Possible energy values given Hamiltonian

  • #1
Rayan
16
1
Homework Statement
The Hamiltonian for a two level system with the orthonormal states |1⟩ and |2⟩ is given by:
Relevant Equations
H = a|1⟩⟨1| + b|1⟩⟨2| + b|2⟩⟨1| + c|2⟩⟨2| ,

where a,b and c are real constants with energy unit.
So first I rewrote H as a matrix:

$$ H =
\begin{pmatrix}
a & b \\
b & c
\end{pmatrix} $$

And tried to find the eigenvalues/energies of H, so I solved

$$ det (H - \lambda I ) =
\begin{vmatrix}
a-\lambda & b \\
b & c-\lambda
\end{vmatrix} = (a-\lambda)(c-\lambda) - b^2 = ac - a\lambda - c\lambda + \lambda^2 - b^2 = 0
$$

but got a complicated solution

$$ \lambda = \frac{a+c}{2} \pm \sqrt{ ( \frac{a+c}{2} )^2 - (ac-b^2) } $$

What am I doing wrong here?
 
Last edited:
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  • #2
Rayan said:
So first I rewrote H as a matrix:

$$ H =
\begin{pmatrix}
a & b \\
b & c
\end{pmatrix} $$
Ok.

Rayan said:
And tried to find the eigenvalues/energies of

$$ H - \lambda I $$
Terminology: You want the eigenvalues of ##H##, not ##H - \lambda I##.

Rayan said:
but got a complicated solution
$$ \lambda = \frac{a-c}{2} \pm \sqrt{ ( \frac{a-c}{2} )^2 - (ac-b^2) } $$
What am I doing wrong here?
It's hard to tell where you made the mistakes. Please show the steps in getting the quadratic equation for ##\lambda## and then the steps in solving for ##\lambda##.
 
  • #3
TSny said:
Ok.Terminology: You want the eigenvalues of ##H##, not ##H - \lambda I##.It's hard to tell where you made the mistakes. Please show the steps in getting the quadratic equation for ##\lambda## and then the steps in solving for ##\lambda##.
You're right! I just updated my question with the steps!:)
 
  • #4
Rayan said:
$$ det (H - \lambda I ) =
\begin{vmatrix}
a-\lambda & b \\
b & c-\lambda
\end{vmatrix} = (a-\lambda)(c-\lambda) - b^2 = ac - a\lambda - c\lambda + \lambda^2 - b^2 = 0
$$
This looks good.

Rayan said:
but got a complicated solution

$$ \lambda = \frac{a+c}{2} \pm \sqrt{ ( \frac{a+c}{2} )^2 - (ac-b^2) } $$
This looks correct.

My preference would be to write what you have as $$ \lambda = \frac{a+c}{2} \pm \frac1 2 \sqrt{ ((a+c)^2 - 4(ac-b^2) } $$
You should be able to simplify the expression inside the square root a little. (Work with the terms involving ##a## and ##c##.)
 

1. What is a Hamiltonian?

A Hamiltonian is a mathematical operator in quantum mechanics that represents the total energy of a system. It is used to calculate the possible energy values of a system.

2. How are possible energy values determined given a Hamiltonian?

Possible energy values are determined by solving the Schrödinger equation for the system, which involves applying the Hamiltonian operator to the wave function of the system. The solutions to the Schrödinger equation represent the possible energy values of the system.

3. Can a Hamiltonian have multiple possible energy values?

Yes, a Hamiltonian can have multiple possible energy values. The number of possible energy values depends on the complexity of the system and the specific form of the Hamiltonian operator.

4. Are all possible energy values equally likely?

No, not all possible energy values are equally likely. The probability of a system having a particular energy value is determined by the wave function of the system, which is influenced by the potential energy of the system as well as other factors.

5. How are possible energy values experimentally determined?

Possible energy values can be experimentally determined through techniques such as spectroscopy, which involves measuring the absorption or emission of light by a system. The energy levels observed in spectroscopy correspond to the possible energy values of the system.

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