Past exam of the theoretical physics II course of the University of Cambridge / 2016 / Paper 1 / 4

A Klein-Gordon solution is second order in time: both $\Psi$ and ${\partial }_t\Psi$ are needed as initial data. Its free modes have both signs of frequency,This differs from the first-order Schrödinger evolution of a fixed one-particle wavefunction.

Moreover, $|\Psi {|}^2$ does not generally obey the required probability continuity equation. The conserved Klein-Gordon density is proportional toup to a positive conventional constant. A pure positive-frequency mode gives $+2\omega |\Psi {|}^2$ and a negative-frequency mode gives $-2\omega |\Psi {|}^2$. This density is not positive definite and cannot be a universal position probability density.

For a free field, one can choose a positive-frequency subspace with a positive integrated inner product. Time-dependent backgrounds generally mix it with the negative-frequency subspace, so simply deleting the latter is not dynamically consistent. In quantum field theory the complex Klein-Gordon field instead has particle and antiparticle sectors; its conserved current is interpreted as charge current, and creation or annihilation of pairs need not fit a fixed-particle-number wavefunction. These are limitations of the naive single-particle interpretation, not a failure of the relativistic field equation.

Work in temporal gauge, with zero scalar potential, and use the charge sign exactly as in the given covariant spatial derivative. A spatially uniform vector potential acts on a Fourier mode asThere are no spatial derivatives of $\mathbf{A}$. Substitution into the field equation and multiplication by $c^2$ therefore givesThe canonical wavevector $\mathbf{k}$ is conserved by spatial translation symmetry; the kinetic momentum is $\hslash \mathbf{k}-e\mathbf{A}(t)$. The absence of a term involving $\dot{\mathbf{A}}$ in this mode equation follows from temporal gauge and the spatial minimal-coupling operator. The electric field still affects evolution through the changing oscillator frequency.

For this mode, abbreviateThe electric pulse is $\mathbf{E}(t)=-{\mathbf{A}}_0\delta (t)$. In temporal gauge the mode equation has a finite step in its frequency squared, with no delta multiplying the mode. Thus both ${\psi }_{\mathbf{k}}$ and its first time derivative are continuous at zero.

Writing the positive-frequency incident amplitude as $a_{\mathbf{k}}^{\mathrm{in}}$, the matched solution isIts initial value is $a_{\mathbf{k}}^{\mathrm{in}}/\sqrt{2\omega }$ and its initial derivative is $-i\omega a_{\mathbf{k}}^{\mathrm{in}}/\sqrt{2\omega }$, as required. In the normalized outgoing-frequency basis it isThe negative-frequency contribution appears even though only positive frequency was incident. These are temporal-mode matching coefficients, so their conserved difference is hyperbolic rather than the sum-one rule of ordinary spatial reflection and transmission probabilities.

Distinguish in-operators and out-operators before identifying the symbols in the final vacuum condition. The full charged-field Fourier component before the pulse containsThe conjugate incoming mode has the conjugate matched solution. Collecting the coefficients of $e^{-i\Omega t}$ and $e^{i\Omega t}$ after the pulse gives the real Bogoliubov mapSince ${\alpha }^2-{\beta }^2=1$, the inverse map isThe Inverse Bogoliubov vacuum condition therefore yieldsHere the $a,b$ inside $(u_{\mathbf{k}}{a}_{\mathbf{k}}+v_{\mathbf{k}}{b}_{-\mathbf{k}}^{†})$ mean out-operators. Using the forward-map coefficient $\beta$ with the same plus sign would be inconsistent with this convention. Mode rephasings can change a displayed coefficient sign, but the map and squeezed-state exponent must change together.

The complete in-vacuum also obeys the antiparticle annihilation condition. In out-operators this is $(u_{\mathbf{k}}{b}_{-\mathbf{k}}^{\mathrm{out}}+v_{\mathbf{k}}{a}_{\mathbf{k}}^{\mathrm{out}†})|0_{\mathrm{in}}⟩=0$. The two conditions jointly select the vacuum sector used below; the particle condition alone would not exclude arbitrary incoming antiparticle excitations.

Suppress the mode label temporarily and let $a,b$ be out-operators. With $q=v/u$, the commuting-mode identityshows that $(a+q{b}^{†})e^{-q{a}^{†}{b}^{†}}|0_{\mathrm{out}}⟩=0$. The analogous relation for $b$ gives the antiparticle condition. Thus in a finite regulated collection of modes,For one particle-antiparticle mode pair, $(a^{†}{b}^{†}{)}^n|0,0⟩=n!|n,n⟩$. The exponential's factorial therefore cancels:The resulting particle-antiparticle two-mode squeezed vacuum has the geometric pair-number distributionHere one pair means one particle in canonical mode $\mathbf{k}$ and one antiparticle in mode $-\mathbf{k}$. They are distinct species, and the state has equal particle and antiparticle occupation. In particular,For a finite set, $|\mathcal{N}|={\prod }_{\mathbf{k}}\sqrt{1-r_{\mathbf{k}}}={\prod }_{\mathbf{k}}|u_{\mathbf{k}}{|}^{-1}$. An overall phase is arbitrary. With no pulse, $\Omega =\omega$, so $v=0$ and only the zero-pair state remains.

The ultraviolet divergence of an instantaneous vector-potential quench qualifies this formal product. For an exactly instantaneous vector-potential step in three spatial dimensions, at large $k$ the mixing behaves as $v_{\mathbf{k}}\sim e{\mathbf{A}}_0\cdot \hat{\mathbf{k}}/(2\hslash k)$. Thus $\int {\mathrm{d}}^{3}k|v_{\mathbf{k}}{|}^2$ diverges in the ultraviolet. In a finite quantization box, a finite momentum cutoff makes the product a normalized Fock-space vector; a sufficiently smooth finite-duration pulse removes this particular instantaneous-quench pathology. The per-mode distribution remains the result above, while the unregulated infinite product should not be asserted to have a finite nonzero global normalization.