Can this PDE problem be reduced from 3D to 1D, by integrating the equation over two of the dimensions? I am looking for a little help with the 3-dimensional (potential-free) Schrodinger equation, which is a partial differential equation for the evolution of some complex function $\psi(x,y,z,t)$ over time $t$, in cartesian coordinates. The equation is
$$
\frac{\partial\psi}{\partial t} = \frac{i\hbar}{2m}\bigg( \frac{\partial^2\psi}{\partial x^2} + \frac{\partial^2\psi}{\partial y^2} + \frac{\partial^2\psi}{\partial z^2} \bigg),
$$
and I will assume that $\psi$ and its first derivative are zero at $\pm\infty.$
Now, in the end I am only interested in the time-evolution of the function after integration over two of the dimensions $x$ and $y$. Therefore I would like to know what is the equation for the time-dependence of the integrated wavefunction
$$
\Phi(z,t) = \int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty} \psi(x,y,z,t)\;\textrm{d}x\;\textrm{d}y
$$
i.e.
$$
\frac{\partial\Phi(z,t)}{\partial t} = \;?
$$
Basically, the problem is really 3D, but since ultimately I am interested only in the integrated field, I am wondering if this actually reduces to a 1D problem?
Is the approach to try to integrate the entire equation with respect to $x$ and $y$, like this
$$
\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty} \frac{\partial\psi}{\partial t} \;\textrm{d}x\;\textrm{d}y = \frac{i\hbar}{2m} \int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty} \bigg( \frac{\partial^2\psi}{\partial x^2} + \frac{\partial^2\psi}{\partial y^2} + \frac{\partial^2\psi}{\partial z^2} \bigg) \;\textrm{d}x\;\textrm{d}y
$$
If so, how can I proceed to write it in terms of $\Phi(z,t)$?
Thank you!

EDIT
After a comment from Sal, I should clarify that what I am really interested in ultimately is not the wavefunction $\Phi$, but the time evolution of the integrated density / probability, given by
$$
n_{\textrm{1d}}(z,t) = \int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty} |\psi(x,y,z,t)|^2\;\textrm{d}x\;\textrm{d}y
$$
I am not sure if this makes a difference to how the equation can be simplified.
 A: As requested, but please see comment above. Given
$$
\Phi(z,t)=\int dx  dy \ \psi(x,y,z,t)
$$
The time derivative goes right through the integral, and $\partial_t\psi$ can be substituted from the Schrodinger equation (with $\hbar=2m$ and $\nabla=\nabla_{x,y}$)
$$
\partial_t \Phi(z,t)=\int dx dy \ i \left[\partial_{zz}\psi +\nabla\cdot\nabla \psi \right]
$$
The last term on the RHS may be evaluated by the divergence theorem (and $S$ is the region over which we integrate)
$$
\partial_t \Phi(z,t)=i\int\limits_S dx dy \ \partial_{zz}\psi \ + \ i\int\limits_{\partial S} dl \ \nabla \psi \cdot \hat{n}
$$
If the boundary terms vanish and if it is permissible to move the partial derivatives past the integral, we are left with
$$ \tag{*}
\partial_t\Phi(z,t)=i\partial_{zz} \Phi(z,t)
$$
Edit
If we are instead interested in
$$
n(z,t)=\int dxdy \ |\psi|^2
$$
then I do not think any simplification occurs. By taking the time derivative and substituting $\partial_t \psi$ from the Schrodinger equation, we find (again with $\nabla=\nabla_{x,y}$)
$$
\partial_t n(z,t)=i\int dxdy \ \left[\psi^* \nabla^2 \psi-\psi \nabla^2\psi^*  \right]+i\int dxdy \ \left[\psi^* \partial_{zz} \psi-\psi \partial_{zz}\psi^*  \right]
$$
The first term on the RHS may be dealt with as before, but the second does not look like it admits a nice expression in terms of $n(z,t)$.
For completeness, let me mention that the full time evolution of $\psi$ is simple to obtain in this case. Let $\psi_0(x,y,z)$ be the initial state, and $\tilde{\psi_0}(\mathbf{k})$ be its Fourier transform. Then we have
$$
\psi(x,y,z,t)=\int \frac{d^3 \mathbf{k}}{(2\pi)^3} \ \tilde{\psi_0}(\mathbf{k}) \exp \left(i \mathbf{k} \cdot \mathbf{r}-ik^2 t \right)
$$
Where $\mathbf{k}=(k_x,k_y,k_z)$ and $\mathbf{r}=(x,y,z)$. Then we have
$$
n(z,t)=\int dx dy \ \int \frac{d^3 \mathbf{k}}{(2\pi)^3} \frac{d^3 \mathbf{k'}}{(2\pi)^3} \ \tilde{\psi_0}(\mathbf{k}) \tilde{\psi_0}^*(\mathbf{k'}) \exp \left(i (\mathbf{k}-\mathbf{k'}) \cdot \mathbf{r} \right) e^{-it(k^2-k'^2)}
$$
The $x$ and $y$ integrals yield deltas, which collapse two $k'$ integrals, and we are left with
$$
n(z,t)= \int \frac{d^3 \mathbf{k}}{(2\pi)^3} \frac{d k_z'}{2\pi} \ \tilde{\psi_0}(\mathbf{k}) \tilde{\psi_0}^*(k_x,k_y,k_z') e^{i (k_z-k_z')z}e^{-it(k_z^2-k_z'^2)}
$$
Which looks a bit messy but contains the full evolution in terms of the initial condition. The Fourier transform of $\psi^*$ can be related to the Fourier transform of $\psi$
If we make the assumption$^\dagger$ that $\psi_0=X(x)Y(y)Z(z)$, then this simplifies considerably using Parseval's theorem
$$
n(z,t)= \int \frac{d k_z}{2\pi} \ \frac{d k_z'}{2\pi} \ \tilde{Z}(k_z) \tilde{Z}^*(k_z') e^{i (k_z-k_z')z}e^{-it(k_z^2-k_z'^2)}
$$
$\dagger$ This is a very restrictive assumption here. We end up with a completely 1D problem by virtue of rejecting any initial state that could not be represented as such.
