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?]

  • $\begingroup$ Is really the second form as you wrote? Isn't there an imaginary unit in the exponential arguments? $\endgroup$ – GiorgioP Jun 9 at 7:06
  • $\begingroup$ @GiorgioP You know most of the time potential energy is less than total energy. So there is an imaginary unit hidden in k'. $\endgroup$ – Dipankar Mitra Jun 9 at 7:18
  • $\begingroup$ Now I understand. But this implies that the squared modulus should appear also in the first case. $\endgroup$ – GiorgioP Jun 9 at 16:02
  • $\begingroup$ @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. $\endgroup$ – Dipankar Mitra Jun 10 at 7:40
  • $\begingroup$ 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. $\endgroup$ – 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.

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.

  • $\begingroup$ Then what are the boundary conditions of this equation? Does boundary conditions depends on systems? Or there is any general boundary condition for Schoedinger's equation? $\endgroup$ – Dipankar Mitra Jun 10 at 7:26
  • $\begingroup$ As you said both are the solutions to the TISE and P1(x) and P2(x) will be same after applying the boundary conditions How it can Be? Say for example, I have choosen two arbitrary values for C and D as C = 2 and D = 3. Now no real values of A and B can equalize P1(x) and P2(x). so how can P1(x) and P2(x) be same? $\endgroup$ – Dipankar Mitra Jun 10 at 7:33
  • $\begingroup$ @DipankarMitra why should A and B be real-valued? $\endgroup$ – J. Murray Jun 10 at 11:13
  • $\begingroup$ @J.Murray I am not sure A and B real-valued or not. Can A and B take complex values? $\endgroup$ – Dipankar Mitra Jun 10 at 11:35
  • $\begingroup$ @DipankarMitra Yes. Wavefunctions are generally complex valued, which shouldn’t be surprising because the two complex exponentials are also solutions. $\endgroup$ – J. Murray Jun 10 at 11:39

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