Position wave function of energy eigenstates in 1D box

AI Thread Summary
The discussion focuses on finding the position-space wave functions ψn(x) for a particle in a one-dimensional box of size L, where the wave function must vanish at the boundaries. The energy operator is defined, and the potential is zero within the box. Participants clarify that the task involves solving the Schrödinger equation for the system and applying boundary conditions to determine the wave functions. The correct form of the wave function includes sine functions, and normalization must be ensured. Ultimately, the goal is to express ψn(x) for discrete energy levels by solving for specific values of k, A, and B.
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Homework Statement


Consider a particle which is confined in a one-dimensional box of size L, so that the position space wave function ψ(x) has to vanish at x = 0 and x = L. The energy operator is H = p2/2m + V (x), where the potential is V (x) = 0 for 0 < x < L, and V (x) = ∞ otherwise.

Find the position-space wave functions ψn(x) ≡ <x|ψn> of the energy eigenstates |ψn>. Make sure that your wave functions are normalized, so that <ψnn> = 1

Homework Equations



maybe

<x'|x|ψ> = x'<x'|ψ>
1 =

The Attempt at a Solution



I'm not 100% sure what the question is asking for but I'm guessing it's asking me to find the RHS of

ψn(x) ≡ <x|ψn>

I can write the RHS as a sum since energy eigenstates are discrete

<x|ψn> = <x|1|ψn> = ∑ψni2

I'm not sure if this is the right step but it's all I could come up with.
 
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Are you familiar with the Schrodinger equation for such a system?
 
Chandra Prayaga said:
Are you familiar with the Schrodinger equation for such a system?

Yes, for a mass moving in 1D, the Schrodinger equation gives

n(x) + Vψn(x) = Eψn(x)

H = Hamiltonian

So if I solve the equation for a 1D box I would get something like

ψn(x) = Asin(kx)+Bsin(kx)

Do I then solve for k,A, and B?
 
I am unsure of what it means by position wave functions of energy eigenstates. Does it mean Solve for ψn(x) for the definite levels of energy levels?
 
Yes. You have the solution, now apply the boundary conditions and normalization. Only specific ##k## do the trick and they can be numbered from n=0 to infinity.

(I assumed you wrote A sin .. + B cos .. ?)
 
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