05 Jun 2026
II: Exact exceptional-Hermite preparation before the angular quench
This chapter constructs exact non-Gaussian preparation states from the one-dimensional Schrödinger equation. The construction is algebraic: a non-normalizable oscillator solution is used to factor the Hamiltonian and produce a regular rational potential with explicitly normalized eigenfunctions.
The derivation fixes the energy shift, normalization, parity, and scale dependence needed to use these states as controlled initial data rather than as a formal ansatz.
Prerequisite and result carried forward
The magnetic-oscillator reduction established the following ingredients; they are taken as input here rather than rederived:
- fixed canonical pairs $(q_+,p_+)$ and $(q_-,p_-)$;
- the unit convention $\hbar=2M=1$;
- positive coordinate scales $\Omega_+\geq\Omega_->0$; and
- for a stable magnetic realization, $\Omega_\pm^2>\omega_c^2$.
This chapter addresses only state preparation. The angular quench and its post-quench dynamics belong to the next stage.
1. Why go beyond an ordinary oscillator?
A harmonic-oscillator eigenstate is analytically convenient, but its ground state is Gaussian. To study dynamics whose initial non-Gaussianity is known exactly, one wants a potential that satisfies four requirements:
- it is nonsingular for every real coordinate;
- its bound states are available in closed form;
- those states can be normalized exactly; and
- the harmonic scale $\Omega$ remains adjustable.
A one-step Darboux factorization of the oscillator meets all four. The resulting wavefunctions contain exceptional-Hermite polynomials divided by a nodeless pseudo-Hermite polynomial. “Exceptional” here does not mean approximate or anomalous: it means that the polynomial sequence is complete with some low polynomial degrees absent.
2. Ordinary oscillator in the present convention
Start with one canonical pair,
\[[q,p_q]=\mathrm i, \qquad p_q=-\mathrm i\frac{\mathrm d}{\mathrm dq},\]and a positive scale $\Omega$. The ordinary Hamiltonian is
\[H_{\mathrm{osc}}=p_q^2+\frac{\Omega^2q^2}{4}.\]Introduce the dimensionless coordinate
\[z=\sqrt{\frac{\Omega}{2}}\,q.\]Then
\[\frac{\mathrm d}{\mathrm dq} =\sqrt{\frac{\Omega}{2}}\frac{\mathrm d}{\mathrm dz}, \qquad H_{\mathrm{osc}} =\frac{\Omega}{2} \left(-\frac{\mathrm d^2}{\mathrm dz^2}+z^2\right).\]The physicists’ Hermite polynomials are defined by
\[H_n(z)=(-1)^ne^{z^2} \frac{\mathrm d^n}{\mathrm dz^n}e^{-z^2},\]and satisfy
\[H_n''-2zH_n'+2nH_n=0.\]The normalized oscillator eigenfunctions and energies are
\[\psi_n(q;\Omega) =\left(\frac{\Omega}{2}\right)^{1/4} \frac{H_n(z)e^{-z^2/2}} {\sqrt{\sqrt\pi\,2^n n!}},\] \[H_{\mathrm{osc}}\psi_n =E_n^{\mathrm{osc}}\psi_n, \qquad E_n^{\mathrm{osc}}=\Omega\left(n+\frac12\right).\]The factor $(\Omega/2)^{1/4}$ is required because $\mathrm dq=\sqrt{2/\Omega}\,\mathrm dz$. Leaving it out normalizes the function in $z$, not in the physical coordinate $q$.
3. Constructing a nodeless seed
Define the pseudo-Hermite polynomial
\[\mathcal H_m(z)=(-\mathrm i)^mH_m(\mathrm iz).\]It obeys
\[\mathcal H_m''+2z\mathcal H_m'-2m\mathcal H_m=0.\]Consequently,
\[u_m(q)=\mathcal H_m(z)e^{z^2/2}\]solves the ordinary oscillator equation at the negative factorization energy
\[H_{\mathrm{osc}}u_m=\epsilon_m u_m, \qquad \boxed{ \epsilon_m(\Omega) =-\left(m+\frac12\right)\Omega. }\]The seed grows at infinity and is not a physical oscillator state. That is intentional: its reciprocal will decay.
Why $m$ must be even
For $m=2r$, the explicit Hermite sum gives
\[\mathcal H_{2r}(z) =(2r)!\sum_{k=0}^{r} \frac{(2z)^{2r-2k}}{k!(2r-2k)!}.\]Every coefficient is positive, including the constant term. Therefore
\[\mathcal H_{2r}(z)>0 \qquad\text{for all real }z.\]Its reciprocal has no real pole. For odd $m$, $\mathcal H_m$ is odd and vanishes at $z=0$, producing a singular potential. We therefore use
\[m=0,2,4,\ldots.\]This is a regularity condition, not a cosmetic restriction.
4. Darboux factorization step by step
Let
\[f_m(q)=\frac{\mathrm d}{\mathrm dq}\ln u_m(q),\]and define
\[A_m=\frac{\mathrm d}{\mathrm dq}-f_m(q), \qquad A_m^\dagger=-\frac{\mathrm d}{\mathrm dq}-f_m(q).\]Because $u_m$ solves the Schrödinger equation,
\[H_{\mathrm{osc}}=A_m^\dagger A_m+\epsilon_m.\]Reverse the operator order to define the partner:
\[H_m^-=A_mA_m^\dagger+\epsilon_m =p_q^2+V_m^-(q;\Omega).\]The partner potential differs by a logarithmic second derivative,
\[V_m^-= \frac{\Omega^2q^2}{4} -2\frac{\mathrm d^2}{\mathrm dq^2}\ln u_m.\]Since
\[\frac{\mathrm d^2}{\mathrm dq^2}\ln u_m =\frac{\Omega}{2} \left[ 1+\frac{\mathcal H_m''}{\mathcal H_m} -\left(\frac{\mathcal H_m'}{\mathcal H_m}\right)^2 \right],\]where primes now mean derivatives with respect to $z$, the regular rational extension is
\[\boxed{ V_m^-(q;\Omega) =\frac{\Omega^2q^2}{4} -\Omega\left[ 1+\frac{\mathcal H_m''(z)}{\mathcal H_m(z)} -\left(\frac{\mathcal H_m'(z)} {\mathcal H_m(z)}\right)^2 \right]. }\]It is convenient to shift the factorization energy to zero:
\[h_m(q;\Omega) =H_m^--\epsilon_m =A_mA_m^\dagger.\]This identity immediately proves
\[\langle\chi|h_m|\chi\rangle =\|A_m^\dagger\chi\|^2\geq0.\]Thus the shifted Hamiltonian is nonnegative without requiring a numerical spectrum.
5. Deriving the exceptional eigenfunctions
The added ground state
The zero-energy equation is
\[A_m^\dagger\phi_{m,0}=0.\]It has the solution
\[\phi_{m,0}\propto\frac{1}{u_m} =\frac{e^{-z^2/2}}{\mathcal H_m(z)}.\]For even $m$ it is regular and square-integrable. This state was absent from the original oscillator spectrum because the seed $u_m$ itself was not normalizable.
The excited states
The intertwining identity
\[H_m^-A_m=A_mH_{\mathrm{osc}}\]maps each oscillator state $\psi_{\ell-1}$, $\ell\geq1$, into a partner state. Applying $A_m$ and using
\[H_n'=2nH_{n-1}, \qquad H_{n+1}=2zH_n-2nH_{n-1},\]produces the numerator
\[\mathcal P_{m,\ell}(z)= \begin{cases} 1, & \ell=0,\\[4pt] \mathcal H_m(z)H_\ell(z) +\mathcal H_m'(z)H_{\ell-1}(z), & \ell\geq1. \end{cases}\]The overall minus sign generated by $A_m$ is an irrelevant phase and has been dropped. The normalized eigenfunction is
\[\boxed{ \phi_{m,\ell}(q;\Omega) =\left(\frac{\Omega}{2}\right)^{1/4} \mathcal N_{m,\ell} \frac{\mathcal P_{m,\ell}(z)}{\mathcal H_m(z)} e^{-z^2/2}, }\]where
\[\mathcal N_{m,\ell}= \begin{cases} \left(\dfrac{2^m m!}{\sqrt\pi}\right)^{1/2}, &\ell=0,\\[9pt] \left[\sqrt\pi\,2^\ell(\ell-1)!(m+\ell)\right]^{-1/2}, &\ell\geq1. \end{cases}\]The excited-state normalization can be checked without doing a new rational integral:
\[\|A_m\psi_{\ell-1}\|^2 =\langle\psi_{\ell-1}| A_m^\dagger A_m|\psi_{\ell-1}\rangle =E_{\ell-1}^{\mathrm{osc}}-\epsilon_m =\Omega(m+\ell).\]The shifted spectrum is
\[\boxed{ \varepsilon_{m,\ell}(\Omega)= \begin{cases} 0, & \ell=0,\\ \Omega(m+\ell), & \ell\geq1. \end{cases} }\]The isolated zero-energy state followed by the level $\Omega(m+1)$ is a characteristic state-adding spectrum.
The undeformed limit
Setting $m=0$ gives $\mathcal H_0=1$. The rational term disappears, $h_0=H_{\mathrm{osc}}-\Omega/2$, and
\[\phi_{0,\ell}=\psi_\ell, \qquad \varepsilon_{0,\ell}=\Omega\ell.\]This limit is a powerful convention check: every factor, index, and energy shift returns to the familiar oscillator.
6. Parity and exact selection rules
For even $m$, $\mathcal H_m$ is even and $\mathcal H_m’$ is odd. Both terms in $\mathcal P_{m,\ell}$ have parity $(-1)^\ell$, hence
\[\boxed{ \phi_{m,\ell}(-q;\Omega) =(-1)^\ell\phi_{m,\ell}(q;\Omega). }\]The eigenfunctions may be chosen real. Parity and reality then imply
\[\langle q\rangle=0, \qquad \langle p_q\rangle=0, \qquad \frac12\langle qp_q+p_qq\rangle=0.\]They also give matrix-element selection rules:
\[\langle m,r|q|m,s\rangle =\langle m,r|p_q|m,s\rangle=0 \quad\text{when }r+s\text{ is even}.\]Both $q$ and $p_q$ are odd under parity, so they connect only states of opposite parity. This halves the number of entries that need numerical quadrature and provides a stringent check on a computed basis.
7. Fully explicit example: $m=2$
The first nontrivial regular seed is
\[\mathcal H_2(z)=4z^2+2=2(2z^2+1).\]It is positive on the real line. Its rational potential is
\[\boxed{ V_2^-(q;\Omega) =\frac{\Omega z^2}{2} -\Omega\left[ 1+\frac{4(1-2z^2)}{(2z^2+1)^2} \right], \qquad z=\sqrt{\frac{\Omega}{2}}q. }\]The factorization energy is
\[\epsilon_2=-\frac52\Omega.\]The first three exceptional numerators are
\[\begin{aligned} \mathcal P_{2,0}(z)&=1,\\ \mathcal P_{2,1}(z)&=4z(2z^2+3),\\ \mathcal P_{2,2}(z)&=4(4z^4+4z^2-1). \end{aligned}\]Their shifted energies are
\[0,\qquad 3\Omega,\qquad 4\Omega.\]For example, the normalized ground state is
\[\boxed{ \phi_{2,0}(q;\Omega) =\left(\frac{\Omega}{2}\right)^{1/4} \sqrt{\frac{2}{\sqrt\pi}}\, \frac{e^{-z^2/2}}{2z^2+1}. }\]It is visibly a rationally dressed Gaussian and has no node or pole.
Symbolic Schrödinger check in Maxima
kill(all)$
Hcal:4*z^2+2$
eps:-5*Omega/2$
Vminus:Omega*z^2/2-Omega*(1+diff(log(Hcal),z,2))$
P0:1$
P1:Hcal*(2*z)+diff(Hcal,z)$
phi(P):=P/Hcal*exp(-z^2/2)$
residual(P,E):=radcan(
(-Omega/2*diff(phi(P),z,2)
+(Vminus-eps-E)*phi(P))*exp(z^2/2)
)$
[residual(P0,0),residual(P1,3*Omega)];
The verified output is [0, 0]. Multiplication by $e^{z^2/2}$ in the
definition of residual merely exposes the rational factor to exact
simplification; it does not change whether the residual vanishes.
8. Exact moments of the $m=2$, $\ell=0$ state
The state is even and real, so its means and symmetrized $qp$ covariance vanish. Define
\[\mathcal M= \sqrt{\frac\pi2}\,e^{1/2} \operatorname{erfc}\!\left(\frac1{\sqrt2}\right) =0.6556795424\ldots.\]The needed rational Gaussian integrals follow from
\[\int_{-\infty}^{\infty} \frac{e^{-z^2}}{z^2+a^2}\,\mathrm dz =\frac{\pi}{a}e^{a^2}\operatorname{erfc}(a), \qquad a>0,\]and differentiation with respect to $a$. Substituting $a=1/\sqrt2$ gives
\[\boxed{ V_q=\langle q^2\rangle =\frac{2(\mathcal M-1/2)}{\Omega} =\frac{0.3113590848\ldots}{\Omega}, }\]and direct differentiation of $\phi_{2,0}$ gives
\[\boxed{ V_p=\langle p_q^2\rangle =\frac{\Omega}{2} \left(\frac32+\frac{\mathcal M}{3}\right) =0.8592799237\ldots\,\Omega. }\]The scale cancels from their product:
\[\Delta q\,\Delta p_q =\sqrt{V_qV_p} =0.5172471466\ldots>\frac12.\]This is consistent with the uncertainty principle. Since the state is pure, the value above the minimum also rules out a pure minimum-uncertainty Gaussian. The rational factor has changed the state in a measurable way even before any dynamics begins.
A warning about momentum moments
It is often useful to write a rationally extended state schematically as
\[|\Psi_{\mathrm{RE}}\rangle =\frac{\mathcal R|\Psi_G\rangle} {\sqrt{\langle\Psi_G|\mathcal R^\dagger\mathcal R|\Psi_G\rangle}}.\]For an operator $\mathcal O$,
\[\langle\mathcal O\rangle_{\mathrm{RE}} =\frac{ \langle\Psi_G|\mathcal R^\dagger\mathcal O\mathcal R|\Psi_G\rangle }{ \langle\Psi_G|\mathcal R^\dagger\mathcal R|\Psi_G\rangle }.\]Coordinate moments reduce to rationally weighted Gaussian integrals. Momentum operators, however, differentiate $\mathcal R$ as well as the Gaussian. Reweighting only \(\lvert\Psi_G\rvert^2\) gives an incorrect momentum variance.
9. Exact two-axis preparation
Apply the one-dimensional construction independently to the two canonical coordinates carried forward from Chapter I:
\[H_{\mathrm{sep}}^{\mathrm{RE}} =h_{m_+}(q_+;\Omega_+) +h_{m_-}(q_-;\Omega_-).\]For even $m_\pm$ and $\ell_\pm=0,1,2,\ldots$, the normalized product state is
\[\boxed{ \Phi_{\boldsymbol m,\boldsymbol\ell}^{\mathrm{RE}}(q_+,q_-) =\phi_{m_+,\ell_+}(q_+;\Omega_+) \phi_{m_-,\ell_-}(q_-;\Omega_-). }\]Its exact preparation energy is
\[\boxed{ E_{\boldsymbol m,\boldsymbol\ell}^{\mathrm{sep}} =\varepsilon_{m_+,\ell_+}(\Omega_+) +\varepsilon_{m_-,\ell_-}(\Omega_-). }\]For a concrete mixed preparation, choose
\[(m_+,\ell_+)=(2,0), \qquad (m_-,\ell_-)=(0,1),\]with
\[\Omega_+=\frac54, \qquad \Omega_-=\frac{17}{20}.\]The $+$ factor is a rationally extended non-Gaussian ground state, while the $-$ factor is the first ordinary oscillator excitation. Their total shifted energy is
\[E^{\mathrm{sep}}=0+\Omega_-=\frac{17}{20}.\]The state is exactly separable at preparation; no finite basis or optimization is involved in defining it.
10. Operator matrices for later dynamics
For later operator calculations, choose cutoffs $0\leq r<N_+$ and $0\leq s<N_-$ and compute the one-axis matrices
\[(Q_\alpha)_{rs} =\int_{-\infty}^{\infty} \phi_{m_\alpha,r}(q;\Omega_\alpha)\, q\, \phi_{m_\alpha,s}(q;\Omega_\alpha)\,\mathrm dq,\] \[(P_\alpha)_{rs} =-\mathrm i\int_{-\infty}^{\infty} \phi_{m_\alpha,r}(q;\Omega_\alpha) \frac{\mathrm d}{\mathrm dq} \phi_{m_\alpha,s}(q;\Omega_\alpha)\,\mathrm dq.\]Use parity to set forbidden entries to exact zero before numerical quadrature. Then verify
\[Q_\alpha^\dagger=Q_\alpha, \qquad P_\alpha^\dagger=P_\alpha.\]At finite cutoff, the canonical commutator cannot equal $\mathrm iI$ in every matrix element because the trace of a finite commutator is zero. Convergence should therefore be tested in the low-energy block actually used, while increasing the total cutoff. Parity-forbidden entries, Hermiticity, normalization, orthogonality, and exact low-level Schrödinger residuals provide independent diagnostics.
Result passed to the next chapter
The preparation stage supplies:
- an exact additive Hamiltonian $H_{\mathrm{sep}}^{\mathrm{RE}}$;
- an exact normalized product eigenstate $\Phi_{\boldsymbol m,\boldsymbol\ell}^{\mathrm{RE}}$;
- its exact energy, parity, and low moments; and
- converged one-axis coordinate and momentum matrices.
Nothing in this construction assumes that the product remains stationary after a coupling is applied. The fixed-confinement angular quench starts from precisely these prepared data.
Discussion