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Separate variables schrodinger equation

  1. Jun 1, 2013 #1
    1. The problem statement, all variables and given/known data
    [2 Dimensional infinite square well]
    Show that you can separate variables such that the solution to the time independent schrodinger equation, ## \hat{H} \psi (x,y) = E \psi (x,y) ## can be written as a product state ## \psi (x,y) = \phi (x) \chi (y) ## where ## \phi (x)## is a function of only the x coordinate and ##\chi(y)## is a function of only the y coordinate.

    2. Relevant equations

    [tex]\hat{H} = -\dfrac{\hbar^2}{2m} \left( \dfrac{\partial^2}{\partial x^2} + \dfrac{\partial^2}{\partial y^2} \right) + V(x,y)[/tex]

    3. The attempt at a solution

    So... putting in ## \psi (x,y) = \phi (x) \chi (y) ## I'll get:

    [tex]\hat{H} \phi (x) \chi (y) = -\dfrac{\hbar^2}{2m} \left( \dfrac{\partial^2 \phi (x)}{\partial x^2} \chi (y) + \dfrac{\partial^2 \chi (y)}{\partial y^2} \phi (x) \right) + V(x,y) \phi (x) \chi (y) [/tex]

    And then dividing both sides by ## \phi (x) \chi (y) ## I'll get:

    [tex]\hat{H} = -\dfrac{\hbar^2}{2m} \left( \dfrac{1}{\phi (x)} \dfrac{\partial^2 \phi (x)}{\partial x^2} + \dfrac{1}{\chi (y)} \dfrac{\partial^2 \chi (y)}{\partial y^2} \right) + V(x,y)[/tex]

    And then... I don't really know where to go from here. I don't get what I'm supposed to do next.
     
  2. jcsd
  3. Jun 1, 2013 #2
    What is the explicit form of V(x, y)? Can it be written as a sum X(x) + Y(y)?
     
  4. Jun 1, 2013 #3
    For a 2D potential in the infinite square well:

    $$
    V(x,y) =
    \begin{cases}
    0, & \text{if } 0 \le x \le a & \text{and} & 0 \le y \le b \\
    \infty, & \text{otherwise}
    \end{cases}
    $$

    Does that mean V(x,y) can be neglected?

    I just don't get what it's asking me to do.
     
  5. Jun 2, 2013 #4

    ehild

    User Avatar
    Homework Helper
    Gold Member

    V=0 in the potential well, if 0≤x≤a and 0≤y≤b, and infinite outside. Solve the equation inside the well with V=0, and outside where V is infinite. Is it possible that the wavefunction of a particle with finite energy differs from zero when the potential is infinite?

    ehild
     
  6. Jun 2, 2013 #5
    Of course not. Think about $$
    V_c(z) =
    \begin{cases}
    0, & \text{if } 0 \le z \le c\\
    \infty, & \text{otherwise}
    \end{cases}
    $$
     
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