Eigenvectors of the hamiltonian

[captainjack2000]

Homework Statement

The Hamiltonian of a system has the matrix representation

H=Vo*(1-e , 0 , 0
0 , 1 , e
0 , e , 2)

Write down the eigenvalues and eigenvectors of the unperturbed Hamiltonian (e=0)

Homework Equations

when unperturbed the Hamiltonian will reduce to Vo* the 3x3 matrix with 1,1,2 along the diagonal. the eigenvalues are therefore Vo,Vo,2Vo (right??)

I am a bit confused about how to calculate the eigenvectors. I have tried looking this up but still get confused. Would they not all be zero since if you sub the eigenvalue Vo back into matrix you would get for the first row

Vo(1-Vo,0,0) * (x,y,z) = (0,0,0) where (x,y,z) is a vertical matrix?

 

[morphemera]

What is the problem then? You should know that the 0 vector is the trivial solution, so it is not count for the eigenvector cos it cannot span any solution space.

 

[captainjack2000]

So how would you calculate the other eigenvectors. Sorry I am still confused.

 

[morphemera]

for e=0,
H=
(Vo , 0 , 0
0 , Vo , 0
0 , 0 , 2Vo)
eigenvalue is Vo,Vo,2Vo as you said. So what is the standard procedure to find the eigenvector? I assume that you should take at least one linear algebra before you take QM.

 

[paranormal]

Yes, you are correct about the eigenvalues for the unperturbed Hamiltonian. The eigenvalues will indeed be Vo, Vo, and 2Vo, as you have correctly calculated.

To calculate the eigenvectors, you will need to solve the following equation for each eigenvalue:

(H - λI)v = 0

Where H is the Hamiltonian matrix, λ is the eigenvalue, I is the identity matrix, and v is the eigenvector. This equation essentially represents the condition for an eigenvector, where the matrix multiplied by the eigenvector is equal to the eigenvalue multiplied by the same eigenvector.

For the first eigenvalue Vo, the equation will be:

(Vo - VoI)v = 0

Solving this equation will give you the eigenvector (1,0,0). Similarly, for the second eigenvalue Vo, the equation will be:

(Vo - VoI)v = 0

Solving this equation will give you the eigenvector (0,1,0). And for the third eigenvalue 2Vo, the equation will be:

(2Vo - 2VoI)v = 0

Solving this equation will give you the eigenvector (0,0,1).

So, the eigenvectors for the unperturbed Hamiltonian are (1,0,0), (0,1,0), and (0,0,1).

I hope this helps clarify the process for calculating eigenvectors for the given Hamiltonian. Let me know if you have any further questions.