Physics 2D Lecture Slides Lecture 27: March 2nd 2005 Vivek Sharma - PDF document

Physics 2D Lecture Slides Lecture 27: March 2nd 2005 Vivek Sharma UCSD Physics Quantum Mechanics In 3D: Particle in 3D Box z Extension of a Particle In a Box with rigid walls 1D → 3D ⇒ Box with Rigid Walls (U= ∞ ) in X,Y,Z dimensions z=L U(r)=0 for (0<x,y,z,<L) Ask same questions: • Location of particle in 3d Box • Momentum • Kinetic Energy, Total Energy • Expectation values in 3D y y=0 y=L To find the Wavefunction and various expectation values, we must first set up the appropriate TDSE & TISE x 1

The Schrodinger Equation in 3 Dimensions: Cartesian Coordinates Time Dependent Schrodinger Eqn: �� � 2 ( , , , ) x y z t � � � 2 + � = ( , , , ) x y z t U x y z ( , , ) ( , ) x t i � .....In 3D � 2 m t � 2 � 2 � 2 � = + + 2 � 2 � 2 � 2 x y z z � � � � � � � � � = � 2 � 2 2 � 2 2 � 2 2 � � = 2 � + � + � So [ ] K � � � � � � � � � 2 m 2 m x 2 2 m y 2 2 m z 2 � � � � � � y = [K ] + [K ] + [K ] x x x � = � so [ H ] ( , ) x t [ ] ( , ) is still the Energy Conservation Eq E x t x Stationary states are those for which all proba bilities are constant in time and are given by the solution of the TDSE in seperable form: = � � � � � -i t � ( x y z t , , , ) ( , ) r t = (r)e This statement is simply an ext ension of what we derive d in case of 1D time-independent potential Particle in 3D Rigid Box : Separation of Orthogonal Spatial (x,y,z) Variables � 2 � 2 � + � = � TISE in 3D: - ( , , ) x y z U x y z ( , , ) ( , , x y z ) E ( , x y , ) z 2m � = � � � x,y,z independent of each other , wr ite ( , , ) x y z ( ) x ( ) y ( ) z 1 2 3 � � � � and substitute in the master TISE, after dividing thruout by = ( ) x ( ) y ( z ) 1 2 3 � and n oting that U(r)=0 fo r (0<x,y,z,<L) � � � � � � 2 � 2 � 2 � 2 � 2 � 2 � � 1 ( ) x � 1 ( ) y � 1 ( ) z � + � + � = = 1 2 3 E Const � � � � � � � � � � � � 2 m ( ) x x 2 2 m ( y ) y 2 2 m ( ) z z 2 � � � � � � 1 2 3 � This can only be true if each term is c onstant for all x,y,z 2 � 2 � 2 � 2 � 2 � 2 � � ( x ) � ( ) y � ( ) z � = � � = � � = � 1 E ( ) ; x 2 E ( ) ; y 3 E ( ) z � 1 1 � 2 2 � 3 3 2 m x 2 2 m y 2 2 m z 2 + + = With E E E E=Constan t (Total Energy of 3D system) 1 2 3 Each term looks like particle in 1D box (just a different dimension) � � � � � � So wavefunctions must be like ( ) x sin k x , ( ) y s in k y , ( z ) s n i k z 1 1 2 2 3 3 2

Particle in 3D Rigid Box : Separation of Orthogonal Variables � � � � � � Wavefunctions are like ( ) x sin k x , ( ) y sin k y , ( ) z sin k z 1 1 2 2 3 3 � � � = Continuity Conditions for and its fi rst spatial derivative s n k L i i i � � � Leads to usual Quantization of Linear Momentum p= k .....in 3D � � � � � � � � � � � � = = � = = � p n ; p n ; p n (n ,n ,n 1,2,3,.. ) � � � � � x 1 y 2 z 3 1 2 3 � L � � L � � L � = Note: by usual Uncertainty Principle argumen t neither of n ,n ,n 0! ( why ?) 1 2 3 � 2 2 1 � 2 + 2 + 2 = 2 + 2 + 2 Particle Energy E = K+U = K +0 = ( p p p ) ( n n n ) x y z 1 2 3 2 2 m 2 mL Energy is again quantized and brought to you by integers n ,n ,n (independent) 1 2 3 � � and (r)=A sin k x sin k y sin k z (A = Overall Normalization Co nstant) 1 2 3 E E � � - i t -i t � � = (r,t)= (r) e A [ sin k x si n k ys in k z ] e � � 1 2 3 Particle in 3D Box :Wave function Normalization Condition E E � � -i t -i t � � = (r,t)= (r) e A [ sin k x sin k y sin k z ] e � � 1 2 3 E E � � i t i t � � = * * (r,t)= (r) e � A [ sin k x s in k y sin k z ] e � 1 2 3 � � � � * 2 2 2 2 (r,t) (r,t)= A [ sin k x si n k y sin k z ] 1 2 3 ��� � Normalization Co ndition : 1 = P(r)dx dyd z x,y, z � �� �� � L L L L L L � � � = 2 2 2 2 2 1 A sin k x dx sin k y dy sin k z dz = A � �� �� � � �� �� � 1 2 3 2 2 2 � �� �� � x=0 y= 0 z =0 3 3 E � 2 � � � 2 � 2 2 - i t � = � � A an d ( r,t)= [ s i n k x s i n k y sin k z ] e � � � � 1 2 3 � L � � L � 3

Particle in 3D Box : Energy Spectrum & Degeneracy � 2 � 2 = 2 + 2 + 2 = � � E ( n n n ); n 1,2,3... , n 0 n ,n ,n 1 2 3 i i 2 2 mL 1 2 3 � 2 2 3 � = Ground State Energy E 111 2 mL 2 � 2 2 6 � � = = Next level 3 Ex cited states E = E E 211 121 112 2 2 mL � � � Different configurations of (r)= (x,y,z) have s ame energy d egeneracy z z=L y y=L x=L x Degenerate States � 6 2 � 2 = = E = E E 211 121 112 2 mL 2 Ground State E 111 E 121 E 112 E 211 ψ ψ z z y y x x 4

Probability Density Functions for Particle in 3D Box Same Energy  Degenerate States Cant tell by measuring energy if particle is in 211, 121, 112 quantum State Source of Degeneracy: How to “Lift” Degeneracy • Degeneracy came from the threefold symmetry of a Energy CUBICAL Box (L x = L y = L z =L) • To Lift (remove) degeneracy  change each dimension such that CUBICAL box  Rectangular Box • (L x ≠ L y ≠ L z ) • Then � � � � � � 2 � 2 2 � 2 2 � 2 n n n = + + E 1 � 2 � 3 � � � � � � 2 2 2 2 mL 2 mL 2 mL � � � � � � x y z 5

The Coulomb Attractive Potential That Binds the electron and Nucleus (charge +Ze) into a Hydrogenic atom -e m e F V +e r kZe 2 = U r ( ) r The Hydrogen Atom In Its Full Quantum Mechanical Glory 1 1 � = � U r ( ) M ore compli cated form of U than bo x r 2 + 2 + 2 x y z By example of particle in 3D box, need to use seperation of r variables(x,y,z) to derive 3 in dependent d iffer ential. eq ns . This approach will get very ugly since we have a "conjoined triplet" To simplify the situation, use appropriate variables � � � Independent Cartesian (x,y,z) Inde. Spherical Polar (r, , ) � 2 � 2 � 2 � = 2 + + � Instead of writing Laplacian , write � � x 2 y 2 z 2 � � � � � 1 � � 1 � � 1 2 � + � + 2 = r 2 s in � � � � 2 � � 2 � � � � � 2 2 � � 2 � r r � r � r si n � � r sin � � � � TISE for (x,y,z)= (r, , ) become s � � � � � � � � � � kZe 2 � � � � 1 (r, , ) 1 (r, , ) = U r ( ) 2 + � + r sin � � � � r � � � � � � � r 2 r � r � r 2 s in � � !!!! fun!!! � � � � 2 1 (r, , ) 2m � � � + (E-U(r)) (r, , ) =0 � � � r 2 si n 2 2 � 2 6