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IainH

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Consider a particle of mass m subject to the following potential function (taking Vo and L

to be positive):

V (x) =

40 Vo if x < 0;

0 if 0 < x < L/2;

2 Vo if L/2 < x < L;

40 Vo if x > L.

(a) Derive the transcendental equation for energy eigenstates having an energy

2 Vo < E < 40 Vo.

To simplify the mathematics and numerics, for the rest of this question take h = 1, L = 1,

m = 1, and choose (in this system of units) Vo = 1.

I've found the solution for the wavefunction in all areas.

For x<0,

## \psi_1(x) = A exp(k_1 x)+B exp(-k_1 x) ##

Using the condition that ##psi(x)##=0 when x approaches infinity,

##\psi_1(x) for x<0 = A exp(k_1 x)##

For region 2, (0<x<L/2), solution is oscillatory as E>V(x), so

## \psi_2(x) = \alpha sin(k_2*x) + \beta cos(k_2*x).##

Applying boundary conditions: at x=0, ## \psi_1(x)=\psi_2(x)##, and the derivatives of## (\psi_1)/\psi_2 ## must also be equal at x=0.

##A exp(0) = \alpha sin(0)+\beta cos(0)

k_1 A exp(0) = \alpha cos(0) -k_2 \beta sin(0)##

##\Rightarrow \beta = A,\alpha = {\frac{k_1}{k_2}}A##

## \Rightarrow \psi_2(x) = {\frac{k_1}{k_2}}A sin(k_2 x) + A cos(k_2 x).##

For region 3, (L/2<x<L), psi is also oscillatory as E>V(x), so

##\psi_3(x) = C sin(k_3 x)+D cos(k_3 x)##

Again, I applied the boundary conditions, where ##\psi_2(x=L/2) = psi_3(x=L/2)##, and the respective derivatives at x=L/2, which produced two equations. Dividing one equation by the other allows the 'A' coefficients to be cancelled.

## \psi_2(L/2) = \psi_3(L/2)\Rightarrow {\frac{k_1}{k_2}}A sin({\frac{k_2 L}{2}}) + A cos({\frac{k_2 L}{2}}) = C sin({\frac{k_3 L}{2}}) + D cos({\frac{k_3 L}{2}})##(1)

Taking derivatives and letting x = L/2;

##{\frac{k_1}{2}} A cos({\frac{k_2 L}{2}}) - {\frac{k_2}{2}} A sin({\frac{k_2 L}{2}) = {\frac{k_3}{2}} C cos({\frac{k_3 L}{2}}) +{\frac{k_3}{2}} D sin({\frac{k_3 L}{2}})}## (2)

Dividing ##{\frac{(1)}{(2)}} \Rightarrow {\frac{{\frac{k_1}{k_2}} A sin({\frac{k_2 L}{2}}) + A cos({\frac{k_2 L}{2}})}{{\frac{k_1}{2}} A cos({\frac{k_2 L}{2}}) - {\frac{k_2}{2}} A sin({\frac{k_2 L}{2})}}} = {\frac{C sin({\frac{k_3 L}{2}}) + D cos({\frac{k_3 L}{2}})}{{\frac{k_3}{2}} C cos({\frac{k_3 L}{2}}) -{\frac{k_3}{2}} D sin({\frac{k_3 L}{2}})}}(3)##

For region 4, x>L, psi_4(x) is decaying, and applying the boundary condition as in region 1,

##\psi_4(x) = G exp(-k_1 x).##

I let ##psi_3(x=L) = psi_4(x=L)##, and the respective derivatives equal each other when x=L. This produced another two equations.

##C sin(k_3 L) + D cos(k_3 L) = G exp(-k_1 L)## (4)

##k_3 C cos (k_3 L) - k_3 D sin(k_3 L) = -k_1 G exp(-k_1 L)## (5)

Dividing (5) by (4) gives:

##{\frac{k_3 C cos (k_3 L) - k_3 D sin(k_3 L)}{C sin(k_3 L) + D cos(k_3 L)}} = -k_1## (6)

Dividing one by the other allows the 'G' coefficients to be removed as above.

I'm then left with a mess of sines and cosines with the factors C and D. I can't seem to get rid of the C and D coefficients to get the required transcendental energy solution. Most of my efforts has been trying to plug in (6) into (3), to cancel C and D, but I've had no luck so far.

Am I missing something here? Any help would be much appreciated. I was thinking that perhaps as ## 2 V_0 < E< 40 V_0,## the wave function won't be found below ## E = 2 V_0##, but I don't think this is correct.

Thanks very much.

to be positive):

V (x) =

40 Vo if x < 0;

0 if 0 < x < L/2;

2 Vo if L/2 < x < L;

40 Vo if x > L.

(a) Derive the transcendental equation for energy eigenstates having an energy

2 Vo < E < 40 Vo.

To simplify the mathematics and numerics, for the rest of this question take h = 1, L = 1,

m = 1, and choose (in this system of units) Vo = 1.

## Homework Equations

## The Attempt at a Solution

I've found the solution for the wavefunction in all areas.

For x<0,

## \psi_1(x) = A exp(k_1 x)+B exp(-k_1 x) ##

Using the condition that ##psi(x)##=0 when x approaches infinity,

##\psi_1(x) for x<0 = A exp(k_1 x)##

For region 2, (0<x<L/2), solution is oscillatory as E>V(x), so

## \psi_2(x) = \alpha sin(k_2*x) + \beta cos(k_2*x).##

Applying boundary conditions: at x=0, ## \psi_1(x)=\psi_2(x)##, and the derivatives of## (\psi_1)/\psi_2 ## must also be equal at x=0.

##A exp(0) = \alpha sin(0)+\beta cos(0)

k_1 A exp(0) = \alpha cos(0) -k_2 \beta sin(0)##

##\Rightarrow \beta = A,\alpha = {\frac{k_1}{k_2}}A##

## \Rightarrow \psi_2(x) = {\frac{k_1}{k_2}}A sin(k_2 x) + A cos(k_2 x).##

For region 3, (L/2<x<L), psi is also oscillatory as E>V(x), so

##\psi_3(x) = C sin(k_3 x)+D cos(k_3 x)##

Again, I applied the boundary conditions, where ##\psi_2(x=L/2) = psi_3(x=L/2)##, and the respective derivatives at x=L/2, which produced two equations. Dividing one equation by the other allows the 'A' coefficients to be cancelled.

## \psi_2(L/2) = \psi_3(L/2)\Rightarrow {\frac{k_1}{k_2}}A sin({\frac{k_2 L}{2}}) + A cos({\frac{k_2 L}{2}}) = C sin({\frac{k_3 L}{2}}) + D cos({\frac{k_3 L}{2}})##(1)

Taking derivatives and letting x = L/2;

##{\frac{k_1}{2}} A cos({\frac{k_2 L}{2}}) - {\frac{k_2}{2}} A sin({\frac{k_2 L}{2}) = {\frac{k_3}{2}} C cos({\frac{k_3 L}{2}}) +{\frac{k_3}{2}} D sin({\frac{k_3 L}{2}})}## (2)

Dividing ##{\frac{(1)}{(2)}} \Rightarrow {\frac{{\frac{k_1}{k_2}} A sin({\frac{k_2 L}{2}}) + A cos({\frac{k_2 L}{2}})}{{\frac{k_1}{2}} A cos({\frac{k_2 L}{2}}) - {\frac{k_2}{2}} A sin({\frac{k_2 L}{2})}}} = {\frac{C sin({\frac{k_3 L}{2}}) + D cos({\frac{k_3 L}{2}})}{{\frac{k_3}{2}} C cos({\frac{k_3 L}{2}}) -{\frac{k_3}{2}} D sin({\frac{k_3 L}{2}})}}(3)##

For region 4, x>L, psi_4(x) is decaying, and applying the boundary condition as in region 1,

##\psi_4(x) = G exp(-k_1 x).##

I let ##psi_3(x=L) = psi_4(x=L)##, and the respective derivatives equal each other when x=L. This produced another two equations.

##C sin(k_3 L) + D cos(k_3 L) = G exp(-k_1 L)## (4)

##k_3 C cos (k_3 L) - k_3 D sin(k_3 L) = -k_1 G exp(-k_1 L)## (5)

Dividing (5) by (4) gives:

##{\frac{k_3 C cos (k_3 L) - k_3 D sin(k_3 L)}{C sin(k_3 L) + D cos(k_3 L)}} = -k_1## (6)

Dividing one by the other allows the 'G' coefficients to be removed as above.

I'm then left with a mess of sines and cosines with the factors C and D. I can't seem to get rid of the C and D coefficients to get the required transcendental energy solution. Most of my efforts has been trying to plug in (6) into (3), to cancel C and D, but I've had no luck so far.

Am I missing something here? Any help would be much appreciated. I was thinking that perhaps as ## 2 V_0 < E< 40 V_0,## the wave function won't be found below ## E = 2 V_0##, but I don't think this is correct.

Thanks very much.

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