Intuition behind Linked Cluster Theorem: connected vs. non-connected diagrams

Question

Within statistical physics and quantum field theory, the linked cluster theorem is widely used to simplify things in the calculation of the partition function among other things.

My question has the following parts:

1. Is there a physical reason to justify the theorem without delving into the math/diagrams?

2. Is there an intuitive reason the theorem holds in general graph theory?

3. Can you sketch out a simple proof for the theorem?

Answer 1

Linked-cluster theorem. Let $Z$ be the partition function and $W_c$ is the generating functional of connected diagrams. Let them be normalized such that for the free theory$^1$ $$ \left.Z\right|_{g=0,J=0}~=~1\qquad\text{and}\qquad\left.W_c\right|_{g=0,J=0}~=~0. \tag{1}$$ In other words, $1$ is the value of the empty diagram, which by definition is not connected. Then $$\ln Z~=~\frac{i}{\hbar} W_c. \tag{2}$$

Proof. We will use a replica trick, cf. Ref. 1.

1. Recall that if a theory consists of $n$ independent sub-theories with partition functions $Z_1, \ldots, Z_n$ (i.e. interactions are only allowed within each sub-theories), then the partition function for the full theory is the product $Z_1 \cdots Z_n$.

2. Introduce $n$ copies of the original theory under investigation, where $n\in\mathbb{N}$ is a positive integer. The replica partition function becomes just a power $$\sum\left\{\text{all replica diagrams}\right\} ~=~Z^n,\tag{3}$$ because different copies do not interact. Each field $\phi^{\alpha}_{(i)}(x)$ in the replica theory now carries a copy label $i\in\{1, \ldots, n\}$, and doesn't talk to other copies.

3. Given a Feynman diagram $D$ in the original theory, the contributions to the corresponding replica Feynman diagram should be multiplied with a factor $n^{\#(D)}$, where $\#(D)$ denotes the number of connected components of $D$. In other words, $$\begin{align} \sum & \left\{\text{all replica diagrams}\right\}\cr &~=~ 1 + n\sum\left\{\text{connected original diagrams}\right\} +{\cal O}(n^2).\end{align} \tag{4}$$ In eq. (4) we have used the normalization (1).

4. Possibly illuminating example. If an original diagram $D^2/2!$ consists of the same connected diagram $D$ twice, the corresponding replica contributions $$(\sum_{i=1}^n D_{(i)})^2/2!~=~ n^2 D^2/2!\tag{5}$$ scale as $n^2$. Here $2!$ is a symmetry factor. The fact that the corresponding diagonal replica contributions $$(\sum_{i=1}^n D_{(i)}^2)/2!~=~ n D^2/2!\tag{6}$$ only scale with a lower power $n$ is not relevant/important because it is (implicitly) assumed that the RHS of eq. (4) is organized according to original (rather than replica) diagrams. End of example.

5. Now let's continue the proof. Equivalent to eq. (4), by Taylor expansion, $$\begin{align} \ln\sum &\left\{\text{all replica diagrams}\right\}\cr ~\stackrel{(4)}{=}~& n\sum\left\{\text{connected original diagrams}\right\} +{\cal O}(n^2). \end{align} \tag{7}$$

6. Combining eqs. (3) & (7) yield $$\ln Z - \sum\left\{\text{connected original diagrams}\right\} ~\stackrel{(3)+(7)}{=}~{\cal O}(n^1) .\tag{8}$$ The LHS. of eq. (8) is independent of $n$, i.e. it is a constant wrt. $n$. But since the RHS. of eq. (8) has no ${\cal O}(n^0)$ terms, the constant must be zero. (Alternatively, we may formally treat the integer $n$ as a real number, and take the limit $n\to 0^{+}.$) This yields the linked-cluster theorem (2). $\Box$

See also this and this related Phys.SE posts.

References:

1. X.G. Wen, QFT of many-body systems, (2004); p. 143.

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$^1$NB: Conventionally in QFT, one allows for a multiplicative normalization factor in the partition function $Z$, which hence corresponds to an additive constant in $W_c$, cf. eq. (2).

Answer 2

1. No
2. Not to my knowledge
3. The idea is thus: diagrams represent transition amplitudes (complex numbers). The structure of the diagram tells you how to write down an integral. So a diagram might be an integral like $$\int f(x) dx.$$ Imagine if we have two copies of the same diagram. It's still a diagram. In that case we'll have something like $$\int f(x)f(y) dx dy = \int f(x)dx \int f(y)dy.$$ I've ignored symmetry factors.

On the other hand, if I were to connect the two disconnected copies with a line, then we'd have a single connected diagram, and the integral couldn't be separated into the product of two integrals, like above. We'd have $$\int f(x,y) dx dy.$$

Basically, a diagram with many disconnected components can be broken up into a product of integrals, which represent connected diagrams.

Since the rules of perturbation theory tell us to write down every possible diagram, we can factorise the disconnected diagrams into connected diagrams.

E.g. if I asked you to expand the expression $(a+b+c+...)^n$, then you'd write down every possible combination of $a,b,c,...$ containing $n$ terms and insert the correct binomial coefficients.

So then we could say the perturbation series is something like

$$\sum_n (n\text{ connected components}) = \sum_n \frac{1}{n!}(\text{sum of connected graphs})^n \\ = \exp(\text{sum of connected graphs})$$

The 1/n! prefactor basically comes out as the correct symmetry factor from the Feynman rules.

One can also separate the connected diagrams into 1PI diagrams and obtain Dyson's series, as well as a 1PI generating functional via a Legendre transformation.