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The potential for a free particle in a potential field is given by \begin{equation} V(x) = V_0\theta(x) - w\delta(x) \end{equation} in which $\theta(x)$ is the unit step function, $\delta(x)$ is the Dirac delta function, and $V_0$ and $w$ are strictly positive constants. A particle of mass $m$ evolves in such a potential.

How do I determine the wave function at all parts in position?

Let the wavefunction be represented by $\psi(x,t)$, The time independent Schrodinger's equation is given by,

$$ \frac{-\hbar^2}{2m} \frac{\delta ^2\psi }{\delta x^2} + V \psi = E \psi $$ The potential function could be written as, $$ V(x) \begin{cases} 0 & x<0 \\ -\infty & x= 0 \\ V_0 & x>0 \end{cases} $$The wave function must be continuous and differential at the $x=0$. This provides us with the boundary conditions,

So solving the general equation for wave function we obtain

$$ \psi(x) \begin{cases} 0 & x<0 \\ -\infty & x= 0 \\ V_0 & x>0 \end{cases} $$ The general solution for a fixed energy fixed potential Time independent schrodinger equation for a free particle is given by,

$$ \psi(x) \begin{cases} A_1 e^{ik_1x} + B_1 e^{-ik_1x} & x<0 & k _1=\sqrt{2m \frac{-E}{\hbar^2}}\\ C & x= 0 \\ A_3 e^{ik_3x} + B_3 e^{-ik_3x}& x>0 & k_3 =\sqrt{2m \frac{V_0-E}{\hbar^2}}\\ \end{cases} $$ On checking the boundary conditions we obtain, (assuming that E is not zero) \begin{align} A_1 + B_1 = A_3 + B_3 = C \\ A_1 k_1 - B_1 k _1 = A_3k _3 - B_3 k_3 \end{align} For normalization and probability constraint we need the following to hold true,

\begin{align} \int_{-\infty}^\infty |\psi(x)|^2 dx &= 1 \\ \int_{-\infty}^0 |\psi(x)|^2 dx +\int_{0}^\infty |\psi(x)|^2 dx = 1 \\ \end{align}

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    $\begingroup$ Suggestion to the post (v2): Avoid the word complex unless it refers to complex numbers. $\endgroup$
    – Qmechanic
    Dec 8, 2021 at 15:27
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    $\begingroup$ I was thinking how we possibly have a complex physical potential. It seems the word you were searching is "complicated". In physics, "complex" almost always has something to do with imaginary numbers. $\endgroup$
    – Shing
    Dec 8, 2021 at 15:42

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the relation you got for the second continuity equation is not quite right.Since you are considering a delta potential at the origin, there is no reason for the derivatives to be continuous at 0. If you look for example into the Wiki page on the usual dirac delta potential, https://en.wikipedia.org/wiki/Delta_potential you'll see that the condition to be imposed is not

$$A_1 k_1 - B_1 k _1 = A_3k _3 - B_3 k_3$$

but

$$-\frac{\hbar^2}{2m}\left(\Psi'(0^+)-\Psi'(0^-)\right)-\omega\Psi(0)=0$$

or in other words

$$-i\frac{\hbar^2}{2m}\left(A_3 k_3 - B_3 k _3 - A_1k_1 +B_1 k_1\right)-\omega C=0$$

You then need to solve these in the 3 possible cases:

$E>V_0$, $0<E<V_0$ and $E<0$.

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  • $\begingroup$ thanks for the info, so for some fields wavefunction does not need to continous? I feel that doesn't make sense. $\endgroup$
    – Mal Mo
    Dec 9, 2021 at 10:12
  • $\begingroup$ It is very specific to the delta potential. If you look through the short demonstration in the Wikipedia page this seems clear: The function is continuous but not its derivative. $\endgroup$ Dec 9, 2021 at 11:30

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