Past exam of the quantum field theory course of the University of Oxford / 2016 / 2

Write ${\int }_{\mathbf{p}}=\int d^{3}p/(2\pi {)}^3$ and $E_{\mathbf{p}}=\sqrt{{\mathbf{p}}^2+m^2}$. Insert the mode expansion of a free field for each quantity into the commutator. Only the mixed creation operator and annihilation operator terms survive. The Dirac delta function in their canonical commutation relation sets the two momenta equal, so the square-root factors multiply to $1/2$:In the second equality, each Fourier transform is the same Dirac delta function, since the integration domain is invariant under $\mathbf{p}\mapsto -\mathbf{p}$. This is the equal-time canonical commutation relation. The analogous calculation gives $[\varphi (\mathbf{x}),\varphi (\mathbf{y})]=[\pi (\mathbf{x}),\pi (\mathbf{y})]=0$, because their two mixed terms have opposite signs.

The unit-normalized Fock vacuum is characterized byChoose the relativistic normalization of a one-particle state from the outset:Since any two creation operators commute, any permutation of their order gives exactly the same state vector. Thus these multiparticle quantum states are symmetric under exchange and describe bosons, without a minus sign or any additional exchange phase.

The Hilbert space is the bosonic Fock space, the direct sum of the symmetric $n$-particle Hilbert spaces, including the zero-particle sector. Sharp-momentum kets are distributionally normalized; square-integrable wave packets provide ordinary Hilbert space vectors. For repeated occupation of a discrete mode, a unit-normalized number state uses $(a^{†}{)}^n|0⟩/\sqrt{n!}$. That factorial does not belong in the above labelled continuum-ket convention.

The normal ordering of both linear maps puts every annihilation operator to the right, soIn particular the stated Hamiltonian has already subtracted the free vacuum energy. For a one-particle state, evaluate the commutators using $[AB,C]=A[B,C]+[A,C]B$:Acting on the Fock vacuum therefore givesThe extra factor $\sqrt{2E_{\mathbf{k}}}$ in the chosen relativistic normalization of a one-particle state is a scalar multiplier and does not change either eigenvalue. Applying the same argument to a multiparticle quantum state gives additive total energy and momentum.

The Fock vacuum condition and the creation operator commutator giveEquality of the square-root product and $2E_{\mathbf{p}}$ holds on the support of the Dirac delta function.

The positive-energy mass shell has the Lorentz invariant measureHere both $p^2$ and the four-dimensional volume element are invariant, and positive energy is preserved by a proper orthochronous Lorentz transformation. Consequently the three-dimensional Jacobian determinant is $J=E_{\mathbf{\Lambda }p}/E_{\mathbf{p}}$. Transforming the Dirac delta function yieldsso its factor exactly cancels that of $2E_{\mathbf{\Lambda }p}$. Thus the displayed inner product is invariant under simultaneous transformation of both momenta. This is also why the one-particle completeness measure is $d^{3}p/((2\pi {)}^{3}2E_{\mathbf{p}})$.^[ref]

The Heisenberg picture evolution of the annihilation operator follows from $[H,a_{\mathbf{p}}]=-E_{\mathbf{p}}{a}_{\mathbf{p}}$. With $p^0=E_{\mathbf{p}}$, this gives the mode expansion of a free fieldDefine ${\Delta }^{+}(z)={\int }_{\mathbf{p}}{e}^{-ip\cdot z}/(2E_{\mathbf{p}})$. The time-ordered product then gives the Feynman propagatorThe last expression follows by closing the $p^0$ contour integral below for $z^0>0$ and above for $z^0<0$; the scalar Feynman propagator pole prescription supplies the required positive- and negative-frequency terms.

For $z^0\ne 0$ every term obeys the homogeneous Klein-Gordon equation. At $z^0=0$, the equal-time commutator of the fields vanishes, so the propagator is continuous and there is no derivative of a Dirac delta function. Its first time derivative has the jumpThe second time derivative therefore supplies precisely the source:Thus the defined Feynman propagator is a Green's function with source normalization $-i$; $i{\Delta }_F$ has source normalization $+1$. Keeping this factor avoids confusing the vacuum expectation value convention with conventions which define the propagator after dividing by $i$.^[ref]