Eigenvalues and eigenvectors of a Hamiltonian

Click For Summary
The discussion focuses on finding the eigenvalues and eigenvectors of a Hamiltonian for a two-level system, represented by a specific matrix form. The initial matrix representation of the Hamiltonian is incorrectly calculated, leading to confusion about the eigenvalues. After realizing a miscalculation in the determinant, the correct eigenvalues are identified as ±√2 ε. The participants emphasize the importance of careful calculation and notation in deriving the correct results. Accurate understanding of the Hamiltonian's matrix representation is crucial for solving similar problems in quantum mechanics.
astrocytosis
Messages
51
Reaction score
2

Homework Statement



The Hamiltonian of a certain two-level system is:

$$\hat H = \epsilon (|1 \rangle \langle 1 | - |2 \rangle \langle 2 | + |1 \rangle \langle 2 | + |2 \rangle \langle 1 |)$$

Where ##|1 \rangle, |2 \rangle## is an orthonormal basis and ##\epsilon## is a number with units of energy. Find its eigenvalues and eigenvectors (as linear combinations of ##|1 \rangle, |2 \rangle##). What is the matrix H representing ##\hat H## with respect to this basis?

Homework Equations



N/A?

The Attempt at a Solution


[/B]
This problem seems like it should be simple but I think I'm having trouble internalizing the notation. I know each bracket pair represents the 11, 12, 21, 22 components of the matrix, so I thought H should be\begin{bmatrix}
\epsilon & \epsilon \\
\epsilon & -\epsilon
\end{bmatrix}

then I tried to find the eigenvalues the usual way by subtracting ##\lambda I## and taking the determinant of\begin{bmatrix}
\epsilon - \lambda & \epsilon \\
\epsilon & -\epsilon- \lambda
\end{bmatrix}
but I ended up with an expression that implies ##\lambda = 0##

$$-\epsilon^2 + \lambda^2 + \epsilon^2 = 0$$

so I must be misunderstanding something.

EDIT: I miscalculated the determinant, so my answer is actually plausible, but would still appreciate confirmation that this is correct

$$\lambda = \pm \frac{1}{\sqrt{2}} \epsilon$$
 
Last edited:
Physics news on Phys.org
Are you sure you wrote that last equation as you intended?
 
I lost a negative sign in the determinant. That will teach me not to skip steps :P Also, it's just ##\sqrt{2} \epsilon##
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 13 ·
Replies
13
Views
3K
Replies
4
Views
2K
  • · Replies 8 ·
Replies
8
Views
1K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
2
Views
4K
  • · Replies 33 ·
2
Replies
33
Views
2K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K