Solving for Eigenvalues in a Finite Square Well with Both Walls Finite

AI Thread Summary
The discussion focuses on deriving the eigenvalue solutions for a one-dimensional finite square well with both walls finite. The established equation for a single finite wall, k cot(kl) = -α, is used as a foundation. The user attempts to find the solution using boundary conditions and the time-independent Schrödinger equation, ultimately leading to the equation tan(kl) = 2αk / (k^2 - α^2). They explore changing the coordinate system to simplify the problem but find that it leads to different equations. The user expresses frustration at not reaching the desired solution, indicating a need for further assistance.
chris_avfc
Messages
83
Reaction score
0

Homework Statement


Already defined that for a 1D well with one finite wall the eigenvalue solutions are given by
k cot(kl) = -α

Show the eigenvalue solutions to well with both walls finite is given by

tan(kl) = 2αk / (k^2 - α^2)

Well is width L (goes from 0 to L) with height V_0

Homework Equations



k cot(kl) = -α

tan(kl) = 2αk / (k^2 - α^2)

Time Independent Schrödinger Equation.

The Attempt at a Solution



I will explain briefly as there is a lot of equations and it will look a mess, I can always upload a picture of my work if needs be.

General Solution:
u(x) = Ce^-αx + De^αx

For x < 0
C = 0, so exponential doesn't go to ∞.
For x > L
D = 0 so exponential doesn't go to ∞.

Using boundary conditions where the function and it's derivative must be continuous
With the solution between 0 and L being

u(x) = A sin(kx)

Substituting in L for x

A sin(kL) = Ce^-αL
Ak cos(kL) = -αCe^-αL

Dividing the second by the first

k cot(kL) = -α (As given in question)

This breaks down when substituting in 0 for x.

A sin(k0) = De^α0
Ak cos(k0) = αDe^α0
0 = αD

I thought about changing the coordinates so the well runs from -L/2 to L/2, but that just appears to give

k cot(kL/2) = -α

And

k cot(kl/2) = α

I was/am looking in one textbook, and that manages to get it down to

k tan (kL/2)= α
k cot (kL/2)= -α

Still not what I need, but might be closer?
 
Physics news on Phys.org
Guess nobody can help then. :(
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Thread 'A cylinder connected to a hanging 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