Past exam of the laser science and quantum information processing course of the University of Oxford / 2022 / 3

For one electromagnetic normal mode, $a$ and $a^{†}$ are the annihilation operator and creation operator, with commutator $[a,a^{†}]=I$. Their action on normalized number states isThey remove or add one photon, whose energy is $\hslash \omega$. The vacuum state $|0⟩$ is normalized and obeys $a|0⟩=0$. It has no photons, but retains zero-point energy $\hslash \omega /2$ for the quantum harmonic oscillator Hamiltonian $H=\hslash \omega (a^{†}a+1/2)$. The coherent state $|\alpha ⟩=D(\alpha )|0⟩$ is a displaced vacuum state.

Set $A=\alpha a^{†}$ and $B=-{\alpha }^{\ast }a$. Their commutator isIt commutes with both operators, so the stated Baker--Campbell--Hausdorff formula terminates:In particular $e^{-{\alpha }^{\ast }a}|0⟩=|0⟩$. Expanding the remaining exponential and using $(a^{†}{)}^n|0⟩=\sqrt{n!}|n⟩$ gives the number state expansionThe orthonormality of the number states givesThis is a Poisson distribution with parameter $\mu =|\alpha {|}^2$. Its probabilities sum to $e^{-\mu }{e}^{\mu }=1$. The phase of $\alpha$ affects the coherent state's relative probability amplitudes but not its photon number probabilities. At $\alpha =0$ the distribution is concentrated at $n=0$.

The number operator is $\hat{N}=a^{†}a$. The number state expansion gives $a|\alpha ⟩=\alpha |\alpha ⟩$ by shifting the summation index; taking the adjoint gives $⟨\alpha |a^{†}={\alpha }^{\ast }⟨\alpha |$. Thus the expectation value isFor the second moment, use $a{a}^{†}=a^{†}a+I$:Its expectation value in the coherent state is $⟨{\hat{N}}^2⟩=|\alpha {|}^4+|\alpha {|}^2$. ThereforeThe variance equals the mean, exactly as for the Poisson distribution found in b. For $\mu >0$ the fractional standard deviation is $1/\sqrt{\mu }$, the usual photon shot noise scaling. For the vacuum state, both the mean photon number and its variance vanish; this does not imply zero optical quadrature fluctuations. At $\mu =0$ the fractional fluctuation is undefined.

For the normalized coherent state, $⟨a⟩=\alpha$ and $⟨a^{†}⟩={\alpha }^{\ast }$. The dimensionless electric field and magnetic field optical quadratures consequently have expectation valuesThe physical electric field and magnetic field strengths have the corresponding proportionality factors and mode functions. In free evolution replace $\alpha$ by $\alpha e^{-i\omega t}$ to recover the classical oscillation.

Expanding the squared operators cancels the $a^2$ and $(a^{†}{)}^2$ terms:On the other hand, $(\alpha +{\alpha }^{\ast }{)}^2+[-i(\alpha -{\alpha }^{\ast }){]}^2=4|\alpha {|}^2$. HenceThe difference is the sum of the variances, not an extra mean field. Indeed $\mathrm{Var}(\mathcal{E})=\mathrm{Var}(\mathcal{B})=1$, independent of $\alpha$. The commutator $[\mathcal{E},\mathcal{B}]=2iI$ requires ${\sigma }_{\mathcal{E}}{\sigma }_{\mathcal{B}}\ge 1$, and the coherent state attains equality. The residual two units are the vacuum fluctuations of the two optical quadratures.

Zero-point fields are the irreducible electromagnetic field fluctuations in the vacuum state. Each normal mode has ground energy $\hslash \omega /2$. Although $⟨\mathcal{E}⟩=⟨\mathcal{B}⟩=0$ in the vacuum state, d gives $⟨{\mathcal{E}}^2⟩=⟨{\mathcal{B}}^2⟩=1$. These are quantum fluctuations, not a classical random coherent field with a known instantaneous value.

Two consequences are the Lamb shift, where coupling to the quantized electromagnetic field changes atomic energy levels, and spontaneous emission, where an excited chemical atom can emit a photon into an initially empty normal mode. In the latter process the creation operator has a nonzero matrix element $⟨1|a^{†}|0⟩=1$; stimulated-plus-spontaneous factors have the form $n+1$, which does not vanish at $n=0$. Descriptions splitting spontaneous emission into vacuum fluctuations and radiation reaction depend on operator-ordering conventions; the measurable total rate does not. Zero-point fields do not mean that the vacuum state contains a nonzero mean photon number.