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Particle in a box, stuck at normalizing the wave function

  1. Apr 28, 2013 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    A particle of mass m is inside a 1 dimensional "box" of length L such that it's restricted to move between ##x=-L/2## and ##x=L/2## where the potential vanishes.
    1)Determine the eigenvalues ##E_n## and the eigenfunctions ##\psi _n## of the Hamiltonian imposing that the eigenfunctions vanish at the extrema of the box.
    2)Graph the first 3 eigenfunctions and analyze their nodes.
    The problem continues up to part 5) but I'll post the rest only if I get stuck later.

    2. Relevant equations
    Schrödinger's equation.
    Normalization of the wave function.


    3. The attempt at a solution
    1)I solved Schrödinger's equation ##\psi ''+ \varepsilon \psi =0## where ##\varepsilon = \frac{2mE}{\hbar ^2}##.
    I've reached that the only possible physical solution is when ##\varepsilon >0## in which case ##\psi (x)=A\cos (\sqrt \varepsilon x )+B \sin (\sqrt \varepsilon x )##.
    I applied the boundary conditions to that psi of x, it gave me a system of 2 equations:
    [tex]A\cos \left ( \frac{L\sqrt \varepsilon }{2} \right )- B \sin \left ( \frac{L\sqrt \varepsilon }{2} \right )=0[/tex] (*)
    [tex]A\cos \left ( \frac{L\sqrt \varepsilon }{2} \right )+ B \sin \left ( \frac{L\sqrt \varepsilon }{2} \right )=0[/tex] (**)
    I wrote it under matricial form, I've seen that the matrix of the system is not invertible for some values of epsilon, so I've checked out what condition ##\det M =0## gives me, and it gives me the condition that ##\varepsilon _n=\left ( \frac{n\pi}{L} \right ) ^2##. For such values of epsilon, the matrix is not invertible and those epsilon's describe are directly related to the allowed energies such that the psi_n's are not 0. I should mention that ##n \in \mathbb{N}##, ##n>0##.
    This gave me that ##E_n= \frac{\hbar^2 n^2 \pi ^2}{2mL^2}##. So far, so good.
    Coming back to (*) (or (**) for that matter) and plugging the values of ##\varepsilon _n## I've found, I've reached that for when n is odd, ##\psi _n(x) =A_n \cos \left ( \frac{n\pi x}{L} \right )## while for when n is even, ##\psi _n (x)=B_n \sin \left ( \frac{n\pi x}{L} \right )##.
    So I've answered part 1).
    Now for part 2), I must normalize, else I can't graph the first ##3 \psi _n##'s. Normalizing implies finding the constants ##A_n##'s and ##B_n##'s.
    I've tried to use ##\int _{-L/2}^{L/2} |\Psi(x)|^2dx=1## where ##\Psi (x)## is an infinite linear combinations of the eigenfunctions, i.e. ##\Psi (x)=\sum _{\text{n odd}} A_n \cos \left ( \frac{n\pi x}{L} \right ) - B_{n+1} \sin \left ( \frac{n\pi x}{L} \right ) ## but this lead me to an integral of an infinite series with 3 terms and I see absolutely no simplications.
    I'm stuck at finding those constants, i.e. normalizing the wave function. Am I missing something?
    Thanks!
     
  2. jcsd
  3. Apr 28, 2013 #2

    cepheid

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    This expression of the wavefunction is, in fact, just a Fourier series expansion, right? I.e. the basis functions (what you call eigenfunctions) are already orthogonal, even if they aren't yet orthonormal. So I think you'll find that, for each term in ψψ* = ψ2, the integral of that term vanishes unless it happens to be a product of the same two basis functions i.e. m = n.
     
  4. Apr 28, 2013 #3

    fluidistic

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    Thanks for the reply. I don't know what m is. If that's related to product of the sine and cosine functions I think I get your point. The integral of the terms with those product will vanish because they are orthogonal with respect to the dot product.
    Using ψψ* = ψ2, I get that ##|\Psi (x)|^2=\sum _{\text{odd n}} A_nA^* _n \cos ^2 \left ( \frac{n\pi x}{L} \right ) +B_{n+1}B^* _{n+1} \sin ^2 \left ( \frac{n\pi x}{L} \right )=\sum _{\text{odd n}} |A_n|^2 \cos ^2 \left ( \frac{n\pi x}{L} \right ) + |B_{n+1}|^2 \sin ^2 \left ( \frac{n\pi x}{L} \right )##. I must integrate this with respect to x, from -L/2 to L/2. Does this look good so far?

    Edit: Integrating and skipping several steps, I get that ##\int _{-L/2} ^{L/2} \Psi (x) \Psi ^* (x)dx=1 \Rightarrow \sum _{odd n} |A_n|^2 \left [ x+\sin \left ( \frac{n\pi x}{L} \right ) \cos \left ( \frac{n\pi x}{L} \right ) \right ] +|B_{n+1}|^2 \left [ x-\sin \left ( \frac{n\pi x}{L} \right ) \cos \left ( \frac{n\pi x}{L} \right ) \right ]=1##.
    Apparently at best I could get ##|A_n|## in terms of ##|B_{n+1}|## but I don't even know how to get that. I'm missing something for sure.
     
    Last edited: Apr 28, 2013
  5. Apr 28, 2013 #4

    vela

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    You want to normalize each eigenfunction ##\psi_n(x)## separately.

    Once you have the normalized eigenstates, you can expand an arbitrary state ##\Psi(x)## in terms of them:
    $$\Psi(x) = \sum_n c_n \psi_n.$$ Without knowing what ##\Psi(x)## is specifically, the best you can do in this case is show that ##\displaystyle\sum_n |c_n|^2 = 1##.
     
  6. Apr 28, 2013 #5

    fluidistic

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    Thanks for the help vela!
    Well if that's the case then I reach that both ##|A_n|^2## and ##|B_n|^2## are worth ##2/L##.
    I've got a question. How can I get ##A_n## and ##B_n## from it? In principle they are complex numbers so I can't just say that they are worth, let's say, ##\sqrt {\frac{2}{L}}##.

    Also it's not clear to me as why each eigenfunction should be normalized. I understand that the general solution (which is a linear combinations of the eigenfunctions) must be normalized if we want to interpret its squared modulus as a probability density. I guess it's because if the eigenfunctions are not normalized, the general Psi is not either.
     
  7. Apr 28, 2013 #6

    vela

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    You can arbitrarily choose the phase of ##A_n## and ##B_n##, so just make them real and positive.

    Normalized eigenfunctions make calculations easier. The inner product of two eigenstates becomes the Kronecker delta. You don't have to worry about some weird factor being introduced from unnormalized eigenstates.
     
  8. Apr 28, 2013 #7

    fluidistic

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    Okay thank you very much!
    I'm going to proceed further alone, for now at least.
     
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