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Schrodinger equation, normalizing

  1. Mar 8, 2012 #1
    1. The problem statement, all variables and given/known data
    Consider the infinite well, a particle with mass m in the potential

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

    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: Mar 8, 2012
  2. jcsd
  3. Mar 8, 2012 #2


    User Avatar
    Homework Helper

    It looks correct.

  4. Mar 8, 2012 #3
  5. Mar 8, 2012 #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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