Non-crossing approximation (NCA) in 'Large-$N$ expansion' (Altland & Simons CMFT)

Question

I have a question about non-crossing approximation (NCA) stated in Condensed Matter Field Theory by Altland & Simons1. It is on page 200-203 for the 3rd edition (on page 223-227 for the 2nd edition). In short, I believe the expression for the first-order contribution of the self-energy operator in the book has two problems; one is quite important while the other might be unnecessary.

Below are the equation for the action (1) and for the self-energy (2) and the diagrammatic expression for each term; the first(second) term of the LHS of (2) for the first(second) diagram in the figure. $\phi = \{\phi^a\}$ is an N-component vector field with $a=1,...,N$. $G_{0,\textbf{p}}$ is the Fourier component of the free Green's function, $G^{ab}_0(x-y)\equiv\langle\phi^a(x)\phi^b(y)\rangle_0$.

$(1) \quad S[\phi] \equiv \int d^dx \,(\frac{1}{2}\partial\phi\cdot\partial\phi +\frac{r}{2}\phi\cdot\phi+\frac{g}{4N}(\phi\cdot\phi)^2)$
$(2) \quad\Sigma_{\textbf{p}}^{(1)}=-\delta^{ab}\frac{g}{L^d}(\frac{1}{N}\sum\limits_{\textbf{p}'}G_{0,\textbf{p}'}+\sum\limits_{\textbf{p}'}G_{0,\textbf{p-p}'})$

Here are the two points I doubt.

1. The $1/N$ coefficient should be in front of the second term of Eq(2) rather than the first one.

2. The Green's function in the second summation should be the same as the first one since $g$ does not carry any momentum.(Actually, I cannot understand why we should introduce such 'wavy-line' unless we adopt some auxiliary field[2] and write Eq(1) as $$S[\phi, A] = \int d^dx \, (\frac{1}{2}\partial\phi\cdot\partial\phi +\frac{r}{2}\phi\cdot\phi -\frac{N}{4g}A^2 +\frac{1}{2}A(\phi\cdot\phi)).$$ Integrating out for $A$ gives the same result with Eq(1). However, also in this case, the $A$ propagator does not carry any momentum due to the absence of a derivative term.)

To explain in more detail,

1. For $\phi$ loop, it contains index summation which gives the factor $N$. For example, for the first order of the Green function we should evaluate the term like (with Einstein's summation)

$\langle\phi^a(x)\phi^b(y)\int d^dz \,\, \phi^c(z)\phi^c(z)\phi^d(z)\phi^c(z)\,\rangle_0$.

The first diagram represents the case when the contraction occurs for (a,c), (b,c) & (d,d); (d,d) contraction gives the $N$ factor. On the other hand the second diagram is the case for (a,c), (b,d) & (c,d). That is why I believe the expression is wrong.

1. I know that $\sum\limits_{\textbf{p}'}G_{0,\textbf{p}'}=\sum\limits_{\textbf{p}'}G_{0,\textbf{p-p}'}$ for this case and I think that it could be just pedagogical expression; to write similarly as the case for Coulomb interaction where bosonic field propagator carries momentum. Is it right to think in this way?

$1/N$ coefficient is important in NCA since it decides which term to ignore under the limit of $N\rightarrow \inf$. Considering the book's context and specified term such as "rainbow diagrams", I know I am wrong. But I cannot find what's the problem. Please help me.

1 Altland A, Simons BD. Condensed Matter Field Theory. 3rd ed.

[2] Hooft, G. 'T ., 2002. LARGE N, in: .. https://doi.org/10.1142/9789812776914_0001

This post imported from StackExchange Physics at 2024-07-19 15:15 (UTC), posted by SE-user Jinu.P

Comments

The wavy line serves a useful notational purpose: O(N) is conserved on each side of the vertex. Any graphical notation that distinguishes between direct and exchange channels necessarily splits the 4-point vertex into two sides.

This post imported from StackExchange Physics at 2024-07-19 15:15 (UTC), posted by SE-user T.P. Ho

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Aha! I understand the point. But even if we distinguish such channels, doesn't that mean the wavy line carries momentum?

This post imported from StackExchange Physics at 2024-07-19 15:15 (UTC), posted by SE-user Jinu.P

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First of all, the diagrammatic notation is human-defined. As long as it is self-consistent, a wavy line can be made to represent anything. And then to answer your question: of course there is momentum exchange across the interaction vertex. You might say that the wavy "propagator" is a momentum-independent constant if that pleases you.

This post imported from StackExchange Physics at 2024-07-19 15:15 (UTC), posted by SE-user T.P. Ho

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In a previous answer of mine, I motivated the diagrammatic notation. Essentially, the large $N$ expansion can be seen as loop expansion in terms of the wavy lines, which makes the accounting simpler

This post imported from StackExchange Physics at 2024-07-19 15:15 (UTC), posted by SE-user LPZ

Answer 1

Let me just write down a proper answer. I will address the questions in reverse order.

The wavy line is a notational device that splits the four-point vertex into two sides. Given the interaction $$ \frac{g}{4N}\sum_{a,b}(\phi_a \phi_a) (\phi_b \phi_b), $$ it is obvious that the $O(N)$ index is separately conserved within each pair. It is diagrammatically useful to indicate the pairing; thus the wavy line that makes a distinction of two sides. If you think this resembles a propagating photon too much, feel free to change the notation to anything else that pleases you. The corresponding Feynman rule for the vertex

is $ \frac{g}{N} \delta_{ab}\delta_{cd} $ (up to sign and prefactor). There is obviously no momentum dependence, but momentum can "flow" across the wavy line: it is, after all, a glorified 4-point vertex that facilitates momentum exchange. But the wavy line does not "carry the O(N) charge", so to speak.

Now let's look at the one-loop self energy diagrams:

The first diagram is rearranged to make it resemble the second. Those $G_{0,p}$ appearing in $\Sigma_{p}^{(1)}$ in OP's question refer to the looping boson lines, not the wavy lines.

It should now be obvious that only the second diagram, that has a complete boson loop in this notation, contains a trace over O(N) flavors resulting in a factor of N. (I have no idea which term is which in OP's expression (2).)

A&S grew out of the lecture notes from Ben Simons' course. I was fortunate to attend those really great lectures. It's a pity that the book contains so many typos it's a pain to read...

This post imported from StackExchange Physics at 2024-07-19 15:15 (UTC), posted by SE-user T.P. Ho

Comments

So you are saying that the second diagram with a complete boson loop has factor a N, right? I also thought in that way and realized the book was telling me the first diagram contribution has a factor N and therefore becomes significant under $N\rightarrow \inf$ limit.

This post imported from StackExchange Physics at 2024-07-19 15:15 (UTC), posted by SE-user Jinu.P

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According to Kogan, Eugene. (2018). QUANTUM FIELD THEORY IN CONDENSED MATTER PHYSICS Course by E. Kogan.'s Fig.30, the Feynman rule for the vertex is defined as $\frac{g}{N}\delta_{ac}\delta_{bd}$, not $\frac{g}{N}\delta_{ab}\delta_{cd}$. Then the rainbow diagrams have a factor N so the book's description makes sense. However, I don't know why I have to define it that way; the way you did is more natural considering $O(N)$.

This post imported from StackExchange Physics at 2024-07-19 15:15 (UTC), posted by SE-user Jinu.P

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@Jinu.P Definition is man made; if you flip it the other way, then the N goes the other way. But honestly I have never seen it defined the other way until your claim here.

This post imported from StackExchange Physics at 2024-07-19 15:15 (UTC), posted by SE-user T.P. Ho

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I see. Thanks for your help.

This post imported from StackExchange Physics at 2024-07-19 15:15 (UTC), posted by SE-user Jinu.P