# Quantum field theory 1, lecture 20

Third, the insertion of this commutation relation yields \begin{equation*} \delta (\bar{\psi }\gamma ^\rho \psi )= -\tfrac{i}{2}\bar{\psi }\epsilon _{\mu \nu }(-i)(\eta ^{\nu \rho }\gamma ^\mu -\eta ^{\mu \rho }\gamma ^\nu )\psi =-\tfrac{1}{2} \bar{\psi }(\epsilon _\mu ^{~\rho }\gamma ^\mu -\epsilon ^{\rho }_{~\nu }\gamma ^{\nu })\psi = \epsilon ^{\rho }_{~\nu }\bar{\psi }\gamma ^\nu \psi . \end{equation*} Since we also know the transformation properties of $$\partial _\rho ,$$ we can easily check that $$\bar{\psi }\gamma ^\rho \partial _{\rho }\psi$$ transforms as a scalar field, \begin{equation*} \delta (\bar{\psi }\gamma ^\rho \partial _\rho \psi ) =\epsilon ^\rho _{~\nu }\bar{\psi } \gamma ^\nu \partial _\rho \psi + \bar{\psi }\gamma ^\rho \epsilon _\rho ^{\;\nu }\partial _\nu \psi =\epsilon_{\rho \nu }\bar{\psi }\gamma ^\nu \partial ^\rho \psi +\epsilon_{\nu\rho}\bar{\psi }\gamma ^\nu \partial ^\rho \psi = 0. \end{equation*}

Electrons with mass $$m$$. Free electrons are massive particles. A mass term in the action involves a fermion bilinear without derivatives that transforms as a scalar field. We have already established that $$\bar{\psi }\psi$$ has these properties. Extending the action by a mass term the action for free electrons involves \begin{equation*} \mathscr{L} = i\bar{\psi }\gamma ^\mu \partial _\mu \psi + im \bar{\psi }\psi . \end{equation*} We will see that this Lorentz invariant action describes a quantum field theory for free electrons and positrons.

#### 10.3 Dirac equation.

Dirac Equation. The functional variation of the associated action $$S$$ with regard to $$\bar{\psi }$$ leads to the famous Dirac equation \begin{equation} \frac{\delta S}{\delta \bar{\psi }} = 0 \Rightarrow (\gamma ^\mu \partial _\mu +m)\psi = 0. \label{eq:dirac} \end{equation} Since $$\mathscr{L}$$ is invariant this is a covariant equation. For a single particle state, this is also the Schrödinger equation, with $$\psi$$ interpreted as a wave function. In this case $$\psi$$ is a complex function (not a Grassmann variable). This interpretation of the Dirac equation as a relativistic Schrödinger equation for a one particle state does not hold in the presence interactions. A generalisation by adding an external electromagnetic is possible, however. In the following we will first concentrate on the interpretation of the Dirac equation as a relativistic Schrödinger equation. The physical properties found there will be useful once we later turn to the interacting quantum field theory for electromagnetism.

Energy-Momentum Relation. A free electron should obey the energy momentum relation for a relativistic particle. This can be established if we square the Dirac equation \begin{equation*} \gamma ^\nu \partial _\nu \gamma ^\mu \partial _\mu \psi = m^2\psi . \end{equation*} Using the anticommutator relation for the $$\gamma$$ matrices \begin{equation*} \tfrac{1}{2}\{\gamma ^\nu ,\gamma ^\mu \}\partial _\nu \partial _\mu \psi = \eta ^{\nu \mu }\partial _\nu \partial _\mu \psi =\partial ^\mu \partial _\mu \psi = m^2\psi , \end{equation*} we find the Klein-Gordon equation $$(\partial ^\mu \partial _\mu -m^2) \psi =0$$. All solutions of the Dirac equation (??) have to solve this equation. The general solution of the Klein-Gordon equation is a superposition of plane waves $$\psi = \psi _0\;e^{ip_\mu x^\mu }=\psi _0\;e^{-i(Et-\mathbf{px})}$$. This implies indeed the relativistic energy-momentum relation, \begin{equation*} (E^2 - \textbf{p}^2 -m^2)\psi = 0 \Rightarrow E^2 = \textbf{p}^2 + m^2. \end{equation*} We observe the existence of solutions for both signs of the energy, $$E=\pm \sqrt{p^2+m^2}$$ and we have to interpret the meaning of the solution with negative energy.

Hamiltonian Formulation. In order to obtain the usual form of the Schrödinger equation we multiply eq. (??) with $$-i\gamma ^0$$, \begin{equation*} -i\gamma ^0\gamma ^\mu \partial _\mu \psi = -i(\gamma ^0)^2 \partial _0\psi - i\gamma ^0\gamma ^k\partial _k\psi = i\gamma ^0 m\psi , \end{equation*} and introduce \begin{equation*} \alpha ^k = -\gamma ^0\gamma ^k = \gamma ^k\gamma ^0 =\begin{pmatrix} -\tau _k & 0 \\ 0 & \tau _k \end{pmatrix}, \quad \quad \quad \beta =i\gamma ^0 = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}. \end{equation*} This leads to the standard form of the Schrödinger equation \begin{equation*} i\dot{\psi } = -i\alpha ^k \partial _k \psi + m\beta \psi =H\psi . \end{equation*} One can check that the Hamiltonian $$H$$ is indeed a hermitian operator. For a free fermion the different momentum modes evolve independently, and one finds in the momentum basis, \begin{equation*} i\dot{\psi } = H\psi \quad \quad \text{with}\quad \quad H=\alpha ^k p_k +m\beta . \end{equation*} It is instructive to consider the rest frame of the particle ($$\mathbf{p}=0$$). For the Hamiltonian one gets \begin{equation*} H = m\begin{pmatrix} 0 & \mathbf{1} \\ \mathbf{1} & 0 \end{pmatrix}. \end{equation*} This matrix mixes the Weyl spinors $$\psi _L$$ and $$\psi _R$$ \begin{equation*} i\partial _t\begin{pmatrix} \psi _L \\ \psi _R \end{pmatrix} =m\beta \begin{pmatrix} \psi _L \\ \psi _R \end{pmatrix} =m \begin{pmatrix} \psi _R \\ \psi _L \end{pmatrix}. \end{equation*} We can verify that $$H$$ has two eigenvectors with positive energy ($$E= +m$$), and another two with negative energy ($$E= -m$$).

Positrons. We can construct linear combinations of $$\psi _L$$ and $$\psi _R$$, which are mass eigenstates \begin{equation} \psi _{\pm } = \tfrac{1}{\sqrt{2}}(\psi _L \pm \psi _R) \quad \quad \quad \text{and} \quad \quad \quad i\dot{\psi }_{\pm } = \pm m\psi _{\pm }. \label{eq:linCombinations} \end{equation} By conjugating the equation for $$\psi _{-}$$, \begin{equation*} -i\dot{\psi }^{*}_{-} = -m\psi ^{*}_{-} \Rightarrow i\dot{\psi }^{*}_{-} = m\psi ^{*}_{-}, \end{equation*} one finds that $$\psi ^{*}_{-}$$ is an eigenstate of the Hamiltonian with positive eigenvalue $$E = +m.$$ This field can be interpreted as the field for a new particle, called the positron. The positron is the antiparticle to the electron. We will see that $$\psi ^{*}_{-}$$ has electric charge $$-e$$, while $$\psi _{+}$$ has charge $$e$$. We use $$\psi _{+}$$ for electrons and therefore $$e<0$$. The existence of antiparticles is a direct consequence of the Dirac equation, which has predicted the positron before its experimental discovery. In turn, it is a consequence of Lorentz symmetry, combined with conserved electric charge. The latter makes it impossible to describe the electron by a Weyl spinor. The complex conjugation in the definition of $$\psi ^*_-$$ is part of a so called ”charge conjugation“ operation $$C$$, which is a discrete symmetry similar to parity $$P$$ and time reversal $$T$$. General arguments that we will not discuss here have shown that Lorentz symmetry implies the invariance of the action under the combined symmetry $$CPT$$. Individual discrete symmetries can be violated, as for neutrinos which violate $$P$$ and $$C$$.

Electrons and Positrons in the Electromagnetic Field. We next investigate the dynamics of electrons and positrons in an electromagnetic field. For this purpose we first construct the piece of the action which describes the interaction of electrons with photons or electromagnetic fields. Taking fixed ”external“ electromagnetic fields the action for the spinor field remains quadratic. We can therefore interpret the Dirac equation as a Schrödinger equation for the one-particle wave function also for this case. In particular, we will see that the positron has the opposite charge of the electron.

As for classical electrodynamics, the electromagnetic field is given by $$A_\mu = (-\phi ,\textbf{A}),$$ and the covariant Lagrangian by \begin{equation} \mathscr{L} = i\bar{\psi }\gamma ^\mu (\partial _\mu -ieA_\mu )\psi + im \bar{\psi }\psi . \label{eq:LagrangianQED} \end{equation} Since $$A_\mu$$ transforms as a covariant vector this action is Lorentz invariant.
The interaction term $$\sim \bar{\psi }\gamma ^\mu \psi A_\mu$$ is dictated by the gauge symmetry of the electromagnetic interactions. This gauge symmetry replaces partial derivatives by covariant derivatives, according to \begin{equation*} \begin{split} \partial _t \psi &\to (\partial _0 + ie\phi )\psi ,\\ \partial _k\psi &\to (\partial _k - ieA_k)\psi , \end{split} \end{equation*} Varying the action with respect to $$\bar{\psi }$$ yields the Dirac equation in an electromagnetic field, \begin{equation*} i\dot{\psi } = \left (\alpha ^k (\hat{p}_k -eA_k) +e\phi + \begin{pmatrix} 0 & m \\ m & 0 \end{pmatrix} \right )\psi , \quad \hat{p}_k=-i\partial _k. \label{eq:diracEqModified} \end{equation*}

With \begin{equation*} \alpha ^k \begin{pmatrix} \psi _L \\ \psi _R \end{pmatrix} = \begin{pmatrix} -\tau _k & 0 \\ 0 & \tau _k \end{pmatrix} \begin{pmatrix} \psi _L \\ \psi _R \end{pmatrix} = \begin{pmatrix} -\tau _k \psi _L \\ \tau _k \psi _R \end{pmatrix}, \end{equation*} the action of $$\alpha ^k$$ on the linear combinations (??) reads \begin{equation*} \begin{split} \alpha ^k \psi _+ &= -\tau _k \psi _{-},\\ \alpha ^k \psi _{-} &= -\tau _k\psi _{+}. \end{split} \end{equation*} For the mass eigenstates $$\psi _{+},\psi _{-}$$ the Dirac equation becomes \begin{equation*} \begin{split} i\dot{\psi }_{+} &= (m + e\phi )\psi _{+} + i(\partial _k - ieA_k)\tau _k \psi _{-},\\ i\dot{\psi }_{-} &= (-m + e\phi )\psi _{-} + i(\partial _k - ieA_k)\tau _k \psi _{+}. \end{split} \end{equation*} The Schrödinger equation for the positron obtains by complex conjugation of the equation for $$\psi _-$$, \begin{equation*} i\dot{\psi }_{-}^{*} = (m - e\phi )\psi _{-}^{*} + i(\partial _k + ieA_k)\tau _k^{*} \psi _{+}^{*}. \end{equation*} The positrons described by $$\psi _{-}^{*}$$ have indeed the opposite charge as the electrons described by $$\psi _{+}$$. One observes that for non-zero momentum $$\hat{p}_k$$ or non-zero vector potential $$A_k$$ the components $$\psi _+$$ and $$\psi _-$$ mix. The eigenfunctions for the electrons and positrons are therefore not simply $$\psi _+$$ and $$\psi ^*_-$$ as for the free fermions at rest. In the absence of interactions the eigenfunctions of the Hamiltonian for non-zero momentum can be obtained from the ones in the rest frame by applying a suitable Lorentz boost. In the presence of electromagnetic fields the eigenfunctions are more complicated. The one-particle description ceases to be valid if the fields are strong enough such that the creation of electron-positron pairs becomes possible.

Quantum electrodynamics. For quantum electrodynamics (QED) the electromagnetic field is treated on the same level as the other fields in the functional integral for a quantum field theory. Thus the functional measure includes an integration over the fields $$A_\mu (x)$$. This contrasts to the fixed ”external fields“ discussed before. The excitations of the electromagnetic field are the photons. Similar to the case of scalar fields, photons are propagating particles. Their number is not conserved. We can discuss processes as photon-electron scattering, or the production of electron-positron pairs by the annihilation of a photon pair. In the presence of the electromagnetic interaction the electric charge remains a conserved quantity. This is guaranteed by the gauge symmetry. The number of electrons is no longer conserved, however.

A Lorentz invariant action for the electromagnetic field employs the field strength $$F_{\mu \nu }$$, \begin{equation} \mathscr{L}_F = \tfrac{1}{4} F^{\mu \nu }F_{\mu \nu },\quad \quad \quad F_{\mu \nu } = \partial _\mu A_\nu - \partial _\nu A_\mu . \label{eq:LagrangianQEDFmunu} \end{equation} Since $$F_{\mu \nu }$$ is gauge invariant, this action also preserves the gauge symmetry. Combining eq. \eqref{eq:LagrangianQED} and \eqref{eq:LagrangianQEDFmunu} yields a Lorentz- and gauge invariant action for electrons and photons. (In the following we often understand by ”electrons“ the description of both electrons and positrons.)

In summary, QED is defined by the functional integral for the partition function \begin{equation*} \begin{split} Z &= \int D\varphi ~ \exp \left (-i\int _x \mathscr{L}_{QED}\right ),\\ \int D\varphi &= \int D\psi D\bar{\psi } DA_\mu , \\ \mathscr{L}_{QED} &= i\bar{\psi }\gamma ^\mu (\partial _\mu - ieA_\mu )\psi +i m \bar{\psi }\psi + \tfrac{1}{4}F^{\mu \nu }F_{\mu \nu }. \end{split} \end{equation*} The integral over the fermion fields is a Grassmann functional integral. The integral over the gauge fields $$A_\mu$$ should preserve both the Lorentz symmetry and the gauge symmetry. It needs a regularisation, to which later parts of the QFT-lecture will turn in detail. For our purpose it is sufficient to know that a suitable functional measure exists.

From the functional integral all correlation functions can, in principle, be computed. They can be compared with precise measurements for many processes. In the case of QED the fine structure constant $$\alpha =e^2/(4\pi ) \sim 1/137$$ is a small parameter. A perturbative computation amounts to an expansion in $$\alpha$$. It can be performed to rather high order. Precise computations with many decimal places agree perfectly with observation.

Gauge symmetry. The action of QED is invariant under local gauge transformations. \begin{equation*} \psi '(x) = e^{i\alpha (x) \psi (x)}, \end{equation*} \begin{equation*}{A'}_\mu (x) = A_\mu (x) + \tfrac{1}{e} \partial _\mu \alpha (x). \end{equation*}
For a local transformation the transformation parameter $$\alpha (x)$$ depends on $$x$$. This contrasts to the Lorentz-transformations which are ”global“ symmetry transformations. The local gauge transformations change $$\psi$$ at every $$x$$ independently. If the action (and measure) is invariant under a local transformation one speaks about a “local symmetry”, and often simply about a ”gauge symmetry“.

In order to verify the gauge invariance of the action for QED we first note that the free fermion kinetic term is not gauge invariant, \begin{equation*} \bar{\psi } \gamma ^\mu \partial _\mu \psi \to \bar{\psi } \gamma ^\mu \partial _\mu \psi + i\partial _\mu \alpha \bar{\psi }\gamma ^\mu \psi . \end{equation*} Also the interaction term $$\sim \bar{\psi } \gamma ^\mu \psi A_\mu$$ alone is not gauge invariant \begin{equation*} -ie\bar{\psi }\gamma ^\mu A_\mu \psi \to -ie\bar{\psi }\gamma ^\mu A_\mu \psi - i\partial _\mu \alpha \bar{\psi }\gamma ^\mu \psi . \end{equation*} Only the combination into a covariant derivative $$D_\mu =\partial _\mu -ie A_\mu$$ yields an invariant expression \begin{equation*} i \bar{\psi } \gamma ^\mu D_\mu \psi = i\bar{\psi }\gamma ^\mu (\partial _\mu -ieA_\mu )\psi . \end{equation*} The gauge invariance of the field strength $$F_{\mu \nu }$$ follows from the commutativity of partial derivatives. \begin{equation*} F_{\mu \nu }= \partial _\mu A_\nu -\partial _\nu A_\mu \to \partial _\mu A_\nu -\partial _\nu A_\mu + \tfrac{1}{e} \partial _\mu \partial _\nu \alpha -\tfrac{1}{e}\partial _\nu \partial _\mu \alpha = F_{\mu \nu }. \end{equation*} Local gauge invariance is an important principle for constructing the action of a quantum field theory. It is closely related to renormalizability.

Renomalizability. Gauge symmetry is a powerfull restriction for the choice of the action. Is it sufficient? Consider a possible term \begin{equation*} \Delta \mathscr{L} = \tfrac{b}{m} \bar{\psi }[\gamma ^\mu ,\gamma ^\nu ] \psi F_{\mu \nu }. \end{equation*} This term is Lorentz invariant and gauge invariant. If we add it with an unknown coefficient $$b$$, predictions will depend on this coefficient. Predictivity of QED, which only involves $$m$$ and $$\alpha = e^2/ 4\pi$$, would be lost. The reason for the absence of such a term will be discussed in later parts of the QFT-lecture related to renormalizability.

Non-relativistic limit of Dirac equation. For the quantum mechanics of atoms one uses the Schrödinger equation for a complex two-component electron wave function $$\chi (x)$$. This equation is not covariant under Lorentz-transformations – it is a non-relativistic equation. It contains various terms, as a coupling between spin and angular momentum. A fundamental theory as QED should fix all such couplings. Since QED contains only two parameters, namely $$m^2$$ and $$\alpha$$, all aspects of the non-relativistic Schrödinger equation should follow from the Dirac equation. The way of deriving the usual Schrödinger equation for electrons proceeds by taking the non-relativistic limit of the Dirac equation.

The Schrödinger equation for the two-component spinor $$\chi$$ for the electron is given by \begin{equation} i\partial _t \chi =H \chi = \frac{1}{2m}(\vec{p}-e\vec{A})^2 +e\varphi -\frac{e}{m}\vec{S}\vec{B},\quad \quad \vec{S} = \tfrac{1}{2}\vec{\tau }, \label{eq:SchoedingerChi} \end{equation} with spin operator $$\vec{S}$$, momentum operator $$\vec{p}_k=-i\vec{\nabla }_k$$ and magnetic field $$\vec{B}$$. We recall that we use units with $$\hbar =1$$. One usually deals with small electromagnetic fields for which one linearizes in $$\vec{A}$$. In particular, for a constant magnetic field $$\vec{B}$$ one takes $$\vec{A} = -\tfrac{1}{2}\vec{r}\times \vec{B}$$ and obtains \begin{equation*} \tfrac{1}{2m} (\vec{p}-e\vec{A})^2 = \frac{\vec{p}^2}{2m}-\frac{e}{2m}\vec{L}\vec{B}, \end{equation*} with $$\vec{L}$$ the angular momentum operator. One recognises the Schrödinger equation for atomic physics. There are indeed no new free couplings.

The magnetic field couples to a linear combination of angular momentum and spin different from the total angular momentum, with a term \begin{equation*} (\vec{L} + g\vec{S})\vec{B},\quad \quad g=2. \end{equation*} The Dirac equation predicts the relative coupling $$g=2$$. For QED, the Dirac equation is not an exact equation. As a one-particle equation it assumes implicitly that the particle number is conserved. We have discussed before that the number of electrons is not conserved in QED since electron-positron pairs can be created or annihilated without violation of charge conservation. One expects possible corrections to the prediction of the Dirac equation due to fluctuation effects. Indeed, the QED corrections from fluctuations yield a small correction to $$g-2$$, which is computed to many decimal places. For the electron one finds \begin{equation*} \frac{g_e}{2}=1.001159652181643(764), \end{equation*} to be compared with the experimental values \begin{equation*} \frac{g_e^{exp}}{2}=1.00115965218073(28), \end{equation*} with errors on the last digits in brackets. The impressive agreement on the level $$10^{-12}$$ is a strong indication for the correctness of QED.

The derivation of the non-relativistic limit of the Dirac equation can be done in several steps.

• Step 1: Square the Dirac equation, \begin{equation*} \gamma ^\nu (\partial _\nu -ieA_\nu )\gamma ^\mu (\partial _\mu -ieA_\mu )\psi = m^2 \psi . \end{equation*}
• Step 2: From the Dirac algebra we use $$[\gamma ^\mu ,\gamma ^\nu ] = 4iT^{\mu \nu }$$ and obtain \begin{equation*} \left ((\partial ^\mu -ieA^\mu )(\partial _\mu -ieA_\mu ) + eT^{\mu \nu }F_{\mu \nu } -m^2\right )\psi =0. \end{equation*}
• Step 3: We use $$T^{\mu \nu }F_{\mu \nu } = \tfrac{1}{2}B_k \tau _k + \tfrac{i}{2} E_k \tau _k \gamma ^5$$, with $$\tau _k =\begin{pmatrix} \tau _k & 0 \\ 0 & \tau _k \end{pmatrix}.$$ Also using $$\psi _\pm$$, one obtains \begin{equation*} \left \{(\partial ^\mu -ieA^\mu )(\partial _\mu -ieA_\mu ) -m^2 + eB_k \tau _k \right \}\psi _{+} = -ieE_k \tau _k \psi _{-}. \end{equation*}
• Step 4: One neglects the positrons by setting $$\psi _-=0$$. The resulting equation for $$\psi _+$$ has only two independent components. The lower two and upper two components obey an identical Schrödinger equation. We take the upper two components.
• Step 5: We introduce the non-relativistic wave function $$\chi$$ by \begin{equation*} \psi _{+} = e^{-imt}\chi . \end{equation*} We formally write \begin{equation*} \begin{split} i\partial _t \chi = H\chi = (E-m)\chi , \end{split} \end{equation*} by putting all terms that are not linear in the time derivative $$\chi$$ on the right hand side.
• Step 6: The non-relativistic limit is given by $$|H| \ll m$$. In this limit one can neglect \begin{equation*} \frac{\partial ^2_t}{m},\quad \frac{A_0 \partial _t}{m}, \quad \frac{(\partial _t A_0)}{m},\quad \frac{A^2_0}{m}. \end{equation*} Omitting these terms in $$H$$ yields the above non-relativistic result. The quantity $$H$$ becomes the non-relativistic Hamiltonian in eq. \eqref{eq:SchoedingerChi}.

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