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Infinite (and finite) square well potentials Announcements: • There will be a homework set put up this afternoon and due Wednesday after break. • No class on Friday. • HW1-8 and Exam 1 grades are now in CULearn. Please check to make sure they are correct. In case you are curious, the average scores (of those turned in) are below: EX1 HW1 HW2 HW3 HW4 HW5 HW6 HW7 HW8 Avg: 75.0 44.2 43.7 42.2 35.6 32.7 39.2 39.2 37.7 EX1 is out of 100, HW4 is out of 40 and the rest are out of 50 http://www.colorado.edu/physics/phys2170/ Physics 2170 – Spring 2009 1 Some wave function rules y(x) and dy(x)/dx must be continuous These requirements are used to match boundary conditions. |y(x)|2 y ( x) dx 1 2 must be properly normalized This is necessary to be able to interpret |y(x)|2 as the probability density y ( x) 0 as x This is required to be able to normalize y(x) http://www.colorado.edu/physics/phys2170/ Physics 2170 – Spring 2009 2 Infinite square well (particle in a box) solution y ( x) B sin( kx) 2 2 2 2 2 n k n k and which gives us an energy of E n 2 a After applying boundary conditions we found 2m 2ma Things to notice: Energy Energies are quantized. Minimum energy E1 is not zero. Consistent with uncertainty principle. x is between 0 and a so Dx~a/2. Since DxDp≥ħ/2, must be uncertainty in p. But if E=0 then p=0 so Dp=0, violating the uncertainty principle. When a is large, energy levels get closer so energy becomes more like continuum (like classical result). http://www.colorado.edu/physics/phys2170/ 16E1 n=4 9E1 n=3 4E1 n=2 E1 0 n=1 Physics 2170 – Spring 2009 V=0 a 3 x Finishing the infinite square well We need to normalize y(x). That is, make sure that y ( x) 2 dx 1 For the region x<0 and x>a the probability |y(x)|2 is zero so we just need to ensure that a y (x) dx 1 2 0 Putting in y ( x) B sin( kx) and doing the integral we find B 2/a Therefore y ( x) 2/a sin( kx) But we also know k n / a So we can write the solution as y ( x) 2 sin nx a a Adding in the time ( x, t) y ( x) (t ) 2 sin nx e iEt/ a a dependence: Is it still normalized? *(t) (t) is not a function of x so can pull out of the integral and find *(t)(t) eiEt/eiEt/ 1 So the time dependent piece is already normalized. http://www.colorado.edu/physics/phys2170/ Physics 2170 – Spring 2009 4 Clicker question 1 2 2 En n2 2 n2 E1 2ma Set frequency to DA 2/a 2/a a/3 2a/3 a 2/a a/3 2a/3 a a/ 2 a 2/a a/ 2 a 2/a 2/a a a a Am I more likely to find the particle close to a/2 in the n=2 or n=3 state? A. n=2 state B. n=3 state C. No difference For n=2 state: |y 2 (a/2) |2 0 For n=3 state: |y 3(a/2) |2 2/a http://www.colorado.edu/physics/phys2170/ Physics 2170 – Spring 2009 5 Be careful to understand everything we plot… Energy V(x) Total energy E (n=3) E (n=2) V=0 eV E (n=1) 0 a y(x) 0 Sometimes plot three things on the same graph! x Potential Energy V(x) Total Energy E Wave Function y(x) http://phet.colorado.edu/simulations/sims.php?sim=Quantum_Bound_States http://www.colorado.edu/physics/phys2170/ Physics 2170 – Spring 2009 6 Comparing classical and quantum results ( x, t ) 2 sin nx e iEnt / a a 2 2 n2 E En n 1 2 2 2ma Note: time dependence depends on energy a Classical physics Particle can have any energy Lowest kinetic energy is 0 (particle is at rest) Quantum physics Particle can only have particular energies (quantized) Lowest energy state in box has kinetic energy (zero point motion) How small would a box need to be for E1 to be 4.7 eV? (6.58 1016 eVs)(3 108 m/s ) n a 2.8 1010 m 2mEn 2(0.511106 eV)(4.7 eV) About the size of an atom so our model wouldn’t work anyway http://www.colorado.edu/physics/phys2170/ Physics 2170 – Spring 2009 7 Reading Quiz 1 Set frequency to DA Please answer this question on your own. No discussion until after. Q. Classically forbidden regions are where… A. a particle’s total energy is less than its kinetic energy B. a particle’s total energy is greater than its kinetic energy C. a particle’s total energy is less than its potential energy D. a particle’s total energy is greater than its potential energy E. None of the above. This would imply that the kinetic energy is negative which is forbidden (at least classically). http://www.colorado.edu/physics/phys2170/ Physics 2170 – Spring 2009 8 Motivation for a finite square well Infinite square well approximation assumes that electrons never get V(x) out of the well so V(0)=V(a)=∞ and y(0)=y(a)=0. Energy V(x) 0 A more accurate potential function V(x) gives a chance of the electron being outside 0 a x What if the particle energy is higher? E What about two wires very close together? These scenarios require the more accurate potential http://www.colorado.edu/physics/phys2170/ Physics 2170 – Spring 2009 9 Need to solve the finite square well Energy V(x) wire 0 0 Need to solve TISE: 4.7 eV 2 d 2y ( x) V ( x)y ( x) Ey ( x) 2 2m dx x < 0: V(x) = 4.7 eV Work function x > a: V(x) = 4.7 eV 0 < x < a: V(x) =0 x a This will be used to understand quantum tunneling which provides the basis for understanding Radioactive decay Scanning Tunneling Microscope which is used to study surfaces Binding of molecules http://www.colorado.edu/physics/phys2170/ Physics 2170 – Spring 2009 10 Analyzing the finite square well Energy V(x) 2 d 2y ( x) 4.7 eV TISE: 2m dx 2 V ( x)y ( x) Ey ( x) wire We rewrite the TISE as Eparticle 0 0 Region I a Region II d 2y ( x) 2m 2 V ( x) E y ( x) 2 dx x Region III Consider three regions In Region II: total energy E > potential energy V so V − E < 0 2 2 m d 2 V ( x ) E Replace 2 with −k to get y (2x) k 2 y ( x) (k is real) dx Same as infinite square well so sin(kx) and cos(kx) or eikx and e-ikx Region II: y II (x) A cos(kx) B sin( kx) http://www.colorado.edu/physics/phys2170/ Physics 2170 – Spring 2009 11