Construct an ONB for a triatomic molecule

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In summary: The null space of$$\begin{pmatrix} -2V_0 & V_0 & V_0\\V_0 & -2V_0 & V_0\\V_0 & V_0 & -2V_0\end{pmatrix}$$is spanned by$$\begin{pmatrix} 1\\1\\1\end{pmatrix}$$So that's one of your eigenvectors, and you can get the others from the other two equations.
  • #1
rogeralms
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Consider a triatomic molecule with three identical atoms that are bound together with each atom at its own corner of an equilateral triangle of edge length a. An added electron on the molecule can be put in an identical atomic orbital on any of the three atoms. Denote the atomic states in which the electron is on atom i as |i>, and assume that these three states {|1>, |2>, and |3>}, form an orthonormal set. Let the mean energy associated with such a state be the same value ε0=<i|H|i> for each state |i>. Suppose also that the electron on an atom can move to either of its neighbors, with an amplitude <i|H|j>=V0 i≠j

a) Construct a 3X3 matrix [H] that represents the Hamiltonian within the subspace spanned by these 3 atomic states, using the states {\|i>} as an ONB for the subspace. Find the trace of this matrix.

[H] =
$$\begin{pmatrix}
ε_0 & V_0 & V_0\\
V_0 & ε_0 & V_0\\
V_0 & V_0 & ε_0
\end{pmatrix}$$

Tr[H] = 3ε0

b) Find the energy eigenvalues and the degeneracies of the molecule with an added electron.

det(H-ε)=
| ε0-ε V0 V0 |
| V0 ε0-ε V0 |
| V0 V0 ε0-ε |

= (ε0-ε) [(ε0-ε)(ε0-ε) - V0]
- V0 [ V00-ε) - V02 ]
+ V0 [ V02 - V00-ε) ]

= (ε0-ε)3 - 3V020-ε) +2V03=0

ε10-V0
ε20-V0
ε30-2V0

c) Construct an ONB of eigenstates {|εn> |n=1,2,3} of the system, as linear combinations of the atomic states.

I could not get Latex to work according to the directions on this site. I apologize for the crude look to the matrices, etc.

I am stuck on part c as follows. Could someone please help me with this part? I get the following which make no sense to me:

(H-ε1) |χ1>=(H-ε0+V0) |χ1>=

V0 V0 V0
V0 V0 V01>
V0 V0 V0

(H-ε3) |χ1>=(H-ε0+2V0) |χ1>=

2V0 V0 V0
V0 2V0 V01>
V0 V0 2V0
 

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  • #2
Your post had broken (unpaired) [sub]-tags, that made it hard to read. I fixed those.

What did you do to get those equations for (c)?
 
  • #3
mfb said:
Your post had broken (unpaired) [sub]-tags, that made it hard to read. I fixed those.

What did you do to get those equations for (c)?

I took the solutions for the eigenvalues from part b (two of which are degenerate) and subtracted them from the Hamiltonian matrix.

These are supposed to be the eigenvalue equations where the chi's should be the eigenvectors.
 
  • #4
Okay, so ##(H-Iε_1) ~|\chi_1\!>~ = 0## right?
Something that won't have a unique solution as your eigenspace has dimension 2.

I'm a bit surprised by the signs in your eigenvalues, shouldn't their sum be equal to the trace of the Hamiltonian?
 
  • #5
You are correct. I wrote down the third eigenvalue incorrectly. It is correct on the attachment but the matrix is incorrect.

The third eigenvalue should be ε3=ε0+2V0

The third matrix should be

-2V0 V0 V0
V0 -2V0 V0 |χ1> = 0
V0 V0 -2V0
 
  • #6
I can see a non-trivial solution for this equation.
 

Related to Construct an ONB for a triatomic molecule

1. What is an ONB?

An ONB, or orthonormal basis, is a set of mutually perpendicular unit vectors that span a vector space. In the context of a triatomic molecule, an ONB can be used to represent the orientation of the molecule's atoms.

2. Why is it important to construct an ONB for a triatomic molecule?

Constructing an ONB for a triatomic molecule allows us to accurately describe the molecule's orientation and understand its behavior. This is crucial for studying its physical and chemical properties and predicting its reaction with other molecules.

3. How do you construct an ONB for a triatomic molecule?

To construct an ONB for a triatomic molecule, we first determine the position of the molecule's atoms in 3D space. Then, we select one atom as the origin and calculate the unit vectors that point from the origin to each of the other two atoms. Finally, we use the Gram-Schmidt process to orthogonalize and normalize these vectors, resulting in an orthonormal basis.

4. Can an ONB be constructed for any triatomic molecule?

Yes, an ONB can be constructed for any triatomic molecule as long as the positions of its atoms in 3D space are known. However, the process may be more complex for molecules with a non-linear shape or with atoms of different masses.

5. How is an ONB used in the study of triatomic molecules?

An ONB is used to calculate various molecular properties, such as bond angles and dipole moments, which are important for understanding a molecule's behavior. It can also be used in simulations and calculations to predict the behavior of triatomic molecules in different environments.

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