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PH 401 Dr. Cecilia Vogel. Review Outline  Time evolution  Finding Stationary states  barriers  Eigenvalues and physical values  Energy Operator 

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Presentation on theme: "PH 401 Dr. Cecilia Vogel. Review Outline  Time evolution  Finding Stationary states  barriers  Eigenvalues and physical values  Energy Operator "— Presentation transcript:

1 PH 401 Dr. Cecilia Vogel

2 Review Outline  Time evolution  Finding Stationary states  barriers  Eigenvalues and physical values  Energy Operator  Stationary States

3 Time Independent Schrödinger Equation   Solutions are stationary states, energy eigenstates  Any state can be written in terms of stationary states:   (x,0)=  a n  n (x)   (x,t)=  a n  n (x)e -iE n t/   Predict the future!  But first need all the stationary states…

4 Two dichotomies  A region of space can be classically allowed (CA) or classically forbidden (CF)  demo  CA means E>V(x) in that region  CF means E<V(x)in that region  A particle can be bound or unbound  If the regions including +infinity are CF,  particle is bound.  If the regions including +infinity are CA,  particle is unbound.

5 Step barrier   particle with energy E>Vo incident from the left  Solutions to TISE:

6 Step barrier continuity  A 1 + B 1 = A 2  k 1 A 1 - k 1 B 1 =k 2 A 2  So…  A 1 =(1+k 2 /k 1 )(A 2 /2)  B 1 =(1-k 2 /k 1 ) (A 2 /2)  One unknown is undetermined  would be found by normalization  if it were normalizable!

7 Step barrier transmission and reflection  A2 is the amplitude for being transmitted into region 2, compare to amplitude in region 1:  T=|A 2 /A 1 | 2  T=[2k 1 /(k 1 +k 2 )] 2  T=1-R  R=[(k 1 -k 2 )/(k 1 +k 2 )] 2  R=[(sqE-sq(E-V))/(sqE+sq(E-V))] 2  R is not zero. The particle might be REFLECTED! By a CA barrier!! What??

8 Tunneling   particle with energy E<Vo incident from the left  Solutions to TISE:

9 Tunneling continuity  A 1 +B 1 =A 2 +B 2  ik 1 A 1 - ik 1 B 1 = K 2 A 2 - K 2 B 2  A 2 e K2a +B 2 e -K2a = A 3 e ik1a  K 2 A 2 e K2a - K 2 B 2 e -K2a = ik 1 A 3 e ik1a

10 Tunneling probability  Tunneling into region 3:  T=|A3/A1| 2  T=[1+(V 2 /4E(V-E))sinh 2 (K 2 a)] -1   If K 2 a>>1, then sinh(K 2 a) approx e K2a


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