What is the general solution of the one-dimensional time-independent Schrodinger's equation with constant potential?

Question

As I tried to learn quantum mechanics I have found two solutions of one-dimensional time-independent Schrödinger equation with constant potential in various resources.

One is,$$\psi(x) = A\sin(kx)+B\cos(kx)\\\text{where}, k = \frac{2π\sqrt{2m(E-U_0)}}{h}\\ \text{In this case, probability density: }P_1(x) = \psi(x)\cdot\psi^*(x) = |\psi(x)|^2\\$$ And another is, $$\psi(x) = Ce^{k'x} + De^{-k'x}\\\text{where}, k' = \frac{2π\sqrt{2m(U_0-E)}}{h}\\ \text{In this case, probability density: } P_2(x)=\psi(x)\cdot\psi^*(x)=|\psi(x)|^2$$ In both cases,

$A,B,C,D$ are arbitrary constants

$\psi(x) = $ The probability function of a particle

$\psi^*(x) = $ The conjugative function of $\psi(x)$

$m$ = the mass of the particle

$E$ = The total Energy of the particle

$U_0 =$ The potential energy of the particle

Now my question is, does $P_1(x)$ and $P_2(x)$ mean the same thing? If yes, then how? And if not, then which is actually the general solution of time independent Schrödinger's equation?

$$\\$$ [Edit: As far as I understood these two solutions are not same. (ie, if I put $C = 2$ and $D=3$ in the second solution, no value of $A\;and\;B$ can equalize these two solutions. Also if I put $A=2\;and\;B=3$ in the first solution no value of $C$ and $D$ can equalize them) So neither of them seems to be a general solution of TISE. Is there any mistake in this example? Or is there no general solution of TISE?]

Answer 1

Since $e^{ix} = \cos{x} + i \sin{x}$ (which is known as Euler's formula), both of your solutions are indeed the same, despite being written in different notations.

Notice, however, that what is meant by "Schrödinger equation" usually involves specifying a particular potential $V(x)$, and different choices of potential will, quite naturally, lead to different solutions. In this sense, your solutions are not general (but they do correspond to the same choice of potential, which I'm guessing to be $V(x) = U_0$).

Answer 2

Both solutions are the same. And not just the Probability density, but the entire wavefunction of a Quantum system is unique.

I think, you are missing the point that the coefficients can be imaginary. I will answer the question you asked in 'Edit part'.

if I put C=2 and D=3 in the second solution, no value of A and B can equalize these two solutions.

$A = 5$ and $B = -i$

Like @Gandalf73 also answered, after applying the Boundary Condition, both $\psi_1$ and $\psi_2$ will turn out to be same.

Answer 3

Yes both are the solutions to the TISE and $P_1(x)$ and $P_2(x)$ will be same after applying the boundary conditions. Roughly speaking, and without going into too much details: the Schroedinger equation (or any other differential equation) has an infinite number of solutions. The system “chooses” the one specific solution according to the boundary conditions.

You can look at it this way: the system defines all possible wave functions. When a wave enters it in a specific way, it gets to be one of the solutions according to its behaviour as it entered the system. (Yes, I know it is not accurate to say the least, but it’s a convenient way to get your head around it).

And if you are looking at solutions for a scattered state, you don't deal with probability densities actually, but the probability current and the coefficient of reflection and transmission.