How to get the energy eigenvalue of the Hamiltonian: H0+λp/m ?

[Jiangwei Du]

TL;DR Summary

We have already know the energy eigenvalue E0 of initial Hamiltonian H0. So when we add the extra item-λp/m, how the energy eigenvalue will vary?

Someone says we can choose the new eigenstate: exp(-iλx/hbar)*ψ,and let the momentum operator p acts upon this new state. At the same time, so does p^2. Something miraculous will happen afterwards. My question is: how to image this point? Thank you very much.

 

[azntoon]

The idea here is that when the momentum operator p is applied to an eigenstate, it will produce a state with the same energy (eigenvalue) as before. However, when the momentum operator squared, p^2, is applied to this same eigenstate, the result will be a state with a different energy. This is because the momentum operator squared contains additional terms corresponding to higher powers of momentum, which require higher energies to produce states with the same eigenvalue. This is an example of what is known as "quantum tunneling", where particles can pass through "barriers" of energy which would normally be too high to be overcome. In this case, the particle is able to "tunnel" through the barrier by utilizing the energy associated with its momentum.

 

[Jiangwei Du]

The idea here is that when the momentum operator p is applied to an eigenstate, it will produce a state with the same energy (eigenvalue) as before. However, when the momentum operator squared, p^2, is applied to this same eigenstate, the result will be a state with a different energy. This is because the momentum operator squared contains additional terms corresponding to higher powers of momentum, which require higher energies to produce states with the same eigenvalue. This is an example of what is known as "quantum tunneling", where particles can pass through "barriers" of energy which would normally be too high to be overcome. In this case, the particle is able to "tunnel" through the barrier by utilizing the energy associated with its momentum.

Sorry, I can't understand your statement. Maybe you have strayed from the point.

 

[PeterDonis]

Someone says

Where? Please give a reference.

 

[DrDu]

You can try to complete the square.

 

[Morbert]

TL;DR Summary: We have already know the energy eigenvalue E0 of initial Hamiltonian H0. So when we add the extra item-λp/m, how the energy eigenvalue will vary?

Someone says we can choose the new eigenstate: exp(-iλx/hbar)*ψ,and let the momentum operator p acts upon this new state. At the same time, so does p^2. Something miraculous will happen afterwards. My question is: how to image this point? Thank you very much.

You can establish a linear dispersion relation with a term like $v \mathbf{\sigma} \cdot \mathbf{p}$ and you can add it your p^2 term to get some generalised k.p approximation useful for some semiconductors/semimentals. Is this what is motivating your question?