Chemistry 4531 Mathematical Preliminaries Spring 2009

I. A Primer on Differential Equations Order of differential equation Linearity of differential equation Partial vs. Ordinary Differential Equations y q Quantum Mechanics involves primarily 2nd order linear partial differential equations First order linear ordinary

df(x)dx

= h(x)

one arbitrary constant

Second order linear ordinary differential equation 2d f(x)

+ a f(x) = 02dx + a f(x) = 0

There must be two arbitrary constants in the general solution. The solution is

( ) ( )f x A ax B ax

or

( ) sin cos= +

( )

( )

f x C ax Dor

f x E ax F

( ) sin

( ) cos

= +

= +( )

( ) ( )

f x E ax F

Heax i ax

( ) cos= +

+ −

or even

f(x) = Gei

The values of the constants are determined by boundary conditions on the differential equations. How do we know which form to use?? It cannot matter, so we use the one that is most convenient that is most convenient.

Taylor Series expansion of a function f(x) about x = x0 df(x) d f(x) (x x )2

02⎤ ⎤ −

f(x) f(x ) + df(x)dx

(x x ) + d f(x)dx

(x x )2!

+ ...0x=x

0x=x

20

0 0

≅ ⎤⎦⎥

−⎤

⎦⎥

Convergence Radius of Convergence Radius of Convergence Absolute convergence

expand eix about x =0

ix i0

x=0

ix

x=0

2 ix

2

2

x=0

3 ix

3

3

e = e + dedx

x + d edx

x2!

+ d edx

x3!

+ ...⎤

⎦⎥

⎤

⎦⎥

⎤

⎦⎥

sin(x) = cos(x) =

Wh t i th it d f ix?

SO EULER FORMULAixe (x) i (x) The= +cos sin

What is the magnitude of eix?

Odd and Even Functions Odd: F(x) = -F(-x) Even: F(x) = F(-x) This definition refers to the symmetry behavior (parity) about This definition refers to the symmetry behavior (parity) about the origin, x=0. The definition can be readily generalized to describe symmetry properties about any point, say x = a. e.g. Odd: F(x-a) = -F(-[x-a]) Even: F(x-a) = F(-[x-a]) Using Mathcad, let us look at examples and also some integrals. x 10 9 95 10

x ..,10 9.95 10F( )x ( )x 2 2

48

64

16

32

48

The function is even about x=2. It is neither odd nor even about any other point.

6 4 2 0 2 4 6 8 100

Let's look at an odd function about x=2: ( )[ ]G x x( ) sin= −64 2 3

64

32

64

32

0

What can we say about integrals centered at x=2 ? Let us evaluate some integrals involving F and G We do so by inspection!!

6 4 2 0 2 4 6 8 1064

Let us evaluate some integrals involving F and G. We do so by inspection!! 8

2( )F x dx∫

8( )F x dx∫ 4( )

−∫

8

2( )G x dx∫

8

2( )G x dx∫

2

( )∫ 8

4( ) ( )F x G x dx

−∫ 8

2( ) ( )F x G x dx∫

OPERATORS An operator is an instruction telling you what to do with a function, much like a function is

an instruction telling you what to do with a number. Examples of operators include d/dx, *,∑ ,√, an nstruct on tell ng you what to do w th a number. Examples of operators nclude d/dx, ,∑ ,√, etc. We will denote a general symbolic operator with a "hat" over the symbol, e.g.  So if  = √ , then  f(x) = f x( ) A ti t l b b d i d An entire operator algebra can be derived. Notation: ÂĈf(x) ≡ Â(Ĉf(x))

Ĉ2 f(x) ≡ Ĉ [Ĉf(x)] (Â + Ĉ) f(x) ≡ Â f(x) + Ĉ f(x)

h DE f k d ff ! ! The ORDER of operators can make a difference! A very important concept!

Â�Ĉ f(x) ?= Ĉ  f(x)

DEFINITION:  Ĉ - Ĉ  ≡ [ Ĉ] the COMMUTATOR of  and Ĉ DEFINITION:  Ĉ Ĉ  ≡ [ , Ĉ] , the COMMUTATOR of  and Ĉ.

If the commutator is zero for any function f, then the operators are said to commute. Namely, the order of carrying out the operations  and Ĉ does not matter. Operators and commutation are critical concepts in quantum mechanics and are directly related to measurement.

Example: let  = x and Ĉ = d/dx. All Quantum Mechanical Operators are linear.

Definition: Â is linear if and only if

 (c1 f + c2 g) = c1  f + c2  g ( g) gwhere c1 and c2 are constants and f and g are any two functions. Which of these operators is linear?? R = ( ) ( )2

2d2

2ddx P = x

d( )dx

3

2d( )⎡ ⎤M 3 d( ) = x dx⎡ ⎤⎢ ⎥⎣ ⎦

Eigenvalue equations and eigenvalues The operator equation Â(f(x)) = c f(x), where c is a constant, is called an eigenvalue equation. The function f(x) is called the eigenfunction of the p t  ith i n l It t ns t th t in nt m m h ni s ll operator Â, with eigenvalue c. It turns out that in quantum mechanics all

physical properties are represented by operators and that the possible values of that property are related to the eigenvalues of that operator. Thus much of our work will be related to finding eigenvalues and g geigenfunctions of operators. Examples: I. Let Â= d/dx. Then we must find f(x) such that dfdx

= c f(x)

The solution in this case is f(x) = ecx , and the eigenvalue is c, where c is any number number.

II. Find the eigenvalues and eigenfunctions of  = -d2 /dx2 .

That is, find all ai and fi(x) such that− =d f x

dxa f xi

i i

2

2( )

( ) . dx

Rewriting,2

iii2

(x)d f (x) 0a fdx

− = , a 2nd order homogeneous differential

equation with solution: q

( ) ( )i i if (x) = Asin a x + Bcos a x , where ai is real, and A and B are any numbers.

More about eigenvalue equations and eigenvalues

Linear momentum, p

Kinetic energy, TKinetic energy, T

Total energy,  H = T + V

POTENTIAL ENERGY We will consider the case that the force F on a particle depends only on its position p p y p

(coordinate). As an aside, when might the force depend on velocity? Conservative system??

If this is the case then one can prove the following: Let the forces acting on a particle be dV

given by a vector F . If there exists a function V(x) such that FdVdx

= − , then the force field is

said to be describable by the potential energy function, V(x). More generally, if there is a function V(x, y, z) such that y

( )F x, y, z = F i+ F j+ F k = Vx

i Vy

jVz

k V(x, y, z)x y z −∂∂

−∂∂

−∂∂

= − ∇

then the force field F is said to be conservative. Here ∇ is the gradient operator and i j d k it t i th d di tii , j and k are unit vectors in the x, y and z directions.

Let's look at several examples to clarify this concept. Gravitational potential for a particle of mass m:

zm

V(z)

z

Since = mgkF − , then V(z) = mg z

An ideal (Hookes Law) Spring

Since F = kx i− , we have

V(x) = k x2

+ C2

The quadratic potential energy curve also reminds us how we generally can visualize The quadratic potential energy curve also reminds us how we generally can visualize the forces acting on a particle.

How would we modify the potential energy for a diatomic molecule??

For the hydrogen atom, -e

rem

+e e

m p The force is = q q

r = 1 2

2F attractiveer

2

2

d h i l i and the potential energy is

V(r) = er

, the Coulomb Potential2−

attractive r

Schrödinger equation for a particle in one dimension Consider a particle of mass m confined to move in one dimension but with NO forces acting on the particle.

Schrödinger postulated that the amplitude of the matter waves, ψ(x), must satisfy the differential equation, 22 22

2d = E

2m dxwhere E is the total energy

and = h/2

ψ ψ

π

−

More generally, if the particle is in a potential energy field V(x), then 2 2d (x)

+ V(x) (x) = E (x)ψ

ψ ψ− 22 m dx+ V(x) (x) = E (x)ψ ψ

Before developing formal postulates, let’s see what this can mean. A solution to the first equation can be written in terms of complex exponentials or sines and cosines:

(x) = e = cos sin , where k = 2 mEikx

2ψ kx + i kx

This is a sine wave of wavelength λ = 2π/k Since there are no forces (no potential energy) This is a sine wave of wavelength λ = 2π/k. Since there are no forces (no potential energy), the total energy E is also the kinetic energy. Thus E = p2/2m and k= p/ . The wavelength λ associated with the solution is thus h/p, the deBroglie wavelength!!

In the more general second case, the only difference is that E in the solution is

replaced by E-V: ( ) ( )ψ(x) = e = cos kx + i sin kx , where k = 2 m(E V)ikx

2

−

The kinetic energy of the particle is (E V) and the wavelength associated with the matter The kinetic energy of the particle is (E-V), and the wavelength associated with the matter

waves is now given by the relation = λh

2 m(E V)− .

Thus, the slower the particle, the longer (and less wiggly) is the matter wave wavelength. Notice that when the particle is at rest, the wavelength is infinite, and we cannot tell where theparticle is. This is the beginnings of the Uncertainty Principle at work!

In the more general second case, the only difference is that E in the solution is

replaced by E-V: ( ) ( )ψ(x) = e = cos kx + i sin kx , where k = 2 m(E V)ikx

2

−

The kinetic energy of the particle is (E V) and the wavelength associated with the matter The kinetic energy of the particle is (E-V), and the wavelength associated with the matter

waves is now given by the relation = λh

2 m(E V)− .

What happens if the potential is infinitely high V = ∞ ?What happens  if the potential is infinitely high,  V = ∞ ?

What happens  if the potential is just high,  V > E ?

Example: A Potential Step

VE2 > V

E1 < V

V=0

We will come back to this “free particle” shortly,but we first will look at the postulates of quantum mechanics

and then solve the particle in a box problem with renewed understanding

Chemistry 4531 Postulates of Quantum Mechanics Spring 2009 Consider a system with N degrees of freedom. Consider a system with N degrees of freedom. 1. There exists a function Ψ(q1 . . . qN ,t) that.contains all information available for this system. It is

called the state function or wave function. Here q represents coordinate(s). Properties of Ψ: a. Ψ*Ψ is a probability density Born hypothesis b Ψ Ψ Ψ Ψ*z dq dq 1 ( m li d) b. Ψ Ψ Ψ Ψ*z ≡ =dq dq N1 1… (normalized)

c. ΨΨ(q),

q

q

2

2∂Ψ∂

∂∂

and must be continuous.

d. Ψ(q1 . . . qN,t) must be a single valued function of its arguments

Chemistry 4531 Postulates of Quantum Mechanics Spring 2009 Consider a system with N degrees of freedom. 1. There exists a function Ψ(q1 . . . qN ,t) that.contains all information available for this system. It

is called the state function or wave function. Here q represents coordinate(s). Properties of Ψ: a Ψ*Ψ is a probability density Born hypothesis a. Ψ Ψ is a probability density Born hypothesis b. Ψ Ψ Ψ Ψ*z ≡ =dq dq N1 1… (normalized)

c. ΨΨ(q),

q

q

2

2∂Ψ∂

∂∂

and must be continuous.

d ( ) b i l l d f i f i d. Ψ(q1 . . . qN,t) must be a single valued function of its arguments

2. For every physical observable A, there corresponds a linear Hermitian operator Â. More later on Hermitian operators.

h  b d b l ll f d d The operator  is obtained by writing A classically in terms of cartesian coordinates and momenta, and then replacing

x x y z ix

iy

izx y z→ → → → − → − → −, , , , , y z p p p∂

∂∂∂

∂∂ y

3. If  corresponds to physical observable A, then for a system with state function Ψ, every measurement of the observable A must yield one of the eigenvalues of Â. y g

4. The time evolution of Ψ(q, t) is given by (q, t)ˆ (q, t) = it

∂ΨΨ

∂H where H is the Hamiltonian t∂

operator. The classical Hamiltonian H corresponds to the total energy (kinetic and potential) of an isolated system. This equation is the time-dependent Schrödinger equation.

(All symbols and operations should become clear in future lectures) We will see shortly that when we wish to calculate the stationary states of a system, the time-dependent Schrödinger equation for a single particle of

d d h l d d h d mass m moving in one dimension reduces to the simpler time-independent Schrödinger equation,

−2

2m d (x)

dx + V(x) (x) = E

2

2

ψψ ψ(x)

Here V(x) is the potential energy of the particle This expression is identical to the eigenvalue Here, V(x) is the potential energy of the particle. This expression is identical to the eigenvalue

equation for the total energy (postulate 3), ˆ EΨ = ΨH .

PARTICLE IN A BOX PARTICLE IN A BOX

a

x) = ∞ , x < 0 or x > a (Regions I ,III) x) = 0 , 0 < x < a (Region II)

2 2d (x)ψ− 22m

d (x)dx

+ V(x) (x) = E (x)ψ

ψ ψ

he formal QM postulates recognize that the solution to the Schrödinger equation, x) has a connection with physical observables In order to represent a physical x), has a connection with physical observables. In order to represent a physical ality the postulates insist that ψ(x) be a) continuous, and b) single valued. e will solve the equation in regions I, II and III and then put the pieces together to ey the above constraints.

REGIONS I and III In these regions, the Schrödinger equation is given by

2 2d (x)ψ22m

d (x)dx

+ (x) = E (x)− ∞ψ

ψ ψ

which implies ψ(x) = 0 for x<0 or x>a Now consider Region II where V(x) = 0 Now consider Region II, where V(x) = 0

−2 2

22md (x)

dx = E (x)

ψψ

so a solution in this region is

ψ(x) = A2mE

x + B2mE

x2 2sin cos⎛

⎝⎜

⎞

⎠⎟

⎛

⎝⎜

⎞

⎠⎟ψ( ) 2 2⎝ ⎠ ⎝ ⎠

for 0≤x≤a . We now have the total solution, and must apply the boundary conditions. Thestate function ψ(x) is certainly well behaved except possibly at x = 0 or x = a. To be state function ψ(x) is certainly well behaved except possibly at x 0 or x a. To be continuous, we must require that the Region II solution vanish at these points THUS at x = 0,

2 2i mE mE⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟2 2

2 20 sin 0 cos 0mE mEA B⎛ ⎞ ⎛ ⎞

= +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ (1)

THUS at x = 0,

2 2

2 20 sin 0 cos 0mE mEA B⎛ ⎞ ⎛ ⎞

= +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ (1) and at x = a,

2 2mE mE⎛ ⎞ ⎛ ⎞

2 2⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ (1)

2 2

2 20 sin cosmE mEA a B a⎛ ⎞ ⎛ ⎞

= +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ (2)

From the first equation, we have 0 A (0) B(1) B 0 or 0 = A (0) + B(1) so B = 0

and from the second,

⎛ ⎞ ⎛ ⎞ ⎛ ⎞2 2 2

2 2 20 sin cos sinmE mE mEA a B a A a⎛ ⎞ ⎛ ⎞ ⎛ ⎞

= + =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

There can only be TWO possibilities: There can only be TWO possibilities: 1) A = 0 , which says ψ(x)= 0 for all x (no good) Why??

or

⎛ ⎞

2) 2

2sin 0mE a⎛ ⎞

=⎜ ⎟⎜ ⎟⎝ ⎠

Can we make this happen?? What parameters are available to us? ONLY the Energy, E !!!!

HOW? Where are the zeros of the sin function? HOW? Where are the zeros of the sin function? We know that sin z = 0 for z = n π, where n = 1, 2, 3, . . So we must have

2mE2

2 where n = 1,2,3 mE a nπ= …

Solving for E,

2 2 2 2 2

2 2 where n = 1,2,3, 2 8n

n n hEma maπ

= = …

and the state function is and the state function is

( ) sin for 0 and 0 otherwisen nn xx A x a

aπψ ⎛ ⎞= ≤ ≤⎜ ⎟

⎝ ⎠ The imposition of boundary conditions and periodic motion has forced

quantized energy states upon us!!

The Born Postulate says that ψ* ψ is a probability density and thus its integral over all space must equal 1.

a∞ ⎛ ⎞ ⎛ ⎞* 2 2 2

0

( ) ( ) 1 sin2

a

n nn x ax x dx A dx A

aπψ ψ

∞

−∞

⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠∫ ∫

A 2 so An = 2a

What do these functions look like???

Particle in a Box ψn and ψn

2

We plot ψn and ψn2 for the first few states in the following picture.

ψn(x) ψn*(x) ψn(x)

16 4

/2 m

a 2

/2 m

a2

4

units

of

29 9

ts o

f 2 π

2 /

3

n

Ener

gy, u

4 4

Ener

gy, u

nit

2

0 0.25 0.5 0.75 10

1

0 0.25 0.5 0.75 10

1

Classical vs. Quantum Behavior 1 Nodes 4 Localization in part of box?

E 1

1. Nodes 4. Localization in part of box? 2. Non-uniform ψn

2 5. "getting across the zeroes” 3. Zero point energy 6. Classical Limit - Large n

What is very misleading about this picture?? What is very misleading about this picture?? Placing the state functions on a potential energy plot! We need to think and talk about this!

Bracket Notation

More on PostulatesBracket Notation Definitions: Let φ and ψ be arbitrary complex-valued functions of coordinate(s) q.

φ ψ φ ψ φψ φ ψ* *A A dq and dq≡ ≡z z φ ψ φ ψ φψ φ ψA A dq and dqz z The * implies taking the complex conjugate of the function. The integration dq is over all values of all coordinates. φ φ≡ * is often called a bra and ψ ψ≡ is similarly called a ket φ φ≡ is often called a bra and ψ ψ≡ is similarly called a ket.

2. For every physical observable A, there corresponds a linear Hermitian operator Â, whose

mathematical properties tell us about the physical quantity A. mathemat cal propert es tell us about the phys cal quant ty . a. Â is Hermitian if and only if

φ ψ φ ψA A= We will see that the Hermitian property of an operator guarantees that its eigenvalues are real.

This Hermitian property also guarantees that eigenfunctions correspondingThis Hermitian property also guarantees that eigenfunctions correspondingto different eigenvalues are orthogonal, ie, ⟨φi|φj⟩ = 0, if ai ≠ aj (if  is Hermitian).

We will prove this statement shortly.

More on Postulates

3. If  corresponds to physical observable A, then for a system with state function Ψ, every measurement of the observable A must yield one of the eigenvalues of Â. g

3a) If a system is characterized by state function ψ, and ψ is aneigenfunction of the operator  with eigenvalue a, then every

t f th t di t  ill i tlmeasurement of the property corresponding to  will give exactly afor the result. 3b) If a system is characterized by state function ψ, and ψ is an

Ânot an eigenfunction of the operator Â, then repeated measurements of the property corresponding to  will result in a distribution of results. The average value of the measurements, ⟨a⟩, is given by

ˆ ˆ*A Aq

a dq= Ψ Ψ = Ψ Ψ∫

Actually, 3b) encompasses all of the other versions

Examples of Significance of Postulates Let us look at some properties of the particle in a box, using the postulates to interpret the results. Let ψn(x) be the normalized eigenfunction for the particle in a box extending from 0 to a.

1) Suppose the state function is ψs = C(2ψ1+5ψ2). Find C such that ψs is normalized.

N.B., we will soon see that this orthogonality of ψ1 and ψ2 extends to any eigenfunctions with different eigenvalues. See Atkins, 2.1, c & d for more.

2) Let the state function be ψ2 Is it normalized?? Calculate ⟨E⟩, ⟨p⟩and ⟨p2⟩. Interpret the results.

3) Given ψs = C(1ψ1+3ψ2), find ⟨E⟩ . Interpret the result.

Another Example

Consider a function W 5x = 30a

x(a x)ψ b g − for 0<x<a, and 0 otherwise. Is ψW an

acceptable state function? acceptable state function? a) continuous and continuous derivatives? b) single valued? b) single valued? c) normalized?

5 0

30 a

a 22(a x dx = 1)x⎛ ⎞ −⎜ ⎟⎝ ⎠ ∫

Thus ψW fulfills the requirements for a state function, but it isNOT an eigenfunction. Why?? Let's plot ψW.

x x(a-x) 0 0 a 0 a/2 a2/4 / / 2/a/4, 3a/4 3 a2/16

a/10, 9a/10 9 a2/100 a

So ψW is symmetric about a/2 and has no nodes. It is an acceptable state function. So ψW is symmetric about a/2 and has no nodes. It is an acceptable state function. It “looks a lot like” ψ1. Is its energy close to that of ψ1?

2 2

20 0

ˆ* *2

a a

E dx dxm x

ψψ ψ ψ − ∂= =

∂∫ ∫H = 30a

2 m

x(a x) ddx

x(a x)dx5

2

0

a 2

2−LNMOQP − −z

Finally,

5 h2 2

⟨ ⟩E = 5

a =

h7.9 m a2 2m

E h2

Recall for ψ1, Ema1 28

= , very close to that above!! That makes sense!! What will we observe if we make repeated measurements of EW?

A distribution of energies, since ψW is not an eigenfunction of E Each member of the distribution will be one of the energy eigenvalues E n h

man =2 2

28

Fi ll E h2

f d “l k l lik ” Finally Ema

= 27 9. for ψW, expected, as ψW “looks a lot like” ψ1

QUANTUM MECHANICAL THEOREMS

Remember postulate 2For every physical observable A there corresponds a linear Hermitian operator  For every physical observable A, there corresponds a linear Hermitian operator Â, whose mathematical properties tell us about the physical quantity A.

Remember: Â is Hermitian if and only if Bracket Notation Definition:

Many properties of Hermitian operators follow from a single derivation.Thus, let ψn and ψm be eigenfunctions of an arbitrary operator  with eigenvalues an and am. Then

 (1) ψm = am ψm (1)And Âψn = an ψn (2)

PROOF:PROOFMultiply (1) by ψn

* and integrate

ψn* Â ψm = ψn

* am ψm = am ψn* ψm (3)

Take * of (2) , multiply by ψm and integrateÂ*ψn

* = an* ψn

*

ψm Â* ψn* = ψm an

* ψn* = an

* ψm ψn* (4)

Now subtract (4) from (3)ψn

* Â ψm - ψm Â* ψn* = am ψm ψn

* - an* ψm ψn

* (5)

= ( am- an* ) ψm ψn

*

This result is valid for ANY operator Â. Suppose that  is Hermitian. Then the left side of (5) vanishes and we have that 0 = ( am- an

* ) ψm ψn*

There are two possibilities involving the indices n and m:Case I n = m Since ψ ψ * = 1 then ( a - a * ) = 0 Case I. n = m. Since ψn ψn = 1, then ( an- an ) = 0.

Thus : "The eigenvalues of a Hermitian operator are real."

Case II. n ≠ m0 = ( am- an ) ψm ψn

* = ( am- an ) ⟨ψn ⏐ ψm ⟩The an

* became an because it is real. If ( am- an ) ≠ 0 , then I can divide by this term to obtain the following:term to obtain the following:ψm ψn

* = 0 if an ≠ am.

Thus: "Eigenfunctions corresponding to different eigenvalues of a Hermitian poperator are orthogonal". This vanishing of the integral of two functions is the definition of orthogonality, as we shall soon see.