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Je m'appelle
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Homework Statement
Solving the following differential equation with the given boundary conditions:
[tex]\hbar^2 \frac{d^2}{dx^2}\psi (x) = 2mE\psi (x), \ \ \ \ \ \forall \ \hbar^2,\ m,\ E > 0 [/tex]
[tex]\psi(a) = \psi(-a) = 0 [/tex]
Homework Equations
The Attempt at a Solution
[tex]\hbar^2 \frac{d^2}{dx^2}\psi (x) = 2mE\psi (x) [/tex]
[tex]\frac{d^2}{dx^2}\psi (x) = \frac{2mE}{\hbar^2}\psi (x) [/tex]
[tex]\frac{d^2}{dx^2}\psi (x) - \frac{2mE}{\hbar^2}\psi (x) = 0 [/tex]
Let
[tex]\frac{2mE}{\hbar^2} = \rho[/tex]
Then
[tex]\frac{d^2}{dx^2}\psi (x) - \rho \psi (x) = 0 [/tex]
The characteristic polynomial for this ODE shall be
[tex]r^2 - \rho = 0 [/tex]
[tex]r^2 = \rho ,\ r_1 = -\sqrt{\rho},\ r_2 = \sqrt{\rho} [/tex]
Therefore
[tex]\psi (x) = c_1e^{r_1 x} + c_2e^{r_2 x} [/tex]
[tex]\psi (x) = c_1^{-\sqrt{\rho} x} + c_2e^{\sqrt{\rho} x} [/tex]
By using the given conditions the only possible solution I can get is
[tex]c_1 = c_2 = 0,\ \psi (x) = 0 [/tex]
As I get to something like
[tex]e^{4 \sqrt{\rho} a} = 1 [/tex]
And since
[tex]\rho,\ a > 0[/tex]
So
[tex]\psi(x) = 0 [/tex]
seems to be the only solution.
Is this it?
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