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Numerical Solutions for Schrodinger's EquationParticle in a Box with an Internal BarrierNumerical integration of Schrodinger's equation: Potential energy: Vx ( ) := if ( x ≥ lb) ⋅( x ≤ rb), V 0 , 0⎡⎣⎤ ⎦Given−1 2d⋅ Ψ( x)2⋅μ2dx+ Vx ( ) ⋅Ψ( x)= E⋅Ψ( x)Ψ( 0) = 0 Ψ' ( 0) = 0.1( )Ψ := Odesolve x,x maxNormalize wave function: Ψ( x):=Ψ( x)x⌠ max⎮ Ψ( x) 2 dx⌡0Integration limit: x max ≡ 1 Effective mass: μ ≡ 1 Barrier height: V 0 ≡ 100Barrier boundaries: lb ≡ .45 rb ≡ .55 Enter energy guess: E ≡ 15.45150Ψ( x)Vx ( )10050) 2 Ψ( x0 0.5 1x00 0.5 1xCalculate potential energy:rb⌠⎮PE := V 0 ⋅Ψ( x) 2 dxPE = 4.932⎮⌡lbCalculate kinetic energy: KE := E − PEKE = 10.518Ratio of potential energy to total energy:PEE= 0.319Calculate probability in barrier:PE= 0.049V 0rb⌠⎮ Ψ( x) 2 dx= 0.049⌡lb

1. Find the first four energy levels, sketch Ψ 2 for each state, and fill in the table below. KE, PE and the probability inthe barrier are calculated above.⎛⎜⎜⎜⎜⎜⎝E15.4520.3062.2080.80KE10.51819.82747.74578.968PE4.9320.47314.4551.832P0.0490.00470.1450.018⎞ ⎟⎟⎟⎟⎟⎠2. Interpret the results for energy in light of the fact that a 100 E h (2720 eV) potential barrier of finite thicknessexists in the center of the box.This is an excellent example of quantum mechanical tunneling. For the first four energy states the particle hasprobability of being found in the tunnel in spite of the fact that its energy is less than the barrier energy.3. Explain the obvious bunching of energy states in pair in terms of the impact of the internal barrier. In other wordswhy is the probability of being in the potential barrier larger for the n = 1 and 3 states than it is for the n = 2 and 4states.π 2The PIB energy levels without an internal barrier are: En ( ) := ⋅2 n2The bunching can be seen by comparing the two energy manifolds. The n = 2 and 4 states have nodes at themiddle of the box where the internal barrier is situated. Thus their potential energy does not increase as much asthe n = 1 and 3 states which do not have nodes in the barrier.150PIB Energy Levels150PIB - Internal Barrier LevelsE1 ( )100E2 ( )E3 ( )E4 ( )50Vx ( )15.4510020.3062.2080.80 5000 0.5 1x00 0.5 1x4. Find the ground state energy for particle masses of 0.5 and 1.5. Record your results in the table below and interpretthem.⎛⎜⎜⎜⎜⎝Mass0.51.01.5E23.9515.4511.55T14.41110.5188.684V9.5394.9322.866P0.0950.0490.029⎞ ⎟⎟⎟⎟ ⎠The higher the mass the lower the energy because in quantummechanics in E ~ 1/mass. The greater the mass the lower theprobability that tunneling will occur. This is due to the fact thatthe deBroglie wavelength is inversely proportional to mass.

5. Find the ground state energy for an m = 1 particle for barrier heights 50 and 150 Eh. Record your results in the tablebelow and interpret them.⎛⎜⎜⎜⎜⎝V 050100150E11.9715.4517.32T7.20310.51813.024V4.7674.9324.296P.0950.0490.029⎞⎟⎟⎟⎟⎠The higher the barrier energy the higher the ground-stateenergy and the lower the tunneling probability.6. On the basis of your calculations in this exercise describe quantum mechanical tunneling. In your answer youshould consider the importance of barrier height and particle mass.Tunneling is inversely proportional to mass and barrier height. It also is inversely proportional to barrier width,but we didn't look into this.7. Calculate the energy of the particle using the PIB (no internal barrier) ground-state wave function.V 0 := 100n := 1n 2 ⋅π 22rb⌠⎮+ V 0 ⋅( 2⋅sin( n⋅π⋅x)) 2⎮dx= 24.771⌡lb8. Calculate the energy of the particle using the PIB (no internal barrier) first excited-state wave function.n := 2n 2 ⋅π 22rb⌠⎮+ V 0 ⋅( 2⋅sin( n⋅π⋅x)) 2⎮dx= 20.384⌡lb9. Explain these results.The n = 2 state has a lower energy than the n = 1 because its wave function has a node at the x = 0.5 where thepotential barrier is centered.