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Problem
An electron in a one-dimensional box with walls at x=(0,a) is in the quantum state
\psi(x) = A, 0<x<a/2
\psi(x) = -A, a/2 < x < a
Obtain an expression for the normalization constant, A. What is the lowest energy of the electron that will be measured in this state?
Solution
Well, we know that
\int_{0}^A |\psi(x)|^2 dx = 1 \iff A^2 \cdot a = 1 \iff A = \sqrt{\frac{1}{a}}
so this is our normalization constant, A. Now, to find the lowest energy of the electrong that will be measured in this state, we have:
-\frac{\hbar^2}{2m} \frac{\partial^2 \psi}{\partial x^2} = E \psi
but the left hand side evaluates to 0 since \psi is independent of x. So this is the lowest (and only observed) energy in this state... is this right?
An electron in a one-dimensional box with walls at x=(0,a) is in the quantum state
\psi(x) = A, 0<x<a/2
\psi(x) = -A, a/2 < x < a
Obtain an expression for the normalization constant, A. What is the lowest energy of the electron that will be measured in this state?
Solution
Well, we know that
\int_{0}^A |\psi(x)|^2 dx = 1 \iff A^2 \cdot a = 1 \iff A = \sqrt{\frac{1}{a}}
so this is our normalization constant, A. Now, to find the lowest energy of the electrong that will be measured in this state, we have:
-\frac{\hbar^2}{2m} \frac{\partial^2 \psi}{\partial x^2} = E \psi
but the left hand side evaluates to 0 since \psi is independent of x. So this is the lowest (and only observed) energy in this state... is this right?