Eigenvectors of this Hamiltonian

In summary: Edit:In summary, the problem is asking for the eigenvalues and eigenstates of the Hamiltonian for a system consisting of two spin half particles in the magnetic field, but I think the problem is wrong and no eigenstate and eigenvalue exist.
  • #1
ShayanJ
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I've got a problem which is asking for the eigenvalues and eigenstates of the Hamiltonian [itex] H_0=-B_0(a_1 \sigma_z^{(1)}+a_2 \sigma_z^{(2)}) [/itex] for a system consisting of two spin half particles in the magnetic field [itex] \vec{B}=B_0 \hat z [/itex].
But I think the problem is wrong and no eigenstate and eigenvalue exist. Here's my reason:
[itex]
-B_0(a_1 \sigma_z^{(1)}+a_2 \sigma_z^{(2)}) \left(\begin{array}{c}x_1\\y_1 \end{array}\right)\left(\begin{array}{c}x_2\\y_2 \end{array}\right) =\lambda\left(\begin{array}{c}x_1\\y_1 \end{array}\right)\left(\begin{array}{c}x_2\\y_2\end{array}\right) \Rightarrow -B_0(a_1 \left(\begin{array}{c}x_1\\-y_1 \end{array}\right)\left(\begin{array}{c}x_2\\y_2 \end{array}\right)+a_2 \left(\begin{array}{c}x_1\\y_1 \end{array}\right)\left(\begin{array}{c}x_2\\-y_2 \end{array}\right)) =\lambda\left(\begin{array}{c}x_1\\y_1 \end{array}\right)\left(\begin{array}{c}x_2\\y_2\end{array}\right)
[/itex]
But as far as I know, that is impossible. Am I right?
I'm asking this because I'm affraid maybe my knowledge on tensor product states and operators contain some holes because I learned it by reading little things here and there.
 
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  • #2
The final step doesn't look right. I haven't looked at the details, but I would try representing the tensor product state using a 4X1 column vector as in http://www.matfys.lth.se/education/quantinfo/QIlect1.pdf (p10) or http://www.tau.ac.il/~quantum/Reznik/lecture notes in quantum information.pdf (p32). The operators should be 4X4 matrices, and ##\sigma_z^{(1)}## should be ##\sigma_z^{(1)} \otimes \mathbb{I}^{(2)}##.

Edit: rubi's and stevendaryl's posts are correct. The 4X1 vector should be have unknowns [a b c d]T.
 
Last edited:
  • #3
The problem with the calculation is that a general vector in the tensor product space isn't ##(\sum_{i=1}^2 a_i e_i)\otimes(\sum_{j=1}^2 b_j e_j)## but rather ##\sum_{i,j=1}^2 c_{ij} e_i\otimes e_j##.
 
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  • #4
The state [itex]\left( \begin{array}\\ x_1 \\ y_1\end{array} \right) \left( \begin{array}\\ x_2 \\ y_2\end{array} \right) [/itex] is not the most general state for two spin-1/2 particles. The most general state looks something like this:

[itex]\alpha_1 \left( \begin{array}\\ 1 \\ 0\end{array} \right) \left( \begin{array}\\ 1 \\ 0\end{array} \right) + \alpha_2 \left( \begin{array}\\ 1 \\ 0\end{array} \right) \left( \begin{array}\\ 0 \\ 1\end{array} \right) + \alpha_3 \left( \begin{array}\\ 0 \\ 1\end{array} \right) \left( \begin{array}\\ 1 \\ 0\end{array} \right) + \alpha_4 \left( \begin{array}\\ 0 \\ 1\end{array} \right) \left( \begin{array}\\ 0 \\ 1\end{array} \right) [/itex]

Now, operate on this general state with the Hamiltonian to find out what kind of constraints on the [itex]\alpha[/itex]s would make it an eigenstate.
 
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  • #5
Thanks people.
Now another question.
The next part of the problem asks for the first order correction for the perturbation [itex] k \, \vec\sigma^{(1)} \cdot \vec\sigma^{(2)} [/itex]. Is the following correct?
[itex]

\vec\sigma^{(1)} \cdot \vec\sigma^{(2)}=(\sigma_x^{(1)*} \,\,\, \sigma_y^{(1)*} \,\,\, \sigma_z^{(1)*}) \left( \begin{array}{c} \sigma_x^{(2)} \\ \sigma_y^{(2)} \\ \sigma_z^{(2)} \end{array} \right)=\sigma_x^{(1)*}\sigma_x^{(2)}+\sigma_y^{(1)*}\sigma_y^{(2)}+ \sigma _z^{(1)*}\sigma_z^{(2)}

[/itex]
 

1. What are eigenvectors in the context of Hamiltonian?

Eigenvectors of a Hamiltonian are special vectors that represent the possible states of a physical system. They are associated with a specific energy value, called an eigenvalue, and govern the evolution of the system over time.

2. How are eigenvectors of a Hamiltonian determined?

Eigenvectors of a Hamiltonian are determined by solving the eigenvalue equation, which involves finding the eigenvalues and corresponding eigenvectors that satisfy the equation H|ψ⟩=E|ψ⟩, where H is the Hamiltonian operator, |ψ⟩ is the eigenvector, and E is the eigenvalue.

3. What is the significance of eigenvectors in understanding quantum systems?

Eigenvectors play a crucial role in understanding quantum systems because they represent the possible states of a system, and their corresponding eigenvalues provide information about the energy levels of the system. This allows us to make predictions about the behavior of the system and make calculations for physical processes.

4. Can eigenvectors of a Hamiltonian change over time?

Eigenvectors of a Hamiltonian are stationary states, meaning they do not change over time. However, the coefficients of the eigenvectors may change, which can affect the overall state of the system. This is known as superposition, where a system can exist in a combination of different eigenvectors at the same time.

5. How do eigenvectors of a Hamiltonian relate to observable quantities?

Eigenvectors of a Hamiltonian are related to observable quantities through the process of measurement. When a system is measured, it will collapse into one of its eigenvectors, and the corresponding eigenvalue will be the measured value. This allows us to predict and understand the outcomes of measurements in quantum systems.

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