one-dimensional Schrodinger Equation

[Megus]

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...

[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 } .

[Tom Mattson]

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 -(π/λ)2. Just define it to be a literal constant, say, -k2. Then your differential equation is:

ψ"+k2ψ=0

Surely you have the solution of that one somewhere!

[Tom Mattson]

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. :wink:

[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.

[Tom Mattson]

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. :wink:

[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))

[Tom Mattson]

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 φI with A and B as follows:

ψ(x)=Acos(kx)+Bsin(kx),

and applying the boundary conditions from there.

[Megus]

Ok - thanks :biggrin: