Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 44 / 3

Use signature $(+,-,-,-)$ and the free Dirac equation $(i{\gamma }^{\mu }{\partial }_{\mu }-m)\psi =0$. For positive $p^0=E_{\mathbf{p}}=\sqrt{{\mathbf{p}}^2+m^2}$, substitution of the two frequency signs givesMultiplying by the opposite factor shows $p^2=m^2$. Each equation has two independent solutions for a massive field.

In the Dirac basis, ${\gamma }^0=\mathrm{diag}(I,-I)$ and ${\gamma }^i=\left(\begin{array}{cc}0& {\sigma }_i\\ -{\sigma }_i& 0\end{array}\right)$. For orthonormal two-component spin labels ${\xi }_s,{\eta }_s$, convenient solutions areThe two-block equations verify these expressions using $(\mathbf{\sigma }\cdot \mathbf{p}{)}^2={\mathbf{p}}^{2}I$. At rest the positive-frequency solutions have only upper components and the negative-frequency solutions only lower components. Their normalizations are $u_s^{†}{u}_r=v_s^{†}{v}_r=2E_{\mathbf{p}}{\delta }_{sr}$ and ${\bar{u}}_s{u}_r=2m{\delta }_{sr}$, ${\bar{v}}_s{v}_r=-2m{\delta }_{sr}$. Their completeness relations areThe field expansion combines fermionic annihilation operators multiplying the positive-frequency waves with antiparticle fermionic creation operators multiplying the negative-frequency waves. The latter frequency sign is not the energy of a physical negative-energy antiparticle state.

For the specified $1/\sqrt{2E_{\mathbf{p}}}$ mode normalization, the canonical anticommutation relations areand every other anticommutator of the $a,b$ operators vanishes. For the Dirac field, they imply the Equal-time canonical anticommutator of a Dirac fieldTo check the normalization, the two sectors contribute $u(\mathbf{p})u^{†}(\mathbf{p})$ and $v(\mathbf{p})v^{†}(\mathbf{p})$ with opposite spatial Fourier signs. Reverse $\mathbf{p}$ in the second contribution. The completeness relations then givewhere $\tilde{p}=(E_{\mathbf{p}},-\mathbf{p})$. This cancels the mode denominator and leaves the Fourier representation of the delta function. Same-momentum $u^{†}v$ orthogonality must not be used in place of this reversed-momentum identity.

For arbitrary spacetime points,For spacelike $z$, choose a Lorentz frame with $z^0=0$; reversing spatial momentum shows $\Delta (z)=0$. Its derivatives vanish there as well, so the field anticommutators vanish at spacelike separation. This is fermionic microcausality. The anticommutation rules make the occupation of each normalized mode zero or one, implementing the Pauli exclusion principle, and make the normal-ordered free Hamiltonian positive for both particle sectors.

With a vacuum annihilated by all $a_{\mathbf{p}}^s,b_{\mathbf{p}}^s$, define the particle and antiparticle occupation operators of a Dirac fieldEach counts its corresponding occupied modes. The total positive particle count isCommuting it through either creation operator gives $[N_{\mathrm{total}},a^{†}]=a^{†}$ and $[N_{\mathrm{total}},b^{†}]=b^{†}$, so both kinds of one-particle states have eigenvalue one. In a normalized discrete mode, $(a^{†}a{)}^2=a^{†}a$ by the anticommutation relations, so its occupation eigenvalues are $0,1$.

For a Dirac field there is also a distinct conserved signed fermion number. Inserting the mode expansion into the normal-ordered local density givesSpatial integration makes the same-sector momenta equal. The cross terms vanish by $u^{†}(\mathbf{p})v(-\mathbf{p})=0$; normal ordering of $b{b}^{†}$ supplies the minus sign for the antiparticles. Thus the density integral is not the total positive particle count. It generates the global phase symmetry, and for electron charge $-e$ the electric charge is $Q_{\mathrm{em}}=-e{Q}_D$.

The free normal-ordered Hamiltonian isso $N_e,N_{\bar{e}},N_{\mathrm{total}}$ and $Q_D$ are conserved in the free theory. In an interacting theory, pair creation can change the total count while preserving the signed charge. This distinguishes a positive occupation number operator from the charge behind Dirac fermion number conservation.

A Positron is the electron's antiparticle: it has the same mass and spin one-half, opposite electric charge $+e$, and fermionic statistics. A state $b_{\mathbf{p}}^{s†}|0⟩$ has positive energy $E_{\mathbf{p}}$ and physical momentum $\mathbf{p}$, as follows from the Hamiltonian and momentum operatorsThe charge commutator gives $[Q_{\mathrm{em}},b^{†}]=+e{b}^{†}$.

The term $b^{†}v(p)e^{ip\cdot x}$ in the field has negative frequency, but it creates this positive-energy Positron. Treating its coefficient as an annihilator of a physical negative-energy electron would give an unbounded energy interpretation. Fermionic quantization instead reorders that sector, giving positive antiparticle occupation energies and the opposite signed charge. The historical Dirac sea describes a Positron as a hole in filled negative-energy electron levels; the field-theory interpretation uses an antiparticle creation operator and normal ordering, without requiring a literal infinite material sea.

Under charge conjugation of a Dirac field, ${\psi }^C=C{\bar{\psi }}^T$, where $C^{-1}{\gamma }^{\mu }C=-({\gamma }^{\mu }{)}^T$, particle and antiparticle modes interchange. Electron–Positron pair creation and annihilation change total particle number while conserving electric charge. Positrons are positive-energy antiparticles, not physical negative-energy electrons.