Finding eigenvalue and normalized eigenstate of a hamiltonian

noblegas
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Homework Statement




The system described by the Hamiltonian H_0 has just two orthogonal energy eigenstates, |1> and |2> , with

<1|1>=1 , <1|2> =0 and <2|2>=1 . The two eignestates have the same eigenvalue , E_0:

H_0|i>=E_0|i>, for i=1 and 2.

Now suppose the Hamiltonian for the system is changed by the addition of the term V, given H=H_0+V.

The matrix elements of V are

<1|V|1> =0 , <1|V|2>=V_12, <2|V|2>=0.

a) Find the eigenvalues of the new Hamiltonian, H , in terms of the quanties above

b) Find the normalized eigenstates of H in terms of |1> , |2> and the other given expressions.


Homework Equations





The Attempt at a Solution


a) I don't know how to begin this problem but I guess I will start by plugging in the values for H_0 and V: H=H_0+V=<1|1>+<1|V|1>=1+0=1 for the first value of H_0 and the first value of V. don't you find the eigen value by writing out this expression: det(H-I*lambda)=0? I have kno w idea.
 
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I think this problem is more easily seen as a matrix:

<br /> H=\left(\begin{array}{cc}E_0 &amp;0 \\ 0 &amp; E_0\end{array}\right)+\left(\begin{array}{cc}0&amp;V_{12} \\ V_{21} &amp; 0\end{array}\right)=\left(\begin{array}{cc}E_0&amp;V_{12} \\ V_{21} &amp; E_0\end{array}\right)<br />

You can then use linear algebra methods to calculate the eigenvalue \lambda and then the eigenvectors (eigenstates) from the new eigenvalues.
 
jdwood983 said:
I think this problem is more easily seen as a matrix:

<br /> H=\left(\begin{array}{cc}E_0 &amp;0 \\ 0 &amp; E_0\end{array}\right)+\left(\begin{array}{cc}0&amp;V_{12} \\ V_{21} &amp; 0\end{array}\right)=\left(\begin{array}{cc}E_0&amp;V_{12} \\ V_{21} &amp; E_0\end{array}\right)<br />

You can then use linear algebra methods to calculate the eigenvalue \lambda and then the eigenvectors (eigenstates) from the new eigenvalues.

Should I write out a matrix for V_11 an V_22 as well, along with E_1 and E_2?
 
noblegas said:
Should I write out a matrix for V_11 an V_22 as well, along with E_1 and E_2?

You've already shown that V_{11}=V_{22}=0, and I included this in the perturbed potential:


<br /> V=\left(\begin{array}{cc}V_{11}&amp;V_{ 12} \\ V_{21} &amp; V_{22}\end{array}\right)=\left(\begin{array}{cc}0&amp;V_{ 12} \\ V_{21} &amp; 0\end{array}\right)<br />

Likewise, you have already claimed that \langle1|H|1\rangle=\langle2|H|2\rangle=E_0 and I too added them in for H_0.
 
jdwood983 said:
You've already shown that V_{11}=V_{22}=0, and I included this in the perturbed potential:


<br /> V=\left(\begin{array}{cc}V_{11}&amp;V_{ 12} \\ V_{21} &amp; V_{22}\end{array}\right)=\left(\begin{array}{cc}0&amp;V_{ 12} \\ V_{21} &amp; 0\end{array}\right)<br />

Likewise, you have already claimed that \langle1|H|1\rangle=\langle2|H|2\rangle=E_0 and I too added them in for H_0.

For part b, would I go about and normalized a matrix by multiplying V by its complex conjugate, i.e., VV*dr=1?
 
noblegas said:
For part b, would I go about and normalized a matrix by multiplying V by its complex conjugate, i.e., VV*dr=1?

No, you need to find the eigenvectors (eigenstates) from your eigenvalues found using this Hamiltonian:

<br /> H=\left(\begin{array}{cc}E_0&amp;V_{ 12} \\ V_{21} &amp; E_0\end{array}\right)<br />

Small hint: You correctly wrote the equation necessary to find the eigenvalues in your first post. Once you find the eigenvalues, how do you find the associated eigenvectors?
 
jdwood983 said:
No, you need to find the eigenvectors (eigenstates) from your eigenvalues found using this Hamiltonian:

<br /> H=\left(\begin{array}{cc}E_0&amp;V_{ 12} \\ V_{21} &amp; E_0\end{array}\right)<br />

Small hint: You correctly wrote the equation necessary to find the eigenvalues in your first post. Once you find the eigenvalues, how do you find the associated eigenvectors?

You would do the hamiltonian-lambda*I and you would find the determinate of det(hamiltonian-lambda*I )=0 to find the values of lambda right? then I would normalized the hamiltonian?
 
noblegas said:
You would do the hamiltonian-lambda*I and you would find the determinate of det(hamiltonian-lambda*I )=0 to find the values of lambda right? then I would normalized the hamiltonian?

Yes then no. You are correct in finding your eigenvalues (\lambda), but you are not asked to normalize the Hamiltonian, you are asked to find the normalized eigenstates (which are the same thing as eigenvectors). If you are unfamiliar with this process, then I would recommend checking out http://www.tmt.ugal.ro/crios/Support/ANPT/Curs/math/s3/s3eign/s3eign.html"
 
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jdwood983 said:
I think this problem is more easily seen as a matrix:

<br /> H=\left(\begin{array}{cc}E_0 &amp;0 \\ 0 &amp; E_0\end{array}\right)+\left(\begin{array}{cc}0&amp;V_{12} \\ V_{21} &amp; 0\end{array}\right)=\left(\begin{array}{cc}E_0&amp;V_{12} \\ V_{21} &amp; E_0\end{array}\right)<br />

You can then use linear algebra methods to calculate the eigenvalue \lambda and then the eigenvectors (eigenstates) from the new eigenvalues.

<br /> <br /> H=\left(\begin{array}{cc}E_0 &amp;0 \\ 0 &amp; E_0\end{array}\right)+\left(\begin{array}{cc}0&amp;V_{ 11} \\ V_{11} &amp; 0\end{array}\right)=\left(\begin{array}{cc}E_0&amp;V_{ 11} \\ V_{21} &amp; E_0\end{array}\right)<br /> would I write two other matrixes fo and <br /> <br /> H=\left(\begin{array}{cc}E_0 &amp;0 \\ 0 &amp; E_0\end{array}\right)+\left(\begin{array}{cc}0&amp;V_{ 22} \\ V_{22} &amp; 0\end{array}\right)=\left(\begin{array}{cc}E_0&amp;V_{ 22} \\ V_{21} &amp; E_0\end{array}\right)<br /> <br />
 
  • #10
noblegas said:
<br /> <br /> H=\left(\begin{array}{cc}E_0 &amp;0 \\ 0 &amp; E_0\end{array}\right)+\left(\begin{array}{cc}0&amp;V_{ 11} \\ V_{11} &amp; 0\end{array}\right)=\left(\begin{array}{cc}E_0&amp;V_{ 11} \\ V_{21} &amp; E_0\end{array}\right)<br /> would I write two other matrixes fo and <br /> <br /> H=\left(\begin{array}{cc}E_0 &amp;0 \\ 0 &amp; E_0\end{array}\right)+\left(\begin{array}{cc}0&amp;V_{ 22} \\ V_{22} &amp; 0\end{array}\right)=\left(\begin{array}{cc}E_0&amp;V_{ 22} \\ V_{21} &amp; E_0\end{array}\right)<br /> <br />

No...V_{11}=V_{22}=0. These have already been included, as I said in Post #4. You just need to find \lambda from

<br /> \det\left[\left(\begin{array}{cc}E_0&amp;V_{ 12} \\ V_{21} &amp; E_0\end{array}\right)-\left(\begin{array}{cc}\lambda&amp;0 \\ 0 &amp; \lambda\end{array}\right)\right]=0<br />

and then find the associated eigenvectors from this and then normalize them.
 
  • #11
jdwood983 said:
No...V_{11}=V_{22}=0. These have already been included, as I said in Post #4. You just need to find \lambda from

<br /> \det\left[\left(\begin{array}{cc}E_0&amp;V_{ 12} \\ V_{21} &amp; E_0\end{array}\right)-\left(\begin{array}{cc}\lambda&amp;0 \\ 0 &amp; \lambda\end{array}\right)\right]=0<br />

and then find the associated eigenvectors from this and then normalize them.

I would have two values for lambda right? and my eigenvectors would be: <br /> <br /> H=\left(\begin{array}{cc}E_0 &amp;0 \\ 0 &amp; E_0\end{array}\right)+\left(\begin{array}{cc}0&amp;V_{ 12} \\ V_{21} &amp; 0\end{array}\right)=\left(\begin{array}{cc}E_0&amp;V_{ 12} \\ V_{21} &amp; E_0\end{array}\right)<br /> <br />*[x,y]=lambda_1*[x,y], and <br /> <br /> H=\left(\begin{array}{cc}E_0 &amp;0 \\ 0 &amp; E_0\end{array}\right)+\left(\begin{array}{cc}0&amp;V_{ 12} \\ V_{21} &amp; 0\end{array}\right)=\left(\begin{array}{cc}E_0&amp;V_{ 12} \\ V_{21} &amp; E_0\end{array}\right)*[x,y]=lambda_2*[x,y] right
 
  • #12
noblegas said:
I would have two values for lambda right? and my eigenvectors would be: <br /> <br /> H=\left(\begin{array}{cc}E_0 &amp;0 \\ 0 &amp; E_0\end{array}\right)+\left(\begin{array}{cc}0&amp;V_{ 12} \\ V_{21} &amp; 0\end{array}\right)=\left(\begin{array}{cc}E_0&amp;V_{ 12} \\ V_{21} &amp; E_0\end{array}\right)<br /> <br />*[x,y]=lambda_1*[x,y], and <br /> <br /> H=\left(\begin{array}{cc}E_0 &amp;0 \\ 0 &amp; E_0\end{array}\right)+\left(\begin{array}{cc}0&amp;V_{ 12} \\ V_{21} &amp; 0\end{array}\right)=\left(\begin{array}{cc}E_0&amp;V_{ 12} \\ V_{21} &amp; E_0\end{array}\right)<br /> <br /> <br />*[x,y]=lambda_2*[x,y] right

This is correct. I will note that there will be a condition involved on V_{12} and V_{21} in finding your eigenvalues. When finding \lambda, remember that your eigenvalues must be real.
 
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