# What is the general solution of one-dimentional time-independent Schrodinger's equation?

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

One is,$$\psi(x) = Asin(kx)+Bcos(kx)\\\text{where}, k = \frac{2π\sqrt{2m(E-U_0)}}{h}\\ \text{In this case, probability density: }P_1(x) = \psi(x)\cdot\psi^*(x) = |\psi(x)|^2\\$$ 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 schrodingers 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?]

• Is really the second form as you wrote? Isn't there an imaginary unit in the exponential arguments? – GiorgioP Jun 9 at 7:06
• @GiorgioP You know most of the time potential energy is less than total energy. So there is an imaginary unit hidden in k'. – Dipankar Mitra Jun 9 at 7:18
• Now I understand. But this implies that the squared modulus should appear also in the first case. – GiorgioP Jun 9 at 16:02
• @GiorgioP Why? As I have understood (there might be lackings in my understanding) in the first case k is used instead of k' (Actually k' = ik). And the value of k is not imaginary or complex for most of the cases. So there should no need of modulus in the first case. – Dipankar Mitra Jun 10 at 7:40
• for most of the cases means for the values of x such that $E>U_0$. And what about the points such that $E<U_0$? So in general the solution will have a piecewise purely real or purely imaginary ${\bf k}$. The squared modulus is in order. – GiorgioP Jun 10 at 8:17

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.