How to do numerical integration with quantum harmonic oscillator wavefunction?

Question

How to do numerical integration (what numerical method, and what tricks to use) for one-dimensional integration over infinite range, where one or more functions in the integrand are 1d quantum harmonic oscillator wave functions. Among others I want to calculate matrix elements of some function in the harmonic oscillator basis:

phi_n(x) = N_n H_n(x) exp(-x²/2)
where H_n(x) is Hermite polynomial

V_m,n = \int_{-infinity}^{infinity} phi_m(x) V(x) phi_n(x) dx

Also in the case where there are quantum harmonic wavefunctions with different widths.

The problem is that wavefunctions phi_n(x) have oscillatory behaviour, which is a problem for large n, and algorithm like adaptive Gauss-Kronrod quadrature from GSL (GNU Scientific Library) take long to calculate, and have large errors.

Answer 1

The WKB approximation?

Answer 2

An incomplete answer, since I'm a little short on time at the moment; if others can't complete the picture, I can supply more details later.

1. Apply orthogonality of the wavefunctions whenever and wherever possible. This should significantly cut down the amount of computation.

2. Do analytically whatever you can. Lift constants, split integrals by parts, whatever. Isolate the region of interest; most wavefunctions are band-limited, and reducing the area of interest will do a lot to save work.

3. For the quadrature itself, you probably want to split the wavefunctions into three pieces and integrate each separately: the oscillatory bit in the center plus the exponentially-decaying tails on either side. If the wavefunction is odd, you get lucky and the tails will cancel each other, meaning you only have to worry about the center. For even wavefunctions, you only have to integrate one and double it (hooray for symmetry!). Otherwise, integrate the tails using a high order Gauss-Laguerre quadrature rule. You might have to calculate the rules yourself; I don't know if tables list good Gauss-Laguerre rules, as they're not used too often. You probably also want to check the error behavior as the number of nodes in the rule goes up; it's been a long time since I used Gauss-Laguerre rules and I don't remember if they exhibit Runge's phenomenon. Integrate the center part using whatever method you like; Gauss-Kronrod is a solid choice, of course, but there's also Fejer quadrature (which sometimes scales better to high numbers of nodes, which might work nicer on an oscillatory integrand) and even the trapezoidal rule (which exhibits stunning accuracy with certain oscillatory functions). Pick one and try it out; if results are poor, give another method a shot.

Hardest question ever on SO? Hardly :)

Answer 3

I am not going to explain or qualify any of this right now. This code is written as is and probably incorrect. I am not even sure if it is the code I was looking for, I just remember that years ago I did this problem and upon searching my archives I found this. You will need to plot the output yourself, some instruction is provided. I will say that the integration over infinite range is a problem that I addressed and upon execution of the code it states the round off error at 'infinity' (which numerically just means large).

// compile g++ base.cc -lm
#include <iostream>
#include <cstdlib>
#include <fstream>
#include <math.h>

using namespace std;

int main ()
        {
        double xmax,dfx,dx,x,hbar,k,dE,E,E_0,m,psi_0,psi_1,psi_2;
        double w,num;
        int n,temp,parity,order;
        double last;
        double propogator(double E,int parity);
        double eigen(double E,int parity);
         double f(double x, double psi, double dpsi);
        double g(double x, double psi, double dpsi);
        double rk4(double x, double psi, double dpsi, double E);

        ofstream datas ("test.dat");

        E_0= 1.602189*pow(10.0,-19.0);// ev joules conversion
        dE=E_0*.001;
//w^2=k/m                 v=1/2 k x^2             V=??? = E_0/xmax   x^2      k-->
//w=sqrt( (2*E_0)/(m*xmax) );
//E=(0+.5)*hbar*w;

        cout << "Enter what energy level your looking for, as an (0,1,2...) INTEGER: ";
        cin >> order;

        E=0;
        for (n=0; n<=order; n++)
                {
                parity=0;
//if its even parity is 1 (true)
                temp=n;
                if ( (n%2)==0 ) {parity=1; }
                cout << "Energy " << n << " has these parameters: ";
                E=eigen(E,parity);
                if (n==order)
                        {
                        propogator(E,parity);
                        cout <<" The postive values of the wave function were written to sho.dat \n";
                        cout <<" In order to plot the data should be reflected about the y-axis \n";
                        cout <<"  evenly for even energy levels and oddly for odd energy levels\n";
                        }
                E=E+dE;
                }
        }

double propogator(double E,int parity)
        {
        ofstream datas ("sho.dat") ;

        double hbar =1.054*pow(10.0,-34.0);
        double m =9.109534*pow(10.0,-31.0);
        double E_0= 1.602189*pow(10.0,-19.0);
        double dx =pow(10.0,-10);
        double xmax= 100*pow(10.0,-10.0)+dx;
        double dE=E_0*.001;
        double last=1;
        double x=dx;
        double psi_2=0.0;
        double psi_0=0.0;
        double psi_1=1.0;
//      cout <<parity << " parity passsed \n";
        psi_0=0.0;
        psi_1=1.0;
        if (parity==1)
                {
                psi_0=1.0;
                psi_1=m*(1.0/(hbar*hbar))* dx*dx*(0-E)+1 ;
                }

        do
                {
                datas << x << "\t" << psi_0 << "\n";
                psi_2=(2.0*m*(dx/hbar)*(dx/hbar)*(E_0*(x/xmax)*(x/xmax)-E)+2.0)*psi_1-psi_0;
//cout << psi_1 << "=psi_1\n";
                psi_0=psi_1;
                psi_1=psi_2;
                x=x+dx;
                } while ( x<= xmax);
//I return 666 as a dummy value sometimes to check the function has run
        return 666;
        }


   double eigen(double E,int parity)
        {
        double hbar =1.054*pow(10.0,-34.0);
        double m =9.109534*pow(10.0,-31.0);
        double E_0= 1.602189*pow(10.0,-19.0);
        double dx =pow(10.0,-10);
        double xmax= 100*pow(10.0,-10.0)+dx;
        double dE=E_0*.001;
        double last=1;
        double x=dx;
        double psi_2=0.0;
        double psi_0=0.0;
        double psi_1=1.0;
        do
                {
                psi_0=0.0;
                psi_1=1.0;

                if (parity==1)
                        {double psi_0=1.0; double psi_1=m*(1.0/(hbar*hbar))* dx*dx*(0-E)+1 ;}
                x=dx;
                do
                        {
                        psi_2=(2.0*m*(dx/hbar)*(dx/hbar)*(E_0*(x/xmax)*(x/xmax)-E)+2.0)*psi_1-psi_0;
                        psi_0=psi_1;
                        psi_1=psi_2;
                        x=x+dx;
                        } while ( x<= xmax);


                if ( sqrt(psi_2*psi_2)<=1.0*pow(10.0,-3.0))
                        {
                        cout << E << " is an eigen energy and " << psi_2 << " is psi of 'infinity'  \n";
                        return E;
                        }
                else
                        {
                        if ( (last >0.0 && psi_2<0.0) ||( psi_2>0.0 && last<0.0) )
                                {
                                E=E-dE;
                                dE=dE/10.0;
                                }
                        }
                last=psi_2;
                E=E+dE;
                } while (E<=E_0);
        }

If this code seems correct, wrong, interesting or you do have specific questions ask and I will answer them.

Answer 4

I'd recommend a few other things:

1. Try transforming the function onto a finite domain to make the integration more manageable.
2. Use symmetry where possible - break it up into the sum of two integrals from negative infinity to zero and zero to infinity and see if the function is symmetry or anti-symmetric. It could make your computing easier.
3. Look into Gauss-Laguerre quadrature and see if it can help you.