QM - matrix elements of operators

In summary: I don't understand why it is implied that <L|H|L> = E0 in the first place.2) in a similar way - from the fact that the electron can "jump" from |L> to |C> (and vice versa) and that this jump is characterized by kinetic energy a ---> we deduce that <C|H|L> = <L|H|C> = a.Again, I don't understand why this is a valid argument. From what you've said, it seems like the electron can only "jump" between nearest neighbors, and so <C|H|L> would only be equal to <L|H|C> if
  • #1
maria clara
58
0

Homework Statement




we shall describe a simple model for a linear molecule, say, CO2.
the states |L>, |C>,|R> are the eigenstates of D operator (corresponds to dipole moment)
D|L>=-d|L> , D|C>=0 , D|R>= +d|R>.
When the electron is localized exactly on the carbon atom, its energy is E1 and
when it is localized on one of the oxygen atoms, it has energy E0 ( assuming
E1 > E0). In addition, the electron can "jump" from one atom to another, and
this jump is characterized by kinetic energy a. We can assume that the jumping
occurs only between nearest neighbors.

Write matrix representation of the Hamiltonian in the basis of |L>, |C>,|R>.
2. The attempt at a solution

we should build the matrix from the matrix elements - <L|H|L>, <L|H|C> etc.

the matrix should look like that:
E0 a 0
a E1 a
0 a E0

it seems very logical, but I still don't see the direct connection between the matrix elements and the given data. for example, why does the element <L|H|L> equals E0?
what does the element <1|H|2> mean in general?
 
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  • #2
What does H|L> equal?
 
  • #3
well, the problem is that |L> is not an eigenstate of H.
So all I know is that |L> can be represented as a superposition of the Hamiltonian's eigenfunctions...
What I understand from the data is that when I measure the energy in the state |L> I get the result E0... so using H will change the state to one of the eigensates of H and yield the result E0...(?)
 
  • #4
maria clara said:
well, the problem is that |L> is not an eigenstate of H.
So all I know is that |L> can be represented as a superposition of the Hamiltonian's eigenfunctions...
What I understand from the data is that when I measure the energy in the state |L> I get the result E0... so using H will change the state to one of the eigensates of H and yield the result E0...(?)

If the matrix you gave is the answer, then you can write down

[tex]H\left|L\right> = E_0 \left|L\right> + a \left|C\right>.[/tex]

Do you see why?

I also don't find this to be very intuitive. For example, I get the energy eigenvalues of [itex]H[/itex] to be

[tex]\frac{1}{2}\left( E_1 + E_0 + \sqrt{\left( E_1 - E_0\right)^2 + 8a^2}\right)[/tex]

[tex]E_0[/tex]

[tex]\frac{1}{2}\left( E_1 + E_0 - \sqrt{\left( E_1 - E_0\right)^2 + 8a^2}\right)[/tex]

with associated eigenvectors

[tex]\left| L \right> + \frac{1}{2a}\left( E_1 - E_0 + \sqrt{\left( E_1 - E_0\right)^2 + 8a^2}\right) \left| C \right> + \left| R \right>[/tex]

[tex]\left|L\right> - \left|R\right>[/tex]

[tex]\left| L \right> + \frac{1}{2a}\left( E_1 - E_0 - \sqrt{\left( E_1 - E_0\right)^2 + 8a^2}\right) \left| C \right> + \left| R \right>[/tex]
 
  • #5
first of all - thanks for your reply(:

secondly - yes, from the matrix I can see very clearly how H acts on any of the states. All I have to do is to multiply it (from the right) with the appropriate column vector; for instance, (1,0,0)T (T denotes "transpose") will represent |L>.
I also understand how to find the eigenvalues and the eigenstates of H - directly from the matrix.

But these problems are to be solved after you find the matrix. And this is the part that is unclear to me- how to create the matrix according to the given data. And to clarify my point: the problem is not technical. I know how to solve the exercise.

My problem is that I don't understand why the following arguments are correct:

1) from the fact that the energy of the system in the state |L> is E0 ---> we can conclude directly that <L|H|L> = E0.

2) in a similar way - from the fact that the electron can "jump" from |L> to |C> (and vice versa) and that this jump is characterized by kinetic energy a ---> we deduce that
<C|H|L> = <L|H|C> = a.

and so on...
 
  • #6
maria clara said:
first of all - thanks for your reply(:

secondly - yes, from the matrix I can see very clearly how H acts on any of the states. All I have to do is to multiply it (from the right) with the appropriate column vector; for instance, (1,0,0)T (T denotes "transpose") will represent |L>.
I also understand how to find the eigenvalues and the eigenstates of H - directly from the matrix.

But these problems are to be solved after you find the matrix. And this is the part that is unclear to me- how to create the matrix according to the given data. And to clarify my point: the problem is not technical. I know how to solve the exercise.

My problem is that I don't understand why the following arguments are correct:

1) from the fact that the energy of the system in the state |L> is E0 ---> we can conclude directly that <L|H|L> = E0.

The only way we can know with certainty that the energy of the system in state |L> is E0 is if |L> is an eigenstate of H with corresponding eigenvalue E0. This follows directly from the probabilistic rules of quantum theory. Then, <L|H|L> = E0 follows easily.

For the given the matrix representation of H, this argument, however, is incorrect.

I used Maple to solve for the energy eigenvalues and eigenstates, and the solution shows that |L> is a complicated linear combination of three energy eigenstates. So, if the system is in state |L>, then the given matrix implies that there are three possible energies of the system that each have a non-zero probability.

This clashes with the statement of the question, which makes me wonder whether the question is even consistent. Did the question come from a book? If so, which one? From the prof?

2) in a similar way - from the fact that the electron can "jump" from |L> to |C> (and vice versa) and that this jump is characterized by kinetic energy a ---> we deduce that
<C|H|L> = <L|H|C> = a.

I have never seen this interpretation.
 
  • #7
Hi again,

this is exactly what I thought - that |L> should be an eigenstate of H, but it is not..!

the question comes from our tutorial, and we also had a homework assignment with a similar exercise. I believe the question was invented by our prof. or tutor.
It does seem to be illogical and inconsistent.:grumpy:

thanks a lot, anyway:smile:
 

What are matrix elements of operators in quantum mechanics?

In quantum mechanics, operators are mathematical entities that represent physical observables, such as position, momentum, or energy. The matrix elements of operators are the elements of a matrix that represent the action of an operator on a particular state in a quantum system.

How do you calculate matrix elements of operators?

The calculation of matrix elements of operators involves finding the inner products of two states, one acted on by the operator and the other being the state on which the operator is applied. This can be done using the eigenfunctions of the operator or by using the matrix representation of the operator.

What is the significance of matrix elements of operators in quantum mechanics?

Matrix elements of operators play a crucial role in quantum mechanics as they provide information about the behavior and properties of a quantum system. They are used to calculate the probabilities of different outcomes of a measurement and to determine the time evolution of a system.

Can matrix elements of operators be complex numbers?

Yes, matrix elements of operators can be complex numbers. In quantum mechanics, operators are represented by complex matrices, and their matrix elements can also be complex numbers. These complex numbers contain information about the phase and amplitude of the wave function of a quantum system.

How are matrix elements of operators related to uncertainty principle?

The matrix elements of operators are related to the uncertainty principle in quantum mechanics. This principle states that certain pairs of physical observables, such as position and momentum, cannot be simultaneously measured with arbitrary precision. The matrix elements of these operators can be used to calculate the uncertainty in these measurements and provide insights into the fundamental nature of quantum systems.

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