Finding the eigenvectors and eigenvalues of the state of a quantum system is one of the most important concepts in quantum mechanics. And it is here where many students get confused.
In order to learn this by heart, one has to do several exercises. There are many ways that can be employed when we deal with these concepts. Let us have an example problem of determining the eigenvectors and eigenvalues of a perturbed quantum system.
A perturbed quantum system
We consider a quantum system with just three linearly independent states. The Hamiltonian, in matrix form, is
We have learned in quantum mechanics that the perturbed system describes a complicated quantum system but can be expressed in terms of a simpler one. The trick then is to begin with a simpler system for which a solution is known, and add an additional perturbing Hamiltonian that represents a small disturbance to the system. In this problem we are tasked to solve for the eigenvalues and eigenvectors of the perturbed quantum system.
We just solve the characteristic equation in order to get the eigenvalues corresponding to the unperturbed Hamiltonian
From the above matrix we can easily obtain the determinant so that we can get this expression
For, the corresponding matrix equation gives
Since the two are arbitrary we have the freedom to choose what their values are and to make things simple we choose 1 and 0 so that the eigenvectors become
For we have the following matrix,
It is easy to see that
Since it is arbitrary we can let any value for it and the most non-trivial and simplest value would be
The eigenvectors corresponding to the different eigenvalues of the unperturbed hamiltonian are then written as follows
If there is a basis defined in a vector space, the vectors can be expressed in terms of components. If we have finite dimensional vector spaces for example with dimension n, the transformations can be represented with n x n square matrices.
Next we solve for the exact eigenvalues of H. We expand each of them as power series in up to second order.
Using the characteristic equation again for solving now the Hamiltonian for the perturbed system we have
Solving for the determinant of this matrix we can easily arrived to this equation
Now, equating the second factor to zero again we have
Simplifying the right hand side algebraically results to
Therefore the expression results to
The eigenvalue problem simply tells us that under the transformation, the eigenvectors experience only changes in magnitude and sign. The result of the eigenvalue shows the amount of stretch or shrink to which a vector is subjected when transformed.
About the author:
Henrilen is a graduate student of physics at MSU-IIT . She hopes to do many researches someday that could truly benefit the people not only in this country but as well as for the whole world.