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Particle in the Box Fall 2014 Chem 356: Introductory Quantum Mechanics Chapter 3 – Schrodinger Equation, Particle in a Box .................................................................... 34 Introduction to the Schrodinger Equation ................................................................................ 34 Linear Operators ....................................................................................................................... 36 Quantization of energy ............................................................................................................. 39 Interpretation of Wave Function .............................................................................................. 40 DeterMination of Constant C .................................................................................................... 42 Useful integrals for particle in the box ..................................................................................... 44 DeMonstration of Uncertainty Principle ................................................................................... 44 Particle in a 3 diMensional box ................................................................................................. 46 Chapter 3 – Schrodinger Equation, Particle in a Box Introduction to the Schrodinger Equation h De Broglie suggested one can associate a wave with a particle and take p = λ 2π h eikx λ = p = k = !k k 2π Generalization to 3 diMensional wave ! ! ! ! eik i x p k = " In chapter 2 we saw that waves in general satisfy a wave equation. Try to postulate a wave equation for “electron-waves” (a guess) Provide soMe rational for Schrodinger equation: ∂2 u 1 ∂2 u Wave equation 2 = 2 2 ∂x V ∂t ν Choose solution with particular ω = 2π u(x,t) =ψ (x)cos(ωt) 34 Fall 2014 Chem 356: Introductory Quantum Mechanics d 2ψ ω 2 2 + 2 ψ ()x = 0 dx V ω 2πν 2π ω = 2πv νλ = V , = = ν (nu) frequency ; V V νλ λ velocity d 2ψ 4π 2 2 + 2 ψ (x) = 0 dx λ 2 h 4π 2 4π 2 ⎛ p⎞ λ = = ⋅ p2 = p 2 h2 ⎝⎜ ⎠⎟ λ ! ∂2ψ !2 + p2ψ (x) = 0 dx2 2 Now substitute p : Let V = V(x) indicate potential: p2 +V = E 2m !2 ∂2ψ p2 2 + ψ (x) = 0 2m ∂x 2m !2 ∂2ψ 2 + (E −V )ψ (x) = 0 2m ∂x !2 ∂2ψ Or Eψ (x) = − 2 +V (x)ψ (x) 2m ∂x ˆ Hψ (x) We obtain a differential equation for function ψ ()x ˆ Hψ (x) = Eψ (x) E is a constant, the energy Hˆ is “operator” that acts on a function. SuMMarizing: 2 2 2 ∂ ψ 1) p ψ (x) = −! 2 (using de Broglie + classical wave equation) ∂x Chapter 3 – Schrodinger Equation, Particle in a Box 35 Fall 2014 Chem 356: Introductory Quantum Mechanics 2) Substitute pmEVx2 =2(− ()) h 2 ∂2ψ −+Vx()ψψ () x = E () x 2m ∂x2 Hxˆψψ()= Ex () pˆ 2 Hˆ = +Vˆ(x) ‘energy operator’ (see later) 2m We need to discuss 2 MatheMatical iteMs a) Operators pˆ , Hˆ , pˆ 2 ….? b) Eigenvalue equations ˆ ! HEψψ= E, p : numbers ! pˆ p ψ = ψ Operators will be indicated by “^” hat or carot Linear Operators (in 1 diMension first) Afˆ () x= g () x Acting with an operator on a function yields a new function. Aˆ fx() Afˆ () x= g () x d 2 2x 0 dx2 2 ⎛⎞dd 3 23 ⎜⎟2 ++23 x 66xx++ 3 x ⎝⎠dx dx d d x x2 x (x2 ) = x ⋅2x = 2x2 dx dx d d d x x2 (x ⋅ x2 ) = (x3) = 3x2 dx dx dx d ikx ikx −i! e !ke dx ⎛ !2 d 2 ⎞ ⎛ !2k 2 ⎞ − +V (x) cos(kx ) +V (x) cos(kx) ⎝⎜ 2m dx2 ⎠⎟ ⎝⎜ 2m ⎠⎟ Chapter 3 – Schrodinger Equation, Particle in a Box 36 Fall 2014 Chem 356: Introductory Quantum Mechanics The operators we consider are linear operators: ˆ Acf(()11 x+ c 2 f 2 ()) x = c Afˆˆ() x c Af () x 11( ) + 2( 2) Where c1 , c2 are (coMplex) constants Example of operator that is not linear: SQR(fx ( )) (())fx2 SQR(()fx+= gx ()) (()) fx22 + (()) gx + 2()() fxgx =SQR(fx ( ))++ SQR( gx ( )) 2 fxgx ( ) ( ) Not linear therefore We can act with operators in sequence ABfˆˆˆˆ() x= A ( Bf ()) x In general: ABfˆˆˆˆ() x≠ BAf () x d Example Axˆ = , Bˆ = dx ⎛⎞ddf ⎜⎟xfxx()= ⎝⎠dx dx ⎛⎞dd df ⎜⎟xfx()==+ ( xfx ()) fx () x ⎝⎠dx dx dx If ABfˆˆˆˆ() x= BAf () x , for any fx()we write ABˆˆˆˆ− BA =0 [,]ABˆ ˆ = 0 Aˆ and Bˆ coMMute, the order does not Matter. This will play an iMportant role later on. Eigenvalue equations (by example) ˆ Aψ (x) = aψ (x) Acting with Aˆ on a function yields the same function Multiplied by a constant. Example: ∂ −i! eikx = (−i!)(ik)eikx x ∂ ikx = !ke Chapter 3 – Schrodinger Equation, Particle in a Box 37 Fall 2014 Chem 356: Introductory Quantum Mechanics ⎛⎞h 2π ikx =⎜⎟⋅e ⎝⎠2πλ h interpretation: ==epeikx ikx λ 2π ix eeikx = λ periodic with period λ ∂ We say pˆ = −i! x x ∂ ikx ikx peˆ xx= p e Number ikx The wave function e is an eigenfunction of operator pˆ x d pˆ x = −i! , with eigenvalue dx h !k = λ pxˆψψ()= px () A particle with definite MoMentuM px is described by eigenfunction of operator pˆ x Consider kinetic energy operator 2 ⎛ ∂ ⎞ −i! pˆ 2 ⎝⎜ ∂x⎠⎟ !2 d 2 = = − 2 2m 2m 2m dx Eigenfunctions of Kinetic energy: 2 2 2 2 ! d ax ! a − 2 e = − < 0 !! (if a is real) 2m dx 2m Not physical 2 2 2 ! d ! 2 − 2 sin(ax) = + a sin(ax) 2m dx 2m Constant Eigenvalue Or 2 2 2 ! d ! 2 − 2 cos(ax) = a cos(ax) 2m dx 2m Also 2 iax ! 2 iax e a e 2m Chapter 3 – Schrodinger Equation, Particle in a Box 38 Fall 2014 Chem 356: Introductory Quantum Mechanics Or Hamiltonian operator: 2 2 2 ˆ pˆ ! ∂ HVx=+() = − 2 +V (x) 2m 2m ∂x Hxˆψψ()= Ex () : particle described by eigenfunction ψ ()x has the definite energy E , (to be discussed in More detail in chapter 4) Quantization of energy We saw that a fundamental feature of ‘new’ quantuM Mechanics was that energy cannot take on any value, but only certain values. Why is that? Let us consider a particle in a box probleM: Vx()= 0 0 <<xa Vx()=∞ elsewhere We wish to solve !2 d 2 − 2 ψ (x) +V (x)ψ (x) = Eψ (x) 2m dx E is a Constant Outside the box Vx()→∞we want finite values of E , the only possibility is ψ ()x = 0outside the box. We also wish ψ ()x to be continuous: Inside the box we have V = 0 Chapter 3 – Schrodinger Equation, Particle in a Box 39 Fall 2014 Chem 356: Introductory Quantum Mechanics !2 dψ 2 − 2 = Eψ (x) 2m dx Boundary Condition: ψψ(0)== (a ) 0 We considered before this equation General Solution: ckxbkxsin( )+ cos( ) !2k 2 x = 0 ψ = 0 b = 0 E = 2m nπ xa= ckasin( )= 0 kn== , 1,2,3 a Any c , c not equal to 0 ⎛⎞nxπ ψ (xc )= sin ⎜⎟ ⎝⎠a !2n2π 2 h2n2 E = 2 = 2 n = 1,2,3..... 2ma 8ma - Quantization: CoMbination of wave equation + Boundary conditions - n =−1, − 2, − 3 also possible, but yields “same” solutions ⎛⎞nxππ nx ccsin⎜⎟−=− sin ⎝⎠aa - c can be anything (still) ˆ For any operator A , with eigenfunction ψ (x) Acˆˆψψ() x= cA () x = caψ () x = ac(())ψ x If ψ ()x is an eigenfunction of operator Aˆ then also cxψ () is eigenfunction. ( c is constant) Interpretation of Wave Function In Mathchapter B we discussed probability distribution p() x dx: Chapter 3 – Schrodinger Equation, Particle in a Box 40 Fall 2014 Chem 356: Introductory Quantum Mechanics px()≥ 0 ∀x ∞ ∫ pxdx()= 1 −∞ ∞ xxpxdx= ∫ () etc. −∞ 2 The absolute square of the wave function ψψψ()xxx= * () ()is to be interpreted like a probability distribution. 2 pxdx()= ψ () x dx Probability to find particle between x and x+ dx ψ ()xfxigx=+ () () complex fx(), gx() real ψ * ()xfxigx= ()− () ψψ* ()xx ()= [ fxigxfxigx ()− ()][()+ ()] fx()22 gx () ifxgx [ ()() gxfx () ()] =++ − 2 2 = f (x) + g(x) (real always) 2 Also ψ ()x > 0 everywhere Probability distribution ∞ 2 Moreover (we should impose): ∫ ψ ()xdx≡ 1 −∞ NorMalization Multiply ψ ()x by constant c , choose c such that cxψψ()= new () x is norMalized Particle in the box (later) 2 ⎛⎞nxπ ψ n (x )= sin ⎜⎟ aa⎝⎠ Further Interpretation xhigh ∫ ψψ()xxdx* () xlow Probability to find particle between xlow and xhigh And Chapter 3 – Schrodinger Equation, Particle in a Box 41 Fall 2014 Chem 356: Introductory Quantum Mechanics ∞ ∞ x = ∫ xψ *(x)ψ (x)dx = ∫ ψ *(x)xψ (x)dx −∞ −∞ Determination of Constant C We will iMpose that the wave functions are norMalized ∞ ∫ ψψ* ()()xxdx≡ 1 For reasons discussed before −∞ ψ * ()x : coMplex conjugate of functions ψ ()xfxigx=+ () () fx(), gx()real ψ * ()xfxigx= ()− () ψψ* ()xx ()= [ fxigxfxigx ()− ()][()+ ()] =++[()][()]fx22 gx ifxgx [()() − gxfx ()()] 22 =+fx() gx () 2 = ψ ()x ≥ 0 everywhere 2 If ψ ()x is real then ψψ()xx= ()2 Consider particle in the box wave functions: nπ x ψ n ()x = Cn sin 0 ≤ x ≤ a a 0, elsewhere ∞ a 2 2 2 ⎛⎞nxπ ψ ()xdxC= sin dx ∫∫n ⎜⎟ −∞ 0 ⎝⎠a a ==C 2 1 n 2 2 Choose Cn = a 2 SiMplest, eiθ would work too. a We can always choose the function ψ ()x to be norMalized (for Meaningful wave functions) A physically Meaningful wave function would be norMalized If Axˆψψ()= ax () eigenfunction of Aˆ , eigenvalue a Chapter 3 – Schrodinger Equation, Particle in a Box 42 Fall 2014 Chem 356: Introductory Quantum Mechanics Then ψψ* ()xA( ˆ () x) =ψψ* ()xa () x = axxψψ* () () ∞ And: ∫ ψψ* ()xA( ˆ () xdx) −∞ ∞ = axxdx∫ ψψ* () () −∞ =a ⋅1 IF ψ ()x is norMalized We define: ∞ Aˆ = ∫ ψ *(x)Aˆψ (x)dx −∞ Called the expectation value of operation Aˆ , depending on ψ ()x , also called the average value of Aˆ If ψ ()x is norMalized, then Aˆ would be the average value Measured for quantity A If ψ ()x is an eigenfunction of Aˆ , then one would always Measure a , and the average value Aa= IF ψ ()x is norMalized If ψ ()x is not an eigenfunction of Aˆ , then Many values could be obtained if A is Measured. The average value would be Aˆ (Much More discussion later) One More definition: 2 ( AAˆˆ− ) : The standard deviation froM the average. The spread of the Measured values ( AAAAˆˆˆˆ−−)( ) 2 =AAAAˆˆˆˆ2 −2 + 2 =AAAAˆˆˆˆ2 −2 + 2 =AAˆˆ2 − 2 ()x = σ A Depends on wave function ψ Chapter 3 – Schrodinger Equation, Particle in a Box 43 Fall 2014 Chem 356: Introductory Quantum Mechanics ∞ Using definition: Aˆ 2 = ∫ ψ *(x)Aˆ 2ψ (x)dx −∞ Useful integrals for particle in the box x sin2bx ∫ sin2 bx dx = − 2 4b 2 2 x xsin2bx cos2bx xsin bx dx = − − 2 ∫ 4 4b 8b x3 ⎛ x2 1 ⎞ cos2bx x2 sin2 bx dx = − − sin2bx − x ∫ 6 ⎜ 4b 8b3 ⎟ 4b2 ⎝ ⎠ nπ nπ Definite Integrals (Most iMportant).
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