Quantum Field Theory: Understanding Path Integrals and Limit Trick

[latentcorpse]

You have a formula for $$Z[J]$$ so you can work out $$Z[0]$$...

Thanks. I've got it now. However, I also have to show that

$Z[J] = \displaystyle\sum_{n=0}^\infty \frac{1}{n!} \int d^dx_1 \dots d^dx_n \left( J(x_1) \dots J(x_n) \frac{ \delta^n Z[J]}{\delta J(x_1) \dots \delta J(x_n)} \right)_{J=0}$

This seems to be a Taylor expansion of some sort but I can't seem to derive it from the previous expression for $Z[J]$ - any advice?

 

[fzero]

What do you find for $$\delta Z[J]/\delta J(x_1)$$?

 

[latentcorpse]

What do you find for $$\delta Z[J]/\delta J(x_1)$$?

Not completely convinced by what I have here but, $\frac{\delta}{\delta J(x_1)}$ is a differential operator so it will act by chain rule so we get

$\frac{ \delta Z[J]}{\delta J(x_1)} = \int [ d \phi(x) ] e^{i S[\phi] + i \int d^d x J(x) \phi(x)} \frac{\delta J(x)}{\delta J(x_1)}$
$\frac{ \delta Z[J]}{\delta J(x_1)} = \int [ d \phi(x) ] e^{i S[\phi] + i \int d^d x J(x) \phi(x)} \delta^{(d)}(x-x_1)$

Is that right?

 

[fzero]

Not completely convinced by what I have here but, $\frac{\delta}{\delta J(x_1)}$ is a differential operator so it will act by chain rule so we get

$\frac{ \delta Z[J]}{\delta J(x_1)} = \int [ d \phi(x) ] e^{i S[\phi] + i \int d^d x J(x) \phi(x)} \frac{\delta J(x)}{\delta J(x_1)}$
$\frac{ \delta Z[J]}{\delta J(x_1)} = \int [ d \phi(x) ] e^{i S[\phi] + i \int d^d x J(x) \phi(x)} \delta^{(d)}(x-x_1)$

Is that right?

You haven't quite used the chain rule, since you didn't include the term coming from the derivative of

$$e^{i S[\phi] + i \int d^d x J(x) \phi(x)},$$

That's actually the most important part.

 

[latentcorpse]

You haven't quite used the chain rule, since you didn't include the term coming from the derivative of

$$e^{i S[\phi] + i \int d^d x J(x) \phi(x)},$$

That's actually the most important part.

Ah! Missed the integral out. So it should have read

$\frac{ \delta Z[J]}{\delta J(x_1)} = \int [ d \phi(x) ] e^{i S[\phi] + i \int d^d x J(x) \phi(x)} \times i \int d^dx \phi(x) \delta^{(d)}(x-x_1)=i \int [ d \phi(x) ] \phi(x_1) e^{i S[\phi] + i \int d^d x J(x) \phi(x)}$

So extrapolating,

$\frac{\delta^n Z[J]}{\delta J(x_1) \dots \delta J(x_n)} = i^n \int [d \phi(x) ] \phi(x_1) \dots \phi(x_n) e^{i S[\phi] + i \int d^d x J(x) \phi(x)}$

 

[latentcorpse]

You haven't quite used the chain rule, since you didn't include the term coming from the derivative of

$$e^{i S[\phi] + i \int d^d x J(x) \phi(x)},$$

That's actually the most important part.

So, despite having worked all this out for the more copmlicated cases where we have path integrals, I am stumped for the finite dimensional question below

if $Z[V] = \int d^Nx e^{-\frac{1}{2} \vec{x} \cdot A \vec{x} - V( \vec{x} )}$
where $V(0)=0$ and if $V_{i_1 \dots i_n} = \partial_i_1 \dots \partial_i_n V( \vec{x} )_{ \vec{x}=0}$ with $V_i=V_{ij}=0$, use the result
$G( \frac{\partial}{\partial b}) F(b)= F( \frac{\partial}{\partial u}) G(u) e^{ub}|_{u=0}$
to show that
$\frac{Z[V]}{Z[0]}= e^{\frac{1}{2} \frac{\partial}{\partial \vec{x}} \cdot A^{-1} \frac{\partial}{\partial \vec{x}}} e^{-V( \vec{x})}|_{\vec{x}=0}$

I can't figure out what to take as F and what to take as G or why?

 

[fzero]

So, despite having worked all this out for the more copmlicated cases where we have path integrals, I am stumped for the finite dimensional question below

if $Z[V] = \int d^Nx e^{-\frac{1}{2} \vec{x} \cdot A \vec{x} - V( \vec{x} )}$
where $V(0)=0$ and if $V_{i_1 \dots i_n} = \partial_i_1 \dots \partial_i_n V( \vec{x} )_{ \vec{x}=0}$ with $V_i=V_{ij}=0$, use the result
$G( \frac{\partial}{\partial b}) F(b)= F( \frac{\partial}{\partial u}) G(u) e^{ub}|_{u=0}$
to show that
$\frac{Z[V]}{Z[0]}= e^{\frac{1}{2} \frac{\partial}{\partial \vec{x}} \cdot A^{-1} \frac{\partial}{\partial \vec{x}}} e^{-V( \vec{x})}|_{\vec{x}=0}$

I can't figure out what to take as F and what to take as G or why?

As a preliminary result, you'll want to show that

$$\int d^Nx e^{-\frac{1}{2} \vec{x} \cdot A \vec{x} } \left(x_1^{n_1}\cdots x_N^{n_N} \right) \propto \left. \left( \frac{\partial^{n_1}}{\partial j_1^{n_1}} \cdots \frac{\partial^{n_N}}{\partial j_N^{n_N}} \right) e^{\frac{1}{2} \vec{j} \cdot A^{-1} \vec{j} } \right|_{\vec{j}=0}.$$

You do this by coupling sources $$\vec{j}$$ to $$\vec{x}$$ as in the scalar field theory a few posts back. As a result, we can write

$$\frac{Z[V]}{Z[0]} =\left. e^{V(\partial/\partial \vec{j})} e^{\frac{1}{2} \vec{j} \cdot A^{-1} \vec{j} }\right|_{\vec{j}=0}.$$

You're meant to use that change of variables result to simplify this expression, but I haven't worked out the details and probably missed something subtle above to make sure things work cleanly.

 

[latentcorpse]

As a preliminary result, you'll want to show that

$$\int d^Nx e^{-\frac{1}{2} \vec{x} \cdot A \vec{x} } \left(x_1^{n_1}\cdots x_N^{n_N} \right) \propto \left( \frac{\partial^{n_1}}{\partial j_1^{n_1}} \cdots \frac{\partial^{n_N}}{\partial j_N^{n_N}} \right) e^{\frac{1}{2} \vec{j} \cdot A^{-1} \vec{j} }.$$

You do this by coupling sources $$\vec{j}$$ to $$\vec{x}$$ as in the scalar field theory a few posts back. As a result, we can write

$$\frac{Z[V]}{Z[0]} = e^{V(\partial/\partial \vec{j})} e^{\frac{1}{2} \vec{j} \cdot A^{-1} \vec{j} }.$$

You're meant to use that change of variables result to simplify this expression, but I haven't worked out the details and probably missed something subtle above to make sure things work cleanly.

This is annoying me because I can see what you want me to do, I just don't know how to do it!

So when I couple a source won't I get

$Z[J]= \int d^Nx e^{-\frac{1}{2} \vec{x} \cdot A \vec{x} - V(x) + \vec{j} \cdot \vec{x}}$
but now I have this $Z[J]$ floating about and I only want to be working with $Z[V]$ and $Z[0]$

 

[fzero]

There's no need to couple a source inside $$Z[V]$$. The connection between the formulas I wrote is entirely made by expanding $$e^{-V(\vec{x})}$$ as a power series and recognizing the terms inside that power series as things that can be computed from $$Z[V=0,j]$$.

 

[latentcorpse]

There's no need to couple a source inside $$Z[V]$$. The connection between the formulas I wrote is entirely made by expanding $$e^{-V(\vec{x})}$$ as a power series and recognizing the terms inside that power series as things that can be computed from $$Z[V=0,j]$$.

So you mean

$Z[V, \vec{j} = e^{\vec{j} \cdot \vec{x}} \int d^N x e^{-\frac{1}{2} \vec{x} \cdot A \vec{x} - V( \vec{x} )}$
That's coupled the source outside of $Z[V]$ right?

 

[fzero]

So you mean

$Z[V, \vec{j} = e^{\vec{j} \cdot \vec{x}} \int d^N x e^{-\frac{1}{2} \vec{x} \cdot A \vec{x} - V( \vec{x} )}$
That's coupled the source outside of $Z[V]$ right?

No that doesn't make any sense because the LHS is independent of $$\vec{x}$$. What I means is that

$$\frac{1}{Z[0]} \int d^Nx e^{-\frac{1}{2} \vec{x} \cdot A \vec{x} } \left(x_1^{n_1}\cdots x_N^{n_N} \right) = \langle x_1^{n_1}\cdots x_N^{n_N} \rangle$$

and that $$Z[V]$$ can be expressed as a linear combination of these correlators.

 

[latentcorpse]

No that doesn't make any sense because the LHS is independent of $$\vec{x}$$. What I means is that

$$\frac{1}{Z[0]} \int d^Nx e^{-\frac{1}{2} \vec{x} \cdot A \vec{x} } \left(x_1^{n_1}\cdots x_N^{n_N} \right) = \langle x_1^{n_1}\cdots x_N^{n_N} \rangle$$

and that $$Z[V]$$ can be expressed as a linear combination of these correlators.

but that doesn't have any j's in it?

I might leave this for a few hours and hopefully when I come back to it I won't be going round in circles!

 

[fzero]

but that doesn't have any j's in it?

I might leave this for a few hours and hopefully when I come back to it I won't be going round in circles!

You should spend some time trying to make sense of the formulas in post #37. I corrected them to show that we're meant to set $$\vec{j}=0$$ after taking derivatives. Hopefully that clears up some confusion.

I've left out a few steps on purpose for you to fill in, since you've seen the necessary manipulations already. It's not really going to improve your understanding if I tell you how to do every single calculation.