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Schrodinger equation, normalizing

  • #1

Homework Statement


Consider the infinite well, a particle with mass m in the potential

[tex]V(x) =
\begin{cases}
0, & 0 < x < a,\\
\infty, & \text{otherwise,}
\end{cases},
[/tex]

At t = 0 the particle is in the state:

[tex]\Psi(x,0) = B \left[\sin{\left(\frac{l \pi}{a}x\right)} + b\sin{\left(\frac{2l \pi}{a}x\right)}\right][/tex]
with b a real number and l a whole number. Use normalization to show how B depends on b.

[tex]1 = B^2 \int_0^a \left[\sin{\left(\frac{l \pi}{a}x\right)} + b\sin{\left(\frac{2l \pi}{a}x\right)}\right]^2 dx \\
= B^2 \left(\frac{1}{2} \int_0^a (1 - \cos{\left(\frac{2 l \pi}{a}x\right)})dx + \frac{b^2}{2} \int_0^a (1 - \cos{\left(\frac{4 l \pi}{a}x\right)})dx +
b\int_0^a \cos{\left(\frac{l \pi}{a}x\right)}dx + b\int_0^a \cos{\left(\frac{3 l \pi}{a}x\right)}dx\right)\\
= B^2(\frac{a}{2} + \frac{b^2a}{2}) [/tex]

[itex]B = \sqrt{\frac{2}{a(1 + b^2)}}[/itex]

Did I do this correctly?
 
Last edited:

Answers and Replies

  • #2
ehild
Homework Helper
15,492
1,874
It looks correct.

ehild
 
  • #4
Now I have to calculate the expectation value of the energy. [itex]E = \frac{p^2}{2m}[/itex], so [itex]\langle E \rangle = \langle \frac{p^2}{2m} \rangle[/itex] and with [itex]\langle p \rangle = \int_{-\infty}^{\infty} \Psi^{\ast} \frac{\hbar}{i} \frac{\partial}{\partial x} \Psi dx[/itex] we get:

[tex]\langle E \rangle = \frac{-\hbar^2}{2m} \frac{2}{a(1 + b^2)} \int_0^a \Psi^{\ast} \frac{\partial^2}{\partial x^2} \Psi dx[/tex]

Is the integrand simply equal to [itex]\frac{d^2}{dx^2} \left[\sin{\left(\frac{l \pi}{a}x\right)} + b\sin{\left(\frac{2l \pi}{a}x\right)}\right]^2[/itex]? Because that becomes a really long calculation, so I was wondering if there is another way...
 

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