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Continuous Variables and Modes

Position and momentum, and the quadratures of a light mode, have continuous spectra. Their states live in infinite-dimensional Hilbert spaces, but many composite-system ideas remain familiar: choose the subsystems, form their joint state, and trace out the part whose measurements are unavailable. Gaussian states make this especially manageable because their first and second moments determine the entire state. General states require more information.

Helpful background. Partial traces explain local predictions; Schmidt decomposition explains entanglement of pure bipartite states. For oscillator notation, review Fock space.

Question Start here
How do continuous observables change the state space? Continuous-variable systems
How is a two-particle wavefunction reduced? Position-space two-particle states
Which modes count as the subsystems? Mode decompositions and two-mode entanglement
When do a mean vector and covariance matrix suffice? Gaussian states
How can squeezing correlate distant measurements? Squeezed states as entangled modes and the EPR state

Continuous coordinates and oscillator modes

Section titled “Continuous coordinates and oscillator modes”

For two distinguishable particles on a line,

H=L2(R,dx)⊗L2(R,dy)≅L2(R2,dx dy).\mathcal H=L^2(\mathbb R,dx)\otimes L^2(\mathbb R,dy) \cong L^2(\mathbb R^2,dx\,dy).

A normalized wavefunction satisfies ∫dx dy ∣Ψ(x,y)∣2=1\int dx\,dy\,|\Psi(x,y)|^2=1. It is product precisely when Ψ(x,y)=ψ(x)ϕ(y)\Psi(x,y)=\psi(x)\phi(y) almost everywhere for suitable normalized factors. Its reduced density operator has kernel

ρA(x,x′)=∫dy Ψ(x,y)Ψ∗(x′,y).\rho_A(x,x')=\int dy\,\Psi(x,y)\Psi^*(x',y).

The shared value of yy in the two factors matters: the trace sums over the same unobserved basis state in the bra and ket. For a product state, normalization of ϕ\phi gives ρA(x,x′)=ψ(x)ψ∗(x′)\rho_A(x,x')=\psi(x)\psi^*(x'), a rank-one projector. General entangled pure states have mixed reductions. Position eigenkets themselves are distributions, not normalized state vectors.

An oscillator mode also has continuous position-like and momentum-like quadratures, even though its number basis is discrete:

Hmode=span⁡{∣0⟩,∣1⟩,…}‾.\mathcal H_{\mathrm{mode}} =\overline{\operatorname{span}\{|0\rangle,|1\rangle,\ldots\}}.

The bar denotes closure in the Hilbert-space norm, so normalizable infinite superpositions are included. Infinite dimension alone does not specify a continuous-variable model: one must identify the relevant observables and their algebra.

Mode choice determines the entanglement question

Section titled “Mode choice determines the entanglement question”

For bosons, an orthogonal decomposition of the one-particle space gives

h=hA⊕hB,F+(h)≅F+(hA)⊗F+(hB).\mathcal h=\mathcal h_A\oplus\mathcal h_B, \qquad \mathcal F_+(\mathcal h)\cong \mathcal F_+(\mathcal h_A)\otimes\mathcal F_+(\mathcal h_B).

These mode spaces can represent paths, frequencies, or spatial wavepackets. The state (∣1,0⟩+∣0,1⟩)/2(|1,0\rangle+|0,1\rangle)/\sqrt2 is entangled across the indicated modes. Defining a new mode with creation operator b+†=(aA†+aB†)/2b_+^\dagger=(a_A^\dagger+a_B^\dagger)/\sqrt2 instead makes this a single excitation in b+b_+ and vacuum in the orthogonal mode. No contradiction occurs: the transformation mixes the original subsystems and changes the partition being tested. Local operations within a fixed partition do not do this.

Fermionic mode spaces can also be represented by tensor factors after choosing an ordering, but their odd operators anticommute between distinct modes. Ordinary tensor products of independent local lowering matrices would incorrectly commute; parity strings or the graded tensor product implement the correct algebra. Fermionic modes are not oscillator continuous-variable modes. See mode decompositions and creation and annihilation operators for these distinctions.

For identical particles, coordinate slots are not automatically independently controllable parties. Mode and region partitions often give the physically relevant questions.

Choose dimensionless bosonic quadratures with ℏ=1\hbar=1:

q=a+a†2,p=a−a†i2,[q,p]=i.q=\frac{a+a^\dagger}{\sqrt2},\qquad p=\frac{a-a^\dagger}{i\sqrt2},\qquad [q,p]=i.

For mm modes let R=(q1,p1,…,qm,pm)TR=(q_1,p_1,\ldots,q_m,p_m)^{\mathsf T}. For states with finite second moments, define

dj=⟨Rj⟩,Γjk=12⟨ΔRjΔRk+ΔRkΔRj⟩,ΔRj=Rj−dj.\begin{aligned} d_j&=\langle R_j\rangle,\\ \Gamma_{jk}&=\tfrac12\langle\Delta R_j\Delta R_k+\Delta R_k\Delta R_j\rangle,\\ \Delta R_j&=R_j-d_j. \end{aligned}

The uncertainty relation reads

Γ+i2Ω⪰0,Ω=⨁r=1m(01−10).\Gamma+\frac{i}{2}\Omega\succeq0, \qquad \Omega=\bigoplus_{r=1}^m\begin{pmatrix}0&1\\-1&0\end{pmatrix}.

The vacuum has d=0d=0 and Γ=I/2\Gamma=I/2. This is a useful first check when comparing formulas: other common quadrature conventions assign the vacuum variance the value 11 instead.

A Gaussian state has a Gaussian characteristic function (equivalently a Gaussian Wigner function) and is completely specified by dd and Γ\Gamma. A general non-Gaussian state need not be determined by those moments. For example, a one-mode number state and a suitable thermal state can have the same covariance but different higher moments (Exercise 2).

For a two-mode Gaussian state, with one mode on each side, positivity under partial transpose is necessary and sufficient for separability. At the covariance level, transposition on BB reverses pBp_B and leaves qBq_B fixed; test the uncertainty relation for the transformed matrix. A violation also witnesses entanglement in a non-Gaussian state with these moments, but passing this covariance test does not generally establish separability without Gaussianity. The Gaussian-state article develops the criterion. Weedbrook and collaborators review the general methods, using a different quadrature normalization.

Finite squeezing gives a normalizable EPR approximation

Section titled “Finite squeezing gives a normalizable EPR approximation”

For real 0≤r<∞0\leq r<\infty, set λ=tanh⁡r\lambda=\tanh r. A two-mode squeezed vacuum with the phase convention used here is

∣TMSV(r)⟩=1−λ2∑n=0∞λn∣n,n⟩.|\mathrm{TMSV}(r)\rangle =\sqrt{1-\lambda^2}\sum_{n=0}^\infty\lambda^n|n,n\rangle.

The two occupation numbers always agree. For r>0r>0, neither mode has a definite number; at r=0r=0, both are in the vacuum. Tracing out BB gives the geometric distribution

ρA=(1−λ2)∑n=0∞λ2n∣n⟩⟨n∣,nˉ=sinh⁡2r.\rho_A=(1-\lambda^2)\sum_{n=0}^\infty\lambda^{2n}|n\rangle\langle n|, \qquad \bar n=\sinh^2r.

For r>0r>0 this marginal is mixed although the joint state is pure, so the modes are entangled. Its entropy in nats is (nˉ+1)ln⁡(nˉ+1)−nˉln⁡nˉ(\bar n+1)\ln(\bar n+1)-\bar n\ln\bar n, with 0ln⁡0=00\ln0=0.

In the same phase convention,

Var⁡(qA−qB)=Var⁡(pA+pB)=e−2r.\operatorname{Var}(q_A-q_B) =\operatorname{Var}(p_A+p_B)=e^{-2r}.

These collective observables commute, so both fluctuations can be small. Their conjugate combinations become noisy. The limit r→∞r\to\infty would give exact EPR correlations, but the energy diverges and no normalized Hilbert-space vector is obtained in that limit. Finite squeezing supplies the physical approximation. Entanglement and small quadrature noise do not alone establish a Bell-inequality violation; that requires a specified measurement test.

Sum the geometric probabilities of ρA\rho_A and compute nˉ\bar n. Check the unsqueezed limit.

Solution

Writing x=λ2<1x=\lambda^2<1, ∑n≥0xn=(1−x)−1\sum_{n\geq0}x^n=(1-x)^{-1} gives normalization. Differentiating yields ∑nxn=x/(1−x)2\sum n x^n=x/(1-x)^2, hence nˉ=x/(1−x)=sinh⁡2r\bar n=x/(1-x)=\sinh^2r. At r=0r=0 the joint state is ∣0,0⟩|0,0\rangle, each marginal is pure vacuum, and the displayed collective variances are 11.

Equal covariance does not mean equal state

Section titled “Equal covariance does not mean equal state”

Compare the number state ∣1⟩|1\rangle with the thermal state having mean occupation nˉ=1\bar n=1. Show that both have d=0d=0 and Γ=3I2/2\Gamma=3I_2/2, but different ⟨N2⟩\langle N^2\rangle.

Solution

For either number-diagonal state, ⟨a⟩=⟨a2⟩=0\langle a\rangle=\langle a^2\rangle=0. Thus each quadrature variance is ⟨N⟩+1/2=3/2\langle N\rangle+1/2=3/2, with zero cross covariance. The number state has ⟨N2⟩=1\langle N^2\rangle=1. The thermal probabilities are pn=2−(n+1)p_n=2^{-(n+1)}, giving ⟨N2⟩=nˉ+2nˉ2=3\langle N^2\rangle=\bar n+2\bar n^2=3. The thermal state is Gaussian; ∣1⟩|1\rangle is not.

Compute [qA−qB,pA+pB][q_A-q_B,p_A+p_B] and explain why small variances do not violate either mode’s uncertainty relation.

Solution

Operators of different bosonic modes commute, so the result is i−i=0i-i=0. The conjugate of qA−qBq_A-q_B is proportional to pA−pBp_A-p_B, not pA+pBp_A+p_B. Its fluctuations grow with squeezing. The local variances also grow: each equals 12cosh⁡2r\tfrac12\cosh2r.

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  • A. Serafini, Quantum Continuous Variables: A Primer of Theoretical Methods, CRC Press, 2017.
  • R. Simon, “Peres–Horodecki separability criterion for continuous variable systems,” Physical Review Letters 84, 2726–2729 (2000). DOI.
  • D. F. Walls and G. J. Milburn, Quantum Optics, 2nd ed., Springer, 2008.
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