chapter 2 (part 1): time-independent schrödinger equation
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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.
The Infinite Square WellV(x)
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
The Infinite Square WellV(x)
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
The Infinite Square WellV(x)
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
( ) 2 sin py æ ö= ç ÷è ø
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
Properties of the SolutionV(x)
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
Properties of the SolutionV(x)
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
Find the wave function at all later times.
Steps to solve:1. Normalize
2. Find cn
3. Build the time dependent wave function
( ) 2 sin py æ ö= ç ÷è ø
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
Find the wave function at all later times.
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
Find the wave function at all later times.
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
( ) 2 sin py æ ö= ç ÷è ø
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
Find the wave function at all later times.
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¥ -
=
æ öY = ç ÷è øS
!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¥ -
=
æ öY = ç ÷è øS
!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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