Negative energy eigenvalues of Hamiltonian

In summary: I'm sorry, I can't remember the word he used. But basically, it's just a model of how the electron moves between the ions. So even though some of the eigenvalues are negative, that doesn't mean that there's actually negative energy in the system.
  • #1
Screwdriver
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0

Homework Statement



If I have a Hamiltonian matrix, [itex]\mathcal{H}[/itex], that only depends on a kinetic energy operator, do the energy eigenvalues have to be non-negative? I have an [itex]\mathcal{H}[/itex] like this, and some of its eigenvalues are negative, so I was wondering if they have any physical significance, or if I should just reject them and their associates eigenstates. Also, one of the eigenstates was just the zero vector, which I think can be ignored since it's a trivial solution.

Homework Equations



[itex]\mathcal{H} = \mathcal{K} + \mathcal{V} = \mathcal{K}[/itex]

The Attempt at a Solution



I don't see any way that the system could have negative energy if the potential is zero everywhere, but I also don't feel entirely comfortable just ignoring eigenstates like that.
 
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  • #2
FYI, I have moved your thread to "Advanced Physics".

Your suspicions seem right to me. The energy eigenvalues are the allowed energies for the system, which really should be positive or zero if there can only be kinetic energy.

Can you post your actual Hamiltonian and describe the system?
 
  • #3
Redbelly98 said:
FYI, I have moved your thread to "Advanced Physics".

Your suspicions seem right to me. The energy eigenvalues are the allowed energies for the system, which really should be positive or zero if there can only be kinetic energy.

Can you post your actual Hamiltonian and describe the system?

Thanks for the reply, I wasn't completely sure if this belonged in the advanced section or not. The system is an ##X_{6}## molecule with a single electron able to move between the different ##X## ions. The basis ket ##|e_{j}\rangle## represents the electron occupying the ##j^{\text{th}}## ion with ##|e_{j}\rangle = |e_{j\pm6}\rangle##. We're to take the kinetic energy operator to be:

$$
\mathcal{K}=-\sum_{i=1}^{6}\left( |e_{j}\rangle\langle e_{j+1}| +|e_{j+1}\rangle\langle e_{j}| \right)
$$

And the potential operator to be zero. The Hamiltonian is then (assuming I did this right):

$$
\mathcal{H} = \mathcal{K}=
\begin{pmatrix}
0& -1 & 0 & 0 & 0 & -1\\
-1& 0& -1& 0& 0& 0\\
0& -1& 0& -1& 0& 0\\
0& 0& -1& 0& -1& 0\\
0& 0& 0& -1& 0& -1\\
-1& 0& 0& 0& -1&0
\end{pmatrix}
$$

With this eigensystem. As you can see, some of the eigenvalues are negative!
 
  • #4
Hmm. I agree that that Hamiltonian has negative eigenvalues, but am puzzled as to why the kinetic energy operator would have the form that you are told to use. I'm used to seeing KE in the form p2/(2m).

I'm going to see if any of the Homework Helpers have an idea here.
 
  • #5
The Hamiltonian looks like it is basically a tight-binding model for one election. You might look up that model in some textbooks to see what they get (here is the relevant section of the wikipedia article in second-quantized notation). I suspect that calling this term the "kinetic energy" is somewhat of a convention and so perhaps one can encounter negative energy eigenvalues. If that's the case then perhaps you can just add a constant energy shift to make the ground state energy zero.
 
  • #6
Mute said:
The Hamiltonian looks like it is basically a tight-binding model for one election. You might look up that model in some textbooks to see what they get (here is the relevant section of the wikipedia article in second-quantized notation). I suspect that calling this term the "kinetic energy" is somewhat of a convention and so perhaps one can encounter negative energy eigenvalues. If that's the case then perhaps you can just add a constant energy shift to make the ground state energy zero.

Hmm yes, that does seem to be of the same form. Honestly though, the material in the article seems way too advanced for the level that we're at (ie. the "do this because it works" stage) so I don't know if I'm even supposed to be worrying about this. You're probably right in that calling it the "kinetic energy" operator is just more of a jargon thing. The only reason I even considered this an issue is because I couldn't solve for the state vector for the given initial condition unless I rejected the zero vector solution and instead chose a different eigenvector that also had eigenvalue 1 (which I had to find manually.) It sort of begs the question as to why Mathematica would report the zero vector as a solution unless it has no other choice, but there's probably some sort of linear algebra argument behind that. So then I started to wonder whether or not I should also reject the negative energies, but I couldn't solve for the initial state that way.
 
  • #7
Redbelly98 said:
I'm going to see if any of the Homework Helpers have an idea here.
One of the Homework Helpers told me -- and I hope I'm not oversimplifying what he said -- that this kinetic energy Hamiltonian is pretty much the change or difference in KE relative to some reference level for an electron that is bound to an atom and does not jump to another atom.

For more details, you can look at Feynman's lectures, in the first section of chapter 13 in Vol III which deals with an electron living on a 1D lattice. See especially Equation (13.3).

Hope that helps.

p.s.: I am about 17 years removed from grad school, and really do not deal with material like this nowadays.
 
  • #8
Redbelly98 said:
One of the Homework Helpers told me -- and I hope I'm not oversimplifying what he said -- that this kinetic energy Hamiltonian is pretty much the change or difference in KE relative to some reference level for an electron that is bound to an atom and does not jump to another atom.

For more details, you can look at Feynman's lectures, in the first section of chapter 13 in Vol III which deals with an electron living on a 1D lattice. See especially Equation (13.3).

Hope that helps.

p.s.: I am about 17 years removed from grad school, and really do not deal with material like this nowadays.

Ah, that does make a lot more sense. Thanks again everyone!
 

1. What are negative energy eigenvalues of Hamiltonian?

Negative energy eigenvalues of Hamiltonian refer to the energy levels of a quantum system that have a negative value. These eigenvalues are obtained from the solutions of the Schrödinger equation, which describes the behavior of particles on a quantum level. Negative energy eigenvalues indicate that the system has a lower energy state than the ground state, and they play an important role in understanding quantum phenomena.

2. How do negative energy eigenvalues affect the behavior of a quantum system?

Negative energy eigenvalues can cause a quantum system to behave in unexpected ways. For example, they can lead to the phenomenon of quantum tunneling, where a particle can pass through a potential barrier without having enough energy to do so classically. They also play a crucial role in particle physics, as they allow for the existence of antimatter and help explain the stability of matter.

3. Can negative energy eigenvalues violate the principle of conservation of energy?

No, the existence of negative energy eigenvalues does not violate the principle of conservation of energy. These eigenvalues only represent the possible energy states of a system and do not have a physical existence. The total energy of a system, including any negative energy eigenvalues, must still remain constant.

4. How are negative energy eigenvalues related to the concept of antimatter?

Negative energy eigenvalues are closely linked to the concept of antimatter. According to quantum field theory, particles and antiparticles are created in pairs from the vacuum. The antiparticle has the opposite charge and spin of the particle, but also has a negative energy eigenvalue. This allows for the creation of particles with negative energy, which can then be converted into positive energy through interactions with other particles.

5. Are negative energy eigenvalues of Hamiltonian always present in quantum systems?

Yes, negative energy eigenvalues are a fundamental aspect of quantum systems and are always present. However, their significance may vary depending on the specific system and the type of interactions involved. In some cases, they may be negligible, while in others they can have a significant impact on the behavior of the system.

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