# One-dimensional Schrodinger Equation

1. Jun 12, 2004

### Megus

one-dimensional Schrodinger's Equation

Hi !
I wonder how to solve one-dimensional Schrodinger's Equation :

$$\frac{d^2 \psi (x)}{dx^2}\ = \ -(\frac{2 \pi}{\lambda})^2 \ * \ \psi (x)$$

I've to obtain $$\psi (x)$$ , when
$$-(\frac{2 \pi}{\lambda})^2$$ is known
Can you solve it as an example, when $$-(\frac{2 \pi}{\lambda})^2 = 2$$ ? Please...

Last edited: Jun 12, 2004
2. Jun 12, 2004

### speeding electron

Based on my rather basic knowledge, making
$$- \left( \frac{2 \pi}{ \lambda} \right)^2 = 2$$
would make the psi function an exponential function rather than having the wave characteristics you want.The solution would be:
$$\psi(x) = e^{ \sqrt{2} x } .$$
But you want a periodic wave function. By taking
$$\frac{2 \pi}{\lambda} = \sqrt{2}$$
(i.e. without the minus sign), the solution would be a wave equation and complex:
$$\psi(x) = e^{ \sqrt{2} i x } .$$

Last edited: Jun 12, 2004
3. Jun 12, 2004

### Tom Mattson

Staff Emeritus
You shouldn't commit to a specific value of -(&pi;/&lambda;)2. Just define it to be a literal constant, say, -k2. Then your differential equation is:

&psi;"+k2&psi;=0

Surely you have the solution of that one somewhere!

4. Jun 12, 2004

### Tom Mattson

Staff Emeritus
No, an exponential function arises when the first derivative is negatively proportional to the wavefunction itself. The problem here involves the second derivative.

The solution to his equation is indeed sinusoidal, but let's let him work it out.

5. Jun 12, 2004

### heardie

In fact $$\frac{{2\pi }} {\lambda } = k$$ when k is the wavenumber, so the k^2 substution can be made even without the knowledge of how it helps solve the DE.

6. Jun 13, 2004

### Tom Mattson

Staff Emeritus
Yes, my choice in symbols was not accidental.

7. Jun 13, 2004

### Megus

Is it the correct answer ?:
$$\psi (x) = e^{ikx + \varphi_R + i \varphi_I} = e^{\varphi_R} \cdot e^{i(kx + \varphi_I)} = A(\cos (kx + \varphi_I) + i \sin (kx + \varphi_I))$$

8. Jun 13, 2004

### Tom Mattson

Staff Emeritus
Yes, though you might find it more convenient to replace the constants A and &phi;I with A and B as follows:

&psi;(x)=Acos(kx)+Bsin(kx),

and applying the boundary conditions from there.

9. Jun 13, 2004

Ok - thanks