Separate variables schrodinger equation

[Cogswell]

Homework Statement

[2 Dimensional infinite square well]
Show that you can separate variables such that the solution to the time independent Schrödinger 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.

Homework Equations

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

The Attempt at a Solution

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

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

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

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

And then... I don't really know where to go from here. I don't get what I'm supposed to do next.

 

[voko]

What is the explicit form of V(x, y)? Can it be written as a sum X(x) + Y(y)?

 

[Cogswell]

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.

 

[ehild]

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.

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

 

[voko]

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?

Of course not. Think about $$
V_c(z) =
\begin{cases}
0, & \text{if } 0 \le z \le c\\
\infty, & \text{otherwise}
\end{cases}
$$