Solutions of the schrödinger eq. for a potential step

1. Apr 6, 2013

71GA

Lets say we have a potential step as in the picture:

In the region I there is a free particle with a wavefunction $\psi_I$ while in the region II the wave function will be $\psi_{II}$. Let me now take the Schrödinger equation and try to derive $\psi_I$ which bugs me:

\begin{align}
&~~W \psi = -\frac{\hbar^2}{2m}\, \frac{d^2 \Psi}{d\, x^2} + W_p \psi ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\nonumber \\
&~~W \psi = -\frac{\hbar^2}{2m}\, \frac{d^2 \Psi}{d\, x^2}\nonumber \\
&\frac{d^2 \Psi}{d\, x^2} = -\frac{2m W}{\hbar^2}\,\psi \nonumber\\
{\scriptsize \text{DE: }} &\boxed{\frac{d^2 \Psi}{d\, x^2} = -\mathcal L\,\psi}~\boxed{\mathcal{L} \equiv \sqrt{\tfrac{2mW}{\hbar^2}}} \nonumber\\
&~\phantom{\line(1,0){18.3}}\Downarrow \nonumber\\
{\scriptsize \text{general solution of DE: }} &\boxed{\psi_{I} = C \sin\left(\mathcal{L}\, x \right) + D \cos \left(\mathcal{L}\, x \right)}\nonumber
\end{align}

I got the general solution for the interval I, but this is nothing like the solution they use in all the books: $\psi_{I} = C e^{i\mathcal L x} + D e^{-i \mathcal L x}$ where $\mathcal L \equiv \sqrt{{\scriptsize 2mW/\hbar^2}}$. I have a personal issue with this because if $x= -\infty$ part $De^{-i \mathcal L x}$ would become infinite and this is impossible for a wavefunction! I know that i would get exponential form if i defined constant $\mathcal L$ a bit differently as i did above:

\begin{align}
{\scriptsize \text{DE: }} &\boxed{\frac{d^2 \Psi}{d\, x^2} = \mathcal L\,\psi}~\boxed{\mathcal{L} \equiv -\sqrt{\tfrac{2mW}{\hbar^2}}} \nonumber\\
&~\phantom{\line(1,0){18.3}}\Downarrow \nonumber\\
{\scriptsize \text{general solution of DE: }} &\boxed{ \psi_{I} = C e^{\mathcal L x } + D^{-\mathcal L x} }\nonumber
\end{align}

This general solution looks more like the one they use in the books but it lacks an imaginary $i$ and $\mathcal L$ is defined with a - while in all the books it is positive. Could anyone tell me what am i missing here so i could connect all this into a solid one piece of knowledge?

2. Apr 6, 2013

bp_psy

The problem is here :
$lim_{x-> \infty} e^{icx} \neq \infty$
Use the fact that:
e^(icx)=cos(cx)+isin(cx)

3. Apr 6, 2013

71GA

That is eulers formula i know and i can ALMOST derive the connection between the example used in books: $\scriptsize \psi_I = Ce^{i\mathcal L x} + De^{-i \mathcal L x}$ and the first general solution to the DE: $\scriptsize \psi_I = C\cos (\mathcal L x) + D \sin (\mathcal L x)$.

Here it goes:

$$\scriptsize \begin{split} \underbrace{Ae^{i\mathcal L x} + B e^{-i \mathcal L x}}_{{\scriptsize \text{used in the books}}} = A \cos(\mathcal L x) +A i \sin (\mathcal L x) + B \cos(\mathcal L x) - B i \sin(\mathcal L x) = \underbrace{(A+B)}_{\equiv C} \cos(\mathcal L x) + \underbrace{(A-B)}_{\equiv D} i \sin (\mathcal L x) \neq \underbrace{C\cos (\mathcal L x) + D \sin (\mathcal L x)}_{{\scriptsize \text{solution to the DE}}} \end{split}$$

I can notice that the function used in books is NOT equal to the solution to the DE. It is diffrent for an imaginary number $i$... Here is allso one small snippet from Griffith where he doesn't use $i$. Take a closer look to the eq. 2.149. I am sorry for posting a snippet, i can remove it if necessary. So my question is why or how does $i$ dissapear???

4. Apr 6, 2013

bp_psy

You are right except that D=(A-B)i, the i is included there and the two forms of the solution are equivalent.

5. Apr 6, 2013

HomogenousCow

The constants don't have to be real.

6. Apr 6, 2013

HomogenousCow

Eulers formula.

7. Apr 7, 2013

71GA

So if i conclude all this (and please correct me if i am in any way wrong). I have a schrödinger equation which for a free particle can be rearanged like this:
$$\frac{d^2 \Psi}{d\, x^2} = -\frac{2m W}{\hbar^2}\,\psi$$
This is ofcourse a differential equation whose general solutions depend on how we define the constant $\mathcal L$.
\begin{align}
\mathcal L \equiv \sqrt{\frac{2mW}{\hbar^2}} \Longrightarrow \underbrace{\psi = C \sin(\mathcal L x) + D \cos (\mathcal L x)}_{\text{1st general solution where $\mathcal L$ is real}}~~~~~~~~\mathcal L \equiv \sqrt{-\frac{2mW}{\hbar^2}} \Longrightarrow \!\!\!\!\!\!\!\!\!\!\!\!\!\!\underbrace{\psi = C e^{\mathcal L x} + D e^{\mathcal L x}}_{\text{2nd general solution where $\mathcal L$ is complex}}
\end{align}
We choose 1st solution which has real $\mathcal L$ and a complex constant $D$ (which can be seen from):
$$\scriptsize \begin{split} \!\!\underbrace{A}_{\text{real}}\!e^{i\mathcal L x} + \!\!\underbrace{B}_{\text{real}}\! e^{-i \mathcal L x}= A \cos(\mathcal L x) +A i \sin (\mathcal L x) + B \cos(\mathcal L x) - B i \sin(\mathcal L x) = \underbrace{(A+B)}_{\equiv C} \cos(\mathcal L x) + \underbrace{(A-B)\,i}_{\equiv D} \sin (\mathcal L x) = C\cos (\mathcal L x) + \!\!\!\!\underbrace{D}_{{\scriptsize \text{complex}}}\!\!\! \sin (\mathcal L x) \end{split}$$
From above equation it can be seen that i can write $\psi = C \sin(\mathcal L x) + D \cos (\mathcal L x)$ in a form $Ae^{i\mathcal L x} + Be^{-i \mathcal L x}$ where $A$ and $B$ are real and $\mathcal L$ is also real ${\scriptsize \mathcal L \equiv \sqrt{2mW/\hbar^2}}$. Please correct me if i am wrong or confirm my assumptions.

8. Apr 7, 2013

bp_psy

Neither constant are (C,D) necessarily real or complex they are determined by your boundary conditions.
The usual argument in a ODE class is that a second order ODE with a characteristic equation with complex roots we have:
$f(x)=e^{(a+bi)x}$
$g(x)=e^{(a-bi)x}$
as solutions. Then by superposition we can get two other solutions

$u(x)=f(x)+g(x)=e^ax(e^{ibx}+e^{-ibx})=2e^{ax} cos(bx)$
$v(x)=f(x)-g(x)=e^ax(e^{ibx}+e^{-ibx})=2i e^{ax} sin(bx)$
Since the a solution multiplied by a real or complex constant is still a solution we can ignore the constants,we get the two solutions
$u(x)=e^{ax} cos(bx)$
$v(x)= e^{ax} sin(bx)$
Then the general solution by superposition will be :
$v(x)=A e^{ax} cos(bx) +Be^{ax}sin(bx)$ where A and B are two constants set by your boundary conditions.
For physics it is usually better to work with the complex exponential form because the integrals are usually more simple.

9. Apr 11, 2013

71GA

Thank you. This solved my issue.