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6.3 Square barrier

A slightly more involved example is the square potential barrier, an inverted square well, see Fig. 6.3.

  
Figure 6.3: The square barrier.
\includegraphics[width=6.0cm]{Figures/square_barrier.eps}

We are interested in the case that the energy is below the barrier height, 0 < E < V0. If we once again assume an incoming beam of particles from the right, it is clear that the solutions in the three regions are

$\displaystyle \phi_{I}^{}$(x) = A1eikx + B1e-ikx,  
$\displaystyle \phi_{II}^{}$(x) = A2cosh($\displaystyle \kappa$x) + B2sinh($\displaystyle \kappa$x),  
$\displaystyle \phi_{III}^{}$(x) = A3eikx. (6.16)

Here

k = $\displaystyle \sqrt{\frac{2m}{\hbar^2}E}$,    $\displaystyle \kappa$ = $\displaystyle \sqrt{\frac{2m}{\hbar^2}(V_0-E)}$. (6.17)

Matching at x = - a and x = a gives (use sinh(- x) = - sinh x and cosh(- x) = cosh x
    
A1e-ika + B1eika = A2cosh$\displaystyle \kappa$a - B2sinh$\displaystyle \kappa$a (6.18)
ik(A1e-ika - B1eika) = $\displaystyle \kappa$(- A2sinh$\displaystyle \kappa$a + B2cosh$\displaystyle \kappa$a) (6.19)
A3eika = A2cosh$\displaystyle \kappa$a + B2sinh$\displaystyle \kappa$a (6.20)
ik(A3eika) = $\displaystyle \kappa$(A2sinh$\displaystyle \kappa$a + B2cosh$\displaystyle \kappa$a) (6.21)

These are four equations with five unknowns. We can express each unknown in A1. The way we proceed is to add eqs. (6.18) and (6.20), subtract eqs. (6.19) from (6.21), subtract (6.20) from (6.18), and add (6.19) and (6.21). We find
    
A1e-ika + B1eika + A3eika = 2A2cosh$\displaystyle \kappa$a (6.22)
ik(- A1e-ika + B1eika + A3eika) = 2$\displaystyle \kappa$A2sinh$\displaystyle \kappa$a (6.23)
A1e-ika + B1eika - A3eika = -2B2sinh$\displaystyle \kappa$a (6.24)
ik(A1e-ika - B1eika + A3eika) = 2$\displaystyle \kappa$B2cosh$\displaystyle \kappa$a (6.25)

We now take the ratio of equations (6.22) and (6.23) and of (6.24) and (6.25), and find (i.e., we take ratios of left- and right hand sides, and equate those)
$\displaystyle {\frac{A_1e^{-ika} + B_1 e^{ika}+A_3 e^{ika}
}{ik(-A_1e^{-ika} + B_1 e^{ika}+A_3 e^{ika})
}}$ = $\displaystyle {\frac{1}{\kappa \tanh \kappa a}}$ (6.26)
$\displaystyle {\frac{A_1e^{-ika} + B_1 e^{ika}-A_3 e^{ika}
}{ik(-A_1e^{-ika} + B_1 e^{ika}+A_3 e^{ika})
}}$ = - $\displaystyle {\frac{\tanh \kappa a}{\kappa }}$ (6.27)

These equations can be rewritten as (multiplying out the denominators, and collecting terms with A1, B3 and A3)
  
A1e-ika($\displaystyle \kappa$tanh$\displaystyle \kappa$a + ik) + B1eika($\displaystyle \kappa$tanh$\displaystyle \kappa$a - ik)      
+ A3eika($\displaystyle \kappa$tanh$\displaystyle \kappa$a - ik) = 0 (6.28)
A1e-ika($\displaystyle \kappa$ - iktanh$\displaystyle \kappa$a) + B1eika($\displaystyle \kappa$ + iktanh$\displaystyle \kappa$a)      
+ A3eika(- $\displaystyle \kappa$ + iktanh$\displaystyle \kappa$a) = 0 (6.29)

Now eliminate A3, add (6.28) ($ \kappa$ - iktanh$ \kappa$a) and (6.29) ($ \kappa$tanh$ \kappa$a - ik), and find
A1e-ika[($\displaystyle \kappa$ - iktanh$\displaystyle \kappa$a)($\displaystyle \kappa$tanh$\displaystyle \kappa$a + ik) +      
($\displaystyle \kappa$tanh$\displaystyle \kappa$a - ik)($\displaystyle \kappa$ - iktanh$\displaystyle \kappa$a)]      
+ B1eika[($\displaystyle \kappa$ - iktanh$\displaystyle \kappa$a)($\displaystyle \kappa$tanh$\displaystyle \kappa$a - ik) +      
($\displaystyle \kappa$tanh$\displaystyle \kappa$a - ik)($\displaystyle \kappa$ + iktanh$\displaystyle \kappa$a)] = 0 (6.30)

Thus we find
B1 = - A1e-2ika$\displaystyle {\frac{\tanh \kappa a (k^2+\kappa^2)}{(\kappa -ik \tanh \kappa a)(\kappa\tanh \kappa a -i k)}}$     (6.31)

and we find, after using some of the angle-doubling formulas for hyperbolic functions, that the absolute value squared, i.e., the reflection coefficient, is

R = $\displaystyle {\frac{\sinh^2 2\kappa a (\kappa^2+k^2)^2}{4 \kappa^2 k^2 + (\kappa^2-k^2)^2 \sinh^2 2\kappa a}}$ (6.32)

In a similar way we can express A3 in terms of A1 (add (6.28) ($ \kappa$ + iktanh$ \kappa$a) and (6.29) (- $ \kappa$tanh$ \kappa$a + ik)), or use T = 1 - R!

We now consider a particle of the mass of a hydrogen atom, m = 1.67 x 10-27 kg, and use a barrier of height meV and of width 10-10m. The picture for reflection and transmission coefficients can seen in Fig. 6.4. We have also evaluated R and T for energies larger than the height of the barrier (the evaluation is straightforward).

  
Figure 6.4: The reflection and transmission coefficients for a square barrier of height 4 meV and width 10-10m.
\includegraphics[width=5.0cm]{Figures/sq_barr_2.ps}

If we heighten the barrier to 50 meV, we find a slightly different picture, see Fig. 6.5.

  
Figure 6.5: The reflection and transmission coefficients for a square barrier of height 50 meV and width 10-10m.
\includegraphics[width=5.0cm]{Figures/sq_barr_5.ps}

Notice the oscillations (resonances) in the reflection. These are related to an integer number of oscillations fitting exactly in the width of the barrier, sin 2$ \kappa$a = 0.


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Next: 7. The Harmonic oscillator Up: 6. Scattering from potential Previous: 6.2 Potential step

© 1998 Niels Walet, UMIST
Email Niels.Walet@umist.ac.uk