# Finding eigenvalue and normalized eigenstate of a hamiltonian

• noblegas
So 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
noblegas

## 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.

## 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.

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.

jdwood983 said:
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.

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:

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

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

jdwood983 said:
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 $H_0$.

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:

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

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?

jdwood983 said:
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?

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"

Last edited by a moderator:
jdwood983 said:
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.

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

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

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.

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

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

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

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

noblegas said:
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

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.

## 1. What is an eigenvalue and why is it important in Hamiltonian systems?

An eigenvalue is a characteristic quantity associated with a particular state in a Hamiltonian system. It represents the energy of that state and is important because it allows us to understand the behavior and properties of the system.

## 2. How do you find the eigenvalues of a Hamiltonian?

The eigenvalues of a Hamiltonian can be determined by solving the Schrödinger equation, which is the fundamental equation of quantum mechanics. This involves using mathematical techniques such as diagonalization or perturbation theory.

## 3. What is a normalized eigenstate and why is it significant?

A normalized eigenstate is a state that has been scaled to have a unit length, meaning its probability amplitude is equal to one. This is significant because it allows us to interpret the probability of finding a particle in that state, making it a fundamental concept in quantum mechanics.

## 4. Can the eigenvalues and eigenstates of a Hamiltonian change over time?

No, the eigenvalues and eigenstates of a Hamiltonian are constant and do not change over time. They represent the fundamental properties of a system and are not affected by external factors or the evolution of the system.

## 5. How are eigenvalues and eigenstates related to the energy levels of a system?

The eigenvalues of a Hamiltonian correspond to the energy levels of a system. Each eigenstate has a unique eigenvalue, which represents the energy of that state. The energy levels of a system can be visualized as a ladder, with each rung corresponding to an eigenvalue and the associated eigenstate being the state at that energy level.

Replies
1
Views
1K
Replies
3
Views
1K
Replies
1
Views
1K
Replies
14
Views
1K
Replies
13
Views
2K
Replies
1
Views
566
Replies
1
Views
1K
Replies
0
Views
954
Replies
2
Views
714
Replies
1
Views
1K