The discussion revolves around the concept of the exponential function in quantum mechanics, specifically in relation to stationary states and wave functions. Participants are exploring the implications of boundary conditions and the relationship between wave functions and energy eigenstates.
The conversation is active, with participants providing insights into the requirements for wave functions and discussing the application of Schrödinger's equation. Some guidance has been offered regarding the steps to take, including plugging the wave function into the equation and considering normalization conditions. There is an acknowledgment of varying levels of understanding among participants.
Participants note that certain concepts, such as the Hamiltonian, have not been covered in their coursework, which may affect their ability to engage fully with the problem. The discussion also references specific solutions learned in previous studies, indicating a gap in foundational knowledge relevant to the current question.
I am not sure what you mean by "no boundaries". The wave function must be zero at x=0 and it must go to zero sufficiently fast at x = infinity, and what is provided fulfills those two conditions.whitegirlandrew said:Homework Statement
The question is here.
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Homework Equations
The Attempt at a Solution
I don't even understand what it's asking because how can a stationary state be an exponential function with no boundaries? I would appreciate any insight, thank you.
We did not go over those concepts.nrqed said:I am not sure what you mean by "no boundaries". The wave function must be zero at x=0 and it must go to zero sufficiently fast at x = infinity, and what is provided fulfills those two conditions.
The first step is to determine if this is a state of definite energy (an eigenstate of the hamiltonian). It is, the eigenvalue will give the energy. Do you see how to check if it's an eigenstate?
Have you learned about Schrödinger's equation or the hamiltonian?whitegirlandrew said:We did not go over those concepts.
nrqed said:Have you learned about Schrödinger's equation or the hamiltonian?
Ok, good. Then all you have to do is to plug the wave function they give into Schrödinger's equation (using for V(x) the potential they give) and then isolate E. That will give you the energy of that state.whitegirlandrew said:We learned about Schrödinger's equation. It's a second year course so it doesn't deal with the pure mathematics of solving the equation but we went through on how it was 'derived', and various solutions such as the infinite and finite potential well.
Then what about the integral given below? What does that have to do with anything.nrqed said:Ok, good. Then all you have to do is to plug the wave function they give into Schrödinger's equation (using for V(x) the potential they give) and then isolate E. That will give you the energy of that state.
The integral will be needed to determine the constant "A", you will need to impose that the integral of ##|\psi|^2 dx ## from 0 to infinity gives 1 and that will fix the value of A. Note that in the first step, when you plug in the wave function in the equation, the A will cancel out everywhere so in that step you will not fix the value of A.whitegirlandrew said:Then what about the integral given below? What does that have to do with anything.
I Think i somewhat understand, I will give it a try, thank you for the help.nrqed said:The integral will be needed to determine the constant "A", you will need to impose that the integral of ##|\psi|^2 dx ## from 0 to infinity gives 1 and that will fix the value of A. Note that in the first step, when you plug in the wave function in the equation, the A will cancel out everywhere so in that step you will not fix the value of A.
I must correct one thing: when you plug in the wave function in Schrödinger's equation, you will also be able to determine the value of "b" that appears in the exponential, so at that stage you will be able to determine both E and b.
You are very welcome. If you get stuck, post here what you tried.whitegirlandrew said:I Think i somewhat understand, I will give it a try, thank you for the help.