How Do You Normalize a Wavefunction and Calculate Particle Location Probability?

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To normalize the wavefunction ψ(x) defined for the range -L/4 to L/4, the normalization constant A is determined by integrating the square of the wavefunction over this range and setting the integral equal to 1, resulting in A = ±2/√L, with the positive value chosen for practical reasons. After normalization, the probability of finding the particle between x=0 and x=L/8 is calculated using the normalized wavefunction, yielding a probability of approximately 40.9%. The discussion confirms that the normalization constant appears as A² in probability calculations, making the sign of A irrelevant. Overall, the process emphasizes the importance of normalization in quantum mechanics for accurate probability assessments.
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Homework Statement


A particle is described by the wavefunction:
\psi (x) = \{ \begin{array}{*{20}{c}}<br /> {A\cos (\frac{{2\pi x}}{L}){\quad\rm{for }} - \frac{L}{4} \le x \le \frac{L}{4}}\\<br /> {0{\quad\rm{otherwise }}}<br /> \end{array}

(a) Determine the normalization constant A
(b) What is the probability that the particle will be found between x=0 and x=L/8 if a measurement is made?

2. Homework Equations

N/A

The Attempt at a Solution



Okay, just verify my logic for me:

(a) We integrate the wave function from -L/4→L/4, where the particle has a 100% chance of appearing, and set this integral equal to 1 (for the probability just mentioned):

1 = \int\limits_{ - \frac{L}{4}}^{\frac{L}{4}} \psi {(x)^*}\psi (x)dx

Integrating this (using Mathematica) and solving for A seems to point at A = \pm \frac{2}{{\sqrt L }}.

I'll only assume the positive value is valid since I'm not sure we can have a negative probability.

(b) Assuming that the normalization constant in part (a) is correct, we use our new normalization constant in our wavefunction and perform the same integration, this time over 0→L/8. Performing this operation seems to yield 40.9% (0.409).
 
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Yes your logic is correct. You're right that we can't have a negative probability, but since the probability comes from ##\psi ^*\psi##, the normalisation constant only ever appears as ##A^2##, so it doesn't matter whether you pick the positive or negative value. Generally we pick the positive one just because positive numbers are easier to think about.
 
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