Unlike the textbook, I was trying to test a new set of boundary condition in step potential where probability density and momentum was continuous at the boundary $x=0$.
Suppose $k_0=\sqrt{2m/\hbar^2E}$ and $k_1=\sqrt{2m/\hbar^2(E-V_0)}$ where $E>V_0$.
$\Psi_1=Ae^{ik_0x}+Be^{-ik_0x}$ for $x<=0$ and $\Psi_2=Ce^{ik_1x}$ for $x>=0$ meet at $x=0.$
Solve the boundary condition equaled to $\Psi_1^*(x)\Psi_1(x)=\Psi_2^*(x)\Psi_2(x)$ (Probability density) and $\Psi_1^*(x)\partial_x\Psi_1(x)=\Psi_2^*(x)\partial_x\Psi_2(x)$(momentum).
Thus $(A+B)^2=C^2$ and $(A+B)(ik_0A-ik_0B)=Cik_1C$.
$(A+B)^2=C^2$ and $A^2-B^2=\frac{k_1}{k_0}C$.
Thus $2AB=\frac{k_0+k_1}{k_0}C^2-2A^2$.
If assuming $A,B,C>0$ and was real.
Eventually I obtained $B=A(k_0-k_1)/(k_0+k_1)$ and $C=2Ak_0/(k_0+k_1)$.
Thus $R=B^2/A^2=(k_0-k_1)^2/(k_0+k_1)^2$ which was consist with the boundary condition of the continuous functions, and $C^2/A^2=4k_0^2/(k_0+k_1)^2$ which was also consist with the usual boundary condition.
Thus we solved the equation without assuming the continuous of the function. Rather, assume the coefficient to be real and momentum and probability amplitude was continuous at the $x=0$.
Could you give me some advice on my solutions?
Answer
Your solution is valid because of your assumptions. If you assume $A,B,C \in \mathbb{R}$ and $A,B,C> 0$ then $\sqrt{|A|^2} = A$, and so in your first boundary condition you could essentially take the square root of both sides to show that continuity of probability density implies continuity of wavefunction.
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