Show that energy E always exists such that junction conditions have a solution

[Penny57]

Homework Statement

Here we explore the physics of “bound states” of definite energy localized due to a potential. Consider a potential which is V (x) = −V0 (V0 ≥ 0) for x ∈ [−W/2, W/2] and zero elsewhere. Physically, this potential corresponds to an attractive potential well of strength −V0 localized near x = 0. It can serve as a simple model of a 1d atom. We look for quantum states of definite energy E < 0 which are localized within the potential well. Assume the wavefunction is reflection symmetric about x = 0, meaning ψ(x) = ψ(−x).

Show that there is always an energy E such that the junction conditions
have a solution, i.e. there is always a bound state.

Relevant Equations

V (x) = −V0 (V0 ≥ 0) for x ∈ [−W/2, W/2] and zero elsewhere.
ψ(x) = ψ(−x)

I know this wavefunction should behave as a symmetric cosine function (possibly as Cos( (k∗x)/(hbar) ?). I also know for a bound state, the wavefunction must decay exponentially outside the well.
Additionally, r = (-β+ik)/(β−ik) .

However, aside from that, I do not know how to get this question started, and am having difficultly finding helpful information. I do not know why there should always be a bound state in the first place. I apologize if I haven't found a lot, but I've been trying to solve this question for the past hour and a half to no success! Any help is appreciated.

 

[kuruman]

You are on the right track with what you know. You need to argue/show that one state is guaranteed. That would be the ground state. What's so special about it? What does it not have that all the other states do have? Draw some wavefunctions.

 

[Penny57]

Could it be related to the fact that it has the lowest possible energy eigenvalue, so the system will naturally fall towards it? The ground state here, to my understanding, is a cosine-like wavefunction. I am assuming that means that, in order to connect to the exponential of the outside, there must an energy value that facilitates this transition because cosine ends at 0.
Due to the energy manipulating the shape of the ground state wavefunction, there must be at least one that creates a smooth transition. Would that be along the right track? Would you recommend any other resources?

 

[kuruman]

Would you recommend any other resources?

https://en.wikipedia.org/wiki/Finite_potential_well#Finding_wavefunctions_for_the_bound_state

 

[vela]

Homework Statement: Here we explore the physics of “bound states” of definite energy localized due to a potential. Consider a potential which is V (x) = −V0 (V0 ≥ 0) for x ∈ [−W/2, W/2] and zero elsewhere. Physically, this potential corresponds to an attractive potential well of strength −V0 localized near x = 0. It can serve as a simple model of a 1d atom. We look for quantum states of definite energy E < 0 which are localized within the potential well. Assume the wavefunction is reflection symmetric about x = 0, meaning ψ(x) = ψ(−x).

Show that there is always an energy E such that the junction conditions
have a solution, i.e. there is always a bound state.
Relevant Equations: V (x) = −V0 (V0 ≥ 0) for x ∈ [−W/2, W/2] and zero elsewhere.
ψ(x) = ψ(−x)

I know this wavefunction should behave as a symmetric cosine function (possibly as Cos( (k∗x)/(hbar) ?). I also know for a bound state, the wavefunction must decay exponentially outside the well.
Additionally, r = (-β+ik)/(β−ik) .

However, aside from that, I do not know how to get this question started, and am having difficultly finding helpful information. I do not know why there should always be a bound state in the first place. I apologize if I haven't found a lot, but I've been trying to solve this question for the past hour and a half to no success! Any help is appreciated.

How are $\beta$ and $k$ defined? What are the conditions that must be met at $x=W/2$?

 

[Rosenthal]

Solve the Schroedinger equation in the three regions. You get a term proportional to cos(k*x) inside the well, a term proportional to exp(-q*x) in the region x>W/2 and a term proportional to exp(q*x) in the region x<W/2. Apply symmetry and match values and first derivatives of wave functions at (either) boundary. You get a transcendental equation linking k and q. The other equation is the energy relationship: the energy in the well is less than the energy outside by an amount V0, giving k^2 - q^2=2*m*V0/hbar^2. Now you need to show that, no matter what the values of W and V0 are, there is always at least one solution of the two equations. For small values of V0, q is practically a linear function of k while for small values of W the transcendental equation is approximately q = (W*k^2)/2, a parabola. There is always one non-trivial intersection of these two curves.