Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 44 / 1 / Solution

Use units $\hslash =c=1$ and the Minkowski metric $g_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$, so $p\cdot x=E_{p}t-\mathbf{p}\cdot \mathbf{x}$. The canonical momentum of the real scalar field is $\pi =\dot{\varphi }$, and the Hamiltonian operator isCanonical quantization of a real scalar field promotes $\varphi$ and $\pi$ to operator-valued distributions satisfying the equal-time canonical commutation relationProducts and operator commutators can first be interpreted after spatial smearing. An additive vacuum constant in $H$ does not affect the following equations. In the Heisenberg picture, $\dot{O}=i[H,O]$. Applying the canonical commutation relation to the quadratic Hamiltonian operator, and integrating the gradient term by parts, givesThus the Klein-Gordon equation is an operator equation. The spatial Fourier transform reduces it to ${\ddot{\varphi }}_{\mathbf{p}}+E_p^2{\varphi }_{\mathbf{p}}=0$, with $E_p=\sqrt{{\mathbf{p}}^2+m^2}$. Its two frequency branches, together with the Hermitian adjoint condition ${\varphi }^{†}=\varphi$, give the mode expansion of a free fieldThe measure $d\mu (p)$ is Lorentz invariant; its normalization fixes that of the creation and annihilation operators rather than imposing any extra physical assumption.

For invariant-normalized scalar mode extraction, take the Klein-Gordon inner product of $f_p$ with the field. The positive-frequency contribution has coefficient $E_p+E_q$ and spatial integral $(2\pi {)}^3{\delta }^3(\mathbf{p}-\mathbf{q})$. The negative-frequency contribution has coefficient $E_p-E_q$ and spatial integral $(2\pi {)}^3{\delta }^3(\mathbf{p}+\mathbf{q})$, so it vanishes because $E_{-p}=E_p$. ConsequentlyTaking the Hermitian adjoint, with the order of the scalar factors immaterial, yieldsEquivalently $a(\mathbf{p})=\int d^{3}x{f}_p^{\ast }(E_p\varphi +i\pi )$ and $a^{†}(\mathbf{q})=\int d^{3}y{f}_q(E_q\varphi -i\pi )$. At a common time their operator commutator therefore isThe same calculation gives $[a(\mathbf{p}),a(\mathbf{q})]=[a^{†}(\mathbf{p}),a^{†}(\mathbf{q})]=0$: the surviving factor is an energy difference supported at opposite spatial momenta. These are the bosonic canonical commutation relations in relativistic normalization of a one-particle state.

Choose the Fock vacuum by $a(\mathbf{p})|0⟩=0$ and $⟨0|0⟩=1$. Only the annihilator-creator product contributes to the Wightman function, givingUsing time ordering and the reverse product for $y^0>x^0$, the Feynman propagator, with the convention $D_F=i{\Delta }_F$ in this calculation, isIn particular the denominator in the spatial measure is $2E_p$ in both terms.

To verify the four-dimensional Fourier transform representation, let $\tau =x^0-y^0$ and integrate the energy variable first:The Feynman i-epsilon prescription puts the positive-energy pole at $E_p-i0$ and the negative-energy pole at $-E_p+i0$. For $\tau >0$, the contour integral closes in the lower half-plane with clockwise orientation. The residue theorem multiplies the residue $i{e}^{-i{E}_p\tau }/(2E_p)$ by $-i$, giving $e^{-i{E}_p\tau }/(2E_p)$. For $\tau <0$, it closes counterclockwise above: multiplying the residue $-i{e}^{i{E}_p\tau }/(2E_p)$ by $i$ gives $e^{i{E}_p\tau }/(2E_p)$. Thus $I(\tau )=e^{-i{E}_p|\tau |}/(2E_p)$. After inserting the spatial exponential, reversing $\mathbf{p}$ in the negative-time term gives exactly the time ordering expression above. HenceThis equality is understood as a distribution limit; the same prescription gives $(\square +m^2)D_F(x-y)=-i{\delta }^4(x-y)$.