lapack tridiag - eigenvectors错误的标志

Question

我使用lapack函数tridiag在无限方形井的模拟中找到哈密顿函数的特征向量，我似乎随机得到一个错误的符号（在这种情况下p = 4和6）：

红色是理论和绿色模拟

（他们也没有正常化，但我认为这很正常吗？）

以下是代码：

#include <iostream>
#include <fstream>
#include <cmath>
using namespace std;


extern "C"
void dsteqr_( char * , int *, double [], double [], double [], int *, double [],
int *  ) ; 

void  tridiag( int n, double d[], double e[], double z[] ) 
/* Appel du sous-programme dsteqr de Lapack : calcul des valeurs et vecteurs
propres d'une matrice tridiagonale symmetrique   */
{
    int info ; // diagnostic
    double work[2*n-2] ; // espace de travail
    char compz='v' ; // mettre 'n' pour n'avoir que les valeurs propres

    dsteqr_( &compz, &n, d, e, z, &n, work, &info ) ; // scalaires pointeurs
    if ( info == 0 ) cout << "La diagonalisation s'est bien passée" << endl ;
    else cout << "Attention ! SSTEQR : info =" << info << endl ;
}

double V(double x){
    return 0;
}

int main() {
    int n=100;
    double d[n], e[n-1], z[n*n], x[n], v[n], L = 5, delta_x = L/n;
    ofstream of1("eigenvalues.dat"), of2("eigenvectors.dat");
    for(int i=0; i<n; i++){
        for (int j=0; j<n; j++){
            if (i == j) z[i+n*j] = 1;
            else z[i+n*j] = 0;
        }
        x[i] = (i+0.5)*delta_x - 0.5*L;
        v[i] = V(x[i]);
        d[i] = 2/pow(delta_x, 2) + v[i];
        if (i<n-2){ 
            e[i] = -1/pow(delta_x, 2);
        }
    }
    tridiag(n,d,e,z);
    for (int i=0; i<n; i++){
        of1 << d[i] << endl;
        for (int j=0; j<n; j++){
            of2 << z[i+n*j] << " ";
        }
        of2 << x[i] << endl;
    }
    of1.close();
    of2.close();
    return 0;
}

所以有人知道发生了什么吗？

Answer 1

The magnitude of your numerical result being larger than the magnitude of the reference result is due to the fact that L=5. The normalization of LAPACK would correspond to L=1. To get the correct vectors, divide all values by L !

Regarding the sign... If you change the signs of vectors in an orthonormal basis, the result is still an orthonormal basis. Moreover if A is an Hermitian matrix and if x is a vector so that A.x=l.x where l is a scalar, then A.(-x)=l.(-x). So the sign of the vector you get is not very significant... Moreover, if the matrix features many times the same eigenvalue, the result can contain any of the orthonormal basis of the subspace associated to this eigenvalue. For instance, if the matrix A is the identity, any orthonormal basis is an acceptable result for dsteqr().