Segal-Bargmann wave function for confined spaces (1D infinite potential well)

I’m trying to calculate the Segal-Bargmann wave function and the Husimi function for a particle in a 1-dimensional infinite potential well of width L (i.e. $V(x) = 0$ if $x in (-L/2,L/2)$). In the Segal-Bargmann space, the Hamiltonian is:
$$
H(z)= frac{p^{2}}{2m} = -frac{1}{4m}left[z^{2}+hbar^{2}frac{partial^{2}}{partial z^{2}}-zhbarfrac{partial}{partial z}-hbarfrac{partial}{partial z}zright]
$$

The solution to the eigenvalue equation $Hpsi^{SB} = Epsi^{SB}$ is:

$$
psi^{SB}(z) = e^{frac{z^{2}}{2hbar}}[Ae^{sqrt{2}ikz} + Be^{-sqrt{2}ikz}], quad A,Bin mathbb{C}, quad k= frac{sqrt{2mE}}{hbar}
$$

Now, if I apply boundary conditions to the Schrödinger wave function, which can be accomplished by calculating the inverse Segal-Bargmann transform of the wave function $psi^{SB}$:

$$
psi^{Sch}(x) = int_{mathbb{C}} dz e^{frac{1}{2hbar}(-overline{z}^{2}+2sqrt{2}overline{z}x-x^2)}psi^{SB}(z)e^{-|z|^{2}}
$$

I obtain the following solution:

$$
psi_{n}^{SB}(z) =
begin{cases}
A,cosleft(frac{sqrt{2}npi}{L}zright)& text{if n odd}
A,sinleft(frac{sqrt{2}npi}{L}zright)& text{if n even}
end{cases}
$$

However, if I calculate the wave function $psi^{SB}$ using the Segal-Bargmann transform from the usual infinite potential well Schrödinger wave function:

$$
begin{cases}
psi_{n}(x) = sqrt{frac{2}{L}}cosleft(frac{npi}{L}xright)& text{if n odd}
psi_{n}(x) = sqrt{frac{2}{L}}sinleft(frac{npi}{L}xright)& text{if n even}
end{cases}
$$

I get the following:

$$
psi^{SB}(z) = frac{1}{pihbar}int_{-L/2}^{L/2} dx e^{(-z^2+2sqrt{2}xz-x^2)/2hbar}psi^{Sch}(x) = frac{1}{sqrt{pihbar L}}e^{frac{1}{2}left(frac{-hbar (npi)^{2}}{L^2}+frac{z^2}{hbar}right)}left[Releft(e^{-frac{sqrt{2}inpi z}{L}}text{erf}left(frac{sqrt{2}L^2-2sqrt{2}ihbar npi+4Lz}{4sqrt{hbar}L}right)right)+ Releft(e^{-frac{sqrt{2}inpi z}{L}}text{erf}left(frac{sqrt{2}L^2+2sqrt{2}ihbar npi-4Lz}{4sqrt{hbar}L}right)right)right]
$$

if n is even, and a similar function if n is odd. As you can see, this expression differs with what I calculated using exclusively the Segal-Bargmann space (although it agrees with the result if $L to infty$). The problem obviously is the fact that the system is confined, i.e. the Schrödinger wave function is zero for every point except for a finite interval. When I apply the Segal-Bargmann transform which is very similar to a Gaussian filter, error function terms appear, which clearly cannot be obtained with the eigenvalue equation. So, how do I fix this? Do I have to change the eigenvalue equation? Can I modifiy the normalization condition somehow so that

$$
int_{-L/2}^{L/2} dx (psi^{Sch})^{*}(x)A^{Sch}(x)psi^{Sch}(x) = int_{mathbb{R}}int_{-L/2}^{L/2} dxdp (psi^{SB})^{*}(z)A^{SB}(z)psi^{SB}(z), e^{-|z^2|}
$$

that is, every expected value of any observable A is he same for both wave functions, each calculated and normalized in their own space?