Can Asymmetric Wave Functions Arise from Symmetric Potentials?

[PhysicsKid0123]

Can the general solution to the Schrodinger equation be asymmetric (has neither even or odd solutions)?

Question (1): I saw somewhere that you cannot have a solution that is both-- it must be either odd or even, and I was wondering: why?

I was working on a problem where the potential function is a delta function of the form

Eq. (1) V(x) = C[d(x-L) + d(x+L)]

and V(x) = 0 otherwise

where "C" denotes a constant and "d" is the Dirac delta function, d(x-L) = infinity at x= L and d(x-L) = 0 when x is not equal to L and similarly for d(x+L). I'll not that the potential function is "even" or symmetric about x = 0. The most general solution is of the form Psi(x) = A sin(k*x) + B cos (k*x) for -L < x < L and | x | > | L |, where k^2 = 2mE/(h_bar)^2. However, the way I approached this problem was to assign different coefficients to the general solution according to which region it was in.

For example, for the region x < -L, the coefficients were as above:
Eq. (2) Psi(x) = A sin(k*x) + B cos (k*x)

For the region -L < x < L, the coefficients were C for sin and D for the cos, or
Eq. (3) Psi(x) = C sin(k*x) + D sin(k*x).

For x > L, the coefficients were E for sin and F for cos or
Eq. (4) Psi(x) = E sin(k*x) + F sin(k*x).

Now when I solved for the boundary conditions, Psi(-L)=0=Psi(+L), using the Psi in the middle region, I got different coefficients and thus two solutions of Psi(x) for the region inside. Namely,

Psi(x= -L) = 0 ---> D = C tan(k*L)
and
Psi(x= +L) = 0 ---> D = -C tan(k*L)

Substituting these into Psi or Eq. (3) to put it in terms of one coefficient implies that we have two Psi(x)'s:

Eq. (5) Psi(x) = C sin(k*x)+ C tan(k*L)*cos(k*x)

and

Eq. (6) Psi(x) = C sin(k*x) - C tan(k*L)*cos(k*x)

Now this is where I get confused.
Question (2): What does it mean two have two Psi's within one region like this?

Question (3): Should I solve the other side of the boundary conditions for each of these equation, namely Eqs (5) and (6)? For example, at the left Psi(x= -L) = Psi(x= -L) on the right or equivalently Eq (2) = Eq (5), and for the other one, Eq (2) = Eq (6). Repeat this for both when Psi(x=+L)=Psi(x=+L)?

Question (4): When I plug in Eq. (5) into the Schrodinger equation, it is NOT a solution UNLESS: x = -L, k = 1, or equivalently E = (1/2m)*(hbar)^2.

For Eq. (6), I get (1+k) = 0 = (1-k) or k = - k for it to satisfy the Schrodinger equation. This implies k = 0 which implies E = 0. What does all this mean?


What I did next to try and find some insights was to set Eq. (5) and Eq. (6) equal to each other. Which seems quite reasonable to do because there should only be one general solution or one only "general" Psi since it is uniquely determined by the boundary conditions, right? Well when I did this, Eq(5)= Psi(x)= Psi(x)= Eq(6), I found that they are equivalent only when C = -C which implies that C = 0, OR when k*x = (pi/2) + n*pi for n = 0,1,2,...

Question (5): IF C=0, then that implies that no wave function exists in the region -L<x<L. What does this mean then? I took this to be a trivial solution, but if it is, then what's left? What is happening in this middle region and why can't I find a solution after using the boundary conditions to determine the coefficients? Before the coefficients are found, the general solution is a solution, but after it is not?

This problem really intrigues me and any insight would help. I'm sure I may be missing some kind of understanding. Again, it isn't entirely clear why must we have either even or odd solutions.

 

[Orodruin]

Question (1): I saw somewhere that you cannot have a solution that is both-- it must be either odd or even, and I was wondering: why?

This is only true if the potential is symmetric. Then you can show that the eigenfunctions of the Hamiltonian are eigenstates of the parity operator with eigenvalues ±1.

Now when I solved for the boundary conditions, Psi(-L)=0=Psi(+L)

These are not the correct boundary conditions. In fact, your problem does not have boundaries. What you need are matching conditions at x=L and x = - L.
You may also want to rethink if there is really a k^2 in your trigonometric functions.

 

[PhysicsKid0123]

This is only true if the potential is symmetric. Then you can show that the eigenfunctions of the Hamiltonian are eigenstates of the parity operator with eigenvalues ±1.These are not the correct boundary conditions. In fact, your problem does not have boundaries. What you need are matching conditions at x=L and x = - L.
You may also want to rethink if there is really a k^2 in your trigonometric functions.

Yes, that's right. Not boundary conditions, but continuity conditions. I've also fixed the the k^2 typo. I suspect there may be more typos and I might have to rework the problem to make sure I've stated everything correctly. So I didn't understand the parity operator part of your response, but just to be clear, you can only have either even functions or odd functions, but not both for a symmetric potential, correct?

 

[Orodruin]

No, you can have both. It is just that all energy eigenstates will be either odd or even.