# Z(J) and W(J)

1. Aug 14, 2010

### NanakiXIII

I apologize for the vague title, I don't know the names for the objects I'm asking about, which also made it hard to search for more information on them.

I'm reading Zee's QFT in a Nutshell and have the feeling I'm missing something. When introducing the path integral formalism, he defines a quantity

$$Z(J) = \langle 0 | e^{-i H T} | 0 \rangle$$

which, if I'm reading it right, should represent the amplitude of "propagating from vacuum to vacuum". He works out what $Z(J)$ looks like and eventually defines

$$Z(J) = e^{i W(J)}.$$

The exponential turns out to be

$$W(J) = -\frac{1}{2} \int \int d^4 x d^4 y J(x) D(x-y) J(y).$$

At first I sort of skipped over all of this, but a good understanding turns out to be important later on. My problem is that I don't fully understand what these quantities represent. What does this vacuum-to-vacuum propagation mean? What does that leave $W(J)$ to mean? Apparently $W(J)$ represents some kind of amplitude for a particle propagating from a disturbance at $x$ to a disturbance at $y$, and though that doesn't sound wrong, I don't understand why it is right either. Zee bases a lot of things on a sentence starting with "We see that $W(J)$ is only large when...", but why does it need to be large?

If anyone could get me on track with these things (and maybe provide some names), I would appreciate it. I'm just looking for the right way to interpret them.

2. Aug 26, 2010

### LAHLH

I recommend Srednicki for a thorough grounding of what these objects actually are, but basically:

One starts by looking at the transition amplitude for a particle to propagate from say q' at time t' to q'' at time t'' (working in 1 spatial dimension), which by the wonders of the path integral formulation turns out to be $$\langle q'',t''\mid q',t'\rangle =\int Dq \, e^{i\int dt L(\dot{q},q)}$$ (see CH6 first couple of pages in Srednicki to see this explicitley worked out).Also note here the reason there is no $$e^{-iHt}$$ sandwhiched is because Srednicki works in Heisenberg pic, as he explains at the start of ch6.

But more often than not we will be interested in initial and final states that are not just position eigenstates. So let's follow what Zee does on p12. So looking at $$\langle F\mid e^{-iHt} \mid I \rangle$$ and inserting complete sets:

$$\int dq'' \int dq' \langle F\mid q'' \rangle \langle q''\mid e^{-iHt}\mid q' \rangle \langle q' \mid I \rangle$$

Now writing this in Schroedinger picture:

$$\int dq'' \int dq' \Psi_{F}(q'')^{*} \langle q''\mid e^{-iHt}\mid q' \rangle \Psi_{I}(q')$$

If we take the I and F state to simply be the ground state then the object is $$\langle 0 \mid e^{-iHt} \mid 0 \rangle$$ then what the above represents is the amplitude for the particle to propagate from q' to q'' and still be in its ground state. This is the object we call Z(0). (Actually there is a slight subtlety here that Srednicki deals with on p47, by employing a trick that allows us to neglect the ground state wavefunctions and just focus on the $$\langle q''\mid e^{-iHt}\mid q' \rangle$$ bit. )

Now it's convenient to define Z(J), by adding a somewhat artificial $$J\phi$$ to the Lagrangian. This can be thought of as a source term, but it essentially lets us pull down factors of $$\phi$$ to form our correlation functions (these are important as we use a formula called the LSZ formula to find amplitudes later which are not just ground state to ground state, and this formula takes correlation functions as input essentially), e.g. $$Z(J)=\int D\phi e^{i\int d^4x[L+J\phi]}$$.

So that now $$\langle 0\mid T\phi(x_1).....\mid0\rangle=\tfrac{1}{i}\tfrac{\delta}{\delta J(x_1)}.....Z(J)\mid_{J=0}$$

(if you're wondering why the RHS of the above is equal to the LHS then it is because the correlation functions are given by
$$\langle 0\mid T\phi(x_1).....\mid0\rangle=\int D\phi \, \phi(x_1)..... e^{i\int d^4x[L+J\phi]}$$ as you can see by going through the same kind of derivation for the path integral at the start of ch6, but this time with a Q operator bunged in at some time in the middle)

The W(J) you have quoted is the value for free field theory, it's derived in ch8 Srednicki, but it's hard to explain what this really means until you get to interacting field theory (ch9 Srednicki). In interacting field theory W is the sum of all connected Feynman diagrams (not including certain types of diagram called vacuum diagrams). If you haven't read up on interacting field theories yet, this prob won't make much sense yet.

Last edited: Aug 26, 2010
3. Aug 26, 2010

### NanakiXIII

Thanks for your reply. Though I don't yet understand everything (I haven't really gone through canonical quantization yet), you cleared some things up I hadn't understood properly.

I've also been reading more since I asked this question, and I think I can ask a more pointed question now. If I understand you, $Z(J)$ is a generalized $Z(0)$. The $J$ can be thought of as sources. But does that mean $Z(0)$ represents all sourceless Feynman diagrams (i.e. quantum fluctuations)? Is this the right way to think about it?

Also, if I understand correctly, $Z(J)$ is the sum of all possible Feynman diagrams, i.e. all possible combinations of propagation and interaction and such, i.e. anything that can possibly happen. I'm guessing somewhere back there I took a wrong turn. It sounds to me like that should just add up to one.

P.S. I'm sorry I didn't go into much detail concerning all the math you took the time to write. I have seen some of it and understand what you're talking about to varying degrees, but what I'm mostly puzzled about for the moment is what $Z(J)$ means, physically.

4. Aug 26, 2010

### LAHLH

OK, $$Z(0)=\langle 0\mid 0 \rangle_{J=0}$$, is the case for no external source, if there is no source it means the system will remain in its ground state with certainty, and so we have the condition $$Z(0)=\langle 0\mid 0 \rangle_{J=0}=1$$. However if there is a source term, the system no longer stays in its ground state with 100% probability, so we need to calculate $$Z(J)=\langle 0\mid 0 \rangle_{J}$$ which turns out to be equal to $$Z(J)=\exp[\sum_{I} C_{I}]$$.

This sum is over all Feynman diagrams (strictly all connected Feynman diagrams, but that is a technicality here), as it stands this sum contains Feynamn diagrams with sources and those without (known as vacuum diagrams). Now we know that Z(0)=1 as discussed above, and thus looking at $$Z(0)=\exp[\sum_{I=\{0\}} C_{I}] =1$$, where now the I={0} indicates that we are only summing over sourceless (vacuum) diagrams, because all the diagrams with sources die when we set J=0. Therefore this part of the sum that forms the general Z(J) is just normalized to 1, i.e. $$Z(J)=\exp[\sum_{I=\{0\}} C_{I}]\exp[\sum_{I\neq\{0\}} C_{I}] =1\times\exp[\sum_{I\neq\{0\}} C_{I}]=\exp[\sum_{I\neq\{0\}} C_{I}]$$, now the $$I\neq\{0\}$$ indicates we are excluding vacuum diagrams.

At this point we define $$iW(J)=\sum_{I\neq\{0\}} C_{I}$$ so iW(J) is the sum of all connected Feynman diagrams with at least one source.

So to answer your question Z(0) is exponential of all sourcless (vacuum) diagrams, which is normalized to 1, since we know that without a source pushing the system, it will certainly remain is its g state. Z(J) is equal to the exponential of diagrams with at least one source on the other hand, and is not generally 1, since there is a chance the system won't end up in its g state.

So what is Z(J)? it is the ground state to ground state transition amplitude in the presence of an external source.

Perhaps someone else could explain what this is in more physical/conceptual terms, I'm not sure if the explaination I'm providing is the one you're looking for?

5. Aug 26, 2010

### NanakiXIII

Thanks for trying to explain. You say $Z(0)$ is normalized to 1. Is that then what is done for each power of $J$? Or is it just done to get rid of it in the definition of $Z(J)$?

You say $Z(J)$ is the amplitude for ground state to ground state transition (I hadn't understood that we were still doing ground state to ground state, or that there are even states at all) in the presence of an external source, but it is the sum of amplitudes for all possible numbers of sources, right? You can expand it in powers of $J$ and $\lambda$ (or $g$, as I think Srednicki calls it), where the power of $J$ indicates the number of sources and $\lambda$ the number of vertices, so $Z(J)$ contains all of these Feynman diagrams.

Maybe what I'm really asking is "why"? What do I need this sum over all possible diagrams for? When dealing with an actual process in particle physics or something, you'll be looking, I presume, at a select number of diagrams that correspond to what you want. What use is there in summing everything together?

6. Aug 26, 2010

### LAHLH

hmm, I'm not really sure what you mean by each power of J here? Z(0) is 1 simply because without a source the ground state will remain the ground state with 100% certainty.

Yes we are doing ground state to ground state (see Zee's paragraph at the bottom of p19) or Srednickis explaination in ch6 and 7. When we have an interacting field, and an interecting part of the Hamiltonian, you do a perturbation (ch9 Srednicki), it then turns out that Z(J) is the exp of the sum of all non vacuum connected diagrams as I said. Why is this the case? hmm, other than it just is, and the fact it follows mathematically( ch9 Srednicki) I struggle to give an explaination on that one, and maybe someone else has some insight. However, the reason why this quantity is useful is because if we know it, we can work out the correlation functions $$\langle 0\mid\phi(x_1)\phi(x_2)....\mid 0 \rangle$$ these turn out to be of great interest, as they appear in a formula known as the LSZ formula. What does the LSZ formula give us? well it allows us to compute $$\langle f \mid i \rangle$$ where now the initial and final states are multiparticle states.

These are the specific processes that you talk about, e.g. if we wanted to look at processes with 2 particles in and 2 out, we would need the so called 4 point correlation functions to plug into the LSZ formula $$\langle 0\mid\phi(x_1)\phi(x_2)phi(x_1')\phi(x_2')\mid 0 \rangle$$. We can get this four point correlation function to various orders in g (or lamba, or whatever you call your coupling constant) directly from Z(J) (by differentiating four times wrt the sources. Note that this picks out of all the diagrams in Z just the ones that have four sources, and then sets those sources to zero.....i.e. all those Feynman diagrams you have prob seen for 2 particle scattering)

Hope that helps, I'm away until Monday now won't be able to reply until then, but perhaps someone else will be able to offer some more insight if you have further questions, if not I will be back online next week. Srednicki 6-9 will give you lots of feel for this stuff, and make it more concrete.

7. Aug 26, 2010

### NanakiXIII

Right, I missed that.

Indeed I have seen how you use $Z(J)$ to get the Green's functions, etc. If $Z(J)$ is used mostly for these kinds of derivations, and it doesn't have any further physical use or meaning itself, that's a good enough answer to me. Thanks for your insights.

If anyone would like to share any other insights, they would be quite welcome, but I think I have a satisfying answer. I'm sure more details will become clear as I continue studying.