# One-dimensional Schrodinger Equation

## Main Question or Discussion Point

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...

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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:
$$\psi(x) = e^{ \sqrt{2} i x } .$$

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Tom Mattson
Staff Emeritus
Gold Member
Megus said:
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...
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!

Tom Mattson
Staff Emeritus
Gold Member
speeding electron said:
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:
$$\psi(x) = e^{ \sqrt{2} x } .$$
No, an exponential function arises when the first derivative is negatively proportional to the wavefunction itself. The problem here involves the second derivative.

But you want a periodic wave function. By taking
The solution to his equation is indeed sinusoidal, but let's let him work it out.

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.

Tom Mattson
Staff Emeritus
Gold Member
heardie said:
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.
Yes, my choice in symbols was not accidental.

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

Tom Mattson
Staff Emeritus
Gold Member
Megus said:
Is it the correct answer ?:
$$\psi (x) = e^{ikx + \varphi_R + i \varphi_I} = e^{\varphi_R} \cdot e^{i(kx + \varphi_I)} = A(\cos (kx + \varphi_I) + i \sin (kx + \varphi_I))$$
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.

Ok - thanks