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!


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.

  • $\begingroup$ You may integrate the entire equation, and use the Liebniz rule to commute $\partial_t$ and $\partial_z$ with the integrals. Note that you get boundary terms appearing since $\int\limits_a^b dx \ \partial_{xx} \psi(x)=\partial_x \psi(b)-\partial_x \psi(a)$. $\endgroup$
    – Sal
    Jul 9, 2021 at 13:21
  • $\begingroup$ Some thoughts: It is clear that Φ contains less information than ψ. What is unclear to me right now is: for given Φ(z,0), if the time evolution given in (*) unique. Are you certain you want the given definition of Φ, rather than say, Φ=∫dxdy ψ∗ψ? Is it necessary that you study an integral of this form, or is it only to try and simplify the problem? $\endgroup$
    – Sal
    Jul 9, 2021 at 22:55
  • $\begingroup$ I've updated my answer. It turns out (as far as I can tell) no nice simplification occurs for $n_{1D}$, but for a free particle it is straightforward to write down the full time evolution and then integrate to get $n$ $\endgroup$
    – Sal
    Jul 14, 2021 at 20:07

1 Answer 1


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

$$ \tag{*} \partial_t\Phi(z,t)=i\partial_{zz} \Phi(z,t) $$


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.

  • $\begingroup$ Thanks for the updated answer Sal. I have a couple of questions: 1) In your edit for n(z,t), is it still okay to apply the divergence theorem now, given that we have a product term like $\psi^*\nabla^2\psi$ ? 2) Do things simplify if we assume, for example, that the distribution in the two directions $x$ and $y$ is gaussian? 3) A comment really, but with regard to the exact evolution in Fourier space, I believe this is simple in principle, but may require a large amount of computer memory in practice if the expansion is large. See this paper: info.ifpan.edu.pl/~deuar/pubs/cpc_208_92.pdf $\endgroup$
    – teeeeee
    Jul 15, 2021 at 15:30
  • $\begingroup$ Also, I am stuggling to understand the difference between $k$ and $k'$. If we have $\psi(k_x,k_y,k_z)$ in Fourier space, then isn't the conjugate $\psi^*(k_x,k_y,k_z)$ ? Why $\psi^*(k_x',k_y',k_z')$ . Thank you! $\endgroup$
    – teeeeee
    Jul 19, 2021 at 11:14
  • $\begingroup$ When you say "the $x$ and $y$ integrals yield deltas which collapse to $k'$ integrals", could you show this step? $\endgroup$
    – teeeeee
    Jul 19, 2021 at 11:16
  • $\begingroup$ In practice, if we take the Fourier transform of both $\psi$ and $\psi^*$, they would be defined on the same $k$ grid (i.e. $k_z=k_z'$), and so all the terms such as $e^{i(k_z-k_z')z}$ would be zero? What am I missing? $\endgroup$
    – teeeeee
    Jul 19, 2021 at 11:48
  • $\begingroup$ @teeeeee 1. Yes you can use vector calc in the first term of the equation for $n(z,t)$, you'll need Green's theorem. 2. If we assume the initial state is separable: $\psi_0=X(x)Y(y)Z(z)$, with eg. $X$ and $Y$ Gaussian, then yes some simplification does occur. You can simply do the $x$ and $y$ integrals in the second term on the right. The resulting equation is in terms of $Z$ and $Z^*$, but I'm not sure this is useful. 3. I'm not at all familiar with numerical methods so can't really comment. $\endgroup$
    – Sal
    Jul 19, 2021 at 22:22

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