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Quantum Physics 2002. Quantum Tunnelling. Recommended Reading: R.Harris, Chapter 5 Sections 1, 2 and 3. Potential Barrier: E < U 0. I. II. III. U = U 0. E = K.E. E = K.E. U. x =L. x. x =0. Region III. Region I. Region II. 1. Potential. where. and. Wavefunctions.
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Quantum Physics 2002 Quantum Tunnelling Recommended Reading: R.Harris, Chapter 5 Sections 1, 2 and 3
Potential Barrier: E < U0 I II III U = U0 E = K.E. E = K.E. U x =L x x =0 Region III Region I Region II 1 Potential where and
Wavefunctions Region I: 2 Incident Reflected Must keep both terms. Do you see why? Region II: 3 Region III: 4 Left Moving Transmitted No term because there is no particle incident from the right.
Boundary Conditions Match wavefunction and derivative at x = 0. 5 6 Match wavefunction and derivative at x = L. 7 8
Boundary Conditions We now have 4 equations and 5 unknowns, Can solve for B, C, D and F in terms of A. This is left as an exercise, A lot of algebra but nothing complicated!! Again we define a Reflection and Transmission coefficient: Since k1 = k3. Substituting for B and F in terms of A gives: Reflection Coefficient R 9
R and T Coefficients: 10 Transmission Coefficient T Recall that sin(i) =sinh(). We can write k1 and k2 in terms of E and U0. This then gives 11
gives Dividing across by 11a similarly we can find an expression for the Reflection coefficient 12 or rearranging 12a
Graph of Transmission Probability 100 100 L = 0.1 nm U0 = 0.1 eV U0 = 1.0 eV L = 0.5 nm 10-5 10-2 T T L = 1.0 nm U0 = 5.0 eV 10-10 10-4 U0 = 10.0 eV 10-15 10-6 0 0.2 0.4 0.6 0.8 1.0 0 0.2 0.4 0.6 0.8 1.0 E/U0 E/U0 Transmission curves for a barrier of constant height 1.0 eV for a series of different widths L. Transmission curves for a barrier of constant width 1.0 nm with different heights U0
Wavefunction U0 E
Optical Analog evanescent wave If reflection angle is greater than the critical angle then the light ray will be totally internally reflected If second prism is brought close to the first there is a small probability for part of the incident wave to couple through the air gap and emerge in the second prism.
and then Limiting Case Tunnelling through wide barriers: Inside the barrier the wavefunction is proportional to exp(-x) or exp(-x/), where = 1/ is the penetration depth (see Potential step lecture). If L then very little of the wavefunction will survive to x = L. The condition for a ‘wide barrier’ is thus 13 The barrier can be considered to be wide if L is large or if E << U0. Making this approximation we see that so for a thick barrier equation 11 reduces to 14 The probability of tunnelling is then dominated by the exponentially decreasing term.
Example An electron (m = 9.11 10-31kg) encounters a potential barrier of height 0.100eV and width 15nm What is the transmission probability if its energy is (a) 0.040eV and (b) 0.060 eV? We first check to see if the barrier is thick (equation 13). for E = 0.04eV = 18.8 >> 1 thick barrier and for E = 0.060: L/ = 15.5 >> 1 thick barrier we can use equation 14 for the transmission probability Very small in both cases!! Can we observe this in a real stuation
+V0 metal Anode Cathode Tunnelling through potential barrier electrons bound by potential step at surface Field Emission
Scanning Tunnelling Microscope (STM) Si (111) Surface Pt Surface
Sample negative Sample positive Pentacene molecules on Silicon
eV0 EF p-type n-type The Tunnel Diode see http://mxp.physics.umn.edu/s98/projects/menz/poster.htm Conduction Band Conduction Band donors EF Valence Band EF acceptors Valence Band - - + + - - + + - - + + p-type n-type
eV0 + Vext eV0- eVext EF EF - + - + - + p-type n-type - + Vext The Tunnel Diode reversed biased forward biased Conduction Band Valence Band - - - ++ + - - - ++ + - - - ++ + p-type n-type - + Vext
Alpha Decay of Nuclei Alpha particle Uranium 238 Strong Nuclear Force Thorium 234 Electrostatic repulsion To escape the nucleus the -particle must tunnel.