The Schrodinger Equation Reading:OGC:(4.6)and (5.1)
The Schrödinger Equation Reading: OGC: (4.6) and (5.1) III-1
The Person Behind The Science Erwin Schrodinger 1887-1961 Highlights Born and educated in Vienna Received Nobel Prize in Physics with Paul Dirac (1933) Moments in a Life In 1927 Schrodinger moved to University of Berlin as Planck's successor Develops his wave equation in 1926 -2
The Person Behind The Science Erwin Schrödinger 1887-1961 In 1927 Schrödinger moved to University of Berlin as Planck's successor Develops his wave equation in 1926 Moments in a Life Highlights Born and educated in Vienna Received Nobel Prize in Physics with Paul Dirac (1933) III-2
The Schrodinger Equation HΨ=EΨ .H is the Hamiltonian Operator;you can't "cancel"the y 一“Cancelling”the平is like“cancelling”the x in f(x)=mx. You just can't do it. Our goal is to operate on (using the H Operator)and get an energy (E)multiplied by the same y. -3
The Schrödinger Equation HΨ = EΨ H is the Hamiltonian Operator; you can’t “cancel” the Ψ — “Cancelling” the Ψ is like “cancelling” the x in f(x) = mx. You just can’t do it. Our goal is to operate on Ψ (using the H Operator) and get an energy (E) multiplied by the same Ψ. III-3
Deriving the Schrodinger Equation .This equation describes the energy of an electron: Total Energy Kinetic Energy Potential Energy E=KE+PE Start with this classical equation. Use classical and quantum mechanical relationships to find the Hamiltonian Operator(H). Find values of that fit the Schrodinger Equation: HΨ=EΨ. -4
Deriving the Schrödinger Equation Total Energy = Kinetic Energy + Potential Energy E = KE + PE This equation describes the energy of an electron: Start with this classical equation. Use classical and quantum mechanical relationships to find the Hamiltonian Operator (H). Find values of Ψ that fit the Schrödinger Equation: HΨ = EΨ. III-4
Describing Kinetic Energy(KE) Classically: Quantum Mechanically: -ihaΨ KE-1mv2 Px= 2πdx 2 Iffx,y,z）=x2+y3+z4, p=mv,so KE= p2 2m then a证=2xand 0 ay 3y2 Combining Equations: KEx p_82y-h2 (close enough for now!) 2m 0x2 8mn2 ,2如 KE-KEx+KEy+KEz-8m2 x2+y2 0z2 g产2 )(V is a form of mathematical shorthand notation -5
∂f ∂ x ∂ f ∂ y — If f(x,y,z) = x2+y3+z4 , Describing Kinetic Energy (KE) Classically: Quantum Mechanically: then = 2x and = 3y2 px = −ih 2π ∂Ψ KE = ∂x 1 2 mv 2 p = mv , so KE = p 2 2 m Combining Equations: (close enough for now!) = - h 2 8 m π 2 ∇ 2 ( Ψ ) ( ∇ is a form of mathematical shorthand notation ) KE x = p x 2 2 m = ∂ 2 Ψ ∂ x 2 -h 2 8 m π 2 So KE =KE x + KE y +KE z = -h 2 8 m π 2 ∂ 2 Ψ ∂ x 2 + ∂ 2 Ψ ∂ y 2 + ∂ 2 Ψ ∂ z 2 III-5
Including Potential Energy (PE) Classically,PE=-e2 Quantum Mechanically,PE=_-e2 4eo「 4J8or We will use the quantum mechanical definition of PE. ·Fm KE--g(2n） So our final equation is: w）-型 Now,we must find the special y's that are solutions to this equation and also satisfy the boundary conditions. Π-6
Classically, PE = -e2 4πεo r Quantum Mechanically, PE = -e2 Ψ • So our final equation is: • Now, we must find the special Ψ’s that are solutions to this equation and also satisfy the boundary conditions. Including Potential Energy (PE) • We will use the quantum mechanical definition of PE. • From the previous slide, KE = -h 2 8 m π 2 ∇ 2 ( Ψ ) 4πεo r 2 -h 8 m π 2 ∇ 2 ( Ψ ) - = E Ψ e 2 Ψ 4πεo r III-6
A Solution to the Schrodinger Equation We can prove that (r)=Ber is a solution,by plugging it into Schrodinger's equation (ignoring angular derivatives): -h2a2平，2aΨ 8mr2ar2+rar厂4e。r eEW EBe or -h2）1 .multiplying through by gives us: 8mat :e2 2 -8m = 4eo「 h2 EBe ar .Now we take the derivatives: aΨ -Be-ar=-aBe-ar 2 102 or or dr2Bear-aBe-wr Ⅲ-7
A Solution to the Schrödinger Equation We can prove that Ψ(r) = Be-αr is a solution, by plugging it into Schrödinger’s equation (ignoring angular derivatives): Now we take the derivatives: multiplying through by −h2 8mπ2 −1 gives us: ∂2 Ψ ∂r 2 = ∂2 ∂r 2 Be−αr = α2 Be−αr ∂Ψ ∂r = ∂ ∂r Be−αr = −αBe−αr -h 2 8 m π 2 ∂ 2 Ψ ∂ r 2 + 2 r ∂ Ψ ∂ r - = E Ψ = EBe e -α r 2 Ψ 4πεo r ∂ 2 Ψ ∂ r 2 + 2 r ∂ Ψ ∂ r + = -8 m π 2 h 2 EBe 8 m π -α r 2 h 2 e 2 Ψ 4πεo r III-7
.Plug the derivatives into Schrodinger's Equation: aΨ =-aBe-ar 2Ψ ar dr2 -a"Be-ar 8mn2 e2 2 a"Be ar-2aBe ar+ h2 Be ar- -8mat h2 EBe-ar r 4eo .Cancel Beur:a2-2a+ 8m2 e2 2 -8mat E h2 h2 .Combining similar terms gives: 2 O+ h2 -2d =0 r These sum to zero for all r only if both terms are zero I-8
Plug the derivatives into Schrödinger’s Equation: Cancel Be-αr : These sum to zero for all r only if both terms are zero Combining similar terms gives: ∂2Ψ ∂r 2 = α2 Be−αr ∂Ψ ∂r = −αBe−αr α 2 Be -α r - 2 r α Be -α r + 8 m π 2 h 2 e 2 Be -α r = -8 m π 2 h 2 EBe -α r 4πεo r α 2 - 2 r α + 8 m π 2 h 2 = -8 m π 2 h 2 E e 2 4πεo r α 2 + 8 m π 2 h 2 E + 1 r - 2 α = 0 2 e 2 m π h 2 εo III-8
.Set both terms to zero: 2 a2+ 8m E 1 2a =0 h2 2 a2+ 8m E=0 2e2mx 2- 2a=0 h2 2 2 8m emat 1 E 0= h2 -a2 nto ao -h2a2 E 8ma2 e'mr 1 .Now set a h2to ao E -e2 -h2 e'm22 -e"m E- 8neoao 8m2 8h2e。2 -9
Set both terms to zero: α2 + 8mπ2 h2 E = 0 E = −h2 α2 8mπ2 8mπ2 h2 E = −α2 h 2 α 2 + 8 m π 2 h 2 E + 1 r 2 e 2 m π - 2 α = 0 εo h 2 2 e 2 m π - 2 α = 0 εo α = e 2 m π h 2 = 1 a εo 0 • Now set α = e 2 m π h 2 = 1 a 0 εo E = -h 2 8 m π 2 e 4 m 2 π 2 h 4 = 4 εo 2 - e m h 2 εo 2 E 8 = -e 2 a 8πεo 0 III-9
So,We Find One Solution (r)=Be-r is a solution to the Schrodinger Equation: =1 -e2 and E= ao 8ne.ao Because the probability of finding the e must be 1.0 if we B look over all space: Other functions of the form (Ar2-Br-C)e-or also work: 1e2 .Enwhere the order of the polynomial is(n+1). Additional work is required to determine the time dependence of any given solution to the Schrodinger equation -10
So, We Find One Solution Ψ(r) = Be-αr is a solution to the Schrödinger Equation: α = 1 a0 and Because the probability of finding the e- must be 1.0 if we look over all space: B = 1 πa0 3 Other functions of the form (Ar2–Br–C)e-αr also work: where the order of the polynomial is (n+1). Additional work is required to determine the time dependence of any given solution to the Schrödinger equation E = -e 2 a 8πεo 0 E = 1 n 2 -e 2 a 8πεo 0 III-10