Squeezed States as Entangled Modes

Squeezed states are oscillator states in which noise in one quadrature is reduced below the vacuum level while the conjugate quadrature becomes noisier. They are central in quantum optics because they can be prepared, transformed by linear optics, measured by homodyne detection, and used to generate continuous-variable entanglement.

This page connects squeezing to mode entanglement. A single-mode squeezed state is a one-mode nonclassical state, not automatically a bipartite entangled state. A two-mode squeezed state is entangled across the two mode factors and is the finite-energy Gaussian relative of the ideal EPR state.

Throughout, quadratures are dimensionless:

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

so the vacuum variances are

$$\operatorname{Var}_0(q) = \operatorname{Var}_0(p) = \frac12.$$

Single-Mode Squeezed State Recap

For a real squeezing parameter $r$, define the single-mode squeeze operator

$$S_1(r) = \exp\!\left[ \frac r2 \left( a^2-a^{\dagger 2} \right) \right].$$

With this phase convention,

$$S_1^\dagger(r)aS_1(r) = a\cosh r-a^\dagger\sinh r.$$

Therefore

$$S_1^\dagger(r)qS_1(r) = e^{-r}q, \qquad S_1^\dagger(r)pS_1(r) = e^{r}p.$$

The squeezed vacuum

$$\lvert\mathrm{sq}(r)\rangle = S_1(r)\lvert0\rangle$$

has covariance matrix

$$V_{\rm sq}(r) = \frac12 \begin{pmatrix} e^{-2r} & 0\\ 0 & e^{2r} \end{pmatrix}.$$

Thus

$$\operatorname{Var}(q) = \frac12e^{-2r}, \qquad \operatorname{Var}(p) = \frac12e^{2r},$$

and the uncertainty product remains minimal:

$$\operatorname{Var}(q)\operatorname{Var}(p) = \frac14.$$

In the number basis, the same state contains only even occupation numbers:

$$\lvert\mathrm{sq}(r)\rangle = \frac{1}{\sqrt{\cosh r}} \sum_{n=0}^{\infty} (-\tanh r)^n \frac{\sqrt{(2n)!}}{2^n n!} \lvert2n\rangle.$$

This even-number structure is a signature of pair creation into the same mode. It is nonclassical, but by itself it is still a state of one mode. Entanglement requires a specified split into at least two subsystems or modes.

Two-Mode Squeezed State

For two modes $A$ and $B$, define the two-mode squeeze operator

$$S_2(r) = \exp\!\left[ r \left( a_A^\dagger a_B^\dagger-a_Aa_B \right) \right].$$

Acting on the two-mode vacuum gives

$$\lvert\mathrm{TMSV}(r)\rangle = S_2(r)\lvert0_A,0_B\rangle = \frac{1}{\cosh r} \sum_{n=0}^{\infty} (\tanh r)^n \lvert n_A,n_B\rangle.$$

This formula shows the physical pair structure: excitations are created in mode pairs. The two modes always have the same occupation number in this ideal state.

The coefficients are normalized because

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

The mean occupation of each mode is

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

The transformation of annihilation operators is

$$S_2^\dagger(r)a_A S_2(r) = a_A\cosh r+a_B^\dagger\sinh r,$$

and similarly with $A$ and $B$ interchanged. This is a Bogoliubov transformation: annihilation operators mix with creation operators, so the vacuum of one mode description becomes a many-particle state in another.

Entanglement of Modes

The two-mode squeezed vacuum is already in Schmidt form:

$$\lvert\mathrm{TMSV}(r)\rangle = \sum_{n=0}^{\infty} \sqrt{p_n}\, \lvert n\rangle_A\lvert n\rangle_B,$$

with

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

For $r=0$, only $p_0$ is nonzero and the state is the product vacuum. For any $r>0$, at least two Schmidt coefficients are nonzero, so the state is entangled across the $A,B$ mode split.

Tracing out mode $B$ gives

$$\rho_A = \sum_{n=0}^{\infty} p_n \lvert n\rangle_A\langle n\rvert.$$

This is a thermal oscillator state with mean occupation $\bar n=\sinh^2 r$. The entanglement entropy is

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

This is the same entropy formula that appears for a thermal oscillator, but here the mixedness is not ordinary ignorance about a single oscillator. It is the reduced-state mixedness obtained from a pure entangled two-mode state.

Single-mode squeezing can also be converted into two-mode entanglement by linear optics. For example, mixing two single-mode squeezed vacua with orthogonal squeezing angles on a balanced beam splitter can produce output modes with EPR-like correlations. This is not a contradiction: the entanglement statement depends on the mode decomposition and the physical transformation applied to the modes.

EPR-Like Correlations

The two-mode squeeze transformation gives

$$S_2^\dagger(r)(q_A-q_B)S_2(r) = e^{-r}(q_A-q_B),$$

and

$$S_2^\dagger(r)(p_A+p_B)S_2(r) = e^{-r}(p_A+p_B).$$

Since the two-mode vacuum has

$$\operatorname{Var}_0(q_A-q_B) = \operatorname{Var}_0(p_A+p_B) = 1,$$

the two-mode squeezed vacuum has

$$\operatorname{Var}(q_A-q_B) = e^{-2r}, \qquad \operatorname{Var}(p_A+p_B) = e^{-2r}.$$

These are the finite-squeezing versions of ideal EPR correlations. The collective observables commute:

$$[q_A-q_B,p_A+p_B] = 0.$$

As $r\to\infty$, these variances tend to zero, but the mean occupation $\bar n=\sinh^2 r$ diverges. Exact EPR correlations are therefore an infinite-energy idealization in this family, not a laboratory state.

The Gaussian States Preview gives the covariance-matrix version of the same facts. The EPR State Preview explains why the exact EPR state is a generalized, non-normalizable limit.

Quantum Optics Applications

Squeezed and two-mode squeezed states appear naturally in systems with parametric interactions. Schematic Hamiltonians include

$$H_{\rm sq} = i\hbar\kappa \left( a^{\dagger 2}-a^2 \right)$$

for degenerate squeezing, and

$$H_{\rm tms} = i\hbar\kappa \left( a_A^\dagger a_B^\dagger-a_Aa_B \right)$$

for two-mode pair creation, after moving to an appropriate interaction picture and absorbing phases into the mode definitions.

Applications include:

• sub-vacuum noise measurements in a chosen quadrature;
• gravitational-wave interferometry and other precision measurements;
• continuous-variable teleportation and dense coding protocols;
• EPR steering and entanglement verification with homodyne detection;
• microwave and optical parametric amplifiers;
• pair generation in parametric down-conversion and four-wave mixing.

These applications are not automatic wins. Loss, phase noise, detector inefficiency, finite squeezing, mode mismatch, and excess thermal noise can quickly degrade the useful correlations. A trustworthy claim should state the modes, the squeezing convention, the measured quadratures, and the relevant noise model.

Common Mistakes

• Calling a single-mode squeezed state bipartite entangled without naming a bipartition.
• Treating finite squeezing as exact EPR correlation.
• Forgetting that the squeezed quadrature depends on phase convention.
• Confusing number correlations $\lvert n_A,n_B\rangle$ with product-state occupation $\lvert n_A\rangle_A\lvert n_B\rangle_B$.
• Assuming a nonzero cross-correlation in a mixed Gaussian state is automatically entanglement.
• Ignoring loss and detector inefficiency when translating ideal formulas to experiments.
• Mixing sign conventions for $S_1(r)$ and $S_2(r)$ without updating quadrature correlations.

Cross-Links

• Squeezed Light for single-mode optical preparation, homodyne verification, sideband modes, and loss-sensitive applications.
• Squeezing for the cross-platform relation among covariance, signal response, readout, loss, and metrological gain.
• Continuous-Variable Systems
• Mode Decompositions
• Two-Mode Entanglement
• EPR State Preview
• Gaussian States Preview
• Occupation-Number Basis
• Creation and Annihilation Operators
• Partial Trace
• Subsystem Entropy
• Entanglement Entropy
• Entanglement in Quantum Optics
• Formula Sheet
• Squeezed States: First Encounter
• Ladder-Operator Solution
• Bogoliubov Quasiparticles — bosonic squeezed vacua as quasiparticle vacua, with the fermionic paired-state contrast.

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, doi:10.1103/RevModPhys.77.513.
• 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, doi:10.1103/RevModPhys.84.621.
• M. D. Reid, P. D. Drummond, W. P. Bowen, E. G. Cavalcanti, P. K. Lam, H. A. Bachor, U. L. Andersen, and G. Leuchs, “Colloquium: The Einstein–Podolsky–Rosen paradox: From concepts to applications”, Reviews of Modern Physics 81, 1727-1751, 2009, doi:10.1103/RevModPhys.81.1727.

Exercises

1. Single-mode uncertainty product. Use the squeezed variances to show that a single-mode squeezed vacuum remains a minimum-uncertainty state.

Solution

The variances are

$$\operatorname{Var}(q) = \frac12e^{-2r}, \qquad \operatorname{Var}(p) = \frac12e^{2r}.$$

Their product is

$$\operatorname{Var}(q)\operatorname{Var}(p) = \frac14e^{-2r}e^{2r} = \frac14.$$

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

Solution

The squared coefficients sum to

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

Using $1-\tanh^2 r=1/\cosh^2 r$, this equals $1$.

3. Reduced state. Trace out mode $B$ from the two-mode squeezed vacuum and show that the result is diagonal in the number basis.

Solution

Start from

$$\rho_{AB} = \sum_{m,n} \sqrt{p_m p_n}\, \lvert m\rangle_A\lvert m\rangle_B \langle n\rvert_A\langle n\rvert_B.$$

Taking the trace over $B$ gives $\langle n\vert m\rangle_B=\delta_{mn}$, so all off-diagonal terms in $m,n$ vanish:

$$\rho_A = \sum_n p_n \lvert n\rangle_A\langle n\rvert.$$

4. Entanglement criterion. Explain why $\lvert\mathrm{TMSV}(r)\rangle$ is entangled for every $r>0$.

Solution

The number-basis expansion is a Schmidt decomposition with probabilities

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

For $r>0$, $\tanh r>0$, so both $p_0$ and $p_1$ are nonzero. The Schmidt rank is therefore larger than one, and the state is entangled.

5. EPR variance. Use

$$S_2^\dagger(r)(q_A-q_B)S_2(r) = e^{-r}(q_A-q_B)$$

to compute $\operatorname{Var}(q_A-q_B)$ in the two-mode squeezed vacuum.

Solution

In the two-mode squeezed state,

$$\operatorname{Var}_{\rm TMSV}(q_A-q_B) = \operatorname{Var}_{0} \left( e^{-r}(q_A-q_B) \right).$$

The two-mode vacuum has $\operatorname{Var}_0(q_A-q_B)=1$, because each vacuum mode contributes $1/2$ and there is no cross-correlation. Hence

$$\operatorname{Var}_{\rm TMSV}(q_A-q_B) = e^{-2r}.$$