Eigenvectors and Eigenvalues of a Perturbed Quantum System

by HENRILEN A. CUBIO

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

where V0 is a constant and

is some small number manifesting the perturbation such that

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.

First we need to write down the eigenvalues and eigenvectors of the unperturbed Hamiltonian.

The unperturbed Hamiltonian in this case is just

For the undisturbed system, it is straightforward to solve the eigenvalue equation

We just solve the characteristic equation in order to get the eigenvalues corresponding to the unperturbed Hamiltonian

In matrix form the above equation is written as

From the above matrix we can easily obtain the determinant so that we can get this expression

The solution to this algebraic equation provides us with the different eigenvalues

of the simpler, unperturbed Hamiltonian.

Now solving for

, we have the solution set as

The eigenvalues now of the simple quantum system are just

For each eigenvalue of a transformation, there is a corresponding eigenvector. The eigenspace of a given transformation for a particular eigenvalue is the set of the eigenvectors associated to this eigenvalue. After we have successfully obtained the eigenvalues, we are now tasked to find the corresponding eigenvectors for each eigenvalue.

For , the corresponding matrix equation gives

Therefore

The remaining two eigenvectors remain arbitrary. The resulting eigenvector for

is then

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

Similarly,

The linear combination of these eigenvectors is the eigenvector for

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

Therefore

The eigenvectors corresponding to the different eigenvalues of the unperturbed hamiltonian are then written as follows

For

we have

For

we have

For

we have

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

We can equate the first factor above to zero giving the expression

This expression yields the first eigenvalue which is

Now, equating the second factor to zero again we have

This would require us to use the quadratic formula to get the desired roots and so by applying we can have this expression

Simplifying the right hand side algebraically results to

The term with the radical sign may be written as

This is because of the power series expansion, up to second order as was asked, given by

Therefore the expression results to

The roots are easily read out separating the + and – signs

We now have the second eigenvalue which is

Solving for the third eigenvalue

This expression results to

Finally, writing down the three desired eigenvalues of the perturbed system

The first one is,

The second eigenvalue results to,

and the third and last eigenvalue is

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.

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20 Responses to “Eigenvectors and Eigenvalues of a Perturbed Quantum System”

Normie Jean Sajor Says:
January 5th, 2010 at 11:38 pm
Hello henrilen. Your article is organized since you have solved the problem step by step that is easy to understand by the readers. You also explain everything. Now, I have a better understanding about this topic. Thanks to you..Keep up the good work!
Sandra L. Manulat Says:
January 6th, 2010 at 9:19 am
Your article is very informative. I suggest though that you label your equations or number them so that it will be easier to follow, especially when you used the power series expansion, the order became confusing. Nevertheless i’ve learned a lot, thank you!
Hananish Joy Odarve Says:
January 6th, 2010 at 12:11 pm
Indeed finding the eigenvectors and eigenvalues of the state of a quantum system is important. You have done a good job in this one. Those who might still be confused and are new to the field will find this one helpful. I find this one really helpful.

More power.
Majvell Kay Odarve Says:
January 6th, 2010 at 2:53 pm
Ms. Henrilen,

The way you presented the process of deriving the eigenvectors and eigenvalues of a perturbed system is very effective. Relating the solution to unperturbed systems makes me understand the procedure clearly. Thaank you and God bless.
Gibson Tipdas Maglasang Says:
January 6th, 2010 at 7:25 pm
Good day to you Ms. Cubio.

The presentation was very impressive in such a way readers from all over the world could easily grasp the ideas presented in this webpage. I like this article since I also is in need of deep understanding regarding this particular topic.

In summary, as i have understood in your article that getting the eigenvalues and eigenvectors both that of unperturbed and perturbed follows the same process only that their eigenvalues differ with some added factor.

Thank you and God bless you and this website.
Lutchie Dyan S. Mendoza Says:
January 6th, 2010 at 8:22 pm
Good Day Henrilen =)
The article is good and the presentation is straight forward. I like the way how you present the topic, very simple yet concise and can be understood easily. The topic seem to be hard at first look yet you still was able to present it as simple as it could be. Good job=]. Keep it up!
Michael Jabines Says:
January 7th, 2010 at 1:19 am
This article is very simple and understandable that even a non-major physics can understand and even I can also understand the concept. Its solution is well organized and you can easily grasp the idea and topic even if without prior knowledge.
Edmar G. Pantohan Says:
January 7th, 2010 at 2:13 am
Your article gives us a way on how to get the eigenvalue and eigenvector of a perturbed system using the information in unperturbed. Its so nice that you used the characteristic equation where I learned already from my instructor Dr. Bacala.
karl patrick casas Says:
January 7th, 2010 at 2:16 am
Thanks for the new knowledge Henrilen! I would like to suggest to put equation numbers like the other authors did so that the readers could site when they comment. I’m just confused with the placement of epsilon in the perturbed Hamiltonian (in matrix form) or its just the addition of the potential of the perturbed and unperturbed. Anyway, its very well presented. Thank you very much.
carlo paul p. morente Says:
January 7th, 2010 at 3:02 am
Its a good and detailed talk about the topic, providing us a good understanding on how to solve and find the eigenvalues and eigenvectors of the system, both perturbed and unperturbed.
Rommel J. Jagus Says:
January 7th, 2010 at 3:15 am
The illustration of the topic is really good. The solutions is well explain. The article is a great help for the beginners of the topic. I really like the way you present the idea of the Perturbed and Unperturbed Quantum Systems.
John Paul Aseniero Says:
January 7th, 2010 at 3:27 am
I have read already about the eigenvectors an eigenvalues of a perturbed system but it amaze me that you have shown a very nice approach regarding this problem.
Eric M. Alcantara Says:
January 7th, 2010 at 5:25 am
Thanx for the very informative article, I really appreciate your effort of giving details clearly. But I think it would be better if you explains the use of employing the power series expansion for a better understanding of the reader. Anyway, its great. and it helps me a lot.
Christine Marie Ceblano Says:
January 7th, 2010 at 7:01 am
Hi Henrilen! You have a nice solution, really organize and understandable. Now this topic seems to be interesting.
Liza Dangkulos Says:
January 7th, 2010 at 8:48 am
The article is being presented in a way
that it could be understood and appreciated
by students of quantum mechanics. It is a good reference for those who want to either learn about or refresh on finding eigenvectors and eigenvalues of a perturbed system.
Good job,ma’am hen!
Marichu T. Miscala Says:
January 8th, 2010 at 8:57 am
Hello Ms. Henrilen. I have read your article and I found it a well-structured one, for it elaborately discusses the method of finding the eigenvectors and eigenvalues of a perturbed system. Also, the manner of presenting the equations and steps is clear and comprehensive especially for those who are still starting to study quantum mechanics. This is indeed one good reference. Keep up the good work.
Patrick Alvin M. Alcantara Says:
January 9th, 2010 at 4:59 am
The article’s quite interesting. It gives the reader a better understanding on the said topic, even newbies could easily get what the author’s trying to imply.
Lotis Racines Says:
January 9th, 2010 at 6:57 am
Your article is easy to understand, that for any beginner, he/she would never need to study higher physics than this just to cope with. The use of simple algebraic expressions never did intimidate them to grasp the physics of a perturbed quantum system. It helped me a lot to refresh my mind on solving for eigenvectors and eigenvalues. Your presentation is great!
Edwin B. Fabillar Says:
January 11th, 2010 at 11:07 pm
I partially encountered this eigenvectors and eigenvalues in math 151 (linear algebra) before but it was not quite challenging. Now as I read the content in your article I really appreciate the beauty of the physical approach and tricks to make the solution of the problem more simpler and understandable. Interesting! Congratulations Ma”am Henrilen and best wishes.
catherine therese quinones Says:
January 12th, 2010 at 2:04 am
Nice post! Honestly speaking, I am not an avid fan of eigenvectors and eigenfunctions ( and …) but the article was worth reading because the solutions were easy to follow. This is a good start to solving quantum mechanical problems involving perturbations. Very helpful indeed! Thanks for that!