Two-Mode Entanglement

Two-mode entanglement is entanglement between two specified modes, such as two optical paths, two cavity modes, two polarization modes, two lattice sites, or two oscillator modes. The tensor factors are the mode Hilbert spaces, not hidden particle labels.

Helpful background. Mode decompositions fixes the subsystem factors, and the Schmidt decomposition turns a pure-state entanglement calculation into an eigenvalue problem. Entropies below use base-two logarithms and are measured in bits.

For two bosonic modes $A$ and $B$, the occupation basis is

$$\lvert n_A,n_B\rangle = \lvert n_A\rangle_A\otimes\lvert n_B\rangle_B, \qquad n_A,n_B=0,1,2,\ldots .$$

A pure two-mode state can be written

$$\lvert\Psi\rangle = \sum_{m,n=0}^{\infty} C_{mn}\lvert m_A,n_B\rangle, \qquad \sum_{m,n} \lvert C_{mn}\rvert^2 = 1.$$

It is a product state across the two-mode split exactly when the coefficient matrix factors as

$$C_{mn} = u_m v_n$$

for two normalized sequences $u_m$ and $v_n$. Otherwise the state is entangled across the chosen mode decomposition.

Pure-State Criterion

The general pure-state criterion is the Schmidt decomposition. If

$$\lvert\Psi\rangle = \sum_k \sqrt{p_k}\, \lvert u_k\rangle_A \lvert v_k\rangle_B, \qquad p_k\ge0, \qquad \sum_k p_k=1,$$

then the state is product if exactly one Schmidt probability is nonzero. It is entangled if at least two Schmidt probabilities are nonzero.

The reduced density operator of mode $A$ is

$$\rho_A = \operatorname{Tr}_B \lvert\Psi\rangle\langle\Psi\rvert.$$

For a pure bipartite state, the nonzero eigenvalues of $\rho_A$ are the Schmidt probabilities $p_k$. The mode entanglement entropy is

$$S_A = -\operatorname{Tr}(\rho_A\log_2\rho_A) = -\sum_k p_k\log_2 p_k.$$

The coefficient matrix is a Hilbert–Schmidt operator, so this countable Schmidt expansion exists for every normalized pure two-mode state. The entropy sum can be infinite; the finite-energy oscillator examples below have finite entropy.

Single Excitation Across Two Modes

The simplest two-mode entangled state has one excitation delocalized over two modes:

$$\lvert\Psi_1\rangle = \alpha\lvert1_A,0_B\rangle + \beta\lvert0_A,1_B\rangle, \qquad \lvert\alpha\rvert^2+\lvert\beta\rvert^2=1.$$

Unless $\alpha=0$ or $\beta=0$, this state is not a product of mode $A$ and mode $B$. Tracing out mode $B$ gives

$$\rho_A = \lvert\alpha\rvert^2 \lvert1_A\rangle\langle1_A\rvert + \lvert\beta\rvert^2 \lvert0_A\rangle\langle0_A\rvert.$$

The entanglement entropy is

$$S_A = -\lvert\alpha\rvert^2\log_2\lvert\alpha\rvert^2 -\lvert\beta\rvert^2\log_2\lvert\beta\rvert^2.$$

For the balanced state

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

the entropy is $1$ bit.

This example is sometimes called “single-particle entanglement,” but that phrase can mislead. The entanglement is between modes. Its operational meaning depends on the available local operations, phase references, and any relevant particle-number superselection constraints.

Two-Mode Number Correlations

Not every two-mode occupation state is entangled. The state

$$\lvert1_A,1_B\rangle = \lvert1_A\rangle_A\otimes\lvert1_B\rangle_B$$

is a product state across the mode split. It has one excitation in each mode, but no mode entanglement.

By contrast, the state

$$\lvert\Psi_2\rangle = \frac{1}{\sqrt2} \left( \lvert2_A,0_B\rangle + \lvert0_A,2_B\rangle \right)$$

is entangled. Its reduced state is

$$\rho_A = \frac12 \lvert2_A\rangle\langle2_A\rvert + \frac12 \lvert0_A\rangle\langle0_A\rvert,$$

so $S_A=\log_2 2$. This state is a two-excitation version of a path-entangled state.

The distinction is structural: definite occupation in each mode can be a product state, while coherent superpositions of different occupation patterns can be entangled.

Two-Mode Squeezed State Preview

The two-mode squeezed vacuum is a central continuous-variable entangled state. In a common phase convention it is

$$\lvert\mathrm{TMSV}(r)\rangle = \frac{1}{\cosh r} \sum_{n=0}^{\infty} (\tanh r)^n \lvert n_A,n_B\rangle, \qquad r\ge0.$$

This is already in Schmidt form. The Schmidt probabilities are

$$p_n = \frac{(\tanh^2 r)^n}{\cosh^2 r}.$$

They sum to one because

$$\sum_{n=0}^{\infty} \frac{(\tanh^2 r)^n}{\cosh^2 r} = \frac{1}{\cosh^2 r} \frac{1}{1-\tanh^2 r} = 1.$$

The mean occupation per mode is

$$\bar n = \sinh^2 r.$$

The entanglement entropy of either mode is the thermal-oscillator entropy

$$S = (\bar n+1)\log_2(\bar n+1) -\bar n\log_2\bar n.$$

For every $r>0$, the two occupation numbers agree perfectly in an ideal number measurement. Increasing $r$ broadens their common distribution and increases the entanglement; it does not improve an already exact equality. Number correlation alone does not certify that entanglement: the dephased state

$$\rho_{\rm deph}=\sum_n p_n\lvert n\rangle\langle n\rvert_A\otimes\lvert n\rangle\langle n\rvert_B$$

is separable and has the same joint number distribution. The coherent amplitudes between different $n$ sectors distinguish the squeezed pure state from this mixture. This state is the normalizable, experimentally meaningful relative of idealized EPR-like position and momentum correlations developed in the EPR State Preview. The broader covariance-matrix analysis begins in the Gaussian States Preview, while Squeezed States as Entangled Modes develops the squeeze-operator viewpoint.

Beam-Splitter Generated States

Mode entanglement often appears when a mode transformation mixes input modes into output modes. A balanced beam-splitter-like transformation can be represented by

$$a_A^\dagger \mapsto \frac{a_C^\dagger+a_D^\dagger}{\sqrt2}, \qquad a_B^\dagger \mapsto \frac{a_C^\dagger-a_D^\dagger}{\sqrt2}.$$

If the input is one excitation in mode $A$ and vacuum in mode $B$,

$$\lvert1_A,0_B\rangle = a_A^\dagger\lvert0\rangle,$$

then the output is

$$\frac{1}{\sqrt2} \left( \lvert1_C,0_D\rangle + \lvert0_C,1_D\rangle \right).$$

With two identical bosonic excitations entering opposite input modes, the same transformation gives

$$\lvert1_A,1_B\rangle \mapsto \frac{1}{\sqrt2} \left( \lvert2_C,0_D\rangle - \lvert0_C,2_D\rangle \right),$$

up to phase conventions. This cancellation assumes the two excitations match in all unresolved degrees of freedom, such as temporal profile and polarization. Distinguishable photons do not have the same two-mode description and generally retain coincidence events.

The word “generated” should be read with care. The physical device changes which output modes are occupied; it does not make entanglement absolute. The entanglement statement is always relative to the output mode split being used.

Operational Caveats

Two-mode entanglement is mathematically precise once the mode tensor factors are fixed. Its operational use depends on additional structure:

• The modes must be physically addressable, at least approximately.
• Loss, detector inefficiency, and mode mismatch turn pure states into mixed states.
• Local particle-number restrictions can limit which operations detect the coherence in single-excitation states.
• A shared phase reference may be needed to access certain superpositions.
• Nonorthogonal or poorly controlled modes can make the intended tensor-product split only approximate.

These are not reasons to avoid mode entanglement. They are the assumptions that make a laboratory claim well-posed.

A beam splitter does not entangle every input

For product coherent input states, the same linear optical device gives

$$\lvert\alpha\rangle_A\lvert\beta\rangle_B \longmapsto \left\lvert\frac{\alpha+\beta}{\sqrt2}\right\rangle_C \left\lvert\frac{\alpha-\beta}{\sqrt2}\right\rangle_D.$$

The output remains product. Mixing modes is therefore not, by itself, proof of entanglement; the input state matters. Photon-number states and suitably oriented squeezed inputs behave differently.

Quantum illumination provides a different application of two-mode squeezed signal–idler correlations: the target-detection task can retain an advantage even when noise destroys the return–idler entanglement. Such a claim requires a specified noise model and comparison resources.

References

• S. D. Bartlett, T. Rudolph, and R. W. Spekkens, “Reference frames, superselection rules, and quantum information,” Reviews of Modern Physics 79, 555–609 (2007), doi:10.1103/RevModPhys.79.555.
• 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.
• M. O. Scully and M. S. Zubairy, Quantum Optics, Cambridge University Press (1997), doi:10.1017/CBO9780511813993.
• 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. Reduced state of a single-excitation state. For $\lvert\Psi_1\rangle=\alpha\lvert1_A,0_B\rangle+\beta\lvert0_A,1_B\rangle$, compute $\rho_A$.

Solution

The density operator contains four terms. The cross terms vanish under the trace over $B$ because $\langle0_B\vert1_B\rangle=0$. Thus

$$\rho_A = \lvert\alpha\rvert^2 \lvert1_A\rangle\langle1_A\rvert + \lvert\beta\rvert^2 \lvert0_A\rangle\langle0_A\rvert.$$

2. Correlation without entanglement. For $\rho_{\rm deph}$ above, compare the joint number probabilities and reduced state with those of the squeezed pure state. Can $S(\rho_A)$ measure their entanglement in both cases?

Solution

Both give $P(n_A,n_B)=p_{n_A}\delta_{n_A,n_B}$ and $\rho_A=\sum_n p_n\lvert n\rangle\langle n\rvert$. The mixture is explicitly separable, while the squeezed pure state has multiple nonzero Schmidt coefficients for $r>0$. Therefore the same positive local entropy measures entanglement only for the globally pure state. Occupation statistics do not reveal the missing coherences.

3. Two-mode squeezed normalization. Verify that the coefficients of $\lvert\mathrm{TMSV}(r)\rangle$ are normalized.

Solution

The norm is

$$\sum_{n=0}^{\infty} \frac{(\tanh^2 r)^n}{\cosh^2 r}.$$

Using the geometric series,

$$\sum_{n=0}^{\infty} (\tanh^2 r)^n = \frac{1}{1-\tanh^2 r}.$$

Since $1-\tanh^2 r=1/\cosh^2 r$, the norm is $1$.

4. Finite occupation cutoff. Put $z=\tanh^2r$. If a numerical calculation retains $n=0,\ldots,N$ in the squeezed-state sum and renormalizes, what is the squared overlap with the exact state?

Solution

The omitted probability is the geometric tail $\delta=\sum_{n=N+1}^\infty(1-z)z^n=z^{N+1}$. Orthogonal projection followed by normalization gives squared overlap $1-\delta$. For $z=1/2$ and $N=9$, the overlap is $1-2^{-10}\approx0.999023$. A chosen cutoff must be checked against $r$; increasing squeezing at fixed cutoff worsens the approximation.

5. Product or entangled? Decide whether $\lvert1_A,1_B\rangle$ and $(\lvert2_A,0_B\rangle+\lvert0_A,2_B\rangle)/\sqrt2$ are product or entangled across the $A,B$ mode split.

Solution

The state $\lvert1_A,1_B\rangle$ is

$$\lvert1_A\rangle_A\otimes\lvert1_B\rangle_B,$$

so it is product. The state

$$\frac{1}{\sqrt2} \left( \lvert2_A,0_B\rangle + \lvert0_A,2_B\rangle \right)$$

has two nonzero Schmidt terms with orthogonal mode states, so it is entangled.