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r1 - 23 Jul 2007 - 19:57:24 - ThanhTruongYou are here: TWiki >  Main Web > LectureNotes > UndergraduatePChem > Chapter5 > Tunneling1D

Tunneling Through a Barrier

Finally, let describe a quantum mechanical phenomenon with translational motion called tunneling.  Let us consider the case of a free particle moving with the total energy E encounters a barrier with the height $V_0$ above E.  In classical mechanics, the particle will bounce back.  What happen to the particle in quantum mechanics?

(insert figure)

The potential energy V(x) in this case is given by

\[V(x) = \left \lbrace \begin{array} {l} 0 \; \; \; \;\; \; (x < 0 ) \\ V_0 \; \; \; \; (0 \leq x \leq a ) \\ 0 \; \; \; \; \; \; (x>a) \end {array} \right.\]

For E < $V_0$, the general solution to the Schrödinger equation in each region is given by

\[\begin{array} {l} \psi_1(x) = Ae^{ik_1x} + Be^{-ik_1x} \; \; \; \; \; (0 <x) \\ \psi_2(x)= Ce^{k_2x} + De^{-k_2x} \; \; \; \; \; \; (0 \leq x \leq a) \\ \psi_3(x)=Fe^{ik_1x} + Ge^{-ik_1x} \; \; \; \; (a < x) \end{array} \]

where

\[k_1 = \frac {\sqrt {2mE}} {\hbar} \; \; \; \; \; and \; \; \; \; \; k_2 = \frac {\sqrt{2m(V_0-E)}} {\hbar}\]

(insert figure)

If we assume the particle originally from the far left (large negative x) moving to the right then we can eliminate the wavefunction representing the particle moving from the far right in Region 3, i.e. G= 0.

Solving for the coefficients A, B, C, D, and F, we rely on the conditions the $\psi$ and $ d\psi / dx$ are continuous at x = 0 and a. These conditions yield

\[ A + B = C + D}\]

\[ ik_1(A-B) = k_2(C-D)\]

\[Ce^{k_2a} + De^{-k_2a} = Fe^{ik_1a}\]

\[k_2Ce^{k_2a} - k_2De^{-k_2a} = ik_1Fe^{ik_1a}\]

There are five unknown but we only have four equations. Fortunately, we are only interested in the transmission coefficient, the probability the particle will tunneling through the barrier, which is give by

\[T = \frac{ \left | F\right |^2 }{ \left | A \right |^2 }\]

To solve for T, the first step is to elimiate B from the first two equations to give A in terms of C and D. Second, solve for C and D in terms of F in the last two equations then substitute them into the resulting equation of the first step to get the equation only involve A and F as follow

\[2ik_1 A = \left[ \left (k_2^2 - k_1^2 + 2ik_1k_2 \right )e^{k_2a} + \left ( k_1^2 - k_2^2 + 2ik_1k_2 \right )e^{-k_2a} \right ] \frac{Fe^{ik_1a }}{ 2k_2} \]

Using the relations $ \sinh x = (e^x - e^{-x})/2$, $\cosh x = (e^x + e^{-x})/2$, and $\cosh^2 x = 1 + \sinh^2 x $ with some algebra manipulations to obtain

\[ T = \frac{ 1}{ 1 + \frac{ \sinh^2 \left (\sqrt { \upsilon_0 \left ( 1- \varepsilon \right ) } \right ) }{ 4 \varepsilon \left ( 1- \varepsilon \right ) } } \]

where

\[ \upsilon_0 = \frac{ 2ma^2V_0}{ \hbar^2} \; \; \; \; \; and \; \; \; \; \; \varepsilon = \frac{ E}{ V_0} \]

(insert figure)

The tunneling transmission coefficient depends on three factors:

  1. Telative energy from the top of the barrier, i.e. the closer to the top of the barrier the larger tunneling probability.
  2. The mass of the particle, i.e. the heavier the mass the smaller the tunneling probability.
  3. The width of the barrier, i.e. the larger the barrier width the smaller the tunneling probability.  

-- ThanhTruong - 23 Jul 2007

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