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.