1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Showing that a wavefunction can be written as a product

  1. Nov 21, 2016 #1
    1. The problem statement, all variables and given/known data
    Let us look at a 3-dimensional potential box. Show, that the wave function in this situation can be written as the product of 3 single-argument functions.

    2. Relevant equations
    The 3D Schrödinger equation:
    \begin{equation}
    -\frac{\hbar^2}{2m} \left( \frac{\partial^2 \Psi}{\partial x^2} + \frac{\partial^2 \Psi}{\partial y^2} + \frac{\partial^2 \Psi}{\partial z^2} \right) + V\Psi = E\Psi,
    \end{equation}
    where ##\Psi = \Psi(x,y,z)##

    3. The attempt at a solution
    I'm not at all sure how I'm supposed to show that this is true. I have not worked with partial differential equations at all, so I'm probably not in the right with what I tried to do:

    I assumed, that ##\Psi(x,y,z) = \Psi_x(x) \Psi_y(y) \Psi_z(z)##, and tried plugging it into the Schrödinger equation as follows:

    \begin{gather*}
    -\frac{\hbar^2}{2m} &\left( \frac{\partial^2 \Psi(x,y,z)}{\partial x^2} + \frac{\partial^2 \Psi(x,y,z)}{\partial y^2} + \frac{\partial^2 \Psi(x,y,z)}{\partial z^2} \right) + V\Psi(x,y,z) = E\Psi(x,y,z)\\
    \\
    -\frac{\hbar^2}{2m} & \left( \frac{\partial^2 \Psi_x(x) \Psi_y(y) \Psi_z(z)}{\partial x^2} + \frac{\partial^2 \Psi_x(x) \Psi_y(y) \Psi_z(z)}{\partial y^2} + \frac{\partial^2 \Psi_x(x) \Psi_y(y) \Psi_z(z)}{\partial z^2} \right)
    + V\Psi_x(x) \Psi_y(y) \Psi_z(z) = E\Psi_x(x) \Psi_y(y) \Psi_z(z)\\
    \\
    -\frac{\hbar^2}{2m} & \left( \Psi_y(y) \Psi_z(z) \frac{\partial^2 \Psi_x(x)}{\partial x^2} + \Psi_x(x) \Psi_z(z) \frac{\partial^2 \Psi_y(y)}{\partial y^2} + \Psi_x(x) \Psi_y(y) \frac{\partial^2 \Psi_z(z)}{\partial z^2} \right)
    + V\Psi_x(x) \Psi_y(y) \Psi_z(z) = E\Psi_x(x) \Psi_y(y) \Psi_z(z)
    \end{gather*}

    Dividing by ## \Psi_x(x) \Psi_y(y) \Psi_z(z)##:
    \begin{gather*}

    -\frac{\hbar^2}{2m} & \left( \frac{1}{\Psi_x(x)} \frac{\partial^2 \Psi_x(x)}{\partial x^2} + \frac{1}{\Psi_y(y)} \frac{\partial^2 \Psi_y(y)}{\partial y^2} + \frac{1}{\Psi_z(z)} \frac{\partial^2 \Psi_z(z)}{\partial z^2} \right)
    + V = E

    \end{gather*}
    But... Now what? Is there a way to proceed from here, or did I not pursue the correct path?
     
    Last edited: Nov 21, 2016
  2. jcsd
  3. Nov 21, 2016 #2

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Looks good. You just need a bit of insight to see what to do now. Can you remember what you did with separation of ##x## and ##t## functions in 1D?
     
  4. Nov 21, 2016 #3
    Basically, we did what I just did: Plugged the product into the original equation, which resulted in a similar expression to what I just wrote. Then the trick was to notice, that each side of the equation is dependent on only a single variable. My book claims that this results in both of the sides of the equation being a separation constant ##C##, which later turned out to be the energy ##E##.

    Are you saying I should have taken the time dependence into account? At a quick glance, it doesn't seem that I could use the same trick here.
     
  5. Nov 21, 2016 #4

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    No, in 1D you had two variables, here you have three. You can use the same trick. To give you a big hint: try setting ##y = z = 0## in your equation.
     
  6. Nov 21, 2016 #5
    Well, if I do that, since we are talking about an infinite potential outside the box, the wave functions ## \Psi_y(y)## and ## \Psi_z(z)## go to zero at ##y = 0## and ##z = 0## respectively, because of continuity. We are of course assuming, that the coordinates are placed as to allow this to happen.

    Our equation then becomes:
    \begin{gather*}
    -\frac{\hbar^2}{2m} & \left( \frac{1}{\Psi_x(x)} \frac{\partial^2 \Psi_x(x)}{\partial x^2} \right) + V = E
    \end{gather*}

    At these specific coordinates of ##y## and ##z## the potential ##V(x,y,z)## is also only a function of ##x##. Therefore our equation becomes:
    \begin{gather*}
    -\frac{\hbar^2}{2m} & \left( \frac{1}{\Psi_x(x)} \frac{\partial^2 \Psi_x(x)}{\partial x^2} \right) + V(x) = E(x)\\
    -\frac{\hbar^2}{2m} & \left( \frac{1}{\Psi_x(x)} \frac{\partial^2 \Psi_x(x)}{\partial x^2} \right) = E(x) - V(x)
    \end{gather*}
    Now the question is, does this hold in general? Could I replace ## -\frac{\hbar^2}{2m} \left( \frac{1}{\Psi_x(x)} \frac{\partial^2 \Psi_x(x)}{\partial x^2} \right)## with ##E(x) - V(x)## in the original equation?
     
  7. Nov 21, 2016 #6

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You misinterpreted my hint! I just choose ##y = z = 0## because it was the easiest to write down. Try setting ##y = y_0, z = z_0##.

    You're missing something much more obvious. There is no complicated maths involved.
     
  8. Nov 21, 2016 #7
    As per usual. :sorry:

    My brain really wants to go the complicated math route for some reason. If ##y## and ##z## constant, that means we are only moving along the ##x##-axis. This means our wave functions don't change in those directions, and their derivatives are zero. Also, our potential ##V = V(x,y_0,z_0)##, and energy ##E = E(x,y_0,z_0)##. The wave equation in this case is:
    [tex]
    -\frac{\hbar^2}{2m} \left( \frac{1}{\Psi_x(x)} \frac{\partial^2 \Psi_x(x)}{\partial x^2} \right)+ V(x,y_0,z_0) = E(x,y_0,z_0)
    [/tex]
    So at certain points ##y=y_0## and ##z = z_0##, our energy only depends on the x-coordinate. If we chose the ##x##-coordinate to be constant, and let ##y## vary, we would get the same result, as with ##z## as well.

    I'm again hesitating to say what this means in general. Does this lead to us being able to write ##V(x,y,z)## as ##V(x) + V(y) + V(z)##, and the same with the energy ##E##? How does this help us?
     
  9. Nov 21, 2016 #8
    Wait, have I just solved it after all? The fact that our problem was reduced to a single variable problem as long as we let the other variables be constant is what I was looking for, wasn't it? Or have I misunderstood this again?
     
  10. Nov 21, 2016 #9

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    That's it. One of your problems I think is that you got confused by the second derivatives. These are just functions of ##x,y,z##. So, you could rewrite your equation with, say, ##f(x) = (\frac{\partial \Psi_x^2}{\partial x^2})/\Psi_x##, and ##g(y), h(z)## similarly.

    Then, it's more obvious that ##g(0), h(0)## or ##g(y_0), h(z_0)## are just some constants, and you have an equation for ##f(x)## involving nothing but constants.
     
  11. Nov 21, 2016 #10
    Thank you so much.
     
  12. Nov 21, 2016 #11
    I also wanted to say, that it wasn't so much the derivatives, but the potential ##V(x,y,z)## and energy ##E(x,y,z)## still being functions of 3 variables even if moved them to the other side of the equation. So even if I solved for ##f(x)##, say, it wouldn't actually be a function of just ##x##, at least in my head.

    It took a lot of thinking to realize, that we chose ##y_0## and ##z_0## at random, which negated any loss of generality. Actually suggesting to let them be equal to zero really threw me off. :confused:
     
  13. Nov 21, 2016 #12

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    This process works because ##V## and ##E## are constant.

    You only needed one value of ##y## and ##z##. They didn't need to be arbitrary. Although, as ##g## and ##h## also turn out to be constant, it didn't matter what values you chose.
     
  14. Nov 21, 2016 #13
    Ah, thanks for clarifying.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted