How to find the eigenvector for a perturbated Hamiltonian?

In summary: This is an isotropic 2D oscillator whose eigenenergies are given by $$E_n=\hbar\omega(n_x+n_y+1)$$ so the first two excited states are $$| 1,0\rangle ~~~and ~| 1,0\rangle ~~~~E_1=2\hbar\omega$$As for the first excited state, is the eigenvector for the matrix in post #1?I'm guessing you are asking if the eigenvector for the first excited state is the eigenvector of the matrix in post #1. The answer is no, the eigenvector for the first excited state is
  • #1
happyparticle
406
20
Homework Statement
Find the eigenvalue (first order) and eigenvector (0 order) for the first and second excited state (degenerate) for a perturbated hamiltonian
Relevant Equations
##H' = kxy##
##H = \frac{p_x^2}{2m} +\frac{p_y^2}{2m} + \frac{1}{2}m \omega^2 x^2 + \frac{1}{2}m \omega^2 y^2 ##
Hi,

I have to find the eigenvalue (first order) and eigenvector (0 order) for the first and second excited state (degenerate) for a perturbated hamiltonian.

However, I don't see how to find the eigenvectors.

To find the eigenvalues for the first excited state I build this matrix

##
\begin{pmatrix}
\langle 01 | H'| \rangle 01 & \langle 10| H'| \rangle 01 \\
\langle 01 | H'| \rangle 10 & \langle 10| H'| \rangle 10
\end{pmatrix}

= \frac{k \hbar}{2 m \omega}
\begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix}
##

Thus, I get the eigenvalue ##\pm \frac{k \hbar}{2 m \omega}##, but now I have no idea how to find the eigenvector for the 0 order.

Any help will me appreciate, thank you.
 
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  • #2
Are you saying that you know how to find the eigenvalues of a 2×2 matrix but you don't know how to find the corresponding eigenvectors? If so, there are plenty of sites that explain how. Just google "How to find the eigenvectors of a two by two matrix" and take your pick.
 
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  • #3
kuruman said:
Are you saying that you know how to find the eigenvalues of a 2×2 matrix but you don't know how to find the corresponding eigenvectors? If so, there are plenty of sites that explain how. Just google "How to find the eigenvectors of a two by two matrix" and take your pick.
I know how to find the eigenvector for a 2x2 matrix.
However, I'm not sure to understand what an eigenvector (0 order) for the first and second excited state really means.
For example is the eigenvector for the first excited state the eigenvector of the matrix in post #1 ?
 
  • #4
This is the nuts and bolts of degenerate perturbation theory. Absent the perturbation (i.e.##H'=0##), $$|n_x,n_y \rangle$$ are exact energy eigenstates so ##|1,0 \rangle## and ##|0,1 \rangle## are exactly degenerate solutions.
With ##H'\neq0## the perturbation theory is an expansion in k solved iteratively. Degenerate solutions can introduce convergence issues unless one selects states with an eye to the next order calculation. This choice is not an exact solution to the full Hamiltonian but it allows uniform convergence for what ensues
 
  • #5
Let's backtrack a bit. The perturbed 2×2 Hamiltonian matrix that you have cannot give you a second excited state. Do you see why? You need to expand the dimensionality of your matrix. How are you going to do this?
 
  • #6
kuruman said:
Let's backtrack a bit. The perturbed 2×2 Hamiltonian matrix that you have cannot give you a second excited state. Do you see why? You need to expand the dimensionality of your matrix. How are you going to do this?
For the second excited state, ##n = 2##, so the possible states are ##\langle 2,0,0| , \langle2,1,0| , \langle 2,1,1| , \langle 2,1,-1|##
 
  • #7
I am a little lost in nomenclature here. This is an isotropic 2D oscillator whose eigenenergies are given by $$E_n=\hbar\omega(n_x+n_y+1)$$ so the first two excited states are $$| 1,0\rangle ~~~and ~| 1,0\rangle ~~~~E_1=2\hbar\omega$$
 
  • #8
Hmm, from my course notes we have ##\langle n,l,m| H' | n,l,m \rangle##. Maybe I misunderstood something somewhere.
Or maybe the expression above is only for 1 particle.
 
  • #9
happyparticle said:
Hmm, from my course notes we have ⟨n,l,m|H′|n,l,m⟩. Maybe I misunderstood something somewhere.
One can do the problem in cylindrical coordinates but I do not know why one would. What are the notes for explicitly (and what course?). Methinks all is not well with you and quantum?
 
  • #10
This is a nice problem applicable in many systems. A few misunderstandings might have happened here.

First, your non-interacting Hamiltonian consists of 2 oscillator
$$ H_{x,y} = p^2_{x,y} + (x,y)^2, $$
where ##(x,y)## can be x or y . All parameters set to 1. Each oscillator has infinite levels. In a perturbation calculation, all levels have to be included. First order perturbation calculation means taking the correlation of all levels with the interaction term ##H^\prime## in to account. Second order perturbation means taking the correlation of the first oder correlations. And so on, applied to all states. What you did (neglecting the correctness) is taking only the first two levels in to account. Hopefully this clarifies the issue.

Second, if you like to use the polar representation ##\langle n,l,m|H^\prime| n^\prime,l^\prime,m^\prime \rangle##, there are probably less than 3 quantum numbers. Your problem lives in 2D space.

Third, a pretty simple thing one can do is to work it out in number space and use that to check the result of your calculation. The Hamiltonian can be written as
$$H = a^{\dagger} a + b^{\dagger} b + h.c. + k x y, $$
where ##a, b## are ladder operators for ##x, y## degrees of freedom. ##x = a^{\dagger} + a##. The interaction term is called dipole coupling. Turning it to quadratic form, you can diagonalize the two coupled oscillator giving ##2 k##-splitting and the collective normal modes.

I hope that helps.
 
Last edited:
  • #11
hutchphd said:
I am a little lost in nomenclature here. This is an isotropic 2D oscillator whose eigenenergies are given by $$E_n=\hbar\omega(n_x+n_y+1)$$ so the first two excited states are $$| 1,0\rangle ~~~and ~| 1,0\rangle ~~~~E_1=2\hbar\omega$$
I would take the "first two excited states" to mean (unperturbed) energy levels ##E_1=2\hbar\omega## and ##E_2=3\hbar\omega##.
 
  • #12
I would certainly not. This seems very bad nomenclature and pedagogy to me. If we solve in cylindrical coordinates do we ignore angular momentum multiplicity? States are states and levels are levels.
 
  • #13
hutchphd said:
Methinks all is not well with you and quantum?
I would say it's an euphemism. We spent the last 3 months deriving equations without further explanation or examples. To be honest I try to apply what I saw without really knowing what I do.

p-waive said:
I hope that helps.
A little bit. However, I thought what I did in post #1 what correct to find the eigenvalue.
 
  • #14
happyparticle said:
A little bit. However, I thought what I did in post #1 what correct to find the eigenvalue.
I stand corrected. My mistake about the first order corrected eigenvectors.

Thanks for pointing out!

I guess you found your 0-th order eigenvectors which are pretty obvious.
 

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