Continuous-Variable Systems

A continuous-variable system has canonical observables such as position and momentum, or oscillator quadratures, whose measurement outcomes range continuously. Its Hilbert space is infinite-dimensional even when another useful observable, such as oscillator occupation number, has a discrete spectrum. For composite systems, the familiar tensor-product and partial-trace rules still apply; normalization and convergence now need care.

Helpful background. Tensor products identify the subsystems, while square-integrable wavefunctions explain which position-space expressions represent normalizable states.

Infinite-Dimensional Tensor Products

For two distinguishable systems, the composition rule is still

$$\mathcal H_{AB} = \mathcal H_A\otimes\mathcal H_B.$$

The new feature is that $\mathcal H_A$ and $\mathcal H_B$ may be infinite-dimensional. For one particle moving on a line,

$$\mathcal H = L^2(\mathbb R).$$

For two distinguishable particles on a line,

$$\mathcal H_{12} = L^2(\mathbb R)\otimes L^2(\mathbb R) \cong L^2(\mathbb R^2).$$

The isomorphism says that a two-particle state can be represented by a square-integrable wavefunction $\Psi(x_1,x_2)$, with normalization

$$\int_{-\infty}^{\infty}dx_1 \int_{-\infty}^{\infty}dx_2\, |\Psi(x_1,x_2)|^2 = 1.$$

This is the continuous-variable version of a coefficient array $\psi_{ij}$ for two finite-dimensional systems. The indices $i,j$ have become continuous coordinates $x_1,x_2$, and sums have become integrals.

For finitely many subsystems, this tensor-product rule is conceptually close to the finite-dimensional case. Infinite tensor products over infinitely many modes, spatial points, or field degrees of freedom are more delicate. In nonrelativistic many-body physics one often works with a finite or countable mode basis and then takes limits carefully. In relativistic QFT, local algebras and operator-valued distributions require additional structure beyond this introductory page.

Two Particles on a Line

A pure product state of two distinguishable particles on a line has the form

$$\Psi(x_1,x_2) = \psi(x_1)\phi(x_2).$$

Here $\psi$ is a normalized state of particle 1 and $\phi$ is a normalized state of particle 2. The probability density factorizes:

$$|\Psi(x_1,x_2)|^2 = |\psi(x_1)|^2|\phi(x_2)|^2.$$

An entangled pure state is a normalized $\Psi(x_1,x_2)$ that cannot be written as one product wavefunction. A common structural form is a Schmidt expansion,

$$\Psi(x_1,x_2) = \sum_n \sqrt{p_n}\, u_n(x_1)v_n(x_2),$$

where $\{u_n\}$ and $\{v_n\}$ are orthonormal sets and $p_n\ge0$ with $\sum_n p_n=1$. For a normalizable vector in these separable Hilbert spaces, the sum is countable and converges in Hilbert-space norm. The state is product exactly when only one nonzero Schmidt coefficient appears. The entropy $-\sum_n p_n\log_2p_n$ may nevertheless diverge.

The reduced density operator of particle 1 has position-space kernel

$$\rho_1(x,x') = \int_{-\infty}^{\infty}dy\, \Psi(x,y)\Psi^*(x',y).$$

This is the partial trace written in a continuous position representation. The formula is useful, but it should not obscure the operator meaning: $\rho_1$ is the unique operator that reproduces all expectations of bounded observables acting only on particle 1. Unbounded observables, such as position or energy, additionally require the relevant moments to exist.

For a product state $\Psi(x,y)=\psi(x)\phi(y)$, the reduced kernel becomes

$$\rho_1(x,x') = \psi(x)\psi^*(x') \int_{-\infty}^{\infty}dy\,|\phi(y)|^2 = \psi(x)\psi^*(x'),$$

so particle 1 remains pure. For an entangled state, $\rho_1$ is mixed, and the subsystem entropy can be nonzero.

The detailed position-space treatment, including phase correlations, reduced kernels, and center-of-mass coordinates, is developed in Position-Space Two-Particle States.

Two Modes of an Oscillator Field

Continuous-variable systems are not limited to particle positions. A harmonic oscillator mode is also an infinite-dimensional quantum system. Its Hilbert space is spanned by number states,

$$\mathcal H_{\rm mode} = \overline{\operatorname{span}} \{\lvert0\rangle,\lvert1\rangle,\lvert2\rangle,\ldots\}.$$

For two modes $A$ and $B$,

$$\mathcal H_{AB} = \mathcal H_A\otimes\mathcal H_B, \qquad \lvert n_A,n_B\rangle = \lvert n_A\rangle_A\otimes\lvert n_B\rangle_B.$$

The state

$$\lvert\Psi\rangle = \frac{1}{\sqrt2} \left( \lvert1_A,0_B\rangle + \lvert0_A,1_B\rangle \right)$$

is entangled with respect to the mode decomposition into subsystem $A$ and subsystem $B$. It is not a product state of mode $A$ and mode $B$. This example is finite in excitation number, but each mode Hilbert space remains infinite-dimensional.

Mode entanglement is decomposition-dependent. If one changes the mode basis, for example by a beam-splitter-like transformation, a state that looks entangled in one mode decomposition may look simple in another. This is not a contradiction; it is the continuous-variable version of the general rule that entanglement is defined relative to a specified tensor-product structure.

Oscillator modes also have quadrature operators. In dimensionless units,

$$q = \frac{a+a^\dagger}{\sqrt2}, \qquad p = \frac{a-a^\dagger}{i\sqrt2},$$

with

$$[q,p]=i.$$

These commutator identities hold on a common invariant domain, for example finite linear combinations of number states. The vacuum has $\operatorname{Var}(q)=\operatorname{Var}(p)=1/2$. Quadratures behave mathematically like position and momentum variables for a mode. They are the natural language of quantum optics, continuous-variable quantum information, and Gaussian-state methods.

Normalizability Caveats

Many useful continuous-variable expressions are distributions, not normalizable states. Position eigenkets obey

$$\langle x\vert x'\rangle = \delta(x-x'),$$

so $\lvert x\rangle$ is not a vector in $L^2(\mathbb R)$. Momentum eigenkets and plane waves have the same status. They are generalized eigenstates used inside integrals, not physical normalizable states by themselves.

The same warning applies to idealized EPR-like expressions such as

$$\int_{-\infty}^{\infty}dx\, \lvert x\rangle_1\lvert x\rangle_2.$$

This expression captures perfect position correlation, but its wavefunction is proportional to $\delta(x_1-x_2)$, which is not square-integrable. Restricting the center position to a finite box does not cure the zero-width relative coordinate: a formal squared norm would involve a product of delta distributions, not just the box length.

A useful replacement has two finite widths. Put $u=(x_1-x_2)/\sqrt2$, $v=(x_1+x_2)/\sqrt2$, and take

$$\Psi_{\sigma,L}(x_1,x_2) =\frac{1}{\sqrt{\pi\sigma L}} \exp\left(-\frac{u^2}{2\sigma^2}-\frac{v^2}{2L^2}\right), \qquad \sigma,L>0.$$

The coordinate transformation has absolute Jacobian one. Each Gaussian integral is finite, and their product makes the norm one. Narrow $\sigma$ gives a small position difference; large $L$ gives a small total momentum spread. Both widths must remain finite and nonzero for this family to describe ordinary normalized vectors. Normalizability by itself does not guarantee finite variance for every possible wavefunction; finite moments are additional assumptions.

The EPR State Preview develops this warning into the standard pair of commuting collective observables, $X_1-X_2$ and $P_1+P_2$.

Normalizability matters because entanglement measures, reduced density operators, and expectation values can fail to be well-defined for ideal distributions. A delta-correlated expression may be an excellent limiting model, but the limit should not be confused with an ordinary Hilbert-space vector.

Gaussian-State Preview

Gaussian states are the most important controlled family of continuous-variable states. They are states whose Wigner functions or characteristic functions are Gaussian in phase-space variables. For $N$ modes, collect the quadratures into

$$R = (q_1,p_1,\ldots,q_N,p_N)^T.$$

A Gaussian state is determined by its first moments

$$d_i = \langle R_i\rangle$$

and covariance matrix

$$V_{ij} = \frac12 \langle \Delta R_i\Delta R_j+\Delta R_j\Delta R_i \rangle, \qquad \Delta R_i=R_i-\langle R_i\rangle.$$

In our convention a physical covariance obeys $V+i\Omega/2\ge0$, where $\Omega=\bigoplus_j\begin{pmatrix}0&1\\-1&0\end{pmatrix}$. This is the uncertainty principle in matrix form. A merely positive classical covariance need not satisfy it.

This finite set of moments can encode an infinite-dimensional state because Gaussianity is a strong restriction. Coherent states, squeezed states, thermal oscillator states, and two-mode squeezed states are central examples.

The preview should be read with two cautions:

• not every continuous-variable state is Gaussian;
• Gaussian entanglement has efficient covariance-matrix criteria, but those criteria do not replace the general definition of entanglement.

The Gaussian States Preview develops this language more carefully. For now, the key point is that continuous-variable entanglement often becomes tractable when the state is Gaussian and the relevant subsystems are modes.

Physical Interpretation

Continuous-variable composition appears in several guises:

• two or more particles with spatial wavefunctions;
• center-of-mass and relative coordinates for composite motion;
• optical modes, microwave modes, phonon modes, and trap modes;
• collective variables in many-body systems;
• field modes that prepare the transition to QFT.

The conceptual question is always the same: what are the subsystems? For particles, the tensor factors may be particle Hilbert spaces. For fields and optics, they may be modes. For spatial regions, the factorization question becomes more subtle, especially in QFT and gauge theories.

Two distinctions to keep in mind

Infinite dimension does not force large entanglement. The normalized state $(u_0v_0+u_1v_1)/\sqrt2$, with orthonormal local wavefunctions, has only two nonzero Schmidt probabilities. Each reduced state has purity $1/2$ and entropy one bit despite the infinite-dimensional ambient spaces.

Continuous labels do not imply a continuum of Schmidt coefficients. Position kets are generalized eigenvectors. The Schmidt vectors of a normalizable bipartite state are ordinary normalized vectors, and its nonzero Schmidt coefficients form a countable sequence.

For applications, continuous-variable computation uses mode registers and Gaussian or non-Gaussian operations; continuous-variable platforms compares sources, control, detection, and noise.

References

• S. L. Braunstein and P. van Loock, “Quantum information with continuous variables,” Reviews of Modern Physics 77, 513–577 (2005), doi:10.1103/RevModPhys.77.513.
• B. C. Hall, Quantum Theory for Mathematicians, Springer (2013), doi:10.1007/978-1-4614-7116-5.
• C. Weedbrook, S. Pirandola, R. García-Patrón, N. J. Cerf, T. C. Ralph, J. H. Shapiro, and S. Lloyd, “Gaussian quantum information,” Reviews of Modern Physics 84, 621–669 (2012), doi:10.1103/RevModPhys.84.621.

Exercises

1. Product wavefunction. Let $\Psi(x_1,x_2)=\psi(x_1)\phi(x_2)$ with $\psi$ and $\phi$ normalized. Show that $\Psi$ is normalized.

Solution

Compute

$$\int dx_1\,dx_2\, |\Psi(x_1,x_2)|^2 = \int dx_1\,|\psi(x_1)|^2 \int dx_2\,|\phi(x_2)|^2.$$

Each factor equals $1$, so the product equals $1$.

2. Reduced kernel for a product state. For $\Psi(x,y)=\psi(x)\phi(y)$, compute $\rho_1(x,x')$.

Solution

By definition,

$$\rho_1(x,x') = \int dy\, \Psi(x,y)\Psi^*(x',y).$$

Substituting the product form gives

$$\rho_1(x,x') = \psi(x)\psi^*(x') \int dy\,|\phi(y)|^2 = \psi(x)\psi^*(x'),$$

because $\phi$ is normalized.

3. Mode entanglement. For

$$\lvert\Psi\rangle = \frac{1}{\sqrt2} \left( \lvert1_A,0_B\rangle + \lvert0_A,1_B\rangle \right),$$

trace out mode $B$ and find the reduced state of mode $A$.

Solution

The density operator is $\rho=\lvert\Psi\rangle\langle\Psi\rvert$. Orthogonality of $\lvert0_B\rangle$ and $\lvert1_B\rangle$ removes the cross terms under the trace over $B$, giving

$$\rho_A = \frac12 \lvert1_A\rangle\langle1_A\rvert + \frac12 \lvert0_A\rangle\langle0_A\rvert.$$

The reduced state is mixed, so the two modes are entangled.

4. Ideal EPR normalization. Explain why $\int dx\,\lvert x\rangle_1\lvert x\rangle_2$ is not normalizable.

Solution

Its wavefunction is a delta distribution in the relative coordinate. A delta distribution is not an $L^2$ function, even if the other coordinate is restricted to a finite interval. The Gaussian family above instead gives

$$\operatorname{Var}(X_1-X_2)=\sigma^2,\qquad \operatorname{Var}(P_1+P_2)=\frac{\hbar^2}{L^2}.$$

The conjugate spreads are $\operatorname{Var}(P_1-P_2)=\hbar^2/\sigma^2$ and $\operatorname{Var}(X_1+X_2)=L^2$. Thus approaching zero EPR spreads makes the complementary spreads unbounded; there is no normalized ideal EPR vector at the endpoint.

5. Quadrature commutator. Using $[a,a^\dagger]=1$, show that $q=(a+a^\dagger)/\sqrt2$ and $p=(a-a^\dagger)/(i\sqrt2)$ obey $[q,p]=i$.

Solution

Compute

$$\begin{aligned} [q,p] &= \frac{1}{2i} [a+a^\dagger,a-a^\dagger] \\ &= \frac{1}{2i} \left( -[a,a^\dagger]+[a^\dagger,a] \right) \\ &= \frac{1}{2i} \left( -1-1 \right) = i. \end{aligned}$$