2
$\begingroup$

The Dirac conjugate is defined as $$ \bar{\psi} = \psi^{\dagger}\gamma^0 $$ where $\psi$ is a spinor (4 dim for spin 1/2) that can describe a mixture (direct sum?) of fermion and anti-fermion states. And $\bar\psi$ is just another form of $\psi$ that can form an Lorentz invariant bilinear.

In other places, however, the bar notation is used for antiparticles, notably, $\bar{u}u$ $\bar{d}d$ to mark antiparticle particle pairs.

These two different ideas can be mixed sometimes, for example in Peskin's QFT Sec. 19.3, the lagrangian and currents used bar symbols to represent the Dirac conjugate,

$$\begin{aligned}j_L^\mu=\bar Q_L\gamma^\mu Q_L, \qquad &j_R^\mu=\bar Q_R\gamma^\mu Q_R, \qquad \\j_L^{\mu a}=\bar Q_L\gamma^\mu\tau^a Q_L, \qquad &j_R^{\mu a}=\bar Q_R\gamma^\mu\tau^a Q_R,\end{aligned}\tag{19.84}$$

while immediately after, it represents antiparticles:

as Fig. 19.5 shows, they must contain net chiral charge, pairing left-handed quarks with the antiparticles of right-handed quarks. The vacuum state with a quark pair condensate is characterized by a nonzero vacuum expectation value for the scalar operator $$ \langle 0|\bar QQ|0\rangle=\langle 0|\bar Q_LQ_R+\bar Q_rQ_L|0\rangle\ne 0,\tag{19.87} $$ [bold mine]

So, is there a connection in the concept of antiparticle and Dirac conjugate, and that the bar notation is an intentional choice? If so, how should I distinguish when spinor represents only particle or a superposition of both?

$\endgroup$

2 Answers 2

2
+50
$\begingroup$

The bar notation is necessary in order to make all of the quantities above Lorentz invariant. In short the relationship between the bar'ed quantitites and charge conjugates of fermions have to do with taking the hermitian conjugate of the creation / annihilation opeartors.

Let us focus for now on a free theory of Fermions. A Dirac Fermion can be written down as the following solution to the Dirac equation

$$ \psi(x) = \sum_{s \pm}\int\tilde{dp}\big[ a_s(\textbf{p})u_s(p)e^{ip\cdot x} + b_s^{\dagger}(\textbf{p})\nu_s(p)e^{-ip\cdot x}\big] \\ \bar{\psi}(x) = \sum_{s \pm}\int\tilde{dp}\big[ a^{\dagger}_s(\textbf{p})\bar{u}_s(p)e^{ip\cdot x} + b_s(\textbf{p})\bar{\nu}_s(p)e^{-ip\cdot x}\big] $$ where $\tilde{dp}$ is a Lorentz invariant measure and $u_s(p)$ and $\nu_s(p)$ are four-component spinors $$ u_s(p) = \begin{pmatrix} \sqrt{p\cdot \sigma} \;\xi^s \\ \sqrt{p\cdot \bar{\sigma}} \;\xi^s \end{pmatrix};\; \nu_s(p) =\begin{pmatrix} \sqrt{p\cdot \sigma} \;\eta^s \\ -\sqrt{p\cdot \bar{\sigma}} \;\eta^s \end{pmatrix};\; \overset{(-)}{\sigma_{\mu}} = (\mathbb{1}, \pm \vec{\sigma}) $$ corresponding to fermion and anti-fermion spinors. $a^{(\dagger)}_s(\textbf{p})$ and $b^{(\dagger)}_s(\textbf{p})$ are annihilation (creation) operators that can act on the Vacuum $$ a_s(\textbf{p})|0\rangle = b_s(\textbf{p})|0\rangle = 0 \\ a^{\dagger}_s(\textbf{p})|0\rangle = \frac{1}{\sqrt{2E_{\textbf{p}}}}|\textbf{p},s\rangle_{\text{ferm}};\; b^{\dagger}_s(\textbf{p})|0\rangle = \frac{1}{\sqrt{2E_{\textbf{p}}}}|\textbf{p},s\rangle_{\text{anti-ferm}}; $$ Charge conjugation is a symmetry that is manifest in some theories of phenomenalogical interest (and some that are not). Therefore Charge conjugation is an operator that commutes with that Hamiltonion and thus, is a quantum number for states in the Hilbert Space.

For a general state one has $\mathcal{C}|\phi\rangle = \eta_C|\phi^c\rangle$ for some field $\phi(x)$. Therefore, if we want to do a charge conjugation on some state, based off of how the creation operators act on the vacuum, what we want is $$ \mathcal{C}^{-1}a_s(\textbf{p})\mathcal{C} = b_s(\textbf{p}) $$ and vice-versa. As it turns out, the charge conjugate of the Dirac fermion field operator is $$ \mathcal{C}^{-1}\psi(x)\mathcal{C} = \psi^c(x) = \eta_cC\bar{\psi}^T(x). $$ Noting the following from Srednickie pg 245 $$ C\bar{u}_s(p)^T = \nu_s(p);\; C\bar{\nu}_s(p)^T = u_s(p) $$ one obtains $$ \psi^c(x) = \sum_{\pm}\int\tilde{dp}\big[ b_s(\textbf{p})u_s(p)e^{ip\cdot x} + a_s^{\dagger}(\textbf{p})\nu_s(p)e^{-ip\cdot x}\big]. $$ Now both $\bar{\psi}$ and $\psi^c$ are different field operators. But think about how both operators act on the vacuum for this free Fermion theory.

To me at least this is how I see it. When people talk about anti-quarks or something in expressions that look like $\bar{u}(x)\gamma^{\mu}u(x)$, I just thing of it as an "abuse of notation" to make physical sense about the quantity we are looking at. In a way they are speaking about the "bar'ed" quantities with respect to how the corresponding free Fermion operator acts on the vacuum.

The reliability of this physical reasoning lies on the reliability of a perturbative expansion of some interacting theory over free fields. Strictly speaking, in strongly coupled theories (or even in something like low-energy QED if you are purist), this kind of physical argument has no rigorous basis.

I hope I answered your question. Hell I hope I am understanding this right. This is just the way I have always understood it.

$\endgroup$
0
$\begingroup$

Confusion may be caused by not paying sufficient attention to the distinction between $\overline{ψ_L}$ and ${\overline{ψ}}_L$, or between $\overline{ψ_R}$ and ${\overline{ψ}}_R$. In fact, $$ \overline{ψ_L} = {ψ_L}^† η = \left(η ψ_L\right)^† = \left((η ψ)_R\right)^† = (η ψ)^†_R = \left(ψ^† η\right)_R = \overline{ψ}_R,\\ \overline{ψ_R} = {ψ_R}^† η = \left(η ψ_R\right)^† = \left((η ψ)_L\right)^† = (η ψ)^†_L = \left(ψ^† η\right)_L = \overline{ψ}_L, $$ where $η = η^†$ is the left-right switcher which - in the usual Dirac matrix algebra representations - is $γ^0$.

Let $\{|←⟩,|→⟩\}$ be the left-right basis, $\{|↑⟩,|↓⟩\}$ the up-down basis, and $\{⟨←|,⟨→|\}$ and $\{⟨↑|,⟨↓|\}$ their respective duals, and for the purposes of algebra, postulate that left-right elements commute with up-down elements. Then $$γ^0 = |←⟩⟨→| + |→⟩⟨←|,\quad γ^i = σ_i \left(|←⟩⟨→| - |→⟩⟨←|\right)\quad (i = 1, 2, 3),$$ where $$σ_1 = |↑⟩⟨↓| + |↓⟩⟨↑|,\quad σ_2 = i(|↓⟩⟨↑| - |↑⟩⟨↓|),\quad σ_3 = |↑⟩⟨↑| - |↓⟩⟨↓|.$$ This is the Weyl representation. It corresponds to the decomposition of the Dirac spinor into Weyl spinors as follows, using van der Waerden notation: $$ ψ = \left(\matrix{ψ_R\\ε ψ_L}\right) = \left(\matrix{ψ_0\\ψ_1\\ψ^\dot0\\ψ^\dot1}\right) = \left(ψ_0|↑⟩ + ψ_1|↓⟩\right)|→⟩ + \left(ψ^\dot0|↑⟩ + ψ^\dot1|↓⟩\right)|←⟩ $$ with indices raised by $$ε ψ_L = \left(\matrix{0&+1\\-1&0}\right)\left(\matrix{ψ_\dot0\\ψ_\dot1}\right) = \left(\matrix{+ψ_\dot1\\-ψ_\dot0}\right) = \left(\matrix{ψ^\dot0\\ψ^\dot1}\right).$$

The Dirac adjoint is given in terms of the left-right switcher $η = |←⟩⟨→| + |→⟩⟨←|$ by $$ \overline{ψ} = ψ^† η = ⟨→|\left(⟨↑|ψ^{\dot0*} + ⟨↓|ψ^{\dot1*}\right) + ⟨←|\left(⟨↑|ψ_0^* + ⟨↓|ψ_1^*\right), $$ where $(⋯)^*$ denotes complex conjugation.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.