Past exam of the quantum field theory course of the University of Oxford / 2021 / 2

Let $A=\mathbf{p}\cdot \mathbf{\sigma }$ and $E=\sqrt{{\mathbf{p}}^2+m^2}$. The Pauli multiplication identity gives $A^2={\mathbf{p}}^2{I}_2$, and the orthonormal two-component spinors obey ${\sum }_s{\xi }^s{\xi }^{s†}=I_2$. Taking the Dirac adjoint of the supplied Dirac spinor therefore gives, directly,For the antiparticle deduction, specify the standard normalization $v^{r†}(-\mathbf{p})v^s(-\mathbf{p})=2E{\delta }^{rs}$, just as $u^{r†}{u}^s=2E{\delta }^{rs}$. Spanning alone does not specify a normalization: replacing every $v$ by $2v$ preserves the spanning property but multiplies its proposed fermion spin sum by four. The standard normalization is the convention implicit in the mode expansion of a Dirac field with the standard canonical anticommutation relations.

To use completeness correctly, introduce the Hermitian Dirac HamiltonianThe Dirac equation makes $u^s(\mathbf{p})$ and $v^s(-\mathbf{p})$ its eigenvectors with eigenvalues $E$ and $-E$. Their mutual orthogonality and spanning property give complementary orthogonal projectorsMultiply the second equality by $2E{\gamma }^0$ on the right and replace $\mathbf{p}$ by $-\mathbf{p}$. For the positive-energy four-momentum $p=(E,\mathbf{p})$,

The canonical anticommutation relations arewith the mixed and equal-type anticommutators zero. Insert the mode expansion of a Dirac field. The phases cancel at equal time, and a givesSince $\bar{\psi }{\gamma }^0={\psi }^{†}$, the requested componentwise answer isThis is the Equal-time canonical anticommutator of a Dirac field: the canonical momentum conjugate to ${\psi }_{\alpha }$ is ${\pi }_{\alpha }=i{\psi }_{\alpha }^{†}$. It is the fermionic counterpart of the canonical field canonical commutation relation, ensuring the correct local canonical quantization. For this first-order Lagrangian density, the conjugate variables are constrained; the displayed relation is the standard result after handling that constraint. It does not assert a nonzero anticommutator between the fields at separated spatial points.

Take a spatial Fourier transform of the mode expansion of a Dirac field:The orthogonality relations $u^{s†}(\mathbf{p})u^r(\mathbf{p})=2E{\delta }^{sr}$ and $u^{s†}(\mathbf{p})v^r(-\mathbf{p})=0$ isolate the fermionic annihilation operator. Projecting onto $v^s(-\mathbf{p})$ instead isolates the fermionic creation operator:The opposite time phases and reversed antiparticle momentum are essential. Both denominators are $\sqrt{2E}$, because the Dirac spinors have norm squared $2E$.

Read the unsuperscripted $a_{\mathbf{p}}$ in the printed Hamiltonian as $a_{\mathbf{p}}^s$: the repeated spin label is otherwise missing. Relabel $\mathbf{p}\mapsto -\mathbf{p}$ in its antiparticle term, and insert both projections from c. Keeping the original operator ordering givesThe difference of the two orthogonal projectors in a is $h(\mathbf{p})/E$, explaining the cancellation of the factors $E$. Under inverse Fourier transformation, $\mathbf{p}$ becomes $-i\nabla$ acting on $\psi$. HenceSpatial indices run from $1$ to $3$. This is the local Hamiltonian density of the free Dirac field. It also follows from $\mathcal{H}=\pi \dot{\psi }-\mathcal{L}$ with $\pi =i{\psi }^{†}$.

The original antiparticle term is $-b{b}^{†}$, so the result retains its infinite additive vacuum energy. If the Hamiltonian is normal-ordered, its occupation-number expression becomes $\int d^{3}p/(2\pi {)}^{3}E{\sum }_s(a^{s†}{a}^s+b^{s†}{b}^s)$ and its local expression is $:\mathcal{H}:$, with the same subtraction. The displayed unsymmetrized Hamiltonian density is Hermitian after spatial integration under vanishing boundary terms; a manifestly Hermitian local form replaces the derivative term by $-\frac{i}{2}{\psi }^{†}\mathbf{\alpha }\cdot \overleftrightarrow{\nabla }\psi$ and differs by a spatial divergence.