3. Although planar waves are not well normalized, one can use them to demonstrate the tunneling...

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3. Although planar waves are not well normalized, one can use them to demonstrate the tunneling phenomenon (a well normalized solution would be a wave packet of this planar waves, namely (x, 0) = dk (k) exp(ikr).) 2T (9) Consider a stationary solution of a free particle with finite energy E and forget for the time being the question of normalization. The momentum of the particle will be given by |hk| E 2m and = with hw = E. (x,t) = A exp(k-wt) +B exp(-kz-wt) (10) Observe that there are two solutions, one with positive (incoming wave) and the other with negative (reflected wave) values of the momentum in the direction. In the previous problems they combined to give sine and cosine functions due to the boundary conditions. Now, imagine that we consider that this particle goes through a potential V(r) = 0 for |x| > a/2, V(x)=V for |x| < a/2. (11) Since we want to consider the transmission of the particle at the right of the potential barrier, consider the case in which at the right of this finite potential barrier, the wave function is described by a particle moving freely in the positive direction of r, (x,t) = C'exp(kr-wt) , for >a/2 (12) while for r < -a/2 we have the function given in Eq. (10). The ratio |B/A2 may be consider as a reflection probability, while |C/A2 as a transmission probability. The solutions for -a/2xa/2, instead, are given by (x,t) = F exp(-wt) +G exp(-kaz-wt) with k = (E Vo)2m/h. Observe that w is unchanged. (13) Using continuity of the wave function and its derivatives at r = ta/2, calculate the reflection and transmission probabilities. Consider both the case of E > Vo and E < V and try to interpret the results. What happens when Vo ? 3. Although planar waves are not well normalized, one can use them to demonstrate the tunneling phenomenon (a well normalized solution would be a wave packet of this planar waves, namely (x, 0) = dk (k) exp(ikr).) 2T (9) Consider a stationary solution of a free particle with finite energy E and forget for the time being the question of normalization. The momentum of the particle will be given by |hk| E 2m and = with hw = E. (x,t) = A exp(k-wt) +B exp(-kz-wt) (10) Observe that there are two solutions, one with positive (incoming wave) and the other with negative (reflected wave) values of the momentum in the direction. In the previous problems they combined to give sine and cosine functions due to the boundary conditions. Now, imagine that we consider that this particle goes through a potential V(r) = 0 for |x| > a/2, V(x)=V for |x| < a/2. (11) Since we want to consider the transmission of the particle at the right of the potential barrier, consider the case in which at the right of this finite potential barrier, the wave function is described by a particle moving freely in the positive direction of r, (x,t) = C'exp(kr-wt) , for >a/2 (12) while for r < -a/2 we have the function given in Eq. (10). The ratio |B/A2 may be consider as a reflection probability, while |C/A2 as a transmission probability. The solutions for -a/2xa/2, instead, are given by (x,t) = F exp(-wt) +G exp(-kaz-wt) with k = (E Vo)2m/h. Observe that w is unchanged. (13) Using continuity of the wave function and its derivatives at r = ta/2, calculate the reflection and transmission probabilities. Consider both the case of E > Vo and E < V and try to interpret the results. What happens when Vo ?