Quantam Griffiths 2.47 Finite Square Well

[gysush]

Consider double finite square well with

-Vo; -(a+b/2) < x < -b/2
V(x) = 0; -b/2 < x < b/2
-Vo; b/2 < x < a + b/2
0; otherwise

Sketch the ground wave function Psi(n=1) and the first excited state Psi(n=2) for
1. b = 0
2. b~a
3. b >> a


So...

for b=0

This is just the normal finite potential well. In particular, we can consider even and odd solutions.

the transcendental equation is

odd: tanz = ((zo/z)^2 - 1)^1/2
even: -cotz = ((zo/z)^2 - 1)^1/2

z=ka...zo=a(2mVo)^(1/2)/h_bar
k=(2m(E+Vo))^(1/2)/h_bar

As a result...for both even and odd...we get E_n + Vo ~ ((n*pi*h_bar)^2)/(2m(2a)^2)

*Note* even though i included the above information, griffiths says "no computations!" "qualitative only."

First Idea

For x< -a...Ae^(kx) (even)...-Ae^(kx) (odd)
For x > a...Ce^(-kx)

i.e. => for these regions shape of Psi (n,x) is very simple, but If I was to graph for n=1 and n=2 on the same graph...would the n=2 state be higher? (i.e. higher y value for the same x values in this region)?

For -a < x < a...

I'm confused on the shape of the graph in this region...

Do i construct for odd and even solutions?
i.e. => like pics. on pg 32 (except for sin and cos as well)
(the pics. on pg 32 are graphs of sin((n*pi/a)x) for n=1,2,3)

then for b~a and b>>a...the results are similar since a well with width a and width 2a have the same physics

Second Idea

This is supposed to just mirror the results of the infinite square well.

Particularly...we have e^(kx) (far left region) and e^(-kx) (far right region)

Also, the graph inside the well would be of form sin((n*pi/2a)x)...
I understand when we did the derivations of the infinite square well we got this result because of the B.C. #1...the continuity of Psi...as a result sin(2ka)=0...=> k=n*pi/2a

I'm confused because griffiths says no computations. Because of that statement, I infer to look back to problems we have done to see how to pursue this problem. As a result, only time we have dealt with finite square well...we didn't even graph Psi...only got information for bound states and scattering states...in particular...no question in the text asks you to graph Psi for finite square well...as a result...only time we graphed Psi for something similar was delta function potential and infinite square well. Obviously, the infinite square well is more similar. So...should I do what my gut tells me and treat each well as Psi(n,x)=sin((n*pi/a)x)

Thank you

 

[Jhero]

for your post. I would approach this problem by first considering the general form of the wave function for a finite square well. In this case, we have a double well potential with three regions: -Vo, 0, and -Vo.

For the first region, -Vo, we can use the even and odd solutions as you have mentioned. For the second region, 0, we can use the general form of the wave function for a finite square well, which is a combination of sine and cosine functions.

Now, for the third region, -Vo, we can use the even and odd solutions again. However, there is a slight difference here because the potential is not symmetric around the middle point. This means that the energy levels will not be the same as in the case of a single finite square well.

For the three cases given, we can sketch the ground state and first excited state as follows:

1. b = 0:
In this case, the potential is just a single finite square well with two regions, -Vo and 0. The ground state will have a wave function that is symmetric around the middle point, and the first excited state will have a wave function that is antisymmetric. The energy levels will be the same as in the case of a single finite square well.

2. b~a:
In this case, the potential is a double well with two regions, -Vo and 0, of equal width. The ground state will have a wave function that is symmetric around the middle point, and the first excited state will have a wave function that is antisymmetric. However, the energy levels will be slightly different due to the asymmetry of the potential.

3. b >> a:
In this case, the potential is a double well with two regions, -Vo and 0, where the width of the second region is much smaller than the first region. This means that the energy levels will be similar to the case of a single finite square well, but the wave function will be slightly modified due to the presence of the second well.

In summary, the ground state and first excited state for a double finite square well can be sketched by considering the general form of the wave function for a finite square well and taking into account the differences in the potential for each case.