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.

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

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

This is the same mathematical structure as finite-dimensional entanglement, but the mode Hilbert spaces are infinite-dimensional.

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

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

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 $\log2$.

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 $S_A=\log2$. 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

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

As $r$ grows, the number correlations become stronger. 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 is the algebraic core behind the familiar two-boson bunching effect.

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.

Common Mistakes

• Calling $\lvert1_A,1_B\rangle$ entangled merely because it contains two modes.
• Treating a single excitation across two modes as particle-label entanglement.
• Ignoring the mode basis in which the state is written.
• Forgetting that a beam splitter changes the relevant input-output mode description.
• Using two-mode squeezed-state formulas without specifying the phase and squeezing convention.
• Treating ideal EPR states as normalizable two-mode squeezed states at finite squeezing.
• Neglecting superselection and reference-frame assumptions when discussing single-excitation entanglement.

Cross-Links

• Continuous-Variable Systems
• Mode Decompositions
• EPR State Preview
• Gaussian States Preview
• Squeezed States as Entangled Modes
• Entanglement Depends on a Decomposition
• Occupation-Number Basis
• Mode Occupations
• Bosonic Fock Space
• Particle-Number Superselection Preview
• Partial Trace
• Subsystem Entropy
• Creation and Annihilation Operators
• Entanglement in Quantum Optics
• Quantum Illumination uses two-mode-squeezed signal–idler correlations for target-channel discrimination even when the returned state is no longer entangled.
• Formula Sheet

References

• D. F. Walls and G. J. Milburn, Quantum Optics, 2nd ed., Springer, 2008.
• M. O. Scully and M. S. Zubairy, Quantum Optics, Cambridge University Press, 1997.
• C. C. Gerry and P. L. Knight, Introductory Quantum Optics, Cambridge University Press, 2005.
• S. L. Braunstein and P. van Loock, “Quantum information with continuous variables”, Reviews of Modern Physics 77, 513-577, 2005.
• C. Weedbrook, S. Pirandola, R. Garcia-Patron, N. J. Cerf, T. C. Ralph, J. H. Shapiro, and S. Lloyd, “Gaussian quantum information”, Reviews of Modern Physics 84, 621-669, 2012.
• S. D. Bartlett, T. Rudolph, and R. W. Spekkens, “Reference frames, superselection rules, and quantum information”, Reviews of Modern Physics 79, 555-609, 2007.

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. Balanced entropy. Show that the balanced single-excitation state has entanglement entropy $\log2$.

Solution

For $\alpha=\beta=1/\sqrt2$, the reduced eigenvalues are $1/2$ and $1/2$. Therefore

$$S_A = -\frac12\log\frac12 -\frac12\log\frac12 = \log2.$$

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. Beam splitter with one input excitation. Apply

$$a_A^\dagger \mapsto \frac{a_C^\dagger+a_D^\dagger}{\sqrt2}$$

to $\lvert1_A,0_B\rangle$ and write the output state.

Solution

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

$$a_A^\dagger\lvert0\rangle \mapsto \frac{1}{\sqrt2} \left( a_C^\dagger\lvert0\rangle + a_D^\dagger\lvert0\rangle \right).$$

Thus

$$\lvert1_A,0_B\rangle \mapsto \frac{1}{\sqrt2} \left( \lvert1_C,0_D\rangle + \lvert0_C,1_D\rangle \right).$$

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.