# Once I have the eigenvalues and the eigenvectors, how do I find the eigenfunctions?

I am using Mathematica to construct a matrix for the Hamiltonian of some system. I have built this matrix already, and I have found the eigenvalues and the eigenvectors, I am uncertain if what I did next is correct: I took the normalized eigenvectors, placed them in matrix form, and did matrix multiplication with the basis set of solutions.

Let me try to be more precise since I am not sure I am using the right language when mentioning the basis solutions. In the problem we are using the set of solutions of the particle in a box model as our basis. I can increase the number of basis elements in the calculation of the matrix of the Hamiltonian (which amounts to doing $<\psi_n|H \psi_k>$ over a specified range of $n$ and $k$) in order for some of my smallest eigenvalues to begin to converge. Once I have this $H$ matrix built, and that I see that my eigenvalues are converging to some degree, I take the eigenvectors of the $H$ matrix, format them to be in matrix form, and multiply them by the set of basis solutions.

I hope that makes things clearer.

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I'm not sure what you are asking; eigenfunctions are a type of eigenvector, in that they satisfy the eigenvalue equation. If your eigenvectors are functions, then you already have your eigenfunctions. –  KDN Feb 21 '13 at 18:26
@KDN, the eigenvectors I find are just numbers, it is my understanding that the eigenfunctions should be a linear combination of the basis solutions with the eigenvectors I found as coefficients. –  user17338 Feb 21 '13 at 18:46
I think I see the confusion; The eigen values are just numbers. The eigenfunctions are the eigenvectors of the operator. In the expression $A \Psi_n = \lambda_n \Psi_n$, the $\lambda_n$ (the numbers) are the eigenvalues, and the $\Psi_n$ are the eigenfunctions, which are also the eigenvectors of $A$. –  KDN Feb 21 '13 at 19:35

If $\bf{v}$ is an eigenvector of the matrix $\bf{H}$ (where the ith row and jth column of $\bf{H}$ is $<\psi_i|H|\psi_j>$) with eigenvalue $\lambda$, i.e.

$$\bf{H} \cdot \bf{v} = \lambda \cdot \bf{v}$$

the function (which is the one you are looking for)

$$\varphi = \bf{v} \cdot \bf{\psi} = \sum_j \bf{v}_j \cdot \psi_j$$

is an 'eigenfunction' (solution of the Schrödinger equation) of the Hamiltonian corresponding to $\bf{H}$ because for all $i$:

$$<\psi_i|H|\varphi> = \sum_j \bf{v}_j \cdot <\psi_i|H|\psi_j> = \sum_j \bf{v}_j \cdot \bf{H}_{ij} = (\bf{H} \cdot \bf{v})_i = (\lambda \cdot \bf{v})_i$$

Assuming your set of basis functions is orthonormal, i.e. $<\psi_i|\psi_j>= \delta_{ij}$ one can rewrite the above expression as:

$$<\psi_i|H|\varphi> = \sum_j \delta_{ij} \cdot \lambda \cdot \bf{v}_j = \lambda \cdot \sum_j \bf{v}_j <\psi_i|\psi_j> = \lambda \cdot <\psi_i|\sum_j \psi_j> = \lambda \cdot <\psi_i|\varphi>$$

because this holds for all $i$

$$H|\varphi> = \lambda \cdot |\varphi>$$

holds.

You say that you put the eigenvector $\bf{v}$ in Matrix form and then multiply it with the vector of basis functions to obtain the function $\varphi$. In fact it should be more like a 'dot product' but if you put the numbers of the eigenvector onto the diagonal (and leave zeros off the diagonal), that should be equivalent.

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