Separate variables schrodinger equation

AI Thread Summary
The discussion focuses on solving the time-independent Schrödinger equation for a two-dimensional infinite square well, specifically addressing the separation of variables. Participants explore how to express the wave function as a product of functions dependent on each coordinate, ## \psi (x,y) = \phi (x) \chi (y) ##. The potential, ## V(x,y) ##, is defined as zero within the well and infinite outside, leading to questions about its role in the equation. It is clarified that the potential cannot be neglected, as it is crucial for determining the behavior of the wave function in the defined regions. Understanding the implications of the potential is essential for solving the equation correctly.
Cogswell
Messages
54
Reaction score
0

Homework Statement


[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.

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.
 
Physics news on Phys.org
What is the explicit form of V(x, y)? Can it be written as a sum X(x) + Y(y)?
 
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.
 
Cogswell said:
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
 
Cogswell said:
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 $$
V_c(z) =
\begin{cases}
0, & \text{if } 0 \le z \le c\\
\infty, & \text{otherwise}
\end{cases}
$$
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top