Guessing trial wave function with variational method

[physicist 12345]

Hellow

i want to ask about guessing the trial wave function at variational method of approximation

usually for example at solving harmonic oscillator or hydrogen atom we have conditions for trial wave function
but in fact i want to ask generally how could i make the guessing .. some problems give a particle of mass m moving at certain potential and want to use variational method to find gs energy now how could i guess the trial wavefunction

 

[A. Neumaier]

how could i guess the trial wavefunction

By having solved a number of problems, or seen how others make the choice.

 

[physicist 12345]

By having solved a number of problems, or seen how others make the choice.

then it some thing come with experience ?

 

[A. Neumaier]

then it some thing come with experience ?

With experience, or with trial and error. Generally one first thinks about the properties the function wanted should have (asymptotic behavior or behavior near distinguished points). Then one selects a class of functions having this property. Often there are paradigmatic exactly solvable cases that show what kind of solution is reasonable, and one can choose similar functions. Normalized wave functions typically decay exponentially. Generic variability is created by polynomial contributions. This suggests an ansatz $e^{-a|x-x_k|} p(x)$ with a polynomial $p(x)$, or (suggested by the linearity of the Schroedinger equation) linear combinations of these. If one wants to have an easy time in calculating inner products, one uses instead Gaussians times polynomials, etc..Unnormalized ones have an asymptotic form reflecting knowledge about scattering, and again one can make up trial functions with this behavior.

 

[marcusl]

Variational methods are often not very sensitive to the choice of trial functions, so long as they're not way off.

 

[A. Neumaier]

Variational methods are often not very sensitive to the choice of trial functions, so long as they're not way off.

They must have qualitatively the correct form, and the approximation cannot be better than what the ansatz allows.