \frac{1}{2\pi\alpha'} \int d^2z (\delta\bar{\Phi}\partial^2\Phi + \partial^2\bar{\Phi}\delta\Phi) = 0 [/tex]

From which follows that we get the equations of motion [itex]\partial^2\Phi = 0[/itex] (2) and [itex] \partial^2 \bar{\Phi} = 0 [/itex] (3). Solving (2) by integrating twice, we get (note that integration constants depend on [itex]\bar{z}[/itex]:

[tex] \Phi(z,\bar{z}) = z\alpha(\bar{z}) + \beta(\bar{z}) [/tex] (4)

and then we must have:

[tex] \bar{\Phi} = \bar{z}\bar{\alpha}(z) + \bar{\beta}(z) [/tex]

Now, here comes my problems. Using (3), we get:

[tex] \partial^2 \bar{\Phi} = \partial^2(\bar{z}\bar{\alpha}(z) + \bar{\beta}(z)) =0 [/tex]

Which once again after integrating twice gives a similar expression (the constants are just complex constants in ℂ this time):

[tex] \bar{z}\bar{\alpha}(z) + \bar{\beta}(z) =az+b [/tex]

Isolating \bar{\beta}(z) and taking the complex conjugate, we have:

[tex] \beta(\bar{z}) = \bar{a}\bar{z}+\bar{b} - z\alpha(\bar{z}) [/tex]

Which if we put this into our expression (4) for [itex] \Phi[/itex] gives:

[tex] \Phi(z,\bar{z}) = z\alpha(\bar{z}) + \beta(\bar{z}) = z\alpha(\bar{z}) + \bar{a}\bar{z}+\bar{b} - z\alpha(\bar{z}) = \bar{a}\bar{z}+\bar{b} [/tex]

Now, if I have done this correctly, I don't see how I am supposed to fit this with criterion (1) in the statement of the problem. As a matter of fact, I'm not sure I understand the criterion since it seems to just cancel the Phis on both sides to give [itex] 0 \equiv 2\pi R[/itex] which I am pretty sure is not the case. I am also later supposed to split [itex]\Phi[/itex] into a holomorphic and anti-holomorphic part, which I should be able to derive left- and right-momentum from. My final solution here seems like it only has an antiholomorphic part to begin with. This is where I am kind of stuck, and not sure if I've done this correctly or not.

--- Thanks to anyone that takes time to read this through!

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