Coherent states for continuous spectrum operators with non-normalizable fiducial states

One of the most simple examples for which one can test the above ideas is the motion of a free particle on a straight line $\mathbb {R}$. After canonical quantization and setting [Q, P] = iℏ for Q and P self-adjoint, the system can be described by the quantum Hamiltonian given by

$$\begin{eqnarray} &&H = \frac{1}{2m} P^2 = \frac{-\hbar ^2 }{2m} \frac{{\rm d}^2}{{\rm d}x^2}. \end{eqnarray} \tag{ 18 }$$

The Hilbert space of the system is simply given by the complex span of the position operator Q eigenbasis denoted by |x〉. H admits the following set of eigenvectors and eigenfunctions:

$$\begin{eqnarray} && H |\Psi _k \rangle = \frac{\hbar ^2 k^2}{2m} |\Psi _k \rangle , \nonumber \\ && \langle x |\Psi _k \rangle =\Psi _k(x) = \frac{1}{\sqrt{2\pi }} \,{\rm e}^{{\rm i} k x }, \qquad \langle x |H |\Psi _k \rangle = \frac{\hbar ^2 k^2}{2m} \Psi _k(x). \end{eqnarray} \tag{ 19 }$$

The properties of the states |Ψ_k〉 are direct: they are not normalizable and obey

$$\begin{eqnarray} &&\langle \Psi _{k} |\Psi _{k^{\prime }}\rangle = \delta (k^{\prime }-k). \end{eqnarray} \tag{ 20 }$$

Thus, for any k, |Ψ_k〉 cannot be chosen as a fiducial vector for defining coherent states for the system.

The next stage is to define two new functions, such that for $\bar{k}\equiv \bar{k}(k_0,k_1) \in (k_0,k_1)$, k₀ ≠ k₁, and for A ≫ 1,

$$\begin{eqnarray} &&{\rm (a)}\quad \Psi _{\bar{k}}(x) = C\int _{k_0}^{k_1} \Psi _k(x) \,{\rm d}k = \frac{ C}{i \sqrt{2\pi }x} [{\rm e}^{{\rm i} k_1 x } - {\rm e}^{{\rm i} k_0 x }], \nonumber \\ &&{\rm (b)}\quad \Psi _{\bar{k};A}(x) = C_A\int _{-\infty }^{\infty } \,{\rm e}^{-A(k-\bar{k})^2} \Psi _k(x) \,{\rm d}k = \frac{C_A}{\sqrt{2A}} \,{\rm e}^{- \frac{x^2}{4A} + {\rm i} \bar{k} x }\;. \end{eqnarray} \tag{ 21 }$$

For any $\bar{k}\in (k_0,k_1)$, we compute the norm of the first function (a):

$$\begin{eqnarray} &&\fl \int _{-\infty }^{\infty } \Psi ^*_{\bar{k}}(x)\Psi _{\bar{k}^{\prime }}(x) \,{\rm d}x = C^2 (k_1 -k_0) = 1 \quad \Leftrightarrow \quad C = \frac{1}{\sqrt{ k_1 -k_0}}. \end{eqnarray} \tag{ 22 }$$

The overlap becomes a finite quantity and is normalized when C is fixed appropriately as done above. On the other hand, for the second type of state (b), we have

$$\begin{eqnarray} &&\fl \int _{-\infty }^{\infty } (\Psi _{\bar{k};A}(x))^* \Psi _{\bar{k};A}(x) \,{\rm d}x = C^2_A \sqrt{\frac{\pi }{2A}} = 1 \quad \Leftrightarrow \quad C_A = \left(\frac{2A}{\pi }\right)^{\frac{1}{4}}. \end{eqnarray} \tag{ 23 }$$

Thus, these states are normalized.

The two vector states $|\Psi _{\bar{k}} \rangle$ and $|\Psi _{\bar{k}} \rangle _A$ can be defined from $\Psi _{\bar{k}}(x) = \langle x |\Psi _{\bar{k}} \rangle$ and $\Psi _{\bar{k};A}(x) = \langle x |\Psi _{\bar{k}} \rangle _A$, respectively,

$$\begin{eqnarray} &&|\Psi _{\bar{k}} \rangle = C\int _{k_0}^{k_1}|\Psi _{ k} \rangle \; {\rm d}k \;,\qquad \; |\Psi _{\bar{k}} \rangle _A = C_A\int _{-\infty }^{\infty }\,{\rm e}^{-A(k-\bar{k})^2} |\Psi _{ k} \rangle \; {\rm d}k\;. \end{eqnarray} \tag{ 24 }$$

These will be considered as our normalized fiducial vectors.

Introducing the unitary operator $U(q,p) = {\rm e}^{-\frac{{\rm i}}{\hbar }q P} \,{\rm e}^{\frac{{\rm i}}{\hbar } p Q}$, $q\in \mathbb {R}$, $p\in \mathbb {R}$, we define two families of states such that

$$\begin{eqnarray} &&\fl {\rm (a)}\quad |q,p;\bar{k} \rangle = U(q,p) |\Psi _{\bar{k}} \rangle , \quad \langle x|q,p;\bar{k} \rangle =\Phi _{\bar{k}}(q,p;x) = {\rm e}^{\frac{{\rm i}}{\hbar } p (x- q)}\Psi _{\bar{k}}( x-q), \end{eqnarray} \tag{ 25 }$$

$$\begin{eqnarray} &&\fl {\rm (b)}\quad |q,p;\bar{k} \rangle _A = U(q,p) |\Psi _{\bar{k}} \rangle _A, \quad \langle x|q,p;\bar{k} \rangle _A =\Phi _{\bar{k};A}(q,p;x) ={\rm e}^{\frac{{\rm i}}{\hbar } p (x- q)}\Psi _{\bar{k};A}( x-q). \nonumber\\ \end{eqnarray} \tag{ 26 }$$

Let us check the GK axioms of coherent states for the set of states $\lbrace |q,p;\bar{k}\rangle \rbrace _{q,p\in \mathbb {R}}$ using the family $\lbrace \Phi _{\bar{k}}(q,p,x)\rbrace _{q,p\in \mathbb {R}}$.

• (i)  
  The continuity in the labels q, p is obvious.
• (ii)  
  The normalizability condition can be checked since it corresponds to the condition that⁴ $\Phi _{\bar{k}; \bullet }(q,p,x) \in L^{2}(\mathbb {R},\,{\rm d}x)$ is of norm 1, the symbol • ∈ {∅, A}. We have

  $$\begin{eqnarray} &&\fl \bullet \langle q,p;\bar{k}|q,p;\bar{k}\rangle _\bullet = \int _{-\infty }^{\infty } (\Phi _{\bar{k};\bullet }(q,p,x))^{*} \Phi _{\bar{k};\bullet }(q,p,x) \;{\rm d}x \nonumber\\ &&= \int _{-\infty }^{\infty } (\Psi _{\bar{k};\bullet }( x-q))^* \Psi _{\bar{k};\bullet }( x-q) \;{\rm d}x \end{eqnarray} \tag{ 27 }$$

  which, by a change of variable $\tilde{x} =x-q$, reduces to the earlier calculations on the L²-normalizability of $\Psi _{\bar{k};\bullet }(x)$. This result can be naturally seen as a consequence of the unitarity of the operator U(q, p).
• (iii)  
  The resolution of the identity has to be verified:

  $$\begin{eqnarray} &&\fl \int _{\mathbb {R}^2} \; \langle x |q,p;\bar{k}\rangle _\bullet \; _\bullet \langle q,p;\bar{k}|x^{\prime }\rangle \frac{{\rm d}p\,{\rm d}q }{2\pi \hbar }= \int _{\mathbb {R}^2} \Phi _{\bar{k};\bullet }(q,p,x)(\Phi _{\bar{k};\bullet }(q,p,x^{\prime }))^{*}\; \frac{{\rm d}p\,{\rm d}q }{2\pi \hbar } \nonumber \\ &&= \int _{-\infty }^{\infty }\left[\int _{-\infty }^{\infty } \Psi _{\bar{k};\bullet }( x-q) \Psi ^*_{\bar{k};\bullet }( x^{\prime }-q)\, {\rm d}q\right] \,{\rm e}^{ \frac{{\rm i}}{\hbar }p(x - x^{\prime }) } \frac{{\rm d}p }{2\pi \hbar } \nonumber\\ &&= \delta (x-x^{\prime }) =\langle x |x^{\prime }\rangle , \end{eqnarray} \tag{ 28 }$$

  where we use the fact that the integrations in q and x are similar to compute the term as performed in (27). Hence, $|q,p;\bar{k}\rangle$ resolves the unity of $L^2(\mathbb {R},\,{\rm d}x)$.
• (iv)  
  The temporal stability condition will be satisfied only at the leading order of a small parameter depending on the two types of coherent states. However, temporal stability can be recovered for expectation values as the last step in a calculation.Let us introduce states endowed with another evolution parameter τ such that

  $$\begin{eqnarray} &&|q,p,\tau ;\bar{k}\rangle _{\bullet } = {\rm e}^{-\frac{{\rm i}}{\hbar } \tau H} |q,p;\bar{k}\rangle _{\bullet } \;; \end{eqnarray} \tag{ 29 }$$

  then of course

  $$\begin{eqnarray} &&{\rm e}^{-\frac{{\rm i}}{\hbar } t H}|q,p,\tau ;\bar{k}\rangle _{\bullet } = |q,p,\tau + t;\bar{k}\rangle _{\bullet } \;. \end{eqnarray} \tag{ 30 }$$

  Hence, the new state $|q,p,\tau ;\bar{k}\rangle _{\bullet }$ implemented with the parameter τ is stable under time evolution. Nevertheless, we could have asked more from our coherent state and have required that without introducing another parameter, ${\rm e}^{-\frac{{\rm i}}{\hbar } \tau H} |q,p;\bar{k}\rangle _{\bullet }$ evolves within the same family of states $\lbrace |q,p;\bar{k}\rangle _{\bullet }\rbrace _{(q,p)\in \mathbb {R}^2}$. This condition should be considered as the true meaning of temporal stability under time evolution. The later statement and further consequences have also been developed in [10, 21] (and a way to generate such states considered in [22]). We seek conditions under which this statement holds.For any coherent state described above, we have

  $$\begin{eqnarray} &&\fl {\rm e}^{-\frac{{\rm i}}{\hbar } \tau H} |q,p;\bar{k}\rangle _{\bullet } = {\rm e}^{-\frac{{\rm i}}{\hbar }q P} \,{\rm e}^{\frac{{\rm i}}{\hbar } \tau \frac{p^2}{2m} } \,{\rm e}^{ -\frac{{\rm i}}{m\hbar } \tau p P } \,{\rm e}^{ \frac{{\rm i}}{\hbar } p Q } |\Psi _{\bar{k}};\tau \rangle _{\bullet } \nonumber\\ &&\,\,\,= {\rm e}^{\frac{{\rm i}}{\hbar } \tau \frac{p^2}{2m} } \,{\rm e}^{-\frac{{\rm i}}{\hbar }(q + \frac{p}{m} \tau ) P} \,{\rm e}^{ \frac{{\rm i}}{\hbar } p Q } |\Psi _{\bar{k}};\tau \rangle _{\bullet }, \end{eqnarray} \tag{ 31 }$$

  where $|\Psi _{\bar{k}};\tau \rangle _{\bullet } = {\rm e}^{-\frac{{\rm i}}{\hbar } \tau H} |\Psi _{\bar{k}}\rangle _{\bullet }$.Computing the first type of shifted state $|\Psi _{\bar{k}};\tau \rangle$, introducing $\tilde{k}(k_1,k_0)$, a function to be specified later, $\delta = k - \tilde{k}$ and at leading orders in δ, one finds

  $$\begin{eqnarray} &&\fl \langle x|\Psi _{\bar{k}};\tau \rangle = {\rm e}^{-{\rm i} \frac{\hbar \tau }{2m}\tilde{k}^2 } \langle x|\Psi _{\bar{k}} \rangle -\frac{C}{\sqrt{2\pi }} \, \left({\rm i}\tilde{k}\frac{\hbar \tau }{m}\right) \,{\rm e}^{-{\rm i} \frac{\hbar \tau }{2m} \tilde{k}^2 } \,{\rm e}^{{\rm i} \tilde{k} x} \int _{k_0-\tilde{k}}^{k_1-\tilde{k}} \delta \,{\rm e}^{{\rm i} \delta x} \,{\rm d}\delta \nonumber\\ &&+\, \frac{C}{\sqrt{2\pi }} \,{\rm e}^{-{\rm i} \frac{\hbar \tau }{2m} \tilde{k}^2 }\,{\rm e}^{{\rm i} \tilde{k} x}\int _{k_0-\tilde{k}}^{k_1-\tilde{k}} O(\delta ^2) \,{\rm e}^{{\rm i} \delta x} \,{\rm d}\delta . \nonumber \end{eqnarray} \tag{ 32 }$$

  Defining Δ = (k₁ − k₀)/2 and $\ell =(k_0+k_1)/2 - \tilde{k}$, and using the new variable $\tilde{\delta }= \delta - \ell$, we can re-express the second term integral above as

  $$\begin{eqnarray*} &&\fl {\rm e}^{{\rm i} \tilde{k} x} \int _{k_0-\tilde{k}}^{k_1-\tilde{k}} \delta \,{\rm e}^{{\rm i} \delta x} \,{\rm d}\delta = 2{\rm i} \,{\rm e}^{{\rm i} (\tilde{k} +\ell ) x} \left[ \frac{ 1}{x^2} \sin ( \Delta x) - \frac{ 1}{x} \Delta \cos ( \Delta x) - \frac{ {\rm i}\ell }{ x} \sin ( \Delta x) \right] \!. \nonumber \end{eqnarray*}$$

  Let us consider the vector $|\Pi _\Delta \rangle$ corresponding to the last quantity and evaluate its norm as

  $$\begin{eqnarray} &&\int _{-\infty }^{\infty } (\Pi _\Delta (x))^* \Pi _\Delta (x) \,{\rm d}x = c \ell ^2 \Delta + O(\Delta ^2), \end{eqnarray} \tag{ 33 }$$

  where c is some constant. Thus, $|\Pi _\Delta \rangle$ is in the Hilbert space and its norm is of the order $\Delta ^{\frac{1}{2}}$. Introducing this result into expression (32) yields

  $$\begin{eqnarray} &&\fl \langle x|\Psi _{\bar{k}};\tau \rangle = {\rm e}^{-{\rm i} \frac{\hbar \tau }{2m}\tilde{k}^2 }\langle x|\Psi _{\bar{k}} \rangle -\frac{{\rm i}}{2\sqrt{\pi \Delta }} \frac{\hbar \tau \tilde{k}}{m} \,{\rm e}^{-{\rm i} \frac{\hbar \tau }{2m}\tilde{k}^2 } \left( \langle x| \Pi _\Delta \rangle + \langle x| O(\Delta ) \rangle \right)\!, \end{eqnarray} \tag{ 34 }$$

  where the notation |O(Δⁿ)〉 stands for a state of norm O(Δⁿ). Thus, all remainder states are of norm at most $O(\Delta ^{\frac{1}{2}})$. By expanding further ${\rm e}^{{\rm i}\tau (2 \bar{k}\delta +\delta ^2)}$ in δ, it can also be checked that higher order terms integrated are of the order of magnitude $O(\Delta ^{\frac{1}{2} + c} )$, c ⩾ 0. Thus, without any further assumption, the temporal stability condition is clearly broken at leading order by the term $| \Pi _\Delta \rangle$ with the order of magnitude O(1) with respect to the window of integration $\sqrt{k_1-k_0}$. Nevertheless, two specific cases of interest may occur: (1) 0 < Δ ⩽ ℓ ≪ 1, i.e. if the position barycenter is close to $\tilde{k}$ (which is the only remaining parameter in the present context and which can be tuned to fulfill that condition), then we infer that the norm of $||\,C| \Pi _\Delta \rangle ||^2 = O(\ell ^2) \ll O(1) = ||\,|\Psi _{\bar{k}}\rangle ||^2$; (2) ℓ = 0; therefore $||\,C| \Pi _\Delta \rangle ||^2 = O(\Delta ) \ll O(1) = ||\,|\Psi _{\bar{k}}\rangle |$. Thus, under these assumptions, we write in shorthand notations

  $$\begin{eqnarray} &&\fl |\Psi _{\bar{k}};\tau \rangle = {\rm e}^{-{\rm i} \frac{\hbar \tau }{2m}\tilde{k}^2 }|\Psi _{\bar{k}} \rangle + |O(\ell )\rangle \qquad {\rm or}\qquad |\Psi _{\bar{k}};\tau \rangle = {\rm e}^{-{\rm i} \frac{\hbar \tau }{2m}\tilde{k}^2 } |\Psi _{\bar{k}} \rangle + |O(\Delta )\rangle , \end{eqnarray} \tag{ 35 }$$

  such that the evolution of the coherent states is given by

  $$\begin{eqnarray} &&\fl {\rm e}^{-\frac{{\rm i}}{\hbar } \tau H} |q,p;\bar{k}\rangle = {\rm e}^{\frac{{\rm i}}{\hbar } \tau (\frac{p^2}{2m} -\frac{\hbar ^2 \tilde{k}^2}{2m} )} |q+ \frac{p}{m} \tau ,p;\bar{k}\rangle +( |O(\ell )\rangle \leftrightarrow |O(\Delta )\rangle ) \end{eqnarray} \tag{ 36 }$$

  with indeed a clear physical meaning: a given coherent state evolves within the same family of states with the parameter q translated to $q+ \textstyle \frac{p}{m} \tau$ at order O(ℓ) or at order O(Δ).For the second type of coherent states, we calculate the evolution of the fiducial state in the same manner as done earlier setting for this time $\tilde{k} = \bar{k}$. We find

  $$\begin{eqnarray} &&|\Psi _{\bar{k}};\tau \rangle _A = {\rm e}^{-{\rm i} \frac{\hbar \tau }{2m}\bar{k}^2 } |\Psi _{\bar{k}}\rangle _A +\left(-{\rm i} \bar{k} \frac{\tau \hbar }{m} \right) \,{\rm e}^{-{\rm i} \frac{\hbar \tau }{2m}\bar{k}^2 } \frac{1}{A} Q |\Psi _{\bar{k}}\rangle _A + \cdots . \end{eqnarray} \tag{ 37 }$$

  A calculation establishes that the norm of the remainder state is

  $$\begin{eqnarray} &&\fl \left( _A \langle \Psi _{\bar{k}} | \left(\bar{k} \frac{\tau \hbar }{m} \right) \,{\rm e}^{+{\rm i} \frac{\hbar \tau }{2m}\bar{k}^2 } \frac{1}{A} Q \right) \left( \left(\bar{k} \frac{\tau \hbar }{m} \right) \,{\rm e}^{-{\rm i} \frac{\hbar \tau }{2m}\bar{k}^2 } \frac{1}{A} Q |\Psi _{\bar{k}}\rangle _A \right) = \frac{1}{A} \left(\frac{\tau \hbar }{m}\bar{k} \right)^2 \end{eqnarray} \tag{ 38 }$$

  which is of the order O(1/A). Hence, we have

  $$\begin{eqnarray} &&{\rm e}^{-\frac{{\rm i}}{\hbar } \tau H} |q,p;\bar{k}\rangle _A = {\rm e}^{\frac{{\rm i}}{\hbar } \tau \left(\frac{p^2}{2m} -\frac{\hbar ^2 \tilde{k}^2}{2m} \right)} |q+ \textstyle \frac{p}{m} \tau ,p;\bar{k}\rangle _A + |O(\textstyle A^{-\frac{1}{2}})\rangle . \end{eqnarray} \tag{ 39 }$$

  In conclusion, this family of coherent states is stable under time evolution at the dominant order, for A sufficiently large.Note that in any situation, interestingly, the inessential phase cancels exactly for $p^2 = \hbar ^2 \tilde{k}^2$, namely when the classical Hamiltonian picks the value of the quantum value $\hbar ^2 \tilde{k}^2$.
• (v)  
  The action identity axiom can be examined by evaluating the mean value of the Hamiltonian.The coherent states of the first type provide the Hamiltonian mean value

  $$\begin{eqnarray} \fl \langle q,p,\tau ;\bar{k}| H |q,p,\tau ;\bar{k}\rangle &=& \frac{1}{2m} \langle \Psi _{\bar{k}} | (P + p)^2 |\Psi _{\bar{k}}\rangle \nonumber \\ &=& \frac{1}{2m} \left[ \frac{1}{3} C^2\hbar ^2\big[k_1^3 \hspace{-0.5pt}\,{-}\,\hspace{-0.5pt}k_0^3\big] \hspace{-0.5pt}\,{+}\,\hspace{-0.5pt} \hbar p C^2 \big[k_1^2 \hspace{-0.5pt}\,{-}\,\hspace{-0.5pt}k_0^2\big] + p^2 \right]= \omega J_{\bar{k}}(p) . \end{eqnarray} \tag{ 40 }$$

  The goal is to find the functions p(J, ω) and q(J, ω), where J will play the role of an action variable and ω is associated with γ which will play the role of an angle variable canonically conjugate to J. This can be achieved by considering

  $$\begin{eqnarray} &&\fl p_\pm (J,\omega ) = - \case{1}{2}\hbar C^2 \big[k_1^2 -k_0^2\big] \pm \sqrt{2m\omega J_{\bar{k}}( p) -\case{1}{3}\hbar ^2C^2 \big[k_1^3 -k_0^3\big] +\case{1}{4} \big[\hbar C^2\big(k_1^2 -k_0^2\big)\big]^2 } \;\nonumber \\ &&\fl \arctan \left(\frac{p}{q}\right) = \omega \quad \Rightarrow \quad q_{\pm } (J,\omega )= p_{\pm }(J,\omega ) \cot \omega . \end{eqnarray} \tag{ 41 }$$

  Hence, setting ω = 1 (omitting henceforth the dependence in ω) and choosing the relevant root p₊(J) = p₊(J, ω = 1), the action identity is given by γ = τ and

  $$\begin{equation} \langle \,q_{+} (J), p_{+}(J),\tau ; \bar{k} |\,H \,| q_{+} (J), p_+(J),\tau ; \bar{k} \,\rangle = J. \end{equation} \tag{ 42 }$$

  Meanwhile, the same calculation for the second type of coherent states yields

  $$\begin{eqnarray} &&\frac{\hbar ^2 \bar{k}^2}{2m} + \frac{\hbar ^2}{2m} \frac{1}{4A} +\frac{\hbar p \bar{k} }{m} +\frac{p^2}{2m} = \omega _A J_A \; \end{eqnarray} \tag{ 43 }$$

  with a similar equation for q as in (41). Again, we invert p in terms of J_A and find

  $$\begin{equation} p_\pm (J_A,\omega _A) =- \hbar \bar{k} \pm \sqrt{ 2m \omega _A J_A - \hbar \frac{1}{4A} }, \end{equation} \tag{ 44 }$$

  such that the action identity reads, choosing ω_A = 1, the root p₊(J_A) = p₊(J_A, ω_A = 1) and γ = τ,

  $$\begin{equation} _{A}\langle \, q_{+} (J_A) ,p_+ (J_A),\tau ;\bar{k}|\, H\, |q_{+}(J_A),p(J_A),\tau ;\bar{k}\,\rangle _{A} = J_A. \end{equation} \tag{ 45 }$$

  Note that, even though τ appears in both (42) and (45), J and J_A do not depend on τ = γ because, in these equations, that dependence explicitly vanishes (complex conjugate phases). Thus, J and J_A are independent and canonically conjugated to γ = τ.

The main statistical properties of the set of states $|q,p,\tau ; \bar{k}\rangle$ cannot be derived since the state $|\Psi _{\bar{k}}\rangle$ does not belong in fact to the domain of Q. Nevertheless, for the second type of coherent states, one can investigate these properties. We will not undertake such a statistical analysis in this paper; however, it is significant to ask for the saturation of the Heisenberg uncertainty relation for the type of states studied here. This is the purpose of this paragraph.

After some algebra, the following identities hold:

$$\begin{eqnarray} && \langle Q \rangle = q , \qquad \!\!\langle Q^2 \rangle = A + q^2 \;,\qquad \!\!(\Delta Q)^2 = A, \nonumber \\ && \langle P \rangle = \hbar \bar{k} + p \;,\qquad \!\!\langle P^2 \rangle = (\hbar \bar{k} + p)^2 + \frac{1}{4A}\hbar ^2 \;,\qquad \!\!(\Delta P)^2 = \frac{1}{4A}\hbar ^2. \end{eqnarray} \tag{ 46 }$$

Hence,

$$\begin{eqnarray} &&\Delta Q \Delta P = \frac{\hbar }{2}, \end{eqnarray} \tag{ 47 }$$

emphasizing the fact that the coherent states defined by (26) saturate the Heisenberg uncertainty relation.

The limits k₁ → k₀ and A → ∞ for the two kinds of coherent states are of interest. The construction at these limit situations rests on the following types of states:

$$\begin{eqnarray} {\rm (a)}\quad |\Psi _{\bar{0}} \rangle &=& \lim _{k_1 \rightarrow k_0} |\Psi _{\bar{k}} \rangle = \lim _{k_1 \rightarrow k_0} \frac{1}{\sqrt{k_1-k_0}}\int _{k_0}^{k_1}|\Psi _{ k} \rangle \; {\rm d}k \rightarrow 0\;; \nonumber \\ {\rm (b)}\quad |\Psi _{\bar{k}} \rangle _\infty &=& \lim _{A \rightarrow \infty } |\Psi _{\bar{k}} \rangle _A = \lim _{A \rightarrow \infty } \left(\frac{2A}{\pi }\right)^{\!\!\frac{1}{4}} \int _{-\infty }^{\infty } \,{\rm e}^{-A(k-\bar{k})^2} |\Psi _{ k} \rangle \; {\rm d}k \rightarrow 0 . \end{eqnarray} \tag{ 48 }$$

Hence, for this limit the entire coherent state construction is vacuous, as it can be checked that their overlap with 〈x| is always vanishing. However, this does not mean that the limits Δ → 0 and A → ∞ are without interest. In fact, such limits should be performed only after taking expectation values. For a general operator T, the following expectations are finite:

$$\begin{eqnarray} &&\lim _{k_1\rightarrow k_0 } \langle q,p; \bar{k}| T| q,p; \bar{k}\rangle , \qquad \lim _{A\rightarrow \infty }\; _A\langle q,p; \bar{k}| T| q,p; \bar{k}\rangle _A. \end{eqnarray} \tag{ 49 }$$

For instance, evaluating the expectations of the identity, one finds

$$\begin{eqnarray} && \lim _{k_1\rightarrow k_0 } \langle q,p; \bar{k}| q,p; \bar{k}\rangle = \lim _{k_1\rightarrow k_0 } \langle \Psi _{\bar{k}}| \Psi _{\bar{k}} \rangle =1,\nonumber \\ && \lim _{A\rightarrow \infty }\; _A\langle q,p; \bar{k}| q,p; \bar{k}\rangle _A =\lim _{A\rightarrow \infty }\; _A\langle \Psi _{\bar{k}}| \Psi _{\bar{k}}\rangle _A=1, \nonumber \\ && \lim _{k_1\rightarrow k_0 } \langle q,p; \bar{k}|H| q,p; \bar{k}\rangle = \frac{p^2}{2m} \;,\nonumber \\ && \lim _{A\rightarrow \infty }\; _A\langle q,p; \bar{k}| H| q,p; \bar{k}\rangle _A =\frac{ (\hbar \bar{k} + p)^2}{2m}. \end{eqnarray} \tag{ 50 }$$

Let us consider the quantum Hamiltonian

$$\begin{eqnarray} &&H = \frac{1}{2m} P^2 - \frac{1}{2}m\omega ^2 Q^2 = \frac{1}{2m}( -\hbar ^2 \partial _{x}^2 - m^2\omega ^2 x^2), \end{eqnarray} \tag{ 51 }$$

where [Q, P] = iℏ. This Hamiltonian has been previously investigated from different perspectives [14–16]. Our goal is to show that the procedure defined above allows us to define coherent states for this system with non-normalizable states with an infinite continuous spectrum. For simplicity in our developments, we will use the notations of [14], such that 2m = 1 = ℏ = ω. The resulting eigenvalue problem

$$\begin{eqnarray} &&H \psi _E = E \psi _E \quad \Leftrightarrow \quad -\left( \partial ^2_x + \frac{1}{4} x^2\right) \psi _E = E \psi _E\; \end{eqnarray} \tag{ 52 }$$

finds solutions in terms of the parabolic cylinder functions [23]

$$\begin{equation} \Psi _{E,1}(x) = \, D_{ -\frac{1}{2}(1+{\rm i}E ) } \big( {\rm e}^{\frac{1}{4} {\rm i}\pi }x \big),\quad \Psi _{E,2}(x) = \, D_{ -\frac{1}{2}(1-{\rm i}E ) }\big( {\rm e}^{\frac{3}{4} {\rm i}\pi } x \big). \end{equation} \tag{ 53 }$$

Both these eigenfunctions are non-normalizable and the system is doubly degenerate. (See the appendix for the basic properties of D_ν(x) and other facts about Hamiltonian (51).) Without loss of generality, we will focus on Ψ_{E, 1}(x). The latter function can be decomposed into a linear combination (with non-trivial, energy-dependent coefficients) of a real and an imaginary part written in terms of ψ_{E, ±}(x) = C_±W(E, ±x) which are again eigenfunctions of the same operator [23]. The constants C_± can be fixed by the orthogonality condition (proofs of the following statements can be found in [14] or in [23])

$$\begin{eqnarray} &&\int _{-\infty }^{\infty } \psi _{E,s}(x) \psi _{E^{\prime },s^{\prime }}(x) \,{\rm d}x = K_{s,E} \delta _{s,s^{\prime }}\delta (E-E^{\prime }), \qquad s,s^{\prime }=\pm , \end{eqnarray} \tag{ 54 }$$

where K_{s, E} is some constant. The functions ψ_{E, s}(x) are however non-normalizable: for x ≫ E, the following approximations hold [23]:

$$\begin{eqnarray} && W( E , x) \sim \sqrt{\frac{2\kappa }{x}} \cos \left[\frac{1}{4} x^2 + E \log x + \frac{1}{4}\pi + \phi (E) \right] \!,\nonumber \\ && W(E, -x) \sim \sqrt{\frac{2}{ \kappa x }} \sin \left[\frac{1}{4} x^2 + E \log x + \frac{1}{4}\pi + \phi (E) \right]\!, \nonumber \\ && \kappa = (1+ {\rm e}^{-2\pi E})^{\frac{1}{2}} - {\rm e}^{-\pi E} , \qquad \phi (E) = \arg \Gamma \left[\frac{1}{2} - {\rm i}E\right]\!. \end{eqnarray} \tag{ 55 }$$

In the following, we will concentrate on the solution ψ_E(x) ≡ ψ_{E, +}(x), such that the following orthogonality relation will be relevant:

$$\begin{eqnarray} &&\int _{-\infty }^{\infty } \psi _{E}(x)\psi _{E^{\prime }}(x) \,{\rm d}x = \delta (E-E^{\prime }); \end{eqnarray} \tag{ 56 }$$

thus, this will fix $C_+ =C_0= (2\pi (1+{\rm e}^{-2\pi E}))^{-\frac{1}{2}} \;.$

We introduce the state family |ψ_{E, +}〉 such that in the |x〉 representation, we have 〈x|ψ_{E, +}〉 =ψ_{E, +}(x), $\langle \psi _{E}|\psi _{E^{\prime }}\rangle$ =δ(E − E') and 〈ψ_E|ψ_E〉 = ∞.

The regularization of the fiducial vector can be done according to our prescription as

$$\begin{eqnarray} &&\fl | \Psi _{\bar{E}} \rangle = C_A \int _{-\infty }^{\infty } \,{\rm e}^{-A(E - \bar{E})^2} |\psi _{E}\rangle \,{\rm d}E ,\nonumber \\ &&\fl \langle x | \Psi _{\bar{E}} \rangle =C_A \int _{-\infty }^{\infty } \,{\rm e}^{-A(E - \bar{E})^2} \psi _{E}(x) \,{\rm d}x = C_A C_0 \int _{-\infty }^{\infty } \,{\rm e}^{-A(E - \bar{E})^2} W(E,x) \,{\rm d}E, \end{eqnarray} \tag{ 57 }$$

where C_A is a normalization constant fixed such that the state $| \Psi _{\bar{E}} \rangle$ is normalized in the same way as (23).

We can check that the set of states defined such that

$$\begin{eqnarray} &&\fl |q,p; \bar{E} \rangle = U(q,p) | \Psi _{\bar{E}} \rangle = {\rm e}^{-{\rm i} q P }\,{\rm e}^{{\rm i} p Q }| \Psi _{\bar{E}} \rangle , \qquad \langle x |q,p; \bar{E} \rangle = {\rm e}^{{\rm i} p (x- q)} \Psi _{\bar{E}}(x-q) \end{eqnarray} \tag{ 58 }$$

are coherent states.

• (i)  
  and (ii): with the continuity in labels being obvious, the normalizability condition is achieved by noticing that

  $$\begin{eqnarray} &&\langle q,p; \bar{E} |q,p; \bar{E} \rangle = \langle \Psi _{\bar{E}} | \Psi _{\bar{E}} \rangle =1. \end{eqnarray} \tag{ 59 }$$
• (iii)  
  Let us determine the resolution of the identity by evaluating the overlap

  $$\begin{eqnarray} \fl \int _{\mathbb {R}^2} \langle x| q,p; \bar{E} \rangle \langle q,p; \bar{E} | x^{\prime } \rangle \frac{{\rm d}q\,{\rm d}p}{2\pi \hbar } &=& \int _{\mathbb {R}^2 } \,{\rm e}^{\frac{{\rm i}}{\hbar } p (x- q)} \Psi _{\bar{E}}(x-q) \,{\rm e}^{-\frac{{\rm i}}{\hbar } p (x^{\prime }- q)} \Psi ^*_{\bar{E}}(x^{\prime }-q) \frac{{\rm d}q\,{\rm d}p}{2\pi \hbar } \nonumber \\ & =& \delta (x\,{-}\,x^{\prime })\int _{\mathbb {R} } \Psi _{\bar{E}}(x\,{-}\,q)\Psi _{\bar{E}}(x\,{-}\,q)\,{\rm d}q =\langle x|x^{\prime }\rangle . \end{eqnarray} \tag{ 60 }$$

  Thus, the states $| q,p; \bar{E} \rangle$ satisfy the resolution of the identity.
• (iv)  
  As we expect, the temporal stability of the states will only be satisfied at some approximation:

  $$\begin{eqnarray} &&{\rm e}^{-\frac{{\rm i}}{\hbar } t H } | q,p; \bar{E} \rangle = {\rm e}^{\frac{{\rm i}}{\hbar } t \frac{p^2}{2m} } \,{\rm e}^{-\frac{{\rm i}}{\hbar }(q + \frac{p}{m} t ) P} \,{\rm e}^{ \frac{{\rm i}}{\hbar } p Q } |\Psi _{\bar{E}};t\rangle , \end{eqnarray} \tag{ 61 }$$

  where we have to expand $|\Psi _{\bar{E}};t\rangle = {\rm e}^{-\frac{{\rm i}}{\hbar } t H } |\Psi _{\bar{E}}\rangle$ as

  $$\begin{eqnarray} &&\fl \langle x| \Psi _{\bar{E}};t\rangle = C_A \int _{-\infty }^{\infty } \,{\rm e}^{-A(E - \bar{E})^2} \,{\rm e}^{-\frac{{\rm i}}{\hbar } t E } \psi _{E}(x)\; {\rm d}E = C_A \int _{-\infty }^{\infty } \,{\rm e}^{-A\delta ^2} \,{\rm e}^{-\frac{{\rm i}}{\hbar } t (\bar{E} + \delta ) } \psi _{\bar{E} + \delta }(x)\; {\rm d}\delta \nonumber \\ &&= C_A \,{\rm e}^{-\frac{{\rm i}}{\hbar } t \bar{E} } \int _{-\infty }^{\infty } \,{\rm e}^{-A\delta ^2} \left(1 - \frac{{\rm i}}{\hbar } t \delta + O(\delta ^2)\right) \psi _{\bar{E} + \delta }(x)\; {\rm d}\delta . \end{eqnarray} \tag{ 62 }$$

  We aim at evaluating the norm of

  $$\begin{eqnarray} &&\langle x | \Upsilon _A \rangle = \Upsilon _A (x) = C_A \frac{{\rm i}}{\hbar } t \int _{-\infty }^{\infty } \,{\rm e}^{-A\delta ^2} \delta \;\psi _{\bar{E} + \delta }(x)\; {\rm d}\delta . \end{eqnarray} \tag{ 63 }$$

  Using the orthogonality of the functions ψ_E(x), one has

  $$\begin{eqnarray} &&\fl \int _{-\infty }^{\infty } \Upsilon ^*_A (x)\Upsilon _A (x) \,{\rm d}x = C^2_A \frac{t^2}{\hbar ^2} \int _{-\infty }^{\infty } \,{\rm e}^{-2A\delta ^2} \delta ^2 \,{\rm d}\delta = \sqrt{\frac{2A}{\pi }} \frac{t^2}{\hbar ^2} \frac{\sqrt{\pi }}{4\sqrt{2} A^{\frac{3}{2}}} = \frac{t^2}{4\hbar ^2}\frac{1}{A}, \end{eqnarray} \tag{ 64 }$$

  which implies that ϒ_A is again of the norm $O(1/\sqrt{A})$. Thus, as claimed, temporal stability will be obeyed at leading order and broken at $O(1/\sqrt{A})$.
• (v)  
  The action identity requires to find (J_A, γ_A) such that, given $H = P^2 -\frac{1}{4} Q^2$ in appropriate units, we have

  $$\begin{eqnarray} &&\fl _A \langle q,p,t;\bar{k}| H | q,p,t;\bar{k} \rangle _A = \; _A\langle \Psi _{\bar{E}}|(P+p)^2| \Psi _{\bar{E}} \rangle _A -\case{1}{4} \; _A \langle \Psi _{\bar{E}} |(Q+q)^2| \Psi _{\bar{E}} \rangle _A \nonumber \\ &&= \; _A\langle \Psi _{\bar{E}}|H| \Psi _{\bar{E}} \rangle _A + 2\; _A\langle \Psi _{\bar{E}}|(pP-\case{1}{4} q Q) | \Psi _{\bar{E}} \rangle _A + p^2 -\case{1}{4} q^2 \nonumber \\ &&= 2\hbar p K_1(A,\bar{E}) - \case{1}{2} q K_2(A,\bar{E}) + p^2 -\case{1}{4} q^2= \omega _A J_A, \end{eqnarray} \tag{ 65 }$$

  where $K_{1,2}(A,\bar{E})$ are functions, yet unknown, obtained after integrations. We will use the other action-angle equation in order to fix (q, p):

  $$\begin{eqnarray} \left\lbrace \begin{array}{@{}ll@{}} 2\hbar p K_1(A,\bar{E}) - \frac{1}{2} q K_2(A,\bar{E}) + p^2 -\frac{1}{4} q^2 = \omega _A J_A & \\ \arctan \left(\frac{p}{q}\right) = \omega _A & \end{array} \right. . \end{eqnarray} \tag{ 66 }$$

  A rapid inspection shows that this system admits solutions in terms of $K_{1,2}(A,\bar{E})$ which should be analyzed in order to fix the last GK axiom.