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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.

square˙barrier


Figure 6.3: The square barrier.

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

\begin{eqnarray} {ϕ}_{I}(x)& =& {A}_{1}{e}^{ikx} + {B}_{ 1}{e}^{−ikx}, %& \\ {ϕ}_{II}(x)& =& {A}_{2}\mathop{ cosh}\nolimits (κx) + {B}_{2}\mathop{ sinh}\nolimits (κx),%& \\ {ϕ}_{III}(x)& =& {A}_{3}{e}^{ikx}. %&(6.16) \\ \end{eqnarray}

Here

k = \sqrt{{2m\over {ℏ}^{2}} E},\kern 2.77695pt \kern 2.77695pt κ = \sqrt{{2m\over {ℏ}^{2}} ({V }_{0} − E)}.
(6.17)

Matching at x = −a and x = a gives (use \mathop{sinh}\nolimits (−x) = −\mathop{sinh}\nolimits x and \mathop{cosh}\nolimits (−x) =\mathop{ cosh}\nolimits x

\begin{eqnarray} {A}_{1}{e}^{−ika} + {B}_{ 1}{e}^{ika}& =& {A}_{ 2}\mathop{ cosh}\nolimits κa − {B}_{2}\mathop{ sinh}\nolimits κa %&(6.18) \\ ik({A}_{1}{e}^{−ika} − {B}_{ 1}{e}^{ika})& =& κ(−{A}_{ 2}\mathop{ sinh}\nolimits κa + {B}_{2}\mathop{ cosh}\nolimits κa)%&(6.19) \\ {A}_{3}{e}^{ika}& =& {A}_{ 2}\mathop{ cosh}\nolimits κa + {B}_{2}\mathop{ sinh}\nolimits κa %&(6.20) \\ ik({A}_{3}{e}^{ika})& =& κ({A}_{ 2}\mathop{ sinh}\nolimits κa + {B}_{2}\mathop{ cosh}\nolimits κa) %&(6.21) \\ \end{eqnarray}

These are four equations with five unknowns. We can thus express for of the unknown quantities in one other. Let us choose that one to be {A}_{1}, since that describes the intensity of the incoming beam. We are not interested in {A}_{2} and {B}_{2}, which describe the wave function in the middle. We can combine the equation above so that they either have {A}_{2} or {B}_{2} on the right hand side, which allows us to eliminate these two variables, leading to two equations with the three interesting unknowns {A}_{3}, {B}_{1} and {A}_{1}. These can then be solved for {A}_{3} and {B}_{1} in terms of {A}_{1}:

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

\begin{eqnarray} {A}_{1}{e}^{−ika} + {B}_{ 1}{e}^{ika} + {A}_{ 3}{e}^{ika}& =& 2{A}_{ 2}\mathop{ cosh}\nolimits κa %&(6.22) \\ ik(−{A}_{1}{e}^{−ika} + {B}_{ 1}{e}^{ika} + {A}_{ 3}{e}^{ika})& =& 2κ{A}_{ 2}\mathop{ sinh}\nolimits κa %&(6.23) \\ {A}_{1}{e}^{−ika} + {B}_{ 1}{e}^{ika} − {A}_{ 3}{e}^{ika}& =& −2{B}_{ 2}\mathop{ sinh}\nolimits κa%&(6.24) \\ ik({A}_{1}{e}^{−ika} − {B}_{ 1}{e}^{ika} + {A}_{ 3}{e}^{ika})& =& 2κ{B}_{ 2}\mathop{ cosh}\nolimits κa%&(6.25) \\ \end{eqnarray}

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)

\begin{eqnarray} {{A}_{1}{e}^{−ika} + {B}_{1}{e}^{ika} + {A}_{3}{e}^{ika}\over ik(−{A}_{1}{e}^{−ika} + {B}_{1}{e}^{ika} + {A}_{3}{e}^{ika})}& =& {1\over κ\mathop{tanh}\nolimits κa}%&(6.26) \\ {{A}_{1}{e}^{−ika} + {B}_{1}{e}^{ika} − {A}_{3}{e}^{ika}\over ik(−{A}_{1}{e}^{−ika} + {B}_{1}{e}^{ika} + {A}_{3}{e}^{ika})}& =& −{\mathop{tanh}\nolimits κa\over κ} %&(6.27) \\ \end{eqnarray}

These equations can be rewritten as (multiplying out the denominators, and collecting terms with {A}_{1}, {B}_{3} and {A}_{3})

\begin{eqnarray}{ A}_{1}{e}^{−ika}(κ\mathop{tanh}\nolimits κa + ik) + {B}_{ 1}{e}^{ika}(κ\mathop{tanh}\nolimits κa − ik)& & %& \\ +{A}_{3}{e}^{ika}(κ\mathop{tanh}\nolimits κa − ik)& =& 0%&(6.28) \\ {A}_{1}{e}^{−ika}(κ − ik\mathop{tanh}\nolimits κa) + {B}_{ 1}{e}^{ika}(κ + ik\mathop{tanh}\nolimits κa)& & %& \\ +{A}_{3}{e}^{ika}(−κ + ik\mathop{tanh}\nolimits κa)& =& 0%&(6.29) \\ \end{eqnarray}

Now eliminate {A}_{3}, add (6.28)(κ − ik\mathop{tanh}\nolimits κa) and (6.29)(κ\mathop{tanh}\nolimits κa − ik), and find

\begin{eqnarray} {A}_{1}{e}^{−ika}\left [(κ − ik\mathop{tanh}\nolimits κa)(κ\mathop{tanh}\nolimits κa + ik)+\right .& & %& \\ \left .(κ\mathop{tanh}\nolimits κa − ik)(κ − ik\mathop{tanh}\nolimits κa)\right ]& & %& \\ +{B}_{1}{e}^{ika}\left [(κ − ik\mathop{tanh}\nolimits κa)(κ\mathop{tanh}\nolimits κa − ik)\right . + & & %& \\ \left .(κ\mathop{tanh}\nolimits κa − ik)(κ + ik\mathop{tanh}\nolimits κa)\right ]& =& 0%&(6.30) \\ \end{eqnarray}

Thus we find

\begin{eqnarray}{ B}_{1} = −{A}_{1}{e}^{−2ika} {\mathop{tanh}\nolimits κa({k}^{2} + {κ}^{2})\over (κ − ik\mathop{tanh}\nolimits κa)(κ\mathop{tanh}\nolimits κa − ik)}& & %&(6.31) \\ \end{eqnarray}

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 = {{\mathop{sinh}\nolimits }^{2}2κa{({κ}^{2} + {k}^{2})}^{2}\over 4{κ}^{2}{k}^{2} + {({κ}^{2} − {k}^{2})}^{2}{\mathop{ sinh}\nolimits }^{2}2κa}
(6.32)

In a similar way we can express {A}_{3} in terms of {A}_{1} (add (6.28) (κ + ik\mathop{tanh}\nolimits κa) and (6.29) (−κ\mathop{tanh}\nolimits κa + ik)), or use T = 1 − R!

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Alternative approach

The equation can be given in matrix form as

\begin{eqnarray} \left (\array{ {e}^{−ika} & {e}^{ika} \cr ik{e}^{−ika}& − ik{e}^{ika}} \right )\left (\array{ {A}_{1} \cr {B}_{1}} \right ) = \left (\array{ \mathop{cosh}\nolimits κa & −\mathop{ sinh}\nolimits κa\cr − κ\mathop{ sinh} \nolimits κa & κ\mathop{ cosh} \nolimits κa} \right )\left (\array{ {A}_{2} \cr {B}_{2}} \right )& & %&(6.33) \\ \left (\array{ {e}^{ika} & {e}^{−ika} \cr ik{e}^{ika}& − ik{e}^{−ika}} \right )\left (\array{ {A}_{3}\cr 0} \right ) = \left (\array{ \mathop{cosh}\nolimits κa & \mathop{sinh}\nolimits κa\cr κ\mathop{ sinh} \nolimits κa &κ\mathop{ cosh} \nolimits κa} \right )\left (\array{ {A}_{2} \cr {B}_{2}} \right ).& & %&(6.34)\\ \end{eqnarray}

Question: Can you invert the matrices and find the same answer as before?

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We now consider a particle of the mass of a hydrogen atom, m = 1.67 × 1{0}^{−27}\ \textrm{kg}, and use a barrier of height 4\ meV and of width 1{0}^{−10}m. The picture for reflection and transmission coefficients can seen in Fig. 6.4a. We have also evaluated R and T for energies larger than the height of the barrier (the evaluation is straightforward).

sq˙barrier˙2   sq˙barrier˙5


Figure 6.4: The reflection and transmission coefficients for a square barrier of height 4 meV (left) amd 50 meV (right) and width 1{0}^{−10}m.

If we heighten the barrier to 50 meV, we find a slightly different picture, see Fig. 6.4b.

Notice the oscillations (resonances) in the reflection. These are related to an integer number of oscillations fitting exactly in the width of the barrier, \mathop{sin}\nolimits 2κa = 0.

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