Epsilon coherent states with polyanalytic coefficients for the harmonic oscillator

Abstract

We construct a new class of coherent states indexed by points z of the complex plane and depending on two positive parameters m and $\varepsilon >0$ by replacing the coefficients $z^{n}/\sqrt{n!}$ of the canonical coherent states by polyanalytic functions. These states solve the identity of the states Hilbert space of the harmonic oscillator at the limit $\varepsilon \rightarrow 0^{+}$ and obey a thermal stability property. Their wavefunctions are obtained in a closed form and their associated Bargmann-type transform is also discussed.

Appendices

Appendix A

Proof

Using the orthogonality relations of the basis elements $\left\{ \varphi _{n}\left( x\right) \right\}$ in (3.7) the scalar product in $L^{2}\left( \mathbb {R}\right)$ beteween two $\varepsilon$-CS can written as

$$\begin{aligned} \langle z;m,\varepsilon \left| w;m,\varepsilon \right\rangle _{L^{2}\left( \mathbb {R}\right) }=\frac{Q_{\varepsilon }\left( z,w\right) }{ \pi m!\sqrt{\mathcal {N}_{m,\varepsilon }\left( z\right) \mathcal {N} _{m,\varepsilon }\left( w\right) }} \end{aligned}$$

(A1)

where

$$\begin{aligned} Q_{\varepsilon }\left( z,w\right) =\sum \limits _{n=0}^{+\infty }\frac{ e^{-n\varepsilon }}{n!}\Phi _{n}^{m}\left( z\right) \overline{\Phi _{n}^{m}\left( w\right) }. \end{aligned}$$

(A2)

Recalling the explicite expression (2.7) of the of the polyanalytic coefficients,  we can split the sum in (A2) into two part as

$$\begin{aligned} Q_{\varepsilon }\left( z,w\right) =\sum \limits _{n=0}^{m-1}e^{-n\varepsilon }n!\left( \left| z\right| \left| w\right| \right) ^{\left( m-n\right) }L_{n}^{\left( m-n\right) }\left( z\overline{z}\right) L_{n}^{\left( m-n\right) }\left( w\overline{w}\right) e^{-i(m-n)\arg z}e^{i(m-n)\arg w}\nonumber \\ +\sum \limits _{n=m}^{+\infty }\frac{e^{-n\varepsilon }}{n!}\left( m!\right) ^{2}\left( \left| z\right| \left| w\right| \right) ^{\left( n-m\right) }L_{m}^{\left( n-m\right) }\left( z\overline{z}\right) L_{m}^{\left( n-m\right) }\left( w\overline{w}\right) e^{-i(m-n)\arg z}e^{i(m-n)\arg w}.\nonumber \\ \end{aligned}$$

(A3)

This quantity can also be decomposed as

$$\begin{aligned} Q_{\varepsilon }\left( z,w\right) =Q_{\varepsilon }^{\left( <\infty \right) }\left( z,w\right) +Q_{\varepsilon }^{\left( \infty \right) }\left( z,w\right) \end{aligned}$$

(A4)

with a finite sum

$$\begin{aligned} Q_{\varepsilon }^{\left( <\infty \right) }\left( z,w\right) :=\sum \limits _{n=0}^{m-1}e^{-n\varepsilon }n!\left( \overline{z}w\right) ^{m-n}L_{n}^{\left( m-n\right) }\left( z\overline{z}\right) L_{n}^{\left( m-n\right) }\left( w\overline{w}\right) \nonumber \\ -\sum \limits _{n=0}^{m-1}\frac{e^{-n\varepsilon }}{n!}\left( m!\right) ^{2}\left( z\overline{w}\right) ^{n-m}L_{m}^{\left( n-m\right) }\left( z \overline{z}\right) L_{m}^{\left( n-m\right) }\left( w\overline{w}\right) \end{aligned}$$

(A5)

and an infinite sum

$$\begin{aligned} Q_{\varepsilon }^{\left( \infty \right) }\left( z,w\right) :=\sum \limits _{n=0}^{+\infty }\frac{e^{-n\varepsilon }}{n!}\left( m!\right) ^{2}\left( z\overline{w}\right) ^{n-m}L_{m}^{\left( n-m\right) }\left( z \overline{z}\right) L_{m}^{\left( n-m\right) }\left( w\overline{w}\right) . \end{aligned}$$

(A6)

Making appeal to the identity ([24], p.98):

$$\begin{aligned} L_{m}^{\left( -k\right) }\left( t\right) =\left( -t\right) ^{k}\frac{\left( m-k\right) !}{m!}L_{m-k}^{\left( k\right) }\left( t\right) ,\quad 1\le k\le m \end{aligned}$$

(A7)

for $k=j-m$ and $t=z\overline{z}$, we can check that the finite sum $Q_{\varepsilon }^{\left( <\infty \right) }\left( z,w\right) =0$. For the infinite sum in (A6) , we rewrite it as

$$\begin{aligned} Q_{\varepsilon }^{\left( \infty \right) }\left( z,w\right) =\frac{\left( m!\right) ^{2}}{\left( z\overline{w}\right) ^{m}}\sum \limits _{n=0}^{+\infty } \frac{1}{n!}\left( z\overline{w}e^{-\varepsilon }\right) ^{n}L_{m}^{\left( n-m\right) }\left( z\overline{z}\right) L_{m}^{\left( n-m\right) }\left( w \overline{w}\right) . \end{aligned}$$

(A8)

We now apply the Wicksell–Campbell–Meixner formula ([25], p.279) : 

$$\begin{aligned} \sum \limits _{n=0}^{+\infty }\frac{\zeta ^{n}}{n!}L_{l}^{\left( n-l\right) }\left( X\right) L_{m}^{\left( n-m\right) }\left( Y\right) =e^{\zeta }\left( \zeta -Y\right) ^{m-l}\frac{\zeta ^{l}}{m!}L_{l}^{\left( m-l\right) }\left( -\left( X-\zeta \right) \left( Y-\zeta \right) \zeta ^{-1}\right) \end{aligned}$$

(A9)

with the notations $\zeta =e^{-\varepsilon }z\overline{w}$, $X=z\overline{z} ,Y=w\overline{w}$ and $l=m.$ With this, Eq.(A8) reduces to

$$\begin{aligned} Q_{\varepsilon }^{\left( \infty \right) }\left( z,w\right) =m!e^{-m\varepsilon }\exp \left( e^{-\varepsilon }z\overline{z}\right) L_{m}^{\left( 0\right) }\left( \left( we^{\varepsilon }-z\right) \left( \overline{w}e^{-\varepsilon }-\overline{z}\right) \right) . \end{aligned}$$

(A10)

Finally, we replace this last expression in the right hand side of (A2) to arrive at the expression (4.3) . We put $z=w$ in (4.3) and we use the condition $\langle z;m,\varepsilon \left| z;m,\varepsilon \right\rangle _{L^{2}\left( \mathbb {R}\right) }=1.$ This allows us to obtain the expression (4.4) of the normalization factor. $\square$

Appendix B

Proof

For $x\in \mathbb {R},$ we can write successively

$$\begin{aligned} \mathcal {O}_{\varepsilon }\left[ \varphi \right] \left( x\right)&=\sum _{n=0}^{+\infty }e^{-n\varepsilon }\langle \varphi \left| \varphi _{n}\right\rangle \langle x|\varphi _{n}\rangle \end{aligned}$$

(B1)

$$\begin{aligned}&=\sum _{n=0}^{+\infty }e^{-n\varepsilon }\left( \int _{-\infty }^{+\infty }\varphi \left( y\right) \overline{\langle y\left| \varphi _{n}\right\rangle }dy\right) \langle x|\varphi _{n}\rangle \end{aligned}$$

(B2)

$$\begin{aligned}&=\int _{-\infty }^{+\infty }\left( \sum \limits _{n=0}^{+\infty }e^{-n\varepsilon }\overline{\langle y\left| \varphi _{n}\right\rangle } \langle x|\varphi _{n}\rangle \right) \varphi \left( y\right) dy. \end{aligned}$$

(B3)

We now look closely at the sum

$$\begin{aligned} \mathcal {G}_{\varepsilon }\left( x,y\right) :=\sum \limits _{n=0}^{+\infty }e^{-n\varepsilon }\overline{\langle y\left| \varphi _{n}\right\rangle } \langle x\left| \varphi _{n}\right\rangle =\sum \limits _{n=0}^{+\infty }e^{-n\varepsilon }\varphi _{n}\left( x\right) \overline{\varphi _{n}\left( y\right) }. \end{aligned}$$

(B4)

Recalling the expression (3.7) of the $\left\{ \varphi _{n}\right\}$ then (4.21) reads

$$\begin{aligned} \mathcal {G}_{\varepsilon }\left( x,y\right) =\frac{1}{\sqrt{\pi }}e^{-\frac{1 }{2}\left( x^{2}+y^{2}\right) }\sum \limits _{n=0}^{+\infty }\left( \frac{1}{2} e^{-\varepsilon }\right) ^{n}\frac{1}{n!}H_{n}\left( x\right) H_{n}\left( y\right) . \end{aligned}$$

(B5)

Equation (B5) can be rewritten as

$$\begin{aligned} \mathcal {G}_{\varepsilon }\left( x,y\right) =e^{-\frac{1}{2}\left( x^{2}+y^{2}\right) }K\left( e^{-\varepsilon };x,y\right) \end{aligned}$$

(B6)

where we have introduced the kernel function

$$\begin{aligned} K\left( \tau ;x,y\right) :=\sum \limits _{j=0}^{+\infty }\tau ^{j}\frac{1}{j!} H_{j}\left( x\right) H_{j}\left( y\right) ;\quad 0<\tau <1. \end{aligned}$$

(B7)

The latter can be written in a closed form by applying the Mehler formula ( [19], p.252):

$$\begin{aligned} K\left( \tau ;x,y\right) =\frac{\pi ^{-\frac{1}{2}}}{\sqrt{1-\tau ^{2}}}\exp \left( \frac{2\tau }{1+\tau }xy-\frac{\tau ^{2}}{1-\tau ^{2}}\left( x-y\right) ^{2}\right) \end{aligned}$$

(B8)

which also is the Poisson kernel for the Hermite polynomials expansion. Taking this into account, Eq. (B3) takes the form

$$\begin{aligned} \mathcal {O}_{\varepsilon }\left[ \varphi \right] \left( x\right) =e^{-\frac{1 }{2}x^{2}}\int _{-\infty }^{+\infty }\varphi \left( y\right) e^{-\frac{1}{2} y^{2}}K\left( e^{-\varepsilon },x,y\right) dy. \end{aligned}$$

(B9)

We can also write the right hand side of (B9) as

$$\begin{aligned} \mathcal {O}_{\varepsilon }\left[ \varphi \right] \left( x\right) =e^{-\frac{1 }{2}x^{2}}M_{\varepsilon }\left[ \varphi \right] \left( x\right) , \end{aligned}$$

(B10)

where

$$\begin{aligned} M_{\varepsilon }\left[ \varphi \right] \left( u\right) =\int _{-\infty }^{+\infty }K\left( e^{-\varepsilon },x,y\right) \varphi \left( y\right) e^{- \frac{1}{2}y^{2}}dy. \end{aligned}$$

(B11)

This suggests us to introduce the function

$$\begin{aligned} f\left( y\right) :=\varphi \left( y\right) e^{-\frac{1}{2}y^{2}},y\in \mathbb {R}. \end{aligned}$$

(B12)

which statisfies

$$\begin{aligned} \left\| f\right\| _{L^{2}\left( \mathbb {\,R},e^{-y^{2}}dy\right) }=\left\| \varphi \right\| _{L^{2}\left( \mathbb {R}\right) }. \end{aligned}$$

(B13)

We now apply the result of B. Muckenhoupt [23] who considered the Poisson integral of Hermite polynomials expansion and proved that for a function $f\in L^{p}\left( \mathbb {R},e^{-y^{2}}dy\right)$ with $1\le p\le +\infty$ the integral defined by

$$\begin{aligned} A\left[ f\right] \left( \tau ,x\right) :=\int _{0}^{+\infty }K\left( \tau ,x,y\right) f\left( y\right) e^{-y^{2}}dy;\quad 0\le \tau <1 \end{aligned}$$

(B14)

with the kernel $K\left( \tau ,\bullet ,\bullet \right)$ defined as given in (B8) satisfies $\lim _{\tau \rightarrow 1^{-}}A\left[ f \right] \left( \tau ,y\right) =f\left( y\right)$ almost everywhere in $[ 0,+\infty [ ,1\le p\le \infty .$ We apply this result in the case $p=2,$$A\equiv M$ and $\tau =e^{-\varepsilon }$ to obtain that $M_{\varepsilon }\left[ \varphi \right] \left( x\right) \rightarrow e^{\frac{1 }{2}x^{2}}\varphi \left( x\right)$, a.e. as $\varepsilon \rightarrow 0^{+}$ , which says that the limit $\mathcal {O}_{\varepsilon }\left[ \varphi \right] \left( x\right) =e^{-\frac{1}{2}x^{2}}M_{\varepsilon }\left[ \varphi \right] \left( x\right) \rightarrow \varphi \left( x\right) ,a.e$. as $\varepsilon \rightarrow 0^{+}$ is valid for every $\varphi \in L^{2}\left( \mathbb {R}\right)$. In other words, we get the limit

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0^{+}}\int \limits _{\mathbb {C}}\left| z;m,\varepsilon \right\rangle \langle z;m,\varepsilon |d\mu _{m,\varepsilon }\left( z\right) =\mathbf {1}_{L^{2}\left( \mathbb {R}\right) .} \end{aligned}$$

(B15)

in terms of Dirac’s bra-ket notation. This completes the proof. $\square$