${\displaystyle i\psi _{t}+\psi _{xx}+2(|\psi |^{2})_{x}\,\psi +|\psi |^{4}\,\psi =0,}$

${\displaystyle i^{2}=-1,}$

${\displaystyle x}$

${\displaystyle t}$

${\displaystyle i\varphi _{t}+\varphi _{xx}=0}$

${\displaystyle \varphi (x,t)}$

${\displaystyle \psi (x,t)}$

${\displaystyle \psi (x,t)={\varphi (x,t)}/{\textstyle (1+2\,\int _{-\infty }^{x}|\varphi (x^{\prime },t)|^{2}\;{\text{d}}x^{\prime })^{1/2}}.}$

The wave-packet solution used here (to the linear Schrödinger equation) is:

${\displaystyle \varphi (x,t)=a\,{\frac {{\text{e}}^{-{\tfrac {1}{8}}v^{2}}}{\sqrt {1+2it}}}\,\exp \left(-{\frac {1}{2}}{\frac {(x-{\tfrac {1}{2}}iv)^{2}}{1+2it}}\right),}$

with ${\displaystyle v}$ the group velocity and ${\displaystyle a}$ the amplitude. This solution has an envelope of Gaussian shape:

${\displaystyle |\varphi (x,t)|={\frac {|a|}{\sqrt[{4}]{1+4t^{2}}}}\,\exp \left(\displaystyle -{\frac {1}{2}}\,{\frac {(x-vt)^{2}}{1+4t^{2}}}\right).}$

So the envelope of the linear solution ${\displaystyle |\varphi (x,t)|}$ widens with time, and becomes lower. Further

${\displaystyle \int _{-\infty }^{x}|\varphi (x^{\prime },t)|^{2}\;{\text{d}}x^{\prime }={\frac {\sqrt {\pi }}{2}}\,|a|^{2}\left[1+{\text{erf}}\left({\frac {x-vt}{\sqrt {1+4t^{2}}}}\right)\right],}$

with erf the error function. The centre of gravity ${\displaystyle x_{c}(t)}$ of the wave packet is computed as ${\displaystyle \textstyle x_{c}=\int x\,|\psi (x,t)|^{2}\,{\text{d}}x/\int |\psi (x,t)|^{2}\,{\text{d}}x.}$

${\displaystyle a=2}$

${\displaystyle v=4.}$

${\displaystyle |\psi (x,t)|}$

${\displaystyle |\varphi (x,t)|}$

${\displaystyle x}$