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)

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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 ! 2 2 ∂ ψ 2 ! 2 + p ψ (x) = 0 dx 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 x = −i! ∂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 ! e a2eiax 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 <

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). Use b = ; bx = a = nπ a x=a a a nxπ a ∫sin2 dx = 0 a 2 a nxπ a2 ∫ xdxsin2 = 0 a 4 a nxπ a33 a x22sin dx =− ∫ 22 0 a 6 4n π a nπ x mπ x sin cos dx = 0, ∀n,m integers ∫ a a 0

Demonstration of Uncertainty Principle

Using the above integrals, we can calculate the following nxπ a) Normalize ψ = C sin nn a a 2 22⎛⎞nxπ a 2 CdxCsin=≡ 1 CC nn∫⎜⎟ n == 0 ⎝⎠a 2 a 2 ⎛⎞nxπ Normalized particle in the box eigen states: sin ⎜⎟ aa⎝⎠

b) Calculate x for normalized ψ n ()x :

Chapter 3 – Schrodinger Equation, Particle in a Box 44

Fall 2014 Chem 356: Introductory Quantum Mechanics

2 a nxππ nx xxdx=∫ sin⋅⋅ sin aa0 a 2 aa2 =⋅= center of the box a 42 c) Calculate x2 2 a nxππ nx x22=∫sin⋅⋅ x sin dx aa0 a 2 ⎛⎞aa33 aa 22 =⋅−⎜⎟22 =− 22 a ⎝⎠6342nnππ d) Standard deviation in x :

22 2 σ x =xx−

22 aa22 aaa 22 an⎡⎤ 22π ⎛⎞ ⎛ ⎞ 2 =−−22⎜⎟=− 22= ⎜ ⎟⎢⎥− 32122322nnππ⎝⎠ ⎝π n ⎠⎣⎦ 2 a nπ x ⎛ d ⎞ nπ x e) P = sin −i! sin dx x a ∫ a ⎝⎜ dx ⎠⎟ a 0 2 ⎛ nπ ⎞ a nπ x nπ x = −i! sin cos dx = 0 a ⎝⎜ a ⎠⎟ ∫ a a 0 2 a nπ x ⎛ d 2 ⎞ nπ x f) P2 = sin −!2 sin x a ∫ a ⎝⎜ dx2 ⎠⎟ a 0 2 2 n2π 2 a ⎛ nπ x⎞ a = !2 ⋅ sin dx a a2 ∫ ⎝⎜ a ⎠⎟ 2 0 !2n2π 2 h2n2 = 2 = 2 ( = 2mEn , of course!) a 4a hn σ ()P = x 2a

We can test the Heisenberg Uncertainty Principle 1 an⎡⎤π 22 2 hn

σσxp=⋅⎢⎥ −⋅2 23π na⎣⎦ 2 1 ! ⎡π 2n2 ⎤ 2 ! = − 2 > 2 ⎢ 3 ⎥ 2 ⎣ ⎦

Chapter 3 – Schrodinger Equation, Particle in a Box 45

Fall 2014 Chem 356: Introductory Quantum Mechanics

a Note 1: σ x → as n →∞ is the same as uncertainty in uniform distribution: 12

a 1 xa2 x == a 220 11a 1 xxa232== a 330 aaa22 2 σ =−= x uniform 3412 σ grows with n. Why? Px n2π 2!2 Pn = ± (2mEn ) = ± 2 a

Spiked distribution Large Uncertainty This represents the classical limit of particle of energy En bouncing back and forth in the box 2 2 Note 2: x , x , Px , Px Can be calculated for any wave function for example: ψ ()xCxax= (− ) also satisfies the boundary conditions

Particle in a 3 dimensional box

Consider rectangular box of length abc,,

Chapter 3 – Schrodinger Equation, Particle in a Box 46

Fall 2014 Chem 356: Introductory Quantum Mechanics

3D Schrodinger Equation: !2 ⎛ ∂2 ∂2 ∂2 ⎞ − + + ψ (x, y,z) = Exyzψ (, ,) 2m ⎝⎜ dx2 dy2 dz 2 ⎠⎟

Boundary Conditions: ψψ(0,yz , )== ( ayz , , ) 0 ∀yz, ψψ(,0,)xz== (,,) xbz 0 ∀xz, ψψ(,xy ,0)== (, xyc ,) 0 ∀xy, “The wave function at the faces of sides of a box is zero” Technique to solve: Separation of variables. Try ψ (,xyz ,)= XxYxZz ()() () Substitute in Schrodinger equation and divide by ψ (,xyz ,) (as we did for vibrating strings) !2 1 d 2 X !2 1 d 2Y !2 1 d 2Z − − − = E 2m X (x) 2 2m Y( y) 2 2m Z(z) 2 dx dy dz

This can only be true if each term itself is constant: EEExyz,, We get 3 equations h22dX a) −=EXx() XXa(0)== ( ) 0 2m dx2 x h22dY b) −=EY() y YYb(0)== ( ) 0 2m dy2 y h22dZ c) −=EZ() z ZZc(0)== ( ) 0 2m dz2 z

EEEExyz++= This is just 3 times the 1D particle in the box equation! We know the (normalized) solution: 2 ⎛⎞kxπ h2 ⎛ k 2 ⎞ Xx( )= sin E = aa⎜⎟ x 8m ⎜ a2 ⎟ ⎝⎠ ⎝ ⎠ 2 ⎛⎞lyπ h2 ⎛ l 2 ⎞ Yy( )= sin E = bb⎜⎟ y 8m ⎜ b2 ⎟ ⎝⎠ ⎝ ⎠ 2 nzπ hn22⎛⎞ Zz( ) sin ⎛⎞ E = ⎜⎟ y = ⎜⎟2 cc⎝⎠ 8m ⎝⎠c Or

Chapter 3 – Schrodinger Equation, Particle in a Box 47

Fall 2014 Chem 356: Introductory Quantum Mechanics

8 ⎛⎞n π ⎛⎞nyyπ ⎛⎞n π ψ =sinxx ⋅⋅ sin sin zz nnnxyz ⎜⎟⎜⎟⎜⎟ abc⎝⎠ a⎝⎠ b⎝⎠ c h2 ⎛⎞n 2 n 2 n 2 E =++x y z nnn, ,= 1,2,3.... ⎜⎟222 xyz 8m ⎝⎠abc Degeneracies for Cubic box Consider the special case of a Cubic box abc==. Then the energy takes the form h2 Ennn=++222 8ma2 ( xyz)

For each triplet nnnxyz,, we get a different wave function, but different values of nnnxyz,, may yield the same energy.

Such energy levels are called degenerate. Eg.for atoms we know there are 1 s-orbital, 3 p- orbitals, 5 d-orbitals.

Table of energies h2 E = (nnnxyz,,) Degeneracy 8ma2 14 (1, 2, 3), (1, 3, 2), (2,1, 3), (2, 3,1), (3,1, 2), (3,2,1) 6 12 (2,2,2) 1 11 (1,1, 3), (1, 3,1), (3,1,1) 3 9 (2,2,1),(2,1,2)(1,2,2) 3 6 (1,1, 2), (1, 2,1), (2,1,1) 3 3 (1,1,1) 1

Chapter 3 – Schrodinger Equation, Particle in a Box 48