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 nontrivial two-mode squeezed vacuum is entangled across the two mode factors and is a finite-energy Gaussian approximation to EPR correlations. Squeezing a sufficiently noisy mixed input need not produce entanglement.

Helpful background. Gaussian covariance matrices describe the variances, and creation and annihilation operators describe the pair amplitudes.

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. To check the sign, write $G=a_A^\dagger a_B^\dagger-a_Aa_B$. Differentiating $A(r)=e^{-rG}a_Ae^{rG}$ gives $A'(r)=e^{-rG}[a_A,G]e^{rG}=B^\dagger(r)$, and a second derivative gives $A''(r)=A(r)$. The initial values $A(0)=a_A$ and $A'(0)=a_B^\dagger$ yield the displayed hyperbolic functions.

This is a Bogoliubov transformation: annihilation operators mix with creation operators while preserving the commutators. Here $S_2$ is an active unitary on two fixed mode factors; it creates excitations from their vacuum.

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

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

The entropy is in bits. Locally this density operator is indistinguishable from a thermal state of the same oscillator: the distinction lies in its correlations with mode $B$, not in any measurement on $A$ alone.

Single-mode squeezing can also be converted into two-mode entanglement by linear optics. For the real balanced transformation $q_C=(q_A+q_B)/\sqrt2$, $q_D=(q_A-q_B)/\sqrt2$ and the same transformation for $p$, choose input $A$ squeezed in $p$ and input $B$ squeezed in $q$, with equal strength $r$. Then $q_C-q_D=\sqrt2q_B$ and $p_C+p_D=\sqrt2p_A$ each have variance $e^{-2r}$: the output is the two-mode squeezed vacuum in our phase convention. If instead both inputs have the same squeeze strength and angle, this beam splitter leaves a product of identical squeezed states. The relative squeezing angle matters.

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

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

for degenerate squeezing, so $U(t)=S_1(\kappa t)$ with the convention above, and

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

for two-mode pair creation, so $U(t)=S_2(\kappa t)$. Here $\kappa$ is real and has units of inverse time. These effective interaction-picture Hamiltonians assume a classical undepleted pump, suitable phase matching or resonance, and omission of rapidly oscillating terms. A different pump phase changes the squeezed quadrature; pump depletion limits indefinitely extrapolating the model.

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.

Loss restores vacuum noise

For a pure-loss channel with transmissivity $\eta$, each output quadrature is $q_{\rm out}=\sqrt\eta q+\sqrt{1-\eta}q_{\rm vac}$, with an independent vacuum environment. Thus a single squeezed variance becomes

$$V_{q,\rm out}=\frac12[\eta e^{-2r}+1-\eta].$$

If both modes of the squeezed vacuum suffer equal independent pure loss, each EPR variance becomes $\eta e^{-2r}+1-\eta$. For $r>0$ and any $\eta>0$ this remains below the separable threshold one per variance, though the violation becomes small. Thermal environmental noise can instead remove the violation and eventually the entanglement. These conclusions assume pure loss; they are not universal for arbitrary noise.

Squeezed light develops optical preparation and homodyne detection. Squeezing in metrology explains why reduced noise must be compared with signal response and readout costs, and Bogoliubov quasiparticles applies related transformations to collective excitations.

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.
• M. O. Scully and M. S. Zubairy, Quantum Optics, Cambridge University Press (1997), doi:10.1017/CBO9780511813993.
• D. F. Walls and G. J. Milburn, Quantum Optics, 2nd ed., Springer (2008).
• 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. 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 squeezing descriptions. Put $b_\pm=(a_A\pm a_B)/\sqrt2$. Express $S_2(r)$ as single-mode squeeze operators on $b_+$ and $b_-$.

Solution

The generator becomes

$$a_A^\dagger a_B^\dagger-a_Aa_B =\frac12[(b_+^{\dagger2}-b_+^2)-(b_-^{\dagger2}-b_-^2)].$$

The two terms commute, so $S_2(r)=S_{1,+}(-r)S_{1,-}(r)$. The same state is product in the $+,-$ mode split and entangled in the $A,B$ split. This equality describes a change of mode factors; implementing a beam splitter on fixed physical ports is an active operation.

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. Loss calculation. Each mode has $e^{-2r}=1/4$ and experiences independent pure loss with $\eta=0.6$. Find the sum of output EPR variances and compare with the separable bound.

Solution

Each variance is $0.6/4+0.4=0.55$. Their sum is $1.10<2$, so entanglement remains certified. For a thermal environment of mean occupation $n_{\rm env}$, the same calculation instead gives each variance $\eta e^{-2r}+(1-\eta)(2n_{\rm env}+1)$; vacuum noise and thermal noise must not be interchanged.

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

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