# Why we often approximate a wave function of a particle to Gaussian wave function?

I was solving problem of two particle system. We were taking wave function generally $$\psi$$. Later we approximated this wavefunction of two-particle system to double Gaussian wave function. My question is that why we often approximate a wave function of a particle to Gaussian wave function ? What is the advantage of Gaussian wave function?

Usually the standard reasons to use (at least didactically) gaussian wave packets are that they allow for exact solutions for the integrals that involve them, instead of relying on approximate solutions. Also gaussian wave packets satisfy the Heisenberg uncertainty relationship between the dispersions of $$\hat X$$ and $$\hat P$$ with the = sign:

$$\langle\Delta \hat X^2\rangle \langle\Delta \hat P^2\rangle = \frac14\left|\langle [\hat X,\hat P]\rangle\right|^2$$

If you can provide additional details on you problem maybe we can find some other reason for which its useful to consider a gaussian function.

• And it is the ground state of harmonic oscillators.
– user137289
Dec 9 '18 at 9:42

Usually wave functions are finite within some volume and disappear outside - due to boundary conditions. There are many wave functions far from Gaussian functions, but for free particles in a collimated beam at some large distance the wave function looks like a Gaussian one.

If you take two beams that do not intersect, you will not get an interference pattern or any other interaction. Thus Gaussian distribution shows the importance of (implied!) inequalities for colliding beams - their widths must me much larger than the region of collision. Then the widths (edge effects) get small and negligible. If so, the Gaussians in the collision region may be approximated further - by plane waves with certain energy-momentum.

To describe a particle by a wave we need a wave packet which has the same velocity,postion and momentum as the particle has.Generally we can specify a region of wave propagation by superposing infinite number of waves .But if we want to specify positon we can find an wave train (a single position)by superposing the wave.And if we want to know the momentum exactly we have to superpose the waves of a finite range of angular frequency.But a guassian wave functions which also can be found from superposition of the waves are the one which can describe both position and momentum well with finite uncertainty satisfied by uncertainty relation.And we are saved in the case of the wave function of a free particle.May be it is the simplest wave function,i.e,$$\psi(x,t)=A ~{\exp}[i(kx-{\omega}t)]$$ $$=A ~{\exp}[-{\frac {i}{\hbar}}(Et-px)]$$