Calculating the Effective Action of a Scalar Field Theory

[latentcorpse]

The effective action Γ[ϕ] for a scalar field theory is a functional of an auxiliary field ϕ(x). Both
Γ and ϕ are defined in terms of the generating functional for connected graphs W[J] as

W[J] + \Gamma[\phi] = \int d^dx J \phi , \quad \frac{\delta}{\delta J(x)} W[J] = \phi(x)

Show

- \int d^dz G_2(x,z) \Gamma_2(z,y) = \delta^{(d)}(x-y)

where G_n(x_1 , \dots , x_n) = (-i)^{n-1} \frac{\delta}{\delta J(x_1)} \dots \frac{\delta}{\delta J(x_n)} W[J]
are the connected n point functions of the theory and
\Gamma_n(x_1 , \dots , x_n) = -i \frac{\delta}{\delta \phi(x_1)} \dots \frac{\delta}{\delta \phi(x_n)} \Gamma[\phi]

So far I have just substituted from the definitions to get
- \int d^dz G_2(x,z) \Gamma_2(z,y) = \int d^dz \frac{\delta}{\delta J(x)} \frac{\delta}{\delta J(z)} W[J] \frac{\delta}{\delta \phi(z)} \frac{\delta}{\delta \phi(y)} \Gamma[\phi]
which becomes
\int d^dz \frac{\delta}{\delta J(x)} \phi(y) \frac{\delta}{\delta \phi(z)} J(y)
But then I am lost...

 

[fzero]

So far I have just substituted from the definitions to get
- \int d^dz G_2(x,z) \Gamma_2(z,y) = \int d^dz \frac{\delta}{\delta J(x)} \frac{\delta}{\delta J(z)} W[J] \frac{\delta}{\delta \phi(z)} \frac{\delta}{\delta \phi(y)} \Gamma[\phi]
which becomes
\int d^dz \frac{\delta}{\delta J(x)} \phi(y) \frac{\delta}{\delta \phi(z)} J(y)
But then I am lost...

From the identity for \phi(z), you should have

\int d^dz \frac{\delta\phi(z)}{\delta J(x)} \frac{\delta J(y)}{\delta \phi(z)} .

 

[latentcorpse]

From the identity for \phi(z), you should have

\int d^dz \frac{\delta\phi(z)}{\delta J(x)} \frac{\delta J(y)}{\delta \phi(z)} .

yeah sry that's what i meant

now can i just cancel the \phi(z)'s or what? and how do i get rid of the integral?

 

[sgd37]

well yeah since from the functional chain rule

<br /> \frac{\delta J(y)}{\delta J(x)} = \int d^dz \frac{\delta\phi(z)}{\delta J(x)} \frac{\delta J(y)}{\delta \phi(z)} = \delta^{(d)}(x-y)<br />

 

[latentcorpse]

well yeah since from the functional chain rule

<br /> \frac{\delta J(y)}{\delta J(x)} = \int d^dz \frac{\delta\phi(z)}{\delta J(x)} \frac{\delta J(y)}{\delta \phi(z)} = \delta^{(d)}(x-y)<br />

Of course! Thanks. As for the last bit, I am asked to show that
G_3(x_1, x_2, x_3) = \int d^dy_1 d^dy_2 d^dy_3<br /> G_2(x_1, y_1)G_2(x_2, y_2)G_2(x_3, y_3) \Gamma_3(y_1, y_2, y_3)

So I substituted for the G's and the Gamma. All the i's cancelled. Then I used the functional chain rule and was left with

\frac{ \delta}{\delta J(x_1)}\frac{ \delta}{\delta J(x_2)}\frac{ \delta}{\delta J(x_3)} \Gamma[\phi]

Then the only constructive thing I could think to do was to sub for Gamma as follows:
\frac{ \delta}{\delta J(x_1)}\frac{ \delta}{\delta J(x_2)}\frac{ \delta}{\delta J(x_3)} \left( \int d^dx J \phi - W[J] \right)
So the 2nd term gives me exactly what I want but I don't know how to get rid of that first one?

 

[fzero]

Then the only constructive thing I could think to do was to sub for Gamma as follows:
\frac{ \delta}{\delta J(x_1)}\frac{ \delta}{\delta J(x_2)}\frac{ \delta}{\delta J(x_3)} \left( \int d^dx J \phi - W[J] \right)
So the 2nd term gives me exactly what I want but I don't know how to get rid of that first one?

You can compute the derivatives of the first term by noting that

<br /> \frac{\delta }{\delta J(x)} = \int d^dz \frac{\delta\phi(z)}{\delta J(x)} \frac{\delta }{\delta \phi(z)}

on functionals of \phi.

 

[latentcorpse]

You can compute the derivatives of the first term by noting that

<br /> \frac{\delta }{\delta J(x)} = \int d^dz \frac{\delta\phi(z)}{\delta J(x)} \frac{\delta }{\delta \phi(z)}

on functionals of \phi.

So we have

<br /> \frac{ \delta}{\delta J(x_1)}\frac{ \delta}{\delta J(x_2)}\frac{ \delta}{\delta J(x_3)} \left( \int d^dx J \phi \right)<br />
= \int d^dz_1 \int d^dz_2 \int d^dz_3 \frac{\delta\phi(z_1)}{\delta J(x_1)} \frac{\delta }{\delta \phi(z_1)} \frac{\delta\phi(z_2)}{\delta J(x_2)} \frac{\delta }{\delta \phi(z_2)} \frac{\delta\phi(z_3)}{\delta J(x_3)} \frac{\delta }{\delta \phi(z_3)} \int d^dx J \phi

But what are the J and the \phi functions of? Without knowing that surely I cannot do the derivatives? And can I just move those functional derivatives inside that last integral so they can act on the J \phi?

Thanks!

 

[fzero]

So we have

<br /> \frac{ \delta}{\delta J(x_1)}\frac{ \delta}{\delta J(x_2)}\frac{ \delta}{\delta J(x_3)} \left( \int d^dx J \phi \right)<br />
= \int d^dz_1 \int d^dz_2 \int d^dz_3 \frac{\delta\phi(z_1)}{\delta J(x_1)} \frac{\delta }{\delta \phi(z_1)} \frac{\delta\phi(z_2)}{\delta J(x_2)} \frac{\delta }{\delta \phi(z_2)} \frac{\delta\phi(z_3)}{\delta J(x_3)} \frac{\delta }{\delta \phi(z_3)} \int d^dx J \phi

This is a bad way to write things, since there's no reason to use the chain rule when the derivatives act on J.

But what are the J and the \phi functions of? Without knowing that surely I cannot do the derivatives? [/tex]

Those are functions of the integration variable x.

And can I just move those functional derivatives inside that last integral so they can act on the J \phi?

Yes.

 

[raviem]

How do i post new topic?

 

[latentcorpse]

This is a bad way to write things, since there's no reason to use the chain rule when the derivatives act on J.

I don't see how else to write it or what the problem is. However, I expanded it out and got:

\int d^dz_1 \int d^d z_2 \int d^dz_3 \frac{\delta \phi(z_1)}{\delta J(x_1)} \frac{\delta \phi(z_2)}{\delta J(x_2)} \frac{\delta \phi(z_3)}{\delta J(x_3)} \int d^dx \frac{\delta}{\delta \phi(z_1)} \frac{\delta}{\delta \phi(z_2)} ( J(x) \delta^{(d)}(x-z_3) )
=\int d^dz_1 \int d^d z_2 \int d^dz_3 \frac{\delta \phi(z_1)}{\delta J(x_1)} \frac{\delta \phi(z_2)}{\delta J(x_2)} \frac{\delta \phi(z_3)}{\delta J(x_3)} \frac{\delta}{\delta \phi(z_1)} \frac{\delta}{\delta \phi(z_2)} J(z_3)
=0 since \frac{\delta J(z_3)}{\delta \phi(z_1)} = \frac{\delta J(z_3)}{\delta \phi(z_2)}=0

How's that? If it's wrong, can you elaborate on what the problem was in the last post please. Thanks.

 

[fzero]

I don't see how else to write it or what the problem is. However, I expanded it out and got:

\int d^dz_1 \int d^d z_2 \int d^dz_3 \frac{\delta \phi(z_1)}{\delta J(x_1)} \frac{\delta \phi(z_2)}{\delta J(x_2)} \frac{\delta \phi(z_3)}{\delta J(x_3)} \int d^dx \frac{\delta}{\delta \phi(z_1)} \frac{\delta}{\delta \phi(z_2)} ( J(x) \delta^{(d)}(x-z_3) )
=\int d^dz_1 \int d^d z_2 \int d^dz_3 \frac{\delta \phi(z_1)}{\delta J(x_1)} \frac{\delta \phi(z_2)}{\delta J(x_2)} \frac{\delta \phi(z_3)}{\delta J(x_3)} \frac{\delta}{\delta \phi(z_1)} \frac{\delta}{\delta \phi(z_2)} J(z_3)
=0 since \frac{\delta J(z_3)}{\delta \phi(z_1)} = \frac{\delta J(z_3)}{\delta \phi(z_2)}=0

How's that? If it's wrong, can you elaborate on what the problem was in the last post please. Thanks.

But

\frac{\delta J(z_3)}{\delta \phi(z_1)} =0 (*)

is not compatible with

<br /> \frac{\delta J(y)}{\delta J(x)} = \int d^dz \frac{\delta\phi(z)}{\delta J(x)} \frac{\delta J(y)}{\delta \phi(z)} = \delta^{(d)}(x-y).<br />

Since we know the 2nd form is correct, the derivative (*) must not be zero.

What I'm telling you is that you need to use a sort of convective derivative here:

\frac{D}{DJ(x)} = \frac{\delta}{\delta J(x)} + \int d^dz \frac{\delta \phi(z) } {\delta J(x)} \frac{\delta}{\delta \phi(z)} .

For instance

\frac{D}{DJ(x)} ( F[J] G[\phi] ) = \frac{\delta F[J]}{\delta J(x)} G[\phi] + F[J] \int d^dz \frac{\delta \phi(z) } {\delta J(x)} \frac{\delta G[\phi]}{\delta \phi(z)}.

 

[fzero]

Disregard my last couple of posts, since I seem to be overcomplicating things.

The first equation you started with


<br /> W[J] + \Gamma[\phi] = \int d^dx J \phi , \quad \frac{\delta}{\delta J(x)} W[J] = \phi(x)<br />

is completely compatible with

\frac{\delta}{\delta J(z) } \left( W[J] + \Gamma[\phi] \right) = \phi(z),

which means that

\frac{\delta \Gamma[\phi] }{\delta J(z) } = 0

so that we should consider

\frac{\delta \phi(x) }{\delta J(z) } = 0.

So there's no reason to use the chain rule to compute derivatives here, since J,\phi are independent variables. You can go back to your post #5 and compute away.

 

[latentcorpse]

Disregard my last couple of posts, since I seem to be overcomplicating things.

The first equation you started with


<br /> W[J] + \Gamma[\phi] = \int d^dx J \phi , \quad \frac{\delta}{\delta J(x)} W[J] = \phi(x)<br />

is completely compatible with

\frac{\delta}{\delta J(z) } \left( W[J] + \Gamma[\phi] \right) = \phi(z),

which means that

\frac{\delta \Gamma[\phi] }{\delta J(z) } = 0

so that we should consider

\frac{\delta \phi(x) }{\delta J(z) } = 0.

So there's no reason to use the chain rule to compute derivatives here, since J,\phi are independent variables. You can go back to your post #5 and compute away.

In post 5 I had <br /> \frac{ \delta}{\delta J(x_1)}\frac{ \delta}{\delta J(x_2)}\frac{ \delta}{\delta J(x_3)} \Gamma[\phi]<br />
which will surely vanish since you have just shown to me that \frac{\delta \Gamma[\phi] }{\delta J(z) } = 0, right?

One other thing, how did you deduce at the end of your last post that \frac{\delta \phi(x) }{\delta J(z) } = 0?

 

[fzero]

Sorry, I think I have to take my last post back. One can't conclude that \delta \phi/\delta J =0 because we also have to take into account that

\frac{\delta \Gamma}{\delta \phi(x)} = J(x),

with a similar equation for W[J]. So we know that

\frac{\delta}{\delta J(y)} \frac{\delta \Gamma}{\delta \phi(x)} \neq 0

and you have to use the chain rule everywhere anyway.

By the way the equation that was confusing me originally works out properly under the chain rule. We had

<br /> W[J] + \Gamma[\phi] = \int d^dx J(x) \phi(x) , ~~~(*)<br />

so the derivative of the LHS with respect to J(z) is

\frac{\delta W[J]}{\delta J(z)} + \int d^dx \frac{\delta \phi(x)}{\delta J(z)}\frac{\delta \Gamma[\phi] }{\delta \phi(x)} = \phi(z) + \int d^dx \frac{\delta \phi(x)}{\delta J(z)} J(x),

where we used \delta \Gamma[\phi]/\delta \phi(x) = J(x), \delta W[J] /\delta J(x) = \phi(x). Whereas on the RHS of (*), we compute directly using the chain rule

\phi(z) + \int d^dx J(x) \frac{\delta\phi(z)}{\delta J(z)}.

This can be shown to be equal to the preceeding expression by integration by parts.I believe, but haven't shown that similar manipulations will lead to the correct result for the 3-point function. Your last post seems to imply that we couldn't derive that result if we ignored the partial derivatives in the chain rule.

Sorry for the confusion, but I think we were on the right track at the beginning.

 

[latentcorpse]

Sorry, I think I have to take my last post back. One can't conclude that \delta \phi/\delta J =0 because we also have to take into account that

\frac{\delta \Gamma}{\delta \phi(x)} = J(x),

with a similar equation for W[J]. So we know that

\frac{\delta}{\delta J(y)} \frac{\delta \Gamma}{\delta \phi(x)} \neq 0

and you have to use the chain rule everywhere anyway.

By the way the equation that was confusing me originally works out properly under the chain rule. We had

<br /> W[J] + \Gamma[\phi] = \int d^dx J(x) \phi(x) , ~~~(*)<br />

so the derivative of the LHS with respect to J(z) is

\frac{\delta W[J]}{\delta J(z)} + \int d^dx \frac{\delta \phi(x)}{\delta J(z)}\frac{\delta \Gamma[\phi] }{\delta \phi(x)} = \phi(z) + \int d^dx \frac{\delta \phi(x)}{\delta J(z)} J(x),

where we used \delta \Gamma[\phi]/\delta \phi(x) = J(x), \delta W[J] /\delta J(x) = \phi(x). Whereas on the RHS of (*), we compute directly using the chain rule

\phi(z) + \int d^dx J(x) \frac{\delta\phi(z)}{\delta J(z)}.

This can be shown to be equal to the preceeding expression by integration by parts.I believe, but haven't shown that similar manipulations will lead to the correct result for the 3-point function. Your last post seems to imply that we couldn't derive that result if we ignored the partial derivatives in the chain rule.

Sorry for the confusion, but I think we were on the right track at the beginning.

That's no problem. I had the right answer and the additional term \frac{\delta}{\delta J(x_1)}\frac{\delta}{\delta J(x_2)}\frac{\delta}{\delta J(x_3)} \int d^dx J(x) \phi(x) that I need to get rid of.

This becomes

= \frac{\delta}{\delta J(x_1)}\frac{\delta}{\delta J(x_2)} \left( \phi(x_3) + \int d^dx J(x) \frac{\delta \phi(x)}{\delta J(x_3)} \right)

Is this looking ok?

 

[fzero]

That's no problem. I had the right answer and the additional term \frac{\delta}{\delta J(x_1)}\frac{\delta}{\delta J(x_2)}\frac{\delta}{\delta J(x_3)} \int d^dx J(x) \phi(x) that I need to get rid of.

This becomes

= \frac{\delta}{\delta J(x_1)}\frac{\delta}{\delta J(x_2)} \left( \phi(x_3) + \int d^dx J(x) \frac{\delta \phi(x)}{\delta J(x_3)} \right)

Is this looking ok?

Yes.

Actually a few more things are coming to me. Since

\phi(x) = \frac{\delta W[J]}{\delta J(x)},

then

\frac{\delta\phi(x)}{\delta J(z)}= \frac{\delta^2 W[J]}{\delta J(z) \delta J(x)}.

Presumably there's a place to use this or a similar expression in your calculation.

 

[latentcorpse]

Yes.

Actually a few more things are coming to me. Since

\phi(x) = \frac{\delta W[J]}{\delta J(x)},

then

\frac{\delta\phi(x)}{\delta J(z)}= \frac{\delta^2 W[J]}{\delta J(z) \delta J(x)}.

Presumably there's a place to use this or a similar expression in your calculation.

Well I can't see how it will vanish though:

\frac{\delta}{\delta J(x_1)} \frac{\delta}{\delta J(x_1)} ( \phi(x_3) + \int d^dx J(x) \frac{ \delta^2 W[J]}{\delta J(x) \delta J(x_3)}
=\frac{\delta}{\delta J(x_1)} \left( \frac{\phi(x_3)}{\delta J(x_2)} + \int d^dx \delta^{(d)}(x-x_2) \frac{delta^2 W[J]}{\delta J(x) \delta J(x_3)} + \int d^dx J(x) \frac{\delta^3 W[J]}{\delta J(x) \delta J(x_2) \delta J(x_3)} \right)
=\frac{\delta}{\delta J(x_1)} \left( 2\frac{\phi(x_3)}{\delta J(x_2)} + \int d^dx J(x) \frac{\delta^3 W[J]}{\delta J(x) \delta J(x_2) \delta J(x_3)} \right)
= \frac{2 \delta^2 W[J]}{\delta J(x_1) \delta J(x_2) \delta J(x_3)} + \int d^dx \delta^{(d)}(x-x_1) \frac{\delta^3W[J]}{\delta J(x) \delta J(x_2) \delta J(x_3)} + \int d^d x J(x) \frac{\delta^4 W[J]}{\delta J(x) \delta J(x_1) \delta J(x_2) \delta J(x_3)}
= \frac{3 \delta^2 W[J]}{\delta J(x_1) \delta J(x_2) \delta J(x_3)} + \int d^d x J(x) \frac{\delta^4 W[J]}{\delta J(x) \delta J(x_1) \delta J(x_2) \delta J(x_3)}

 

[fzero]

Well I can't see how it will vanish though:

\frac{\delta}{\delta J(x_1)} \frac{\delta}{\delta J(x_1)} ( \phi(x_3) + \int d^dx J(x) \frac{ \delta^2 W[J]}{\delta J(x) \delta J(x_3)}
=\frac{\delta}{\delta J(x_1)} \left( \frac{\phi(x_3)}{\delta J(x_2)} + \int d^dx \delta^{(d)}(x-x_2) \frac{delta^2 W[J]}{\delta J(x) \delta J(x_3)} + \int d^dx J(x) \frac{\delta^3 W[J]}{\delta J(x) \delta J(x_2) \delta J(x_3)} \right)
=\frac{\delta}{\delta J(x_1)} \left( 2\frac{\phi(x_3)}{\delta J(x_2)} + \int d^dx J(x) \frac{\delta^3 W[J]}{\delta J(x) \delta J(x_2) \delta J(x_3)} \right)
= \frac{2 \delta^2 W[J]}{\delta J(x_1) \delta J(x_2) \delta J(x_3)} + \int d^dx \delta^{(d)}(x-x_1) \frac{\delta^3W[J]}{\delta J(x) \delta J(x_2) \delta J(x_3)} + \int d^d x J(x) \frac{\delta^4 W[J]}{\delta J(x) \delta J(x_1) \delta J(x_2) \delta J(x_3)}
= \frac{3 \delta^2 W[J]}{\delta J(x_1) \delta J(x_2) \delta J(x_3)} + \int d^d x J(x) \frac{\delta^4 W[J]}{\delta J(x) \delta J(x_1) \delta J(x_2) \delta J(x_3)}

That's a mess and I think it's partly due to trying to use the chain rule to claim that

<br /> \int d^dy_1 d^dy_2 d^dy_3<br /> G_2(x_1, y_1)G_2(x_2, y_2)G_2(x_3, y_3) \Gamma_3(y_1, y_2, y_3) = <br /> \frac{ \delta}{\delta J(x_1)}\frac{ \delta}{\delta J(x_2)}\frac{ \delta}{\delta J(x_3)} \Gamma[\phi].<br />

The problem is that you'll find that

\left[ \frac{\delta \phi(z)}{\delta J(x)} , \frac{\delta}{\delta \phi(y)} \right] \neq 0,

so you can't just commute the factors of G_2 around like you'd want. In any case, the substitution

W[J] + \Gamma[\phi] = \int d^dx J \phi

also adds complications.


It seems like the smallest amount of fuss would be deriving this from the original identity on 2-pt functions:

<br /> \frac{\delta}{\delta J(x_1)} \int d^dy_2 G_2(x_2,y_2) \Gamma_2(y_2,y_3) = 0.<br />

 

[latentcorpse]

That's a mess and I think it's partly due to trying to use the chain rule to claim that

<br /> \int d^dy_1 d^dy_2 d^dy_3<br /> G_2(x_1, y_1)G_2(x_2, y_2)G_2(x_3, y_3) \Gamma_3(y_1, y_2, y_3) = <br /> \frac{ \delta}{\delta J(x_1)}\frac{ \delta}{\delta J(x_2)}\frac{ \delta}{\delta J(x_3)} \Gamma[\phi].<br />

The problem is that you'll find that

\left[ \frac{\delta \phi(z)}{\delta J(x)} , \frac{\delta}{\delta \phi(y)} \right] \neq 0,

so you can't just commute the factors of G_2 around like you'd want. In any case, the substitution

W[J] + \Gamma[\phi] = \int d^dx J \phi

also adds complications.


It seems like the smallest amount of fuss would be deriving this from the original identity on 2-pt functions:

<br /> \frac{\delta}{\delta J(x_1)} \int d^dy_2 G_2(x_2,y_2) \Gamma_2(y_2,y_3) = 0.<br />

Erm sorry but where do I use this identity?

 

[fzero]

Acting with the derivative we get a sum of two terms, one involving G_3 and the other \Gamma_3. You can derive the stated identity without having to worry about any ordering of derivatives.