Physics 2D Lecture Slides UCSD Physics Vivek Sharma Dec 1 QM in - PowerPoint PPT Presentation

Physics 2D Lecture Slides UCSD Physics Vivek Sharma Dec 1

QM in 3 Dimensions • Learn to extend S. Eq and its solutions from “toy” examples in 1-Dimension (x) → three orthogonal dimensions (r ≡ x,y,z) � ˆ = + + ˆ ˆ r ix jy kz • Then transform the systems – Particle in 1D rigid box � 3D rigid box – 1D Harmonic Oscillator � 3D z Harmonic Oscillator • Keep an eye on the number of different integers needed to specify system 1 � 3 (corresponding to 3 available y degrees of freedom x,y,z) x

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

The Schrodinger Equation in 3 Dimensions: Cartesian Coordinates Time Dependent Schrodinger Eqn: ∂Ψ � 2 ( , , , ) x y z t − ∇ Ψ + Ψ = � 2 ( , , , ) ( , , ) ( , ) .....In 3D x y z t U x y z x t i ∂ 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 2 2 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 2 2 2 2 m m x m y m z y = [K ] + [K ] + [K ] x x x Ψ = Ψ [ ] ( , ) [ ] ( , ) is still the Energy Conservation Eq so H x t 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: =Ψ � � ω Ψ ψ -i t ( , , , ) ( , ) = (r)e x y z t r t 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 ∇ ψ + ψ = ψ 2 in 3D: - ( , , ) ( , , ) ( , , ) ( , , ) TISE x y z U x y z x y z E x y z 2m ψ = ψ ψ ψ x,y,z independent of each other , wr ite ( , , ) ( ) ( ) ( ) x y z x y z 1 2 3 ψ ψ ψ ψ 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) ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ∂ ψ ∂ ψ ∂ ψ � 2 � 2 � 2 2 2 2 1 ( ) 1 ( ) 1 ( ) z x y − + − + − = = ⎜ 1 ⎟ ⎜ 2 ⎟ ⎜ 3 ⎟ E Const ψ ∂ ψ ∂ ψ ∂ 2 2 2 ⎝ 2 ( ) ⎠ ⎝ 2 ( ) ⎠ ⎝ 2 ( ) ⎠ m x x m y y m z z 1 2 3 ⇒ This can only be true if each term is c onstant for all x,y,z ∂ ψ ∂ ψ ∂ ψ � 2 2 � 2 2 � 2 2 ( ) ( ) ( ) x y z − = ψ − = ψ − = ψ 1 2 3 ( ) ; ( ) ; ( ) E x E y E z ∂ ∂ ∂ 1 1 2 2 3 3 2 2 2 2 2 2 m x m y m z + + = 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) ψ ∝ ψ ∝ ψ ∝ So wavefunctions must be like ( ) sin x , ( ) s in y , ( ) s n i x k y k z k z 1 1 2 2 3 3

Particle in 3D Rigid Box : Separation of Orthogonal Variables ψ ∝ ψ ∝ ψ ∝ Wavefunctions are like ( ) sin x , ( ) sin y , ( ) sin x k y k z k z 1 1 2 2 3 3 ψ ⇒ π = 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 π π π ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ � � � = = ⎜ = = ∞ ; ; (n ,n ,n 1,2,3,.. ) ⎜ ⎟ ⎟ ⎜ ⎟ p n p n p n 1 2 3 1 2 3 x ⎝ ⎠ y ⎝ ⎠ z ⎝ ⎠ L L L = Note: by usual Uncertainty Principle argumen t neither of n ,n ,n 0! ( ?) why 1 2 3 π � 2 2 1 + + = + + 2 2 2 2 2 2 Particle Energy E = K+U = K +0 = ( ) ( ) p p p n n n 1 2 3 x y z 2 2 m 2 mL Energy is again quantized and brought to you by integers n ,n ,n (independent) 1 2 3 � ψ and (r)=A sin x sin y sin (A = Overall Normalization Co nstant) k k k z 1 2 3 � � E E - i -i t t Ψ ψ = � � (r,t)= (r) e [ sin x si n ys in ] e A k k k z 1 2 3

Particle in 3D Box :Wave function Normalization Condition � � E E -i -i t t Ψ ψ = � � (r,t)= (r) e [ sin x sin y sin ] e A k k k z 1 2 3 � � E E i i t t Ψ ψ = * * � � (r,t)= (r) e [ sin x s in y sin ] e A k k k z 1 2 3 � � Ψ Ψ * 2 2 2 2 (r,t) (r,t)= [ sin x si n y sin ] A k k 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 sin x dx sin y dy sin z dz = A k k k A ⎜ ⎟⎜ ⎟⎜ ⎟ 1 2 3 2 2 2 ⎝ ⎠⎝ ⎠⎝ ⎠ x=0 y= 0 z =0 3 3 ⎡ ⎤ ⎡ ⎤ E � 2 2 2 2 - i t ⇒ = ⎢ ⎥ Ψ � an d ( r,t)= [ s i n x s i n y sin ] e A k k k z ⎢ ⎥ 1 2 3 ⎣ ⎦ ⎣ ⎦ L L

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

Degenerate States π 2 � 2 6 = = E = E E 211 121 112 2 2 mL Ground State E 111 E 121 E 112 E 211 ψ ψ z z y y x x

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 ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ π π π 2 2 2 2 2 2 n n n = + + ⎜ ⎟ ⎜ 1 ⎟ 2 3 ⎜ ⎟ E ⎜ ⎟ 2 2 2 ⎝ 2 ⎠ 2 ⎝ 2 ⎠ mL mL mL ⎝ ⎠ x y z

The Hydrogen Atom In Its Full Quantum Mechanical Glory 1 1 ∝ = ⇒ ( ) M ore compli cated form of U than bo x U r + + 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 → θ φ Independent Cartesian (x,y,z) Inde. Spherical Polar (r, , ) ∂ ∂ ∂ 2 2 2 ∇ = + + ∂ 2 Instead of writing Laplacian , write ∂ ∂ 2 2 2 x y z ∂ ∂ ∂ ∂ ∂ ⎛ ⎞ ⎛ ⎞ 2 1 1 1 ∇ + θ + 2 2 = ⎜ ⎟ ⎜ s in ⎟ r ∂ ∂ θ ∂ θ ∂ θ θ ∂ φ 2 ⎝ ⎠ 2 ⎝ ⎠ 2 2 2 r si n sin r r r r ψ ψ θ φ TISE for (x,y,z)= (r, , ) become s ∂ ∂ ψ θ φ ∂ ∂ ψ θ φ ⎛ ⎞ ⎛ ⎞ 1 (r, , ) 1 (r, , ) + θ + 2 2 sin kZe ⎜ ⎟ ⎜ ⎟ r = ( ) ∂ ∂ θ ∂ θ ∂ θ U r 2 ⎝ ⎠ 2 ⎝ ⎠ r s in r r r r !!!! fun!!! ∂ ψ θ φ 2 1 (r, , ) 2m ψ θ φ + (E-U(r)) (r, , ) =0 θ ∂ φ � 2 2 2 2 si n r

Spherical Polar Coordinate System Vol ume Element dV = θ φ θ ( sin ) ( )( ) dV r d rd dr θ θ φ 2 = r si n drd d