Chapter 2 (part 1): Time-Independent Schrödinger Equation

TodayChapter 2 (part 1): Time-Independent Schrödinger Equation

Time-independent Schrödinger EquationInfinite square well

Stationary solutionsFourier trickGeneral solutionExample

Text: 2.1 – 2.2

Solving the Schrödinger Equation2 2

2–2

Y¶=

¶

¶ Y+ Y

¶!

!t

i Vm x

Consider the case when V is independent of time: V(x)

Can solve Schrödinger equation by the method of separation of variables:

( ) ( ) ( ), y jY =x t x t( ) ( ) ( )

( ) ( ) ( )2 2

2 2

,

,

x t dx tt dtx t d x

tx dx

y j

yj

ì¶Y=ïï ¶

í¶ Yï =ï ¶î

2 2

2–2

j yy j yj= +!

!d

dtdi V

m dx2 2

2

1 1–2

j yj y

= +!

!ddt

di Vm dx

{function of t} = {function of x} = ?

Time-independent Schrödinger Equation2 2

2

1 1–2

j yj y

= + =!

!ddt

di V Em dx

2 2

2

1

1–2

jj

yy

=ìïïíï + =ïî

!

!

dE

dti

d V Em dx

j j= -!

ddt

iE

2 2

22y y y- + =

! d V Em dx

Had: partial differential equationObtained: two ordinary differential equations

( ) iEtt ej -= !

Time-independent Schrödinger Equation

( )2 2

22d V x E

m dxy y y- + =

!

Time dependence is the same for any V(x)!

Properties of Separable Solution( ) ( ), iEtx t x ey -Y = !1. They are stationary states

( ) ( )( )2 * * *, iEt iEtx t e ey y y y+ -Y =Y Y = =! !

( ) ( )2 2, yY =x t x

* probability does not depend on time:

* expectation values don’t depend on time:*( , ) , iEt iEtdQ x p e Q x e dx

i dxy y- +æ ö= ç ÷

è øò ! !!

We might as well drop the j(t) altogether and use y in place of Y

Often y is called the wave functionRemember: true wave function always has a factor iEte- !

*( , ) , y yæ ö= ç ÷è øò! dQ x p Q x dxi dx

Properties of Separable Solution( ) ( ), iEtx t x ey -Y = !2. They are states of definite constant energy E

Total energy (classical Hamiltonian):2

( , ) ( )2

= +pH x p V xm

Total energy (quantum):

¯

¶æ öç ÷¶è ø!

p

i x

2 2

2ˆ ( )

2¶

= - +¶

!H V xm x

2 2

22y y y- + =

! d V Em dx

Compare with time independent SE:

ˆy y=H E

Time-independent SE can be written as:

Properties of Separable Solution( ) ( ), iEtx t x ey -Y = !2. They are states of definite constant energy E

Energy expectation value:

2 2* ˆy y y= = = Y =ò ò òH H dx E dx E dx E

( ) ( ) ( )2 2ˆ ˆ ˆ ˆ ˆy y y y= = = =H H H H E E H E

=H E

2 * 2ˆ ˆ ...y y= =òH H dx * 2 2... y y= =ò E dx EEnergy variance:

22 2 2 2 0s = - = - =H H H E E

2 2

2ˆ ( )

2¶

= - +¶

!H V xm x

ˆy y=H E

Total energy expectation value:

Energy is exactly defined in these states!

Properties of Separable Solution3. General solution is a linear combination of separable solution

2 2

22y y y- + =

! d V Em dx

Time-independent Schrödinger equation:

Yields an infinite collection of solutions: ( ) ( ) ( )1 2 3, , ,...y y yx x x

Each with its own energy: 1 2 3, , ,...E E E

There is an infinite number of wave functions: ( ) ( ), niE tn nx t x ey -Y = !

General solution to the Schrödinger equation:

1( , ) ( )y

¥-

=

Y =å !niE tn n

nx t c x e

Every solution to the Schrödinger equation can be written in this form

H = cn2En

n=1

∞

∑

cn2= 1

n=1

∞

∑

cn2 Is the probability that a measurement of the

energy would return the value En

Coefficients

A measurement will always yield one of the allowed valuesand is the probability of getting value cn

2En

What is the physical interpretation of the coefficients ?cn{ }

Solving ProblemsGiven: time-independent potential V(x)

Solving strategy:

1. Solve the TISE to find 2 2

22y y y- + =

! d V Em dx

( ) ( ) ( )1 2 3, , ,...y y yx x xE1 , E2 , E3 , …

2. Write down general solution for t = 0:1

( ,0) ( )y¥

=

Y =å n nn

x c x

3. Find coefficients cn using initial conditions at t = 0 (Fourier’s trick)Note: you can always match the specified initial conditions by a proper choice of cn

4. Construct the wave function1

( , ) ( )y¥

-

=

Y =å !niE tn n

nx t c x e

5. Enjoy the solution to the problem

Find: wave function of a particle

( )j -= !niE tt e( )0j

( )0 1j =

cn = ψ n∗(x)Ψ(x,0)dx

−∞

+∞

∫

initial condition

Solving ProblemsSuperposition of stationary wave functions (general solution)

is not necessarily a stationary state

http://en.wikipedia.org/wiki/Schr%C3%B6dinger_equation

(animation)

Stationary Solution: Real Part only?

2 2

22y y y- + =

! d V Em dx

In general, the solution of the time-independent Schrödinger equation is a complex function. Can we take just a real part to build the general solution?

Time-independent SE:

® If y is solution then y* is also a solution and so is any linear combinationay + by*

Complex conjugate SE:2 2 *

* *22y y y- + =

! d V Em dx

( ) ( )*1' Re2

y y y y= + =

When solving problems we can use only Re(yn) to construct general solution to the problem:

1( , ) ( )y

¥-

=

Y =å !niE tn n

nx t c x e

Note: cn are complex numbers!

Can pick:

The Infinite Square WellV(x)

a x

( )0, if 0 ,

, otherwise£ £ì

= í¥î

x aV x

1. Outside the well:

2 2

22y y y- + =

! d V Em dx

Solve time-independent SE:

2 2

22y y y- +¥ =

! d Em dx

y = 0 if x < 0 or x > a

A particle in this potential is completely free, except at the two ends, where an infinite force prevents it from escaping.


a x

( )0, if 0 ,

, otherwise£ £ì

= í¥î

x aV x

2 2

22y y y- + =

! d V Em dx

2. Inside the well:2 2

22y y- =

! d Em dx

ψ (x) = Asin kx + Bcoskx

22

2

d kdxy y= - 2

º!

mEk

Continuity: ( ) ( )0 0y y= =a

Solution:

( )0 sin 0 cos0y = + =A B B 0=B

( ) sin 0y = =a A kap=ka n

(A=0 – no particle!)

2 2 2 2 2

22 2p

= =! !n

nk nEm ma

p=nnka

n = 1, 2, 3, …

This is the equation of the simple harmonic oscillator

ψ x( ) = c1eikx + c2e− ikxor


a x

( )0, if 0 ,

, otherwise£ £ì

= í¥î

x aV x

2 2

22y y y- + =

! d V Em dx

2. Inside the well: siny =n n nA k x2 2 2 2 2

22 2p

= =! !n

nk nEm ma

p=nnka

How do we get A?

2y+¥

-¥=ò n dx ( )2 2

0sin =ò

a

n nA k x dx 2

0

sin 22 4

æ ö-ç ÷

è ø

a

nn

n

k xxAk

2 sin 22 4

æ ö= -ç ÷

è øn

nn

k aaAk

( )2 2sin 2 12 4 2

pp

æ ö= - = =ç ÷ç ÷

è øn na n aA A

n a

2=nA a


a x

A particle inside the infinite square well

( ) 2 sin py æ ö= ç ÷è ø

nnx x

a a

2 2 2

22p

=!

nnEma

Note: actual state is a linear combination:

1( , ) ( )y

¥-

=

Y =å !niE tn n

nx t c x e

Q: Do we need to normalize that at t>0? A. YesB. No

for 0 < x < a

n = 1, 2, 3, …

Properties of the SolutionV(x)

a x


nnx x

a a

1. They are alternately even and odd:(in respect to the center of well)

2. Each successive state has one more node:(y crossing zero)

3. They are mutually orthogonal:

( ) ( )*y y d=ò m n mnx x dx δmn =0, if m ≠ n1, if m = n

⎧⎨⎪

⎩⎪

4. They are complete:- any other function f(x) can be represented as:

( ) ( )1 1

2 sinn n nn n

nf x c x c xa a

py¥ ¥

= =

æ ö= = ç ÷è øS S


a x

Prove 3. They are mutually orthogonal:

( ) ( )*y y d=ò m n mnx x dx

( ) ( )*y y =ò m nx x dx( ) 2 sin py æ ö= ç ÷

è øn

nx xa a0

2 sin sinp pæ ö æ ö =ç ÷ ç ÷è ø è øò

a m nx x dxa a a

0

1 cos cosp pé - + ùæ ö æ ö= - =ç ÷ ç ÷ê úè ø è øë ûòa m n m nx x dx

a a a

( ) ( )0

1 sin sinp pp p

ì ü- +ï ïæ ö æ ö= - =í ýç ÷ ç ÷- +è ø è øï ïî þ

aa m n a m nx x

a m n a m n a

( )( )

( )( )

sin sin1 p pp

ì ü- +é ù é ùï ïë û ë û= - =í ý- +ï ïî þ

m n m nm n m n

0

:¹m nCase


a x

Prove 4: They are complete:- any other function f(x) can be represented as:

( ) ( )1 1

2 sinn n nn n

nf x c x c xa a

py¥ ¥

= =

æ ö= = ç ÷è øS S

Fourier series for function between 0 and a

Expansion Coefficients cnV(x)

a x

Fourier’s trick:

( ) ( ) ( ) ( )* *

1y y y

¥

=

= åò òm m n nn

x f x dx x c x dx

( ) ( )*y= òn nc x f x dx

Proof:

( ) ( )*

1y y

¥

=

= =å òn m nnc x x dx

1d

¥

=

= =å n mn mnc dx c

( ) ( )1n n

nf x c xy

¥

=

=S

SolutionsV(x)

a x

( ) ( )2 2 2/22, sin pp -æ öY = ç ÷è ø

!i n ma tn

nx t x ea a

Stationary states: (Expectation values do not depend on time)

General solution: (Expectation values can depend on time)

( ) ( )2 2 2/2

1

2, sin i n ma tn

n

nx t c x ea a

pp¥-

=

æ öY = ç ÷è ø

å !

Stationary

http://en.wikipedia.org/wiki/Particle_in_a_box

Classical

Superposition ofY1, Y2, Y3 and Y4,

Re(Y)Im(Y)

Example 2.2: Infinite WellExample 2.2:A particle in an infinite square well has the initial wave function:

( ,0) ( )Y = -x Ax a x

Find the wave function at all later times.

The figure shows this function plus the ground state wave function. They are very similar. We expect to find the solution looks mostly like the ground state, but with small amounts of other terms mixed in….

Ground

state

Example 2.2: Solution StrategyExample 2.2:A particle in an infinite square well has the initial wave function:

( ,0) ( )Y = -x Ax a x


Steps to solve:1. Normalize

2. Find cn

3. Build the time dependent wave function


nnx x

a a2 2 2

22p

=!

nnEma

( ) ( )*y= òn nc x f x dx

1( , ) ( )y

¥-

=

Y =å !niE tn n

nx t c x e

Stationary states

Example 2.2: NormalizationExample 2.2:A particle in an infinite square well has the initial wave function:

( ,0) ( )Y = -x Ax a x


2

0

2 2 2

0

52

5

1 ( ,0)

( )

30 30

= Y

= -

= ® =

ò

ò

a

a

x dx

A x a x dx

aA Aa

1. Normalize:

Example 2.2: cn CoefficientsExample 2.2:A particle in an infinite square well has the initial wave function:

5

30( ,0) ( )Y = -x x a xa


2. Solve for the coefficients cn:

[ ]

50

26

0 0

3

2 30sin ( )

60 sin sin ...

4 15 = cos(0) cos( )( )

p

p p

pp

æ ö= - =ç ÷è ø

é ùæ ö æ ö= - = =ê úç ÷ ç ÷è ø è øë û

- =

ò

ò ò

a

n

a a

n xc x a x dxa a a

n x n xaxdx x dxa a a

nn


nnx x

a a( ) ( )*y= òn nc x f x dx

Stationary states

Table of integrals

3

0 if n is even=

8 15 / ( ) if n is oddp

ìïíïî

nc n

Example 2.2: The SolutionExample 2.2:A particle in an infinite square well has the initial wave function:

5

30( ,0) ( )Y = -x x a xa


3. The solution:

3

0 if n is even

8 15 / ( ) if n is oddp

ìï= íïî

nc n

( ) ( )2 2 2/2

1

2, sin pp¥ -

=

æ öY = ç ÷è øS

!i n ma tn

n

nx t c x ea a

Review Normalization

Individual wave functions are normalized: ( ) 2 sin py æ ö= ç ÷è ø

nnx x

a a

( ) ( )*y y d=ò m n mnx x dx

Require normalization of general solution:1

( ,0) ( )y¥

=

Y =å n nn

x c x

*2

1 11 ( ,0) ( ) ( )y y

¥ ¥

= =

æ ö æ ö= Y = ç ÷ ç ÷è ø è øå åò ò m m n nm n

x dx c x c x dx

Normalization:2

11

¥

=

=å nnc

- the probability of finding a particle in the nth stationary state

2nc

* * * *

1 1 1

2*

1 1 1

= ( ) ( ) ( ) ( )

=

y y y y

d

¥ ¥ ¥

= = =

¥ ¥ ¥

= = =

=

=

åå åò ò

åå å

m n m n n n n nm n n

m n mn nm n n

c c x x dx c c x x dx

c c c

Example 2.2: - Probabilities

3

0 if n is even

8 15 / ( ) if n is oddp

ìï= íïî

nc n

( ) ( )2 2 2/2

1

2, sin pp¥ -

=


!i n ma tn

n

nx t c x ea a

- the probability of finding a particle in the nth stationary state

2nc

- the probability that the measurement of the energy yields En

2nc

2 2 2

22p

=!

nnEma

2nc

2

11

¥

=

=å nnc

Energy of General State

Expectation value for energy: ˆy y=n n nH E

* ˆ( , ) ( , )= Y YòH x t H x t dx

Notice that this result is independent of time t. Hence the expectation value of the Hamiltonian is a constant, a manifestation of the conservation of energy in quantum mechanics.

( ) ( )* ˆ= ( ) ( ) ( ) ( )y j y jå åò m m m n n nc x t H c x t dx

* * *= ( ) ( ) ( ) ( )y y j jåå òm n n m n m nc c E x x t t dx

2=å n nc E

TI SE

2 2

2ˆ ( )

2¶

= - +¶

!H V xm x

*m n mndx dY Y =ò

H = cn2En

n=1

∞

∑

Example 2.2: Energy (expectation)

3

0 if n is even

8 15 / ( ) if n is oddp

ìï= íïî

nc n

( ) ( )2 2 2/2

1

2, sin pp¥ -

=


!i n ma tn

n

nx t c x ea a

2 2 2

22p

=!

nnEma

2= =å n nH c E

2

4 2 41,3,5,...

480 1p

¥

=

= =å !

n ma n

2

4 2 41,3,5,...

480 1p

¥

=å!

nma n

2

2

5=!Hma

2 2 2

1 2 2

4.9352p

= »! !Ema ma

Compare with:

The solution

2 2 2 2

3 21,3,5,...

8 15( ) 2

pp

¥

=

é ùê úë û

å !

n

nn ma

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Chapter 2 (part 1): Time-Independent Schrödinger Equation