(Quantum Mechanics) Infinite Square Well

[emol1414]

Hi, I'm stuck in this Griffiths' Introduction to QM problem (#2.8)

Homework Statement

A particle in the infinite square well has the initial wave function

\Psi(x,0) = Ax(a-x)

Normalize \Psi(x,0)

Homework Equations

\int_{0}^{a} |\Psi(x)|^2 dx = 1

The Attempt at a Solution

Haha, this is supposed to be the least of my problems but... doing

A^2 \int_{0}^{a} x^2 (a-x)^2 dx = 1
gives us A = \sqrt{\frac{30}{a^5}}.

When the correct answer is A = \sqrt{\frac{2}{a}}. I have no clue what I did wrong...

 

[kuruman]

The normalization constant A = sqrt(2/a) is appropriate for ψ(x) = A sin(nπx/a). Here you have ψ(x) = Ax(a-x). Different wavefunctions have different normalization constants.

 

[emol1414]

Kuruman... indeed A = \sqrt{\frac{2}{a}} is appropriate for \psi as a sin function. But when checking the result i got, i found this solution for the exact same problem:

http://img714.imageshack.us/img714/6186/64732846.jpg

The integral limit is set to 0,a/2... I'm not sure why, and if this is correct. For now i'll stick with my solution

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By the way... (a correlated problem) trying to prove that: \int_{-\infty}^{\infty} \psi_n^*(x) \psi_m (x) dx = \delta_{n,m} (Kronecker delta).

For n = m, and \psi_n (x) = \sqrt{\frac{2}{a}} sin(\frac{n \pi x}{a}),\in \Re, I'm coming up with

\int_{-\infty}^{\infty} \psi_{n}^{2}(x) dx = \frac{-1}{a} cos(\frac{n\pi x}{a}) sin(\frac{n\pi x}{a})|_{-\infty}^{\infty} + \frac{1}{a} \int_{-\infty}^{\infty} dx

The first term goes to 0, but the second would result in 1 for like... \int_{0}^{a}, not \int_{-\infty}^{\infty}. Again... where am I messing up?

 

[kuruman]

Kuruman... indeed A = \sqrt{\frac{2}{a}} is appropriate for \psi as a sin function. But when checking the result i got, i found this solution for the exact same problem:

http://img714.imageshack.us/img714/6186/64732846.jpg

The integral limit is set to 0,a/2... I'm not sure why, and if this is correct. For now i'll stick with my solution

It is not the exact same problem. The problem above has yet a third wavefunction that is constant for 0≤x≤a/2 and zero everywhere else. That is the reason why the upper limit is set at a/2. If you extend the integral beyond a/2, you will be adding a whole bunch of zeroes since ψ(x) is zero, so you might as well use a/2 for the upper limit.

By the way... (a correlated problem) trying to prove that: \int_{-\infty}^{\infty} \psi_n^*(x) \psi_m (x) dx = \delta_{n,m} (Kronecker delta).

For n = m, and \psi_n (x) = \sqrt{\frac{2}{a}} sin(\frac{n \pi x}{a}),\in \Re, I'm coming up with

\int_{-\infty}^{\infty} \psi_{n}^{2}(x) dx = \frac{-1}{a} cos(\frac{n\pi x}{a}) sin(\frac{n\pi x}{a})|_{-\infty}^{\infty} + \frac{1}{a} \int_{-\infty}^{\infty} dx

The first term goes to 0, but the second would result in 1 for like... \int_{0}^{a}, not \int_{-\infty}^{\infty}. Again... where am I messing up?

Here I assume that you are using particle-in-the-box wavefunctions. All of them are zero for x≤0 and x≥a. What do you think your limits of integration ought to be?

 

[emol1414]

Hmm, ok! So it's a different problem

It makes sense now, I forgot the conditions... You're right, it is \int_{0}^{a}!
Thank you