I had the question about the boundary condition of wave equations in quantum mechanics, where I think it was sufficient to show that the probability density and momentum was continuous at the boundary, rather than the function itself and its first derivative was continuous at the boundary condition.
See details: https://physics.stackexchange.com/questions/394788/finite-square-well-boundary-condition
Where the following example was given by solve the step well potential.
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 $x=0$ meet at $x=0.$
Solve the boundary condition equaled to $\Psi_1^*(0)\Psi_1(x)=\Psi_2^*(0)\Psi_2(x)$ (Probability density) and $\Psi_1^*(0)\partial_x\Psi_1(x)=\Psi_2^*(0)\partial_x\Psi_2(x)$(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 $x=0$.
https://physics.stackexchange.com/questions/397127/step-potential-well-solve-the-other-kind-of-given-boundary-condition
My question was that, was those solution valid? Further, was my original statement in the first paragraph correct?
Dear Yucong Cai,
It's impossible to read your equations. Leave the $ symbols, the editor here doesn't process them into formatted text. You are given the options upper case, lower case, and you can copy and paste from soe doc-file, Greek symbols and mathematical symbols. For instance
Suppose
k0= √(2mE/ħ2) and k1=√[2m(E-V0)/ħ2]
where E>V0.
ψ1=A exp(ik0 x) + B exp(-ik0x) for x ≤ 0, and
ψ2=C exp(ik1 x} for x>0
meet at x=0.
Continue as I showed you, and people would be able to read easily your question.
In general, for proving that a set of boundary conditions solve a problem, you should start the solution from the boundary abd continue in small steps until you cover all the doain over which the wave-function is defined. Next, you should also show whether the solution is unique.
By the way I notice your statement
"Thus we solved the equation without assuming the continuous of the function."
It doesn't go like that. The continuity of the wave-function is a requirement that you can't ignore. After solving your equatios you MUST show that the wave-function obtined is continuous and has continuous 1st derivative over all the domain on which it is defined.
Best regards
@Sofia D. Wechsler
Sorry, I had not figured out how to type latex in Researchgate yet. My question was mainly about the second half of the copenhagen interpretation " ... The wavefunction evolves smoothly in time while isolated from other systems.'" see: https://en.wikipedia.org/wiki/Copenhagen_interpretation#Principles
I got some feedback these days, and some people say by implying the probability density and momentum to be continuous, the function and its first derivative had to be continuous as well.(i.e. in step well by implying the coefficients to be real) But I never see a proof for the general case. Do you know if there was any cases exists?
Thanks.
Momentum isn't continuous at boundaries. If the potential changes, momentum changes.
The density is continuous across boundaries, so that isn't anything new. The density does satisfy a continuity equation-it's just that the current isn't a local function of the density-it's a local function of the wavefunction. That's the difference between this continuity equation and the diffusion equation and expresses the fact that the continuity equation, in quantum-and classical-mechanics doesn't describe diffusion.
This is known, since the 19th century, as Liouville's theorem.
So it's just a question of taste, whether one is solving the continuity equation (Liouville's equation) or the Schrödinger equation. It's just much simpler to solve the latter than the former in many cases.
Cf. http://www.feynmanlectures.caltech.edu/III_21.html section 21-2.
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