# Finding eigenvalue and normalized eigenstate of a hamiltonian

1. Nov 29, 2009

### noblegas

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

3. 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.

2. Nov 29, 2009

### jdwood983

I think this problem is more easily seen as a matrix:

$$H=\left(\begin{array}{cc}E_0 &0 \\ 0 & E_0\end{array}\right)+\left(\begin{array}{cc}0&V_{12} \\ V_{21} & 0\end{array}\right)=\left(\begin{array}{cc}E_0&V_{12} \\ V_{21} & E_0\end{array}\right)$$

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

3. Nov 29, 2009

### noblegas

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

4. Nov 29, 2009

### jdwood983

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

$$V=\left(\begin{array}{cc}V_{11}&V_{ 12} \\ V_{21} & V_{22}\end{array}\right)=\left(\begin{array}{cc}0&V_{ 12} \\ V_{21} & 0\end{array}\right)$$

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

5. Nov 30, 2009

### noblegas

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

6. Nov 30, 2009

### jdwood983

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

$$H=\left(\begin{array}{cc}E_0&V_{ 12} \\ V_{21} & E_0\end{array}\right)$$

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?

7. Nov 30, 2009

### noblegas

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?

8. Nov 30, 2009

### jdwood983

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" [Broken]

Last edited by a moderator: May 4, 2017
9. Dec 1, 2009

### noblegas

$$H=\left(\begin{array}{cc}E_0 &0 \\ 0 & E_0\end{array}\right)+\left(\begin{array}{cc}0&V_{ 11} \\ V_{11} & 0\end{array}\right)=\left(\begin{array}{cc}E_0&V_{ 11} \\ V_{21} & E_0\end{array}\right)$$ would I write two other matrixes fo and $$H=\left(\begin{array}{cc}E_0 &0 \\ 0 & E_0\end{array}\right)+\left(\begin{array}{cc}0&V_{ 22} \\ V_{22} & 0\end{array}\right)=\left(\begin{array}{cc}E_0&V_{ 22} \\ V_{21} & E_0\end{array}\right)$$

10. Dec 1, 2009

### jdwood983

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

$$\det\left[\left(\begin{array}{cc}E_0&V_{ 12} \\ V_{21} & E_0\end{array}\right)-\left(\begin{array}{cc}\lambda&0 \\ 0 & \lambda\end{array}\right)\right]=0$$

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

11. Dec 1, 2009

### noblegas

I would have two values for lambda right? and my eigenvectors would be: $$H=\left(\begin{array}{cc}E_0 &0 \\ 0 & E_0\end{array}\right)+\left(\begin{array}{cc}0&V_{ 12} \\ V_{21} & 0\end{array}\right)=\left(\begin{array}{cc}E_0&V_{ 12} \\ V_{21} & E_0\end{array}\right)$$*[x,y]=lambda_1*[x,y], and $$H=\left(\begin{array}{cc}E_0 &0 \\ 0 & E_0\end{array}\right)+\left(\begin{array}{cc}0&V_{ 12} \\ V_{21} & 0\end{array}\right)=\left(\begin{array}{cc}E_0&V_{ 12} \\ V_{21} & E_0\end{array}\right)$$*[x,y]=lambda_2*[x,y] right

12. Dec 1, 2009

### jdwood983

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.