RGApproachToInteractingFermions

Introduction

The renormalization group (RG) is a very powerful tool in physics. Essentially, it is a way to continuously map a given theory onto other theories possessing the same low-energy physics by successively integrating out "fast", or high-energy, modes. This is expressed in terms of differential equations giving the "flows" of different coupling constants that appear in the theory. We are often interested in determining the fixed points of these flows, or points which these flows may end at, since these give us some important information about the system. For example, each fixed point corresponds to a certain phase of the system. Since there are many different sets of initial parameters that all flow to the same fixed point, we have an explanation of universality, or the observation that many different systems all possess similar physical properties, such as critical exponents [1]. Here, we will discuss how to perform RG for fermions, which differs in some respects from the case with bosonic fields. This discussion will follow that in Shankar's paper [1]. The most important difference, one which is true of any fermionic system, is that we have a Fermi sea in the ground state of our system. This means that, unlike in the bosonic case, we cannot simply impose a cutoff on the momentum, restricting it to the inside of a sphere of radius Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Lambda} . We will illustrate the correct procedure as we go along for different cases.

System at half-filling in one dimension

We will begin with a simple system - Fermions on a one-dimensional tight-binding lattice at half filling. The Hamiltonian for such a system without interactions is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H_0=-t\sum_{\langle ij\rangle}\left (c^{\dagger}_{i+1} c_i+c^{\dagger}_i c_{i+1}\right ),}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \langle ij\rangle} means that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle i} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle j} are nearest neighbors, with Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle j>i} . We can diagonalize this Hamiltonian by performing a Fourier transform, upon which we find that the energies are Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \epsilon(k)=-2t\cos{K}} , where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\pi<K<\pi} . If we write the partition function for this Hamiltonian as a path integral, with the action written in momentum and frequency space, we get

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Z_0=\int{D[\psi^{\ast},\psi] e^{-S_0}},}

where

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_0=\int_{-\infty}^{\infty}\frac{d\omega}{2\pi}\int_{-\pi}^{\pi}\frac{dK}{2\pi}\,\psi^{\ast}(K,\omega)(-i\omega+\cos{K})\psi(K,\omega).}

Note that we write an integral over the frequency, rather than a Matsubara sum; we are working at zero temperature, and will do so throughout this article. We have absorbed the constant Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t} into the field and frequency integral to bring our notation into contact with that of Shankar [1].

The RG transformation

We will now discuss how to perform RG for this system. First, we will find an RG transformation that will leave the above "bare" action invariant. We first impose a cutoff on the momentum integral. As noted before, we cannot simply restrict its range to a small region Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\Lambda<K<\Lambda} due to the fact that the system's ground state is a filled Fermi sea. In this case, all states with Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\tfrac{\pi}{2}<K<\tfrac{\pi}{2}} are occupied. This means that our low-energy excitations are not excitations with small Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k} , but with Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k} close to the Fermi surface. Therefore, our cutoff should restrict Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle K} to small regions around Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \pm\tfrac{\pi}{2}} . Let us define a new momentum Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k=|K|-\tfrac{\pi}{2}} . Since there are two regions that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle K} is restricted to, we will introduce a new label to our fields, specifying whether the momentum is near the left (Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\tfrac{\pi}{2}} ) or the right (Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle +\tfrac{\pi}{2}} ) Fermi point. Since we are interested only in small Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k} , we will expand the energy to leading (in this case, linear) order. The action becomes

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_0=\sum_{i=L,R}\int_{-\infty}^{\infty}\frac{d\omega}{2\pi}\int_{-\Lambda}^{\Lambda}\frac{dk}{2\pi}\,\psi^{\ast}_i(k,\omega)(-i\omega+k)\psi_i(k,\omega),}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle i} labels the "branch" of the Fermi surface, and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Lambda} is our cutoff.

We are now ready to begin finding the appropriate RG transformation. This is done in three steps [1]. First, we split the fields into "slow" (Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \displaystyle \psi_{<}} ) and "fast" modes (Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \displaystyle \psi_{>}} ) and integrate out the fast modes. The slow modes are defined over a range Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0\leq |k|\leq\tfrac{\Lambda}{s}} , where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle s>1} , and the fast modes are defined over Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tfrac{\Lambda}{s}\leq |k|\leq\Lambda} . We can then write Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \displaystyle \psi_i(k,\omega)=\psi_{i,<}(k,\omega)+\psi_{i,>}(k,\omega)} . We will split the action into two parts - one containing only slow modes, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_{0,<}} and one containing only fast modes, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_{0,>}} . There are no terms mixing the two types of modes since they are defined on two disjoint intervals and both fields are at the same momentum. We can easily integrate out the fast modes since the integral is over a Gaussian, leaving just the slow modes. The result of the integration is just a constant, which we will drop. The action for the slow modes is thus

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_0=\sum_{i=L,R}\int_{-\infty}^{\infty}\frac{d\omega}{2\pi}\int_{-\Lambda/s}^{\Lambda/s}\frac{dk}{2\pi}\,\psi^{\ast}_{i,<}(k,\omega)(-i\omega+k)\psi_{i,<}(k,\omega).}

In order to compare this theory to other theories with the same cutoff, we must now rescale the momenta and frequencies to restore the momentum integral back to its original range [1]. To this end, let us define a new momentum Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k'=sk} and new frequency Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \omega'=s\omega} . Upon introducing these variables into the action, we obtain

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_0=\frac{1}{s^3}\sum_{i=L,R}\int_{-\infty}^{\infty}\frac{d\omega'}{2\pi}\int_{-\Lambda}^{\Lambda}\frac{dk'}{2\pi}\,\psi^{\ast}_{i,<}(\tfrac{k'}{s},\tfrac{\omega'}{s})(-i\omega'+k')\psi_{i,<}(\tfrac{k'}{s},\tfrac{\omega'}{s}).}

Finally, we rescale the fields so that the coefficient of some quadratic term in the action remains constant [1]. We define a new field Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi'_i(k',\omega')=\tfrac{1}{\zeta}\psi_{i,<}(\tfrac{k'}{s},\tfrac{\omega'}{s})} . We only have one quadratic term in our action, so we will leave its coefficient invariant. In terms of this new field, the action becomes

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_0=\frac{\zeta^2}{s^3}\sum_{i=L,R}\int_{-\infty}^{\infty}\frac{d\omega'}{2\pi}\int_{-\Lambda}^{\Lambda}\frac{dk'}{2\pi}\,{\psi'_i}^{\ast}(k',\omega')(-i\omega'+k')\psi'_i(k',\omega').}

If we define Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \zeta=\tfrac{1}{s^{3/2}}} , then we recover the original action in terms of rescaled variables.

We see that the three steps of the RG transformation we seek are: integrate out the fast modes, introduce rescaled momenta Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k'=sk} and frequencies Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \omega'=s\omega} , and introduce rescaled fields Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi'_i(k',\omega')=s^{3/2}\psi_{i,<}(\tfrac{k'}{s},\tfrac{\omega'}{s})} [1].

Quadratic perturbations

Now that we know the appropriate transformation, we may now look at how different perturbations scale under this transformation. Let us start with a quadratic perturbation,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta S_2=\sum_{i=L,R}\int_{-\infty}^{\infty}\frac{d\omega}{2\pi}\int_{-\Lambda}^{\Lambda}\frac{dk}{2\pi}\,\mu(k,\omega)\psi^{\ast}_i(k,\omega)\psi_i(k,\omega).}

Note that this perturbation preserves the symmetry between the two Fermi points. We now perform our RG transformation on this system. This separates directly into a piece depending only on the slow modes and one depending only on the fast modes, just as in the bare action. Therefore, when we integrate out the fast modes, we only generate a constant term in the action, which we drop. Performing the remaining two steps of the transformation, we obtain

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta S_2=s\sum_{i=L,R}\int_{-\infty}^{\infty}\frac{d\omega'}{2\pi}\int_{-\Lambda}^{\Lambda}\frac{dk'}{2\pi}\,\mu(\tfrac{k'}{s},\tfrac{\omega'}{s}){\psi'_i}^{\ast}(k',\omega')\psi'_i(k',\omega'),}

We now expand the coupling function Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mu} in a Taylor series [1],

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mu(k,\omega)=\mu_{00}+\mu_{10}k+\mu_{01}i\omega+\cdots+\mu_{mn}k^m(i\omega)^n+\cdots.}

We may now determine how each of these terms scales with Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle s} in the transformed action. In general, defining Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mu'(k',\omega')=s\mu(\tfrac{k'}{s},\tfrac{\omega'}{s})} ,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mu'_{mn}=\frac{1}{s^{m+n-1}}\mu_{mn}.}

We see that $\displaystyle \mu _{00}$ scales with a positive power of $s$, so that it grows as we increase $s$, and therefore integrate out more modes. Such a perturbation that grows under the RG transformation is called relevant [1]. As was guaranteed by our definition of the transformation, we find that $\displaystyle \mu _{10}$ and $\displaystyle \mu _{01}$ remain constant. Such perturbations are referred to as marginal [1]. Finally, if $m+n>1$, then $\displaystyle \mu _{mn}$ decreases under the RG transformation; these types of perturbations are referred to as irrelevant [1]. In all of our calculations, we will only retain terms in the action that are not irrelevant under the RG transformation.

Quartic perturbations

The scaling of quartic perturbations is not so trivial; in fact, these will lead to the flow equations for our system. The most general such perturbation, assuming a two-body interaction, is

$\delta S_{4}={\frac {1}{2!2!}}\int _{K\omega }u(4,3,2,1)\psi ^{\ast }(4)\psi ^{\ast }(3)\psi (2)\psi (1),$

where, following Shankar [1], we use the shorthand notation,

$\displaystyle \psi (i)=\psi (K_{i},\omega _{i})$

and

$\int _{K\omega }=\prod _{i=1}^{4}\left(\int _{-\pi }^{\pi }{\frac {dK_{i}}{2\pi }}\int _{-\infty }^{\infty }{\frac {d\omega }{2\pi }}\right)\times 2\pi {\bar {\delta }}(K_{1}+K_{2}-K_{3}-K_{4})\times 2\pi \delta (\omega _{1}+\omega _{2}-\omega _{3}-\omega _{4}),$

where ${\bar {\delta }}$ is a delta function that is infinite if its argument is zero modulo $2\pi$. Because the Grassman fields in the above perturbation are antisymmetric under the exchange of 1 and 2 or of 3 and 4, we require the same to be true of $u(4,3,2,1)$ [1]:

$\displaystyle u(4,3,2,1)=-u(4,3,1,2)=-u(3,4,2,1)=u(3,4,1,2)$

Even if $u$ had a symmetric part, it would be filtered out due to the antisymmetry under the exchange of the Grassman fields. Let us now rewrite this perturbation in terms of $k$ and impose a cutoff on the $k$ integrals. We get

$\delta S_{4}={\frac {1}{2!2!}}\sum _{i_{1},i_{2},i_{3},i_{4}=L,R}\int _{k\omega }^{\Lambda }u_{i_{4}i_{3}i_{2}i_{1}}(4,3,2,1)\psi _{i_{4}}^{\ast }(4)\psi _{i_{3}}^{\ast }(3)\psi _{i_{2}}(2)\psi _{i_{1}}(1),$

where, again following Shankar [1], we write

$\displaystyle \psi _{i_{j}}(j)=\psi _{i_{j}}(k_{j},\omega _{j})$

and

$\int _{k\omega }^{\Lambda }=\prod _{i=1}^{4}\left(\int _{-\Lambda }^{\Lambda }{\frac {dk_{i}}{2\pi }}\int _{-\infty }^{\infty }{\frac {d\omega }{2\pi }}\right)\times 2\pi {\bar {\delta }}(k_{1}+k_{2}-k_{3}-k_{4}+(\epsilon _{i_{1}}+\epsilon _{i_{2}}-\epsilon _{i_{3}}-\epsilon _{i_{4}}){\tfrac {\pi }{2}})\times 2\pi \delta (\omega _{1}+\omega _{2}-\omega _{3}-\omega _{4}),$

where $\epsilon _{i}$ is -1 if $i=L$ or +1 if $i=R$. Let us consider all the possible values of $\epsilon _{i_{1}}+\epsilon _{i_{2}}-\epsilon _{i_{3}}-\epsilon _{i_{4}}$. The only possible values of $\epsilon _{i_{1}}+\epsilon _{i_{2}}$ (and also Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \epsilon_{i_3}+\epsilon_{i_4}} ) are 2 (if both Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \epsilon} quantities are +1), 0 (if one is +1 and the other is -1), and -2 (if both are -1). This means that the only possible values of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \epsilon_{i_1}+\epsilon_{i_2}-\epsilon_{i_3}-\epsilon_{i_4}} are -4, -2, 0, +2, and +4. However, due to the fact that our Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k} variables are restricted to a region Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |k|\leq\Lambda} , with Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Lambda} much smaller than Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tfrac{\pi}{2}} , we will never satisfy the restriction imposed by the delta function if Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \epsilon_{i_1}+\epsilon_{i_2}-\epsilon_{i_3}-\epsilon_{i_4}=\pm 2} . This means that we can only have couplings of the forms Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u_{LRLR}} and all couplings related to it by the above antisymmetry condition, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u_{LLRR}} , or Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u_{RRLL}} . This also means that we can eliminate the term containing the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \epsilon} quantities, which will make performing the RG transformation much easier.

Let us now apply the RG transformation to this term. Upon separating the fields into slow and fast modes, we note that, unlike the bare action and the quadratic perturbations, we cannot simply separate the quartic perturbation into a term with all slow modes and one with all fast modes. This is because, unlike in the quadratic case, our fields are not always at the same momentum because each field's momentum is integrated over separately. This allows for terms that mix slow and fast modes. Because of this, integrating out the fast modes is not a trivial task. One typically employs the cumulant expansion,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \langle e^{-\delta S}\rangle_{0,>} = \exp\left [-\langle \delta S\rangle_{0,>} +\tfrac{1}{2}\left (\langle (\delta S)^2\rangle_{0,>} -\langle \delta S\rangle^2_{0,>}\right )+\cdots\right ],}

and performs the integration perturbatively. Let us first consider the term containing all slow modes; this is known as the tree-level term [1].

Rescaling of the tree-level term

After carrying out the first step of our RG transformation, the tree-level term becomes

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta S_4 = \frac{1}{2!2!}\sum_{i_1,i_2,i_3,i_4=L,R}\int_{k\omega}^{\Lambda/s}u(4,3,2,1)\psi^{\ast}_{i_4}(4)\psi^{\ast}_{i_3}(3)\psi_{i_2}(2)\psi_{i_1}(1).}

We introduce our rescaled variables as usual. We will pick up a factor of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (s^{3/2})^4=s^6} from rescaling the fields, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle s^{-8}} from the rescaling of the integration measures, and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle s^2} from the delta functions, since delta functions scale inversely with their arguments. Overall, this introduces no factors of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle s} , so that the new coupling function Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u'_{i_4 i_3 i_2 i_1}(k'_i,\omega'_i)} is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u'_{i_4 i_3 i_2 i_1}(k'_i,\omega'_i)=u_{i_4 i_3 i_2 i_1}(\tfrac{k'_i}{s},\tfrac{\omega'_i}{s}).}

If we expand the above in powers of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k'} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \omega'} , we will find that the constant term is marginal and all higher-order terms are irrelevant. The constant term has further restrictions on the Fermi point labels due to all Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \omega} variables being zero. To be exact, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u_{LLRR}} and $u_{RRLL}$ must have zero constant terms. All that remains are $u_{LRLR}$ and the couplings related to it by permutations of the labels. We will thus label all of these as $u_{LRLR}=u_{0}$, and we will combine all of these terms together as one:

$\delta S_{4}=u_{0}\int _{k\omega }^{\Lambda }\psi _{L}^{\ast }(4)\psi _{R}^{\ast }(3)\psi _{L}(2)\psi _{R}(1).$

One-loop RG

Now we turn our attention to terms containing fast modes. This is where the above-mentioned cumulant expansion is used [1]. Let us begin with the first-order term in the expansion. The terms containing an odd number of fast modes will be zero, and the term containing all fast modes will simply give a constant, which we ignore. The only terms that give us any interesting non-zero contribution are those containing only two fast modes. There are, in fact, only two such terms. They are

$u_{0}\int _{k\omega }^{\Lambda }\langle \psi _{L,<}^{\ast }(4)\psi _{R,>}^{\ast }(3)\psi _{L,<}(2)\psi _{R,>}(1)\rangle _{0,>}$

and a similar term with both "L" fields being fast modes and both "R" fields being slow modes. These terms would introduce corrections to the quadratic terms,

$\delta S_{2}=\mu \sum _{i=L,R}\int _{-\infty }^{\infty }{\frac {d\omega }{2\pi }}\int _{-\Lambda }^{\Lambda }{\frac {dk}{2\pi }}\,\psi _{i}^{\ast }(k,\omega )\psi _{i}(k,\omega ).$

Note that we dropped the subscript on $\mu$.

Our original action, however, did not include a term of this form, and thus we must introduce one in order for there to be a non-trivial fixed point [1]. Let us suppose that we did have such a term, and see how it would rescale with this contribution added. We may write the above average as

$-u_{0}\int _{k\omega }^{\Lambda }\langle \psi _{R,>}^{\ast }(3)\psi _{R,>}(1)\rangle _{0,>}\psi _{L,<}^{\ast }(4)\psi _{L,<}(2).$

We can pull out the slow modes from the average because it is an average over only the fast modes. The average over two fields that is left is just a Green's function:

$\langle \psi _{R,>}^{\ast }(3)\psi _{R,>}(1)\rangle _{0,>}=-2\pi \delta (\omega _{3}-\omega _{1})\times 2\pi \delta (k_{3}-k_{1})G_{R}(k_{1},\omega _{1})$

Substituting in this result and eliminating all delta functions, we obtain

$=u_{0}\int _{-\infty }^{\infty }{\frac {d\omega _{2}}{2\pi }}\int _{-\Lambda /s}^{\Lambda /s}{\frac {dk_{2}}{2\pi }}\int _{-\infty }^{\infty }{\frac {d\omega _{1}}{2\pi }}\int _{\Lambda /s<|k|<\Lambda }{\frac {dk_{1}}{2\pi }}\,G_{R}(k_{1},\omega _{1})\psi _{L,<}^{\ast }(k_{2},\omega _{2})\psi _{L,<}(k_{2},\omega _{2}).$

We rescale the slow fields and the associated momenta and frequencies, obtaining

$=su_{0}\int _{-\infty }^{\infty }{\frac {d\omega '_{2}}{2\pi }}\int _{-\Lambda }^{\Lambda }{\frac {dk'_{2}}{2\pi }}\int _{-\infty }^{\infty }{\frac {d\omega _{1}}{2\pi }}\int _{\Lambda /s<|k|<\Lambda }{\frac {dk_{1}}{2\pi }}\,G_{R}(k_{1},\omega _{1}){\psi '_{L}}^{\ast }(k'_{2},\omega '_{2})\psi '_{L}(k'_{2},\omega '_{2}).$

Let us define

$\delta \mu =u_{0}\int _{-\infty }^{\infty }{\frac {d\omega _{1}}{2\pi }}\int _{\Lambda /s<|k|<\Lambda }{\frac {dk_{1}}{2\pi }}\,G_{R}(k_{1},\omega _{1}).$

We know that, at tree level, $\mu$ rescales as $\mu \rightarrow s\mu$. Therefore, putting all the pieces together, we find that, at one loop, the new quadratic coupling $\mu '$ is

$\mu '=s(\mu +\delta \mu )=s\left(\mu -u_{0}\int _{-\infty }^{\infty }{\frac {d\omega }{2\pi }}\int _{\Lambda /s<|k|<\Lambda }{\frac {dk}{2\pi }}\,{\frac {1}{i\omega -k}}\right).$

Here, we used the fact that the Green's function is just

$G_{R}(k,\omega )={\frac {1}{-i\omega +k}}.$

We now evaluate the integrals, starting with the frequency integral. To this end, we will introduce a small convergence factor, $\displaystyle e^{i\omega \eta }$, where $\eta$ is a small positive number that we take to zero at the end of our calculation. The frequency integral becomes

$\int _{-\infty }^{\infty }{\frac {d\omega }{2\pi }}\,{\frac {e^{i\omega \eta }}{i\omega -k}}.$

We may now evaluate the integral as a contour integral:

$\oint _{C}{\frac {d\omega }{2\pi }}\,{\frac {e^{i\omega \eta }}{i\omega -k}},$

where the contour $C$ is a semicircle whose straight-line base is on the real axis and the arc is in the half-plane where $\omega$ has a positive imaginary part. We may enlarge this semicircle to infinite radius. The arc gives zero contribution to the integral by Jordan's lemma since it decreases exponentially as the imaginary part of $\omega$ increases. Therefore, only the line contributes. The integrand has a pole at $\omega =-ik$. If $k$ is positive, then the pole lies outside the contour, and the integral is zero. If, on the other hand, $k$ is negative, then the pole is inside the contour, and we get a non-zero integral. The residue of this pole is $-i$, so the integral is given by

$\oint _{C}{\frac {d\omega }{2\pi }}\,{\frac {e^{i\omega \eta }}{i\omega -k}}=\theta (-k).$

Substituting this back into the equation for $\mu '$, we obtain

$\mu '=s(\mu +\delta \mu )=s\left(\mu -u_{0}\int _{\Lambda /s<|k|<\Lambda }{\frac {dk}{2\pi }}\,\theta (-k)\right)=s\left(\mu -u_{0}\int _{-\Lambda }^{-\Lambda /s}{\frac {dk}{2\pi }}\right)=s\left[\mu -{\frac {\Lambda u_{0}}{2\pi }}\left(1-{\frac {1}{s}}\right)\right].$

We may now derive the flow equation for this coupling constant. If we rearrange this equation, we will obtain, in terms of $\delta \mu =\mu '-\mu$,

$\delta \mu =(s-1)\left(\mu -{\frac {\Lambda u_{0}}{2\pi }}\right).$

Now, let $s=1+t$, where $t$ is infinitesimal. Then

${\frac {d\mu }{dt}}=\mu -{\frac {\Lambda u_{0}}{2\pi }}.$

One comment is in order. Typically, we wish to work at a fixed particle density. This means that we must hold the chemical potential constant. However, the above flow equation would imply that, in fact, the chemical potential changes. To counter this, we must change the bare chemical potential in such a way that, upon adding interactions, we are returned to the original value of the chemical potential [1]. The fixed point of this flow tells us how large a "counter term" we will need to add to keep the particle density fixed [1]. Before we can calculate this fixed point, we would need to know the flow equation for $u_{0}$ as well. We will derive this equation in the next section, and it will turn out that $u_{0}$ does not flow [1]. That is, we have a whole family of fixed points as a function of $u_{0}$. Therefore, $\mu$ at the fixed point is

$\mu ={\frac {\Lambda u_{0}}{2\pi }},$

or the point at which ${\tfrac {d\mu }{dt}}$ is zero.

Now we turn our attention to the second-order term in the cumulant expansion. This term will generate the one-loop corrections to the coupling $u_{0}$. We will find it convenient to think, to some degree, in terms of Feynman diagrams. Let us first write down the general form of one of these second-order corrections:

$u_{0}\int _{k\omega (8765)}^{\Lambda }\int _{k\omega (4321)}^{\Lambda }\langle \psi _{i_{8}}^{\ast }(8)\psi _{i_{7}}^{\ast }(7)\psi _{i_{6}}(6)\psi _{i_{5}}(5)\psi _{i_{4}}^{\ast }(4)\psi _{i_{3}}^{\ast }(3)\psi _{i_{2}}(2)\psi _{i_{1}}(1)\rangle _{0,>}$

All possible combinations of fast and slow fields will appear in this integral. However, we will only be interested in those terms that contain four slow modes and four fast modes. The others either give a constant, a two-loop correction to the quadratic coupling, or contribute to higher-order couplings. We will now use diagrams to help us visualize the different terms that will appear.

References

[1] R. Shankar, Rev. Mod. Phys. 66, 129 (1994).