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Heavy quark effective theory (HQET)

Last updated Aug 5, 2023

Heavy Quark Effective Theory (HQET) is an EFT for low energy QCD to describe hadrons containing one heavy quark (charm, bottom)

Bound states of contain one heavy quark $Q$ and many light d.o.f. (light quarks $q$ and gluons)

  1. Meson $Q\bar{q}$: $3\otimes\bar{3}=1\oplus 8$ - “Heavy-light meson”
  2. Barion $Qqq$: $3\otimes3\otimes 3=1\oplus …$

# Classification of Hadrons

Hadrons are classified by different properties

  1. Total angular momentum $j$ $$\vec{J} = \vec{S}_Q + \vec{S}_l$$ Sum of angular momentum of heavy quark $Q$ and light d.o.f. $l$
  1. Parity $P$
  1. Isospin $I_3$
  1. Electric charge

# Heavy-light mesons

If $s_l = s_q = \frac{1}{2}$ we have two possible $j$s: $j_+ = 1$ ($D^,B^$), $j_- = 0$ ($D,B$)

We can build the flavor $SU(3)$ diagram (Isosping-Strangeness) for $0^-$ (Pseudoscalar) and $1^-$ (Vector) states ($j^P$)

# Heavy-light baryons

# HQET model

Physical model: One heavy quark $Q$ ($c$ or $b$) surrounded by “cloud” of light d.o.f.:

  1. One light (anti)quark $q$ (for mesons) or one pair $qq$
  2. Light virtual pairs $q\bar{q}$
  3. gluons Only 1. gives the cloud its quantum numbers (spin, color, …)

The radius of the system is $R\approx \frac{1}{\Lambda_{QCD}}$. The energy $\Lambda_{QCD}$ is the order of the momentum and energy of the light d.o.f. The mass of the heavy quark $M$ $$M \gg \Lambda_{QCD}$$

# Heavy quark symmetry

For $m\to\infty$ the interaction of the heavy quark does not depend on flavor or spin. This leads to the heavy quark symmetry $$SU(2)\times U(N_h).$$

  1. $SU(2)$: Heavy quark Spin symmetry
  2. $U(N_h)$: Heavy quark flavor symmetry with the number of heavy quarks $N_h$ (usually c,b $N_h=2$ ?)

Heavy quark symmetry is only an approximate symmetry for HQET not for the QCD Lagrangian in general. It is broken by $\mathcal{O}(\frac{\Lambda_{QCD}}{m})$ and manifest at order $\mathcal{O}(\frac{1}{m^0})$.

# HQET construction

following the EFT recipe for $\mathcal{O}(\frac{1}{m^0})$ (static limit):

  1. energy scales: $m \gg \Lambda_{QCD}$
  2. degrees of freedom (d.o.f): Heavy quarks, light quarks, “soft” gluons, all momenta $\approx \Lambda_{QCD}$
  3. symmetries: Symmetries of QCD, Lorentz inv. not manifest, for $m\to \infty$ heavy quark symmetry
  4. Lagrangian: Consider hadron moving with velocity $v^\mu$, in the hadron rest frame $v^\mu=(1,0,0,0)$ The momentum of the heavy quark in the hadron $p^\mu = mv^\mu + k^\mu$ with $k^2\approx \Lambda_{QCD}$ due to interaction with gluons Using the momentum we can approximate the feynman rules by “expanding in $\frac{k^2}{m^2}\ll 1$”:

The Lagrangian that results in these rules is the QCD lagrangian with modified term for the heavy quark fields: $$\mathcal{L}=\bar{hv}(iv\cdot D)hv - \frac{1}{4}F^a_{\mu\nu}F^{\mu\nu a}+\sum^{h_l}_l\bar{q}_l(i\mathrlap{/}D - m_l)q_l + \mathcal{O}(\frac{1}{m}),$$ with $iD^\mu = i\partial^\mu - g A^\mu$ The heavy quark field $hv$ satisfies $\mathrlap{/}v hv = hv$ (why ? - because projection on hadron part ?) In the rest frame $\cancel{v} = \gamma_0$, so that from $hv$ only two components really contribute.

Alternative derivation of the HQET Lagrangian

In QCD for the heavy quark: $$\mathcal{L}_\text{heavy} = \bar{\psi}(i\cancel{D} - m)\psi$$ We can define the quark field $$\psi(x) = e^{-imvx}(h_v(x) + H_v(x))$$ with the fields $$h_v(x) = e^{imvx}\frac{1+\cancel{v}}{2}\psi(x) \quad H_v(x) = e^{imvx}\frac{1-\cancel{v}}{2}\psi(x)$$. With $\cancel{v}h_v = h_v$, $\cancel{v}H_v = -H_v$ and $\bar{h}vmH_v=0$ this leads to $$\mathcal{L}\text{heavy} = \bar{h}_v i v\cdot Dh_v + \mathcal{O}(\frac{1}{m}),$$ when applying the equation of motion of the field $H_v$: $(iv\cdot D + 2m)H_v = i(\cancel{D} - \cancel{v} v\cdot D)h_v$

The heavy quark symmetry is satisfied by the static limit HQET lagrangian:

  1. Heavy flavor symmetry: Satisfied since the lagrangian does not depend on the mass $m$
  2. Heavy quark spin symmetry: Satisfied as one can check when applying an infinitesimal transformation using the $SU(2)$ generator

Example: Spectroscopy implications

Heavy quark symmetry can make predictions for differences of hadron masses e.g. $m_{B^*}-m_B=\mathcal{O}(\frac{\Lambda^2_\text{QCD}}{m_b})$ etc..

# Antiquarks

In HQET pair creation is forbidden ($hv$ does not create a heavy antiquark). The Lagrangian for a heavy antiquark is obtained from $v \to -v$ since the decomposition of the momentum of the heavy antiquark is $p^\mu = -mv^\mu + k^\mu$. $$\mathcal{L} = \bar{h}{-v}(-ivD)h{-v}+…$$ where $h_{-v}$ is the antiquark field in HQET.

# States

While “normal” relativistic hadronic states $\ket{H(p)}$ have a relativistic normalization $$\braket{H(q)|H(p)}=2E_p(2\pi)^3\delta^3(\vec{p}-\vec{q}),$$ in HQET the hadron states are labeled by the four velocity of the hadron $v^\mu$ and the residual momentum $k^\mu$: $$\braket{H(v^\prime,k^\prime)|H(v,k)}=2v^0\delta_{vv^\prime}(2\pi)^3\delta^3(\vec{k}-\vec{k}^\prime)$$ These states have mass dimension $-\frac{3}{2}$.

# Heavy meson decay constants

Decay constants parametrize hadronic matrix elements:

Definition: Decay constant

The decay constant for a pseudoscalar heavy-light meson $f_p$ is defined by $$\braket{0|(\bar{q}\gamma^\mu\gamma_5 Q)|P(p)}=f_p p^\mu$$ Mass dimension: $[f_p]=1$

The decay constant for a vector heavy-light meson $f_{p^}$ is defined as $$\braket{0|(\bar{q}\gamma^\mu Q)|P^(p,\epsilon)}=f_{p^} \epsilon^\mu$$ Mass dimension: $[f_{p^}]=1$

These are the most general decomposition due to parity and the conservation of the vector current.

In HQET the decompositions becomes $$\braket{0|(\bar{q}\gamma^\mu\gamma_5h_v)|P(v)}=av^\mu$$ $$\braket{0|(\bar{q}\gamma^\mu h_v)|P^(v,\epsilon)}=\tilde a\epsilon^\mu$$ with mass dimension $[a] = [\tilde a] = \frac{3}{2}$. In the hadron rest frame $v^\mu=(1,0,0,0)$ and $\epsilon^\mu =(0,0,0,1)$, using $S^3_Q\ket{P(v)}=\frac{1}{2}\ket{P^(v,\epsilon)}$ we find

$$\boxed{a=\tilde a}$$

Since $\ket{P(p)}=\sqrt{m_p}\ket{P(v)}$ and $\ket{P^(p,\epsilon)}=\sqrt{m_{p^}}\ket{P(v,\epsilon)}$, we can make predictions for fractions of decay constants, e.g. $$f_P = \frac{a}{\sqrt{m_P}} \Rightarrow \frac{f_B}{f_D}=\sqrt{\frac{m_D}{m_B}}$$