Constructing squeezed states of light with associated Hermite polynomials

Abstract

A new class of states of light is introduced that is complementary to the well-known squeezed states. The construction is based on the general solution of the three-term recurrence relation that arises from the saturation of the Schrödinger inequality for the quadratures of a single-mode quantized electromagnetic field. The new squeezed states are found to be linear superpositions of the photon-number states whose coefficients are determined by the associated Hermite polynomials. These results do not seem to have been noticed before in the literature. As an example, the new class of squeezed states includes superpositions characterized by odd-photon number states only, so they represent the counterpart of the prototypical squeezed-vacuum state which consists entirely of even-photon number states.

Appendices

A Variances associated to the squeezed states

Consider the operators $A=A^{\dagger }$ and $B=B^{\dagger }$, which act on ${\mathcal {H}}={\text {Span}}\{ \vert n\rangle \}_{n=0}^{\infty }$. Let $\vert \beta \rangle \in {\mathcal {H}}$ be a solution of the eigenvalue equation

$$\begin{aligned} (A-i\lambda B)\vert \beta \rangle =\beta \vert \beta \rangle \, , \quad \beta =\langle A\rangle -i\lambda \langle B\rangle \in {\mathbb {C}} \, . \end{aligned}$$

(A-1)

The variances of A and B are easily computed using ${\widetilde{A}}=A-\langle A \rangle$ and ${\widetilde{B}}=B-\langle B \rangle$ through

$$\begin{aligned} (\Delta A)^{2}=\langle {\widetilde{A}}^{2}\rangle \, , \quad (\Delta B)^{2}=\langle {\widetilde{B}}^{2}\rangle . \end{aligned}$$

(A-2)

From Eq. (A-1), it follows ${\widetilde{A}}\vert \beta \rangle =i\lambda {\widetilde{B}}\vert \beta \rangle$, so that $\widetilde{A}^{2}\vert \beta \rangle =i\lambda {\widetilde{A}}{\widetilde{B}}\vert \beta \rangle$. Then,

$$\begin{aligned}&(\Delta A)^{2}\equiv \langle {\widetilde{A}}^{2}\rangle =i\lambda \langle {\widetilde{A}}{\widetilde{B}}\rangle =i\lambda \langle \beta \vert \left( \frac{AB+BA}{2}+\frac{AB-BA}{2}-\langle A\rangle \langle B\rangle \right) \vert \beta \rangle \nonumber \\&\quad =i\lambda \left( \sigma _{A,B}+\frac{i}{2}\vert \langle [A,B]\rangle \vert \right) \, , \end{aligned}$$

(A-3)

where we have used $[A,B]^{\dagger }=-[A,B]$. Considering now the complex number $\lambda =\vert \lambda \vert e^{i\theta _{\lambda }}$ and $(\sigma _{A,B}+\frac{i}{2}\vert \langle [A,B]\rangle \vert )=(\sigma ^{2}_{A,B}+\frac{1}{4}\vert \langle [A,B]\rangle \vert ^{2}) e^{i\theta _{A,B}}$, we obtain

$$\begin{aligned} (\Delta A)^{2}=\vert \lambda \vert \left( \sigma ^{2}_{A,B}+\frac{1}{4}\vert \langle [A,B]\rangle \vert ^{2}\right) ^{1/2} \, e^{i\left( \frac{\pi }{2}+\phi _{\lambda }+\phi _{A,B}\right) } \, . \end{aligned}$$

(A-4)

The complex phase in (A-4) is fixed by recalling $(\Delta A)^{2}\ge 0$, as $A=A^{\dagger }$, and the expectation value is computed through regular vectors. Therefore, we find the following relation between the complex phases:

$$\begin{aligned} \frac{\pi }{2}+\phi _{\lambda }+\phi _{A,B}=2n\pi \, , \quad n=0,1,\ldots , \end{aligned}$$

(A-5)

and consequently, the variance for A is reduced to

$$\begin{aligned} (\Delta A)^{2}_{\lambda }=\vert \lambda \vert \left( \sigma ^{2}_{A,B}+\frac{1}{4}\vert \langle [A,B]\rangle _{\lambda }\vert ^{2}\right) ^{1/2}. \end{aligned}$$

(A-6)

The calculations for $(\Delta B)^{2}$ are carried out in a similar way. We multiply ${\widetilde{B}}$ to the left of the eigenvalue equation ${\widetilde{A}} \vert \beta \rangle =i\lambda {\widetilde{B}}\vert \beta \rangle$, and the straightforward calculation, together with the condition for the complex phases in Eq. (A-5), leads to

$$\begin{aligned} (\Delta B)^{2}_{\lambda }=\frac{1}{\vert \lambda \vert }\left( \sigma ^{2}_{A,B}+\frac{1}{4}\vert \langle [A,B]\rangle _{\lambda }\vert ^{2}\right) ^{1/2} \, . \end{aligned}$$

(A-7)

Using $A=x=\frac{1}{\sqrt{2}}(a^{\dagger }+a)$ and $B=p=\frac{i}{\sqrt{2}}(a^{\dagger }-a)$ in the above results, we recover Eq. (5).

B Solving recurrence relations by the comparison method

In this appendix, we construct the general solution of recurrence relations before considering the initial conditions. Our approach, hereafter referred to as comparison method, is addressed to determine whether the solutions of two recurrence relations can be paired by assuming that one of the recurrence problems is already solved. We focus on three-term recurrence relations, but the procedure is easily adapted to recurrences having a different number of terms.

Consider the recurrences

$$\begin{aligned} a_{n} f_{n+1}+ b_{n}f_{n}+ d_{n}f_{n-1}=0, \quad a_n, b_n, d_n\in {\mathbb {C}}, \quad n=1,2, \ldots \, , \end{aligned}$$

(B-1)

and

$$\begin{aligned} A_{n} F_{n+1} + B_{n} F_{n} + D_{n} F_{n-1}=0, \quad A_n, B_n, D_n\in {\mathbb {C}}, \quad n=1,2, \ldots \end{aligned}$$

(B-2)

Assuming that the set $F_n$ defined by (B-2) is already known, we want to determine whether $f_{n}$ can be written in terms of $F_{n}$. The affirmative answer depends strictly on the profile and properties of $F_{n}$.

Let $h_n$ be a function such that $f_n= h_n F_n$. Introducing it into Eq. (B-1) and comparing the result with (B-2), we arrive at the relationships

$$\begin{aligned} \frac{h_{n+1}}{h_n}=\frac{A_n b_n}{B_n a_{n}}, \qquad \frac{h_{n+1}}{h_{n}}=\frac{B_{n+1} d_{n+1}}{D_{n+1} b_{n+1}}. \end{aligned}$$

(B-3)

Both equations in (B-3) should lead to the same function $h_n$, so we impose the compatibility condition

$$\begin{aligned} \frac{A_{n} D_{n+1} b_{n}b_{n+1}}{B_{n} B_{n+1} a_{n}d_{n+1}}=1, \quad n=0,1,2, \ldots \end{aligned}$$

(B-4)

If (B-4) is fulfilled, the auxiliary function is obtained by solving any of the recurrence relations in (B-3). One finds

$$\begin{aligned} h_{n+1}=h_0 \prod _{k=0}^{n}\frac{A_k b_k}{B_k a_k}= h_0 \prod _{k=0}^{n}\frac{B_{k+1} d_{k+1}}{D_{k+1} b_{k+1}}, \quad n=0,1, \ldots , \end{aligned}$$

(B-5)

where $h_0$ may be determined from the initial conditions through $f_{n}=h_{n} F_{n}$.

1.1 B-2.1 The three-term recurrence relation of Sect. 2.2

We apply the comparison method to solve the recurrence relation (9) of Sect. 2.2. First we show the way in which the conventional results are recovered and then we obtain more general results.

1.1.1 B-2.1.1 Usual solution

Let us rewrite the recurrence relation (9) as follows:

$$\begin{aligned} P_{n+1}- \alpha P_{n}+ \xi n P_{n-1}=0, \quad n=1,2, \ldots \end{aligned}$$

(B-6)

To apply the comparison method we use $f_{n}=P_{n}(\alpha ,\xi )$, with $a_{n}=1$, $b_n=\alpha$ and $d_{n}= \xi n$.

Exploring the well-known recurrence relations for the classical orthogonal polynomials [80], we find that the recurrence relation for the Hermite polynomials

$$\begin{aligned} H_{n+1}(z)-2zH_{n}(z)+2nH_{n-1}(z)=0 \, ,\quad n=1,2, \ldots , \end{aligned}$$

(B-7)

is useful in the present case. That is, taking $F_{n}=H_{n}(z)$, with $A_{n}=1$, $B_{n}=-2z$, and $D_{n}=2n$, the compatibility condition (B-5) is fulfilled with $z=\alpha /\sqrt{2\xi }$. Therefore, we obtain $h_{n}=h_0 \left( \frac{\xi }{2}\right) ^{n/2}$.

Now, taking into account the constraint (10), meaning $P_0 \ne 0$, we may fix $h_0$ by the initial condition $P_{0}(\alpha ,\xi )=1$. Therefore, $f_{n} = h_{n} F_{n}$ yields the well known result $P_{n}(\alpha ,\xi )=(\xi /2)^{n/2}H_{n}(\alpha /\sqrt{2\xi })$. These roots of the recurrence problem (9)–(10) have been used to recover the expression of the conventional squeezed states $\vert \alpha , \xi ; + \rangle$ in Eq. (28) of the main text.

1.1.2 B-2.1.2 General solution

Looking for a general solution, one should recall that the confluent hypergeometric function ${}_{1}F_{1}(a,c;z) \equiv M(a,c; z)$, with $a =-n$ and $c \ne -m$ yields a polynomial of degree n in z [53] (n and m positive integers). Then, we may wonder whether the solutions of (B-6) can be paired with such polynomials. A first insight is obtained by comparing the confluent hypergeometric recurrence relation

$$\begin{aligned} (c-a) M(a-1,c;z) + (2a-c+ z) M(a,c;z) -a M(a+1,c;z)=0, \quad a=-n, c\ne -m, \end{aligned}$$

(B-8)

with Eq. (B-6) since it makes clear that the compatibility condition (B-4) cannot be achieved. Nevertheless, decoupling (B-6) into even and odd values of n, we, respectively, have

$$\begin{aligned} P_{2n+2} (\alpha ,\xi ) + \left( 4 \xi n + \xi -\alpha ^2 \right) P_{2n} (\alpha ,\xi )+ 4 \xi ^2 n \left( n-\tfrac{1}{2} \right) P_{2n-2} ( \alpha , \xi )=0, \end{aligned}$$

(B-9)

and

$$\begin{aligned} P_{2n+3} (\alpha ,\xi ) + \left( 4 \xi n + 3\xi -\alpha ^2 \right) P_{2n+1} (\alpha , \xi )+ 4 \xi ^2 n \left( n + \tfrac{1}{2} \right) P_{2n-1}(\alpha ,\xi )=0. \end{aligned}$$

(B-10)

These results are now compatible with (B-8) for either $c=\tfrac{1}{2}$ or $c= \tfrac{3}{2}$. It is useful to recall the relationship between the confluent hypergeometric function and the Hermite polynomials

$$\begin{aligned} M(-n, \tfrac{1}{2}; x^2)= (-1)^{n} \frac{n!}{(2n)!} H_{2n}(x), \quad M(-n, \tfrac{3}{2}; x^2)= \frac{ (-1)^{n} }{2x} \frac{n!}{(2n+1)!} H_{2n+1}(x). \end{aligned}$$

(B-11)

Thus, in the present case we may consider $F_n = M(-n,c;z)$ with c equal to either 1/2 or 3/2 in order to get the corresponding auxiliary function (B-5). The latter provides a first solution to the problem. A second solution can be obtained by recalling that the confluent hypergeometric equation admits two linearly independent solutions. Given $y_1=M(a,c;z)$, the function $y_4=z^{1-c} e^z M(1-a, 2-c, -z)$ is such that $W(y_1, y_4) =(1-c) z^{-c} e^z$ [53], so that $y_1$ and $y_2$ are linearly independent if $c\ne 1$. Therefore, if $f_{2n} = h_{2n} M(-n, \tfrac{1}{2}, z)$ is our first solution, we may write ${{\widetilde{h}}}_{2n} M(1+n, \tfrac{3}{2}, -z)$ for the second one, with ${{\widetilde{h}}}_{2n}$ absorbing the factors $z^{1/2} e^z$ and being to be determined. In this form, the general solution for the even labels 2n is written as a linear combination of the above functions. The straightforward calculation yields

$$\begin{aligned} P_{2n} = (-2\xi )^{n} \left[ \left( \tfrac{1}{2} \right) _{n} \kappa _{1}\, M \!\left( -n,\tfrac{1}{2}; \tfrac{\alpha ^{2}}{2\xi }\right) + n! \, \widetilde{\kappa }_{1} \, M\! \left( n+1, \tfrac{3}{2};- \tfrac{\alpha ^{2}}{2\xi }\right) \right] , \end{aligned}$$

(B-12)

where the complex-valued coefficients $\kappa _{1} (\alpha ,\xi )$ and $\widetilde{\kappa }_{1} (\alpha ,\xi )$ are fixed by the initial conditions. Equivalently, for the odd labels $2n+1$, we have

$$\begin{aligned} P_{2n+1} = (-2\xi )^{n} \left[ \left( \tfrac{3}{2} \right) _n \kappa _{2} M \! \left( -n,\tfrac{3}{2};\tfrac{\alpha ^{2}}{2\xi }\right) + n! \, \widetilde{\kappa }_{2} M\! \left( n+1,\tfrac{1}{2};-\tfrac{\alpha ^{2}}{2\xi }\right) \right] , \end{aligned}$$

(B-13)

with $\kappa _{2} (\alpha ,\xi )$ and $\widetilde{\kappa }_{2} (\alpha ,\xi )$ defined by the initial conditions.

• Solutions obeying the constraint (10). Taking into account the constraint (10), that is $P_0 \ne 0$, we may take $P_{0}(\alpha ,\xi )=1$. Then, $P_{1}(\alpha ,\xi )=\alpha$, and

  $$\begin{aligned} {\widetilde{\kappa }}_{1} (\alpha ,\xi )= \widetilde{\kappa }_{2} (\alpha ,\xi )=0, \quad \kappa _{1} (\alpha ,\xi )= 1, \quad {\kappa }_{2} (\alpha ,\xi )= \alpha . \end{aligned}$$

  The above results permit to recover the well-known expression of the squeezed states (28).

• Solutions that do not satisfy the constraint (10). Making $P^{(2)}_{0}(\alpha ,\xi )=0$ and $P^{(2)}_{1}(\alpha ,\xi )=1$ we find ${\widetilde{\kappa }}_{1} (\alpha ,\xi )= \widetilde{\kappa }_{2} (\alpha ,\xi )=1$, and

  $$\begin{aligned} \kappa _{1} (\alpha ,\xi )= -M \! \left( 1, \tfrac{3}{2};- \tfrac{\alpha ^{2}}{2\xi } \right) , \quad {\kappa }_{2} (\alpha ,\xi )= \tfrac{\alpha ^2}{\xi } M \! \left( 1, \tfrac{3}{2};- \tfrac{\alpha ^{2}}{2\xi } \right) . \end{aligned}$$

  After some calculations, from the above expressions one arrives at the results presented in Eqs. (32) and (33) of the main text.

C Orthogonal and associated polynomials

Following [81], we consider a set $\{ p_n(x) \}$ of orthogonal polynomials

$$\begin{aligned} p_n (x) = \gamma _n x^n + \gamma _{n-1} x^{n-1} + \cdots + \gamma _0, \quad \gamma _n >0, \end{aligned}$$

that satisfy the three-term recurrence relation

$$\begin{aligned} x p_n(x) = a_{n+1} p_{n+1} (x) + b_n p_n(x) + a_n p_{n-1}(x), \quad n \ge 0, \end{aligned}$$

(C-1)

with initial values

$$\begin{aligned} p_{-1}(x) =0, \quad p_0(x) =1, \end{aligned}$$

(C-2)

and

$$\begin{aligned} a_n = \frac{\gamma _{n-1}}{\gamma _n} >0, \quad b_n = \int x p_n^2(x) d\mu (x) \in {\mathbb {R}}. \end{aligned}$$

(C-3)

The function $\mu (x)$ in the recurrence coefficients (C-3) is a probability measure on the real line such that

$$\begin{aligned} \int p_{n}(x)p_{m}(x) d\mu (x)=\delta _{nm}, \quad m, n \ge 0. \end{aligned}$$

(C-4)

Markedly, except for the classical orthogonal polynomials [53], finding the measure $\mu$ and the solutions $p_n(x)$ of the system (C-1)–(C-4) represents a formidable amount of work in general. In this respect the Favard’s theorem [69] (see also [68]) is very useful since it states that for the recurrence problem defined by (C-1)–(C-2), there exists a probability measure $\mu$ so that the recurrence coefficients acquire the form (C-3) and the orthogonality (C-4) is satisfied [81], and vice versa. Therefore, it is natural to concentrate in solving (C-1)–(C-2) and then to allude the Favard’s theorem to ensure orthogonality.

A slight alteration of the three-term recurrence relation (C-1) produces new results. Namely,

$$\begin{aligned} x p_n^{(k)} (x) = a_{n+k+1} p_{n+1}^{(k)} (x) + b_{n+k} p_n^{(k)} (x) + a_{n+k} p_{n-1}^{(k)} (x), \quad n \ge 0, \end{aligned}$$

(C-5)

with

$$\begin{aligned} p_{-1}^{(k)} (x) =0, \quad p_0^{(k)} (x) =1, \end{aligned}$$

(C-6)

defines the kth associated orthogonal polynomials $p_n^{(k)}(x)$ [81, 82] (associated with the ones with $k=0$), also called numerator polynomials [68]. Given k, the set $\{ p_n^{(k)}(x) \}$ defines a solution of the recurrence relation (C-1) with $\mu ^{(k)}$ the corresponding measure. (Guidelines for determining $\mu$ can be found in [68, 83], and references quoted therein.) The associated recurrence problem (C-5)–(C-6) is very useful for the purposes of this work since it permits to avoid the strong restriction $P_0 \ne 0$ from the constraint (10).

In the main text, we work with (C-1) rewritten in the form: [68, 84]

$$\begin{aligned} p_{n+1}=(x-c_{n})p_{n}-\lambda _{n}p_{n-1}, \quad p_{-1}=0, \quad p_{0}=1, \quad n=0,1,2,\ldots , \end{aligned}$$

(C-7)

where the coefficients $c_n$ and $\lambda _n$ are complex in general.

We identify (C-7) with a second-order difference equation [48], so there are two independent solutions for each value of n. If $p_{n}$ and $g_{n}$ solve a difference equation, they are independent if

$$\begin{aligned} {\mathcal {C}}(p_n,g_n)= {\text {Det}}\left( \begin{array}{ll} p_{n} \quad &{} g_{n} \\ p_{n+1} \quad &{} g_{n+1} \end{array}\right) \ne 0. \end{aligned}$$

(C-8)

The above determinant is known as the Casorati function [48] and is the discrete version of the Wronskian in differential equations.

Given a first solution $p_{n}$, a second solution $g_{n}$ such that ${\mathcal {C}}(p_n,g_n) \ne 0$ may be constructed by reducing the order of the corresponding difference equation [85]. The method is summarized with the algorithm

$$\begin{aligned} \frac{g_{n+1}}{p_{n+1}}=d_{0} + d_1 \sum _{m=0}^{n} \frac{{\mathcal {R}}(m)}{p_{m}p_{m+1}} \, , \quad g_{0}=0 , \quad {\mathcal {R}}(m)=\prod _{k=0}^{m-1}\lambda _k , \end{aligned}$$

(C-9)

with $d_{0}$ and $d_{1}$ arbitrary complex constants. Nevertheless, such a method yields unnecessary complications. We circumvent them by considering the additional difference problem

$$\begin{aligned} p_{n+1}^{(1)}=(x-c_{n+1}) p_{n}^{(1)} -\lambda _{n+1}p_{n-1}^{(1)}, \quad p_{-1}^{(1)}=0 \, , \quad p_{0}^{(1)}=1, \quad n=0,1, \ldots , \end{aligned}$$

(C-10)

where $\lambda _{n}$ and $c_{n}$ are the same as those defining (C-7). Thus, we pay attention to the associated polynomials $p_{n}^{(1)} (x)$ of the $p_n(x)$ that solve (C-7).

Making $g_{n}=p_{n-1}^{(1)}$, and considering the initial condition $p_{-1}^{(1)}=0$, produces $g_{0}=0$, which is the initial condition considered in the algorithm (C-9). Therefore, the difference problem (C-10) acquires the form

$$\begin{aligned} g_{n+1}=(x-c_n)g_{n}-\lambda _n g_{n-1}, \quad g_{0}=0, \quad g_{1}=1. \end{aligned}$$

(C-11)

It is now clear that the associated polynomials $g_n$ do not satisfy the constraint (10) since $g_0 =0$. Besides, the Casorati function between $p_{n}$ and $g_{n}$ is $C(p_n,g_n)=\lambda _1$ [81], so these solutions are independent, provided that $\lambda _1 \ne 0$. That is, the second independent solution $g_{n}$ given by Eq. (C-9) may be also computed from the recurrence relation (C-11).