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An electron in a one-dimensional box with walls at [tex]x=(0,a)[/tex] is in the quantum state

[tex]\psi(x) = A, 0<x<a/2[/tex]

[tex]\psi(x) = -A, a/2 < x < a[/tex]

Obtain an expression for the normalization constant, [tex]A[/tex]. What is the lowest energy of the electron that will be measured in this state?

Solution

Well, we know that

[tex]\int_{0}^A |\psi(x)|^2 dx = 1 \iff A^2 \cdot a = 1 \iff A = \sqrt{\frac{1}{a}}[/tex]

so this is our normalization constant, [tex]A[/tex]. Now, to find the lowest energy of the electrong that will be measured in this state, we have:

[tex]-\frac{\hbar^2}{2m} \frac{\partial^2 \psi}{\partial x^2} = E \psi[/tex]

but the left hand side evaluates to 0 since [tex]\psi[/tex] is independent of [tex]x[/tex]. So this is the lowest (and only observed) energy in this state... is this right?

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# Electron in a One-Dimensional Box

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