Particle in finite well (Schroedinger)

[elninio0397]

Homework Statement

A particle with energy greater than the potential is defined as below:

V(x) = Vo (x<0)

V(x) = 0 (0<x<a)

V(x) = Vo (x>a)

a) Write the complete solutions (time and space) to the S. Eqn for the 3 regions

b) What condition must the width of the potential satisfy for the transmission of a wave from the left to be a maximum?

c) What is the minimum possible value for the transmission? What conditions must the weidth of the potential satisfy for this?

Homework Equations

None, really. It seems more conceptual.

The Attempt at a Solution

If I am not mistaken, this represents a simple harmonic oscillator.
I believe to have figured out the answer to part A.
Region 2 has the Schrodinger equation:
(-h(bar)/2m)(d²\psi/dx²) + 1/2kx²\psi = ih(bar)d\psi/dt
whose bound state solutions are \psi(x,t)=\psi(x)e^(-iEt/h(bar))

The solution for Region 1 = Region 3 is:
(-h(bar)/2m)(d²\psi/dx²) + 1/2kx²\psi = ih(bar)d\psi/dt
\psi''(x)=(alpha)²\psi(x) where (alpha)² = (2m/h(bar)²)(V(x)-E)

For problem b, I have an idea that since the wave function is Asinkx, the length (x) would have to be long enough so that it reaches a maximum at the second barrier (sin(pi/2)).

I could be going about this problem completely wrong. I've read through the textbook many times and it just doesn't reach this level of detail.

Any help would be much appreciated!

 

[vela]

Homework Statement

A particle with energy greater than the potential is defined as below:

V(x) = Vo (x<0)

V(x) = 0 (0<x<a)

V(x) = Vo (x>a)

a) Write the complete solutions (time and space) to the S. Eqn for the 3 regions

b) What condition must the width of the potential satisfy for the transmission of a wave from the left to be a maximum?

c) What is the minimum possible value for the transmission? What conditions must the weidth of the potential satisfy for this?

Homework Equations

None, really. It seems more conceptual.

The Attempt at a Solution

If I am not mistaken, this represents a simple harmonic oscillator.

You are mistaken. You're told what V(x) equals, and it's clearly not V(x)=1/2 kx².

I believe to have figured out the answer to part A.
Region 2 has the Schrodinger equation:
(-h(bar)/2m)(d²\psi/dx²) + 1/2kx²\psi = ih(bar)d\psi/dt
whose bound state solutions are \psi(x,t)=\psi(x)e^(-iEt/h(bar))

The solution for Region 1 = Region 3 is:
(-h(bar)/2m)(d²\psi/dx²) + 1/2kx²\psi = ih(bar)d\psi/dt
\psi''(x)=(alpha)²\psi(x) where (alpha)² = (2m/h(bar)²)(V(x)-E)

For problem b, I have an idea that since the wave function is Asinkx, the length (x) would have to be long enough so that it reaches a maximum at the second barrier (sin(pi/2)).

I could be going about this problem completely wrong. I've read through the textbook many times and it just doesn't reach this level of detail.

Any help would be much appreciated!

Use the time-independent Schrodinger equation. You're given the potential, so just plug it in and solve, with the assumption that E>V₀. The idea is to solve the equation in the three regions and then join the solutions together smoothly at the boundaries. At x=0, you require \psi(0^-)=\psi(0^+) and \psi&#039;(0^-)=\psi&#039;(0^+), and you have similar conditions at x=a.

 

[elninio0397]

I'm afraid you have me confused then. How is it that I can use the time-independant S. Equation when the problem specifically asks for a complete time and space solution?

 

[elninio0397]

I've reworked my solution for part a and have come up with the following solutions:

For the region in the well - \psi(x,t) = Aexp(ik2x)exp(iwt) + Bexp(-ik2x)exp(-iwt) where k2=((2mE)^-1/2)/hbar.

For the region outside the well - d^2\psi/dx^2 + 2m/hbar^2 (E-V0)\psi = 0 which has the solution \psi(x) = Aexp(ik1x)exp(-iwt) + Bexp(-ik1x)exp(-iwt) where k1=((2m(E-V0))^-1/2)/hbar.

Assuming this is correct, how are parts b and c accomplished?

My work for part b)
T=1 occurs when sin^2(k1a)=0 from the equation T= 1 + (sin^2(k1a))/(4(E/V0)((E/V0-1)) = 1

I'm still stumped for part c. How can this equation become lower than 1 and how could I find this minimum value?

 

[vela]

I'm afraid you have me confused then. How is it that I can use the time-independent S. Equation when the problem specifically asks for a complete time and space solution?

Sorry, I missed that. In any case, the time dependence separates out, so you use the time-independent equation to solve for the spatial part and put the two pieces together to get the complete time-and-space solution. Your book should show how that works. Or you can refer to http://scienceworld.wolfram.com/physics/SchroedingerEquation.html.

I've reworked my solution for part a and have come up with the following solutions:

For the region in the well - \psi(x,t) = Aexp(ik2x)exp(iwt) + Bexp(-ik2x)exp(-iwt) where k2=((2mE)^-1/2)/hbar.

For the region outside the well - d^2\psi/dx^2 + 2m/hbar^2 (E-V0)\psi = 0 which has the solution \psi(x) = Aexp(ik1x)exp(-iwt) + Bexp(-ik1x)exp(-iwt) where k1=((2m(E-V0))^-1/2)/hbar.

You have made some algebra errors. The expressions for k1 and k2 are wrong. You need to fix those before you go on.

Assuming this is correct, how are parts b and c accomplished?

My work for part b)
T=1 occurs when sin^2(k1a)=0 from the equation T= 1 + (sin^2(k1a))/(4(E/V0)((E/V0-1)) = 1

I'm still stumped for part c. How can this equation become lower than 1 and how could I find this minimum value?

Your expression for T is wrong. It can't exceed 1.

 

[elninio0397]

...Misread my own notes for T. =1 should read ^-1.

Thank you for all your help thus far. I'll get back at it in the morning..

 

[elninio0397]

How exactly do I find the expressions for k1 and k2? The ones that I took came from a similar example done in class. I can't find any information regarding how k is found, I can only find concrete values for specific examples.

 

[vela]

Well, you really shouldn't be taking them from anywhere. You should be solving the Schrodinger equation and deriving the results on your own.

In the middle region (0<x<a), you have V(x)=0, so the time-independent Schrodinger equation reduces to

-\frac{\hbar^2}{2m}\psi&#039;&#039;(x) = E\psi

After a little rearranging, you get

\psi&#039;&#039;(x)+\frac{2mE}{\hbar^2}\psi(x)=0

This is a simple, second-order linear differential equation with constant coefficients. What are its solutions?