Physics 2D Lecture Slides
Lecture 16: Feb 12th
Sunil Sinha
UCSD Physics
P(x,t)= |Ψ(x,t) |2
Wave Function of “Stuff” & Probability Density
Probability of a particle to
be in an interval a ≤ x ≤b is
area under the curve from
x=a to a=b
x
x=a
x=b
• Although not possible to specify with certainty the location of
particle, its possible to assign probability P(x)dx of finding particle
between x and x+dx
• P(x) dx = | Ψ(x,t)|2 dx
• E.g intensity distribution in light diffraction pattern is a measure of
the probability that a photon will strike a given point within the
pattern
Ψ: The Wave function Of A Particle
•
The particle must be some where
+"
2
|
!
(
x
,
t
)
|
dx = 1
$
#"
•
•
Any Ψ satisfying this condition is
NORMALIZED
Prob of finding particle in finite interval
The Wave Function is a mathematical
function that describes a physical
object  Wave function must have some
rigorous properties :
•
P(a " x " b) = #! * ( x, t ) ! ( x, t )dx •
a
• Fundamental aim of Quantum Mechanics •
– Given the wavefunction at some
•
instant (say t=0) find Ψ at some
b
subsequent time t
– Ψ(x,t=0)  Ψ(x,t) …evolution
– Think of a probabilistic view of
particle’s “newtonian trajectory”
• We are replacing Newton’s
2nd law for subatomic
systems
Ψ must be finite
Ψ must be continuous fn of x,t
Ψ must be single-valued
Ψ must be smooth fn 
d!
must be continuous
dx
WHY ?
Bad (Mathematical) Wave Functions Of a Physical System : You Decide Why
?
A Simple Wave Function : Free Particle
• Imagine a free particle of mass m , momentum p and K=p2/2m
• Under no force , no attractive or repulsive potential to influence it
• Particle is where it wants : can be any where [- ∞ ≤ x ≤ + ∞ ]
– Has No relationship, no mortgage , no quiz, no final exam….its essentially a
bum !
– how to describe a quantum mechanical bum ?
• Ψ(x,t)= Aei(kx-ωt) =A(Cos(kx-ωt)+ i sin (kx-ωt))
p
E
; !=
!
!
For non-relativistic particles
k=
p2
!k 2
E=
" ! (k)=
2m
2m
Has definite momentum
and energy but location
unknown !
X
Wave Function of Different Kind of Free Particle : Wave Packet
Sum of Plane Waves:
$ ( x, 0) =
+"
&
a (k )eikx dk
#"
$ ( x, t ) =
+"
&
Combine many free waves to create a
Localized wave packet (group)
a (k )ei ( kx #! t ) dk
#"
Wave Packet initially localized
in %X, %t undergoes dispersion
The more you know now,
The less you will know
later
Why ?
Spreading is due to DISPERSION resulting
from the fact that phase velocity of individual
waves making up the packet depends on λ (k)
Normalization Condition: Particle Must be Somewhere
Example : ! ( x, 0) = Ce
"
x
x0
, C & x 0 are constants
This is a symmetric wavefunction with diminishing amplitude
The Amplitude is maximum at x =0 # Probability is max too
Normalization Condition: How to figure out C ?
A real particle must be somewhere: Probability of finding
particle is finite P(-$ + x + +$) =
+$
,
2
! ( x, 0) dx =
-$
# 1 = 2C
$
2
,e
0
"2
x
x0
,
-$
% x0 &
dx = 2C ' ( = C 2 x0
)2*
2
# ! ( x, 0) =
+$
1
e
x0
"
x
x0
C 2e
"2
x
x0
dx = 1
Where is the particle within a certain location x ± Δx
Prob |Ψ(x,0)|2
Lets Freeze time (t=0)
P(-x 0 # x # +x 0 ) =
?
+x 0
+
-x 0
x
2
! ( x, 0) dx =
+x 0
+
C 2e
"2
x
x0
-x 0
$ x0 %
"2
"2
$
%
$
= 2C ) * '1 " e ( = '1 " e %( = 0.865 & 87%
'2(
2
dx
Where Do Wave Functions Come From ?
• Are solutions of the time
dependent Schrödinger
Differential Equation (inspired
by Wave Equation seen in 2C)
! 2 ! 2 " ( x, t )
!" ( x, t )
#
+
U
(
x
)
"
(
x
,
t
)
=
i
!
2m !x 2
!t
• Given a potential U(x) 
particle under certain force
– F(x) =
"
!U ( x)
!x
Schrodinger had an interesting life
Schrodinger Wave Equation
Wavefunction " which is a sol. of the Sch. Equation embodies
all modern physics experienced/learnt so far:
E=hf,
h
p= ,
!
#x.#p ! " , #E.#t ! ", quantization etc
Schrodinger Equation is a Dynamical Equation
#
#
much like Newton's Equation F= m a
$
" (x,0) $ Force (potential) $ " (x,t)
Evolves the System as a function of space-time
The Schrodinger Eq. propogates the system
forward & backward in time:
& d" '
" (x,% t) = " (x,0) ± (
%t
)
* dt + t =0
Where does it come from ?? ..."First Principles"..no real derivation exists
Time Independent Sch. Equation
! 2 " 2 #(x,t)
"#(x,t)
!
+ U (x)#(x,t) = i!
2
2m "x
"t
Sometimes (depending on the character of the Potential U(x,t))
The Wave function is factorizable: can be broken up
( )
# x,t = $ (x) % (t)
Example : Plane Wave #(x,t)=e i(kx-& t) = e i(kx)e-i(& t)
In such cases, use separation of variables to get :
-! 2
" 2$ (x)
"% (t)
% (t). 2
+ U (x)$ (x)% (t) = i!$ (x)
2m
"t
"x
Divide Throughout by #(x,t)=$ (x)% (t)
-! 2 1 " 2$ (x)
1 "% (t)
'
. 2
+ U (x) = i!
2m $ (x) " x
% (t) "t
LHS is a function of x; RHS is fn of t
x and t are independent variables, hence :
' RHS = LHS = Constant = E
Factorization Condition For Wave Function Leads to:
-! 2 # 2! ( x)
+ U ( x)! ( x) = E ! ( x)
2
2m # x
#" (t )
i!
= E" (t )
#t
What is the Constant E ? How to Interpret it ?
Back to a Free particle :
# (x,t)= Aeikx e-i! t , " (x)= Aeikx
U(x,t) = 0
Plug it into the Time Independent Schrodinger Equation (TISE) $
%! 2 d 2 ( Ae(ikx ) )
!2k 2 p 2
( ikx )
+ 0 = E Ae
$E=
=
= (NR Energy)
2
2m
dx
2m 2m
Stationary states of the free particle: # (x,t)=" (x)e-i! t
2
$ # ( x, t ) = " ( x )
2
Probability is static in time t, character of wave function depends on " ( x)
A More Interesting Potential : Particle In a Box
U(x)
Write the Form of Potential: Infinite Wall
U(x,t) = !; x " 0, x # L
U(x,t) = 0 ; 0 < X < L
•
Classical Picture:
•Particle dances back and forth
•Constant speed, const KE
•Average <P> = 0
•No restriction on energy value
• E=K+U = K+0
•Particle can not exist outside box
•Can’t get out because needs to borrow
infinite energy to overcome potential of
wall
What happens when the joker
is subatomic in size ??
Example of a Particle Inside a Box With Infinite Potential
(a) Electron placed between 2 set of electrodes C &
grids G experiences no force in the region between
grids, which are held at Ground Potential
However in the regions between each C & G is a
repelling electric field whose strength depends on
the magnitude of V
(b) If V is small, then electron’s potential energy vs x
has low sloping “walls”
(c) If V is large, the “walls”become very high & steep
becoming infinitely high for V→∞
(d) The straight infinite walls are an approximation of
such a situation
U=∞
U(x)
U=∞
Ψ(x) for Particle Inside 1D Box with Infinite Potential Walls
Inside the box, no force " U=0 or constant (same thing)
"
-! d ! ( x )
+ 0 ! ( x) = E ! ( x)
2m dx 2
2
2
d 2! ( x)
2mE
2
2
"
=
#
k
!
(
x
)
;
k
=
dx 2
!2
d 2! ( x)
or
+ k 2! ( x) = 0 $ figure out what ! (x) solves this diff eq.
2
dx
In General the solution is ! ( x) = A sinkx + B coskx (A,B are constants)
Why can’t the
particle exist
Outside the box ?
 E Conservation
∞
∞
Need to figure out values of A, B : How to do that ?
Apply BOUNDARY Conditions on the Physical Wavefunction
We said ! ( x) must be continuous everywhere
So match the wavefunction just outside box to the wavefunction value
just inside the box
" At x = 0 " ! ( x = 0) = 0 & At x = L " ! ( x = L) = 0
% ! ( x = 0) = B = 0 (Continuity condition at x =0)
& ! ( x = L) = 0 " A Sin kL = 0 (Continuity condition at x =L)
" kL = n& " k =
n&
, n = 1, 2,3,...'
L
n 2& 2 ! 2
So what does this say about Energy E ? : E n =
Quantized (not Continuous)!
2mL2
X=0
X=L
Quantized Energy levels of Particle in a Box
The Quantum Mechanics of Christina Aguilera!
Christina at rest between two walls originally at infinity: Uncertainty in her location ΔX = ∞ .
At rest means her momentum P=0 , ΔP=0 (Uncertainty principle)
Slowly two walls move in from infinity on each side, now ΔX = L , so Δp ≠ 0
She is not at rest now, in fact her momentum P ≈ ± (h/2π L)
L
X Axis
0
Christina’s Momentum p
On average, measure <p> = 0
but there are quite large fluctuations!
Width of Distribution = !P
!
!P = ( P 2 ) ave " ( Pave ) 2 ; !P "
L
Bottomline : Christina dances to the tune of Uncertainty Principle!
What About the Wave Function Normalization ?
The particle's Energy and Wavefunction are determined by a number n
We will call n ! Quantum Number , just like in Bohr's Hydrogen atom
What about the wave functions corresponding to each of these energy states?
# n = A sin(kx) = A sin(
n" x
)
L
for 0<x < L
for x % 0, x % L
=0
Normalized Condition :
L
L
0
0
1 = '# n *# n dx = A2 ' Sin 2 (
n" x
)
L
Use 2Sin 2$ = 1 & 2Cos 2$
A2 (
2n" x )
1=
1
&
c
os(
) + and since ' cos $ = sin$
*
'
2 0,
L L
A2
1=
L
2
. A=
So # n =
2
sin(kx) =
L
2
L
2
n" x
sin(
)
L
L
...What does this look like?
Wave Functions : Shapes Depend on Quantum # n
Wave Function
Probability P(x): Where the
particle likely to be
Zero Prob
Where in The World is Carmen San Diego?
• We can only guess the
probability of finding the particle
somewhere in x
– For n=1 (ground state) particle most
likely at x = L/2
– For n=2 (first excited state) particle
most likely at L/4, 3L/4
• Prob. Vanishes at x = L/2 & L
– How does the particle get
from just before x=L/2 to
just after?
» QUIT thinking this way,
particles don’t have
trajectories
Classically, where is the particle most
» Just probabilities of
Likely to be : Equal prob of being
being somewhere
anywhere inside the Box
NOT SO says Quantum Mechanics!
How to Calculate the QM prob of Finding Particle in Some region in Space
Consider n =1 state of the particle
L
3L
Ask : What is P ( # x # ) ?
4
4
3L
4
P=
1"
L
4
2
1
2
dx =
L
3L
4
!x
$2% 1
1L sin L dx = ') L (* . 2
2
4
3L
4
2! x
1L (1 & cos L )dx
4
3L / 4
1 +L ,+ L
2! x ,
1 1
P = - & . - sin
= &
.
L / 2 0 / 2!
L 0 L / 4 2 2!
2! 3L
2! L %
$
. & sin
. (
' sin
L 4
L 4*
)
1 1
P= &
(&1 & 1) = 0.818 2 81.8%
2 2!
Classically 2 50% (equal prob over half the box size)
2 Substantial difference between Classical & Quantum predictions
When The Classical & Quantum Pictures Merge: n→∞
But one issue is irreconcilable:
Quantum Mechanically the particle can not have E = 0
This is a consequence of the Uncertainty Principle
The particle moves around with KE inversely proportional to the Length
Of the 1D Box
Finite Potential Barrier
• There are no Infinite Potentials in the real world
– Imagine the cost of as battery with infinite potential diff
• Will cost infinite $ sum + not available at Radio Shack
• Imagine a realistic potential : Large U compared to KE
but not infinite
Region I
U
Region II
Region III
E=KE
X=0
X=L
Classical Picture : A bound particle (no escape) in 0<x<L
Quantum Mechanical Picture : Use ΔE.Δt ≤ h/2π
Particle can leak out of the Box of finite potential P(|x|>L) ≠0
Finite Potential Well
-! 2 d 2" ( x)
+ U" ( x) = E " ( x)
2
2m dx
d 2" ( x) 2m
$
= 2 (U # E )" ( x)
2
dx
!
2m(U-E)
= ! 2 " ( x); ! =
!2
$General Solutions : " ( x) = Ae +! x + Be #! x
Require finiteness of " ( x)
$ " ( x) = Ae +! x .....x<0
(region I)
" ( x) = Ae #! x .....x>L (region III)
Again, coefficients A & B come from matching conditions
at the edge of the walls (x =0, L)
But note that wave fn at " ( x) at (x =0, L) % 0 !! (why?)
d" ( x)
Further require Continuity of " ( x) and
dx
These lead to rather different wave functions
Finite Potential Well: Particle can Burrow Outside Box
Finite Potential Well: Particle can Burrow Outside Box
Particle can be outside the box but only
for a time Δt ≈ h/ ΔE
ΔE = Energy particle needs to borrow to
Get outside ΔE = U-E + KE
The Cinderella act (of violating E
Conservation cant last very long
Particle must hurry back (cant be
caught with its hand inside the cookie-jar)
1
!
Penetration Length ! = =
"
2m(U-E)
If U>>E # Tiny penetration
If U $ % # ! $ 0
Finite Potential Well: Particle can Burrow Outside Box
1
!
Penetration Length ! = =
"
2m(U-E)
If U>>E # Tiny penetration
If U $ % # ! $ 0
n! !
En =
, n = 1, 2,3, 4...
2
2m( L + 2" )
When E=U then solutions blow up
# Limits to number of bound states(E n < U )
2
2
2
When E>U, particle is not bound and can get either
reflected or transmitted across the potential "barrier"
Simple Harmonic Oscillator:
Quantum and Classical
Spring with
Force Const
k
m
X=0
x
Unstable
a
Stable
U(x)
b
Stable Equilibrium: General Form :
c
Stable
x
1
U(x) =U(a)+ k ( x ! a ) 2
2
1
Rescale " U ( x ) = k ( x ! a ) 2
2
Motion of a Classical Oscillator (ideal)
Ball originally displaced from its equilibirium
position, motion confined between x=0 & x=A
1 2 1
k
2 2
U(x)=
kx
=
m
#
x
;
#
=
= Ang . Freq
Particle of mass m within a potential U(x)
2
2
m
!
dU ( x )
1
F(x)= E = kA2 " Changing A changes E
dx
2
!
E can take any value & if A $ 0, E $ 0
dU ( x )
F(x=a) = = 0,
Max. KE at x = 0, KE= 0 at x= ± A
dx
!
!
F(x=b) = 0 , F(x=c)=0 ...But...
look at the Curvature:
! 2U
! 2U
> 0 (stable),
< 0 (unstable)
2
2
!x
!x
Quantum Picture: Harmonic Oscillator
Find the Ground state Wave Function ! (x)
1
Find the Ground state Energy E when U(x)= m" 2 x 2
2
-! 2 # 2! ( x ) 1
Time Dependent Schrodinger Eqn:
+ m" 2 x 2! ( x ) = E ! ( x )
2
2m # x
2
d 2! ( x ) 2m
1
$
=
(
E
%
m" 2 x 2 )! ( x ) = 0 What ! (x) solves this?
2
2
dx
!
2
Two guesses about the simplest Wavefunction:
1. ! (x) should be symmetric about x 2. ! (x) & 0 as x & '
d! (x)
+ ! (x) should be continuous &
= continuous
dx
2
My guess: ! (x) = C0e %( x ; Need to find C0 & ( :
What does this wavefunction & PDF look like?
Quantum Picture: Harmonic Oscillator
2
P(x) = C 0e
" (x) = C0e
"2! x 2
C2
#! x 2
C0
x
How to Get C0 & α ?? …Try plugging in the wave-function into
the time-independent Schr. Eqn.
Time Independent Sch. Eqn & The Harmonic Oscillator
! 2" (x) 2m 1
2 2
Master Equation is :
=
[
m
#
x $ E]" (x)
2
2
!x
! 2
d" (x)
$% x2
$% x2
Since " (x) = C0 e ,
= C0 ($2% x)e ,
dx
d 2" (x)
d($2% x) $% x 2
2 $% x2
2 2
$% x2
= C0
e
+ C0 ($2% x) e
= C0 [4% x $ 2% ]e
2
dx
dx
& C0 [ 4% x $ 2% ]e
2
2
$% x2
2m 1
2 2
$% x2
= 2 [ m# x $ E]C0 e
! 2
Match the coeff of x 2 and the Constant terms on LHS & RHS
2m 1
m#
2
2
& 4% = 2 m# or % =
2!
! 2
2m
& the other match gives 2% = 2 E, substituing % &
!
1
E= !#
2
!!!!......(Planck's Oscillators)
What about C0 ? We learn about that from the Normalization cond.
SHO: Normalization Condition
+#
+#
" m% x 2
2
!
0
2
|
!
(x)
|
dx = 1 = $ C e
$ 0
"#
"#
+#
Since
dx
$e
" ax 2
dx =
"#
Identifying a=
&
a
(dont memorize this)
m%
and using the identity above
!
1
4
( m% +
'
C0 = *
&
!
)
,
Hence the Complete NORMALIZED wave function is :
1
4
( m% +
! 0 (x) = *
- e
) &! ,
" m% x 2
2!
Ground State Wavefunction
has energy E = h% /2
Planck's Oscillators were electrons tied by the "spring" of the
mutually attractive Coulomb Force
Quantum Oscillator In Pictures
E = KE + U ( x) > 0 for n=0
Quantum Mechanical Prob for particle
To live outside classical turning points
Is finite !
U(x)
C0
U
-A
+A
-A
+A
Classically particle most likely to be at the turning point (velocity=0)
Quantum Mechanically , particle most likely to be at x=x0 for n=0
Classical & Quantum Pictures of SHO compared
• Limits of classical vibration : Turning Points (do on
Board)
• Quantum Probability for particle outside classical turning
points P(|x|>A) =16% !!
– Do it on the board (see Example problems in book)
Excited States of The Quantum Oscillator
m" x 2
2!
m"
x) e
;
!
H n ( y) = Hermite Polynomials
#
! n (x) = Cn H n (
with
H 0 (y)=1
H1 (y)=2y
H 2 (y)=4y2 # 2
H 3 (y)=8y3 # 12 y
H n (y)=(-1) n e y
2
n # y2
d e
dy n
and
1
En = (n + )!"
2
Again n=0,1,2,3...$ Quantum #
Excited States of The Quantum Oscillator
Ground State Energy >0 always
Measurement Expectation: Statistics Lesson
•
Ensemble & probable outcome of a single measurement or the average outcome
of a large # of measurements
"
n
< x >=
n1 x1 + n2 x2 + n3 x3 + ....ni xi
=
n1 + n2 + n3 + ...ni
$n x
i i
i =1
N
=
% xP( x )dx
#"
"
% P( x )dx
#"
For a general Fn f(x)
"
n
< f ( x ) >=
$ ni f ( xi )
i =1
N
=
*
!
% ( x ) f ( x )! ( x )dx
#"
Sharpness of A Distr:
Scatter around average
"
% P( x )dx
#"
!=
!=
2
(x
"
x
)
% i
N
( x 2 ) " ( x )2
! = small # Sharp distr.
Uncertainty $X = !
Particle in the Box, n=1, <x> & Δx ?
" (x)=
2
%! &
sin ' x (
L
)L *
$
2
%! &
sin ' x ( x
L
)L *
<x>= 0
-$
2
%! &
sin ' x ( dx
L
)L *
2
%! &
%! &
= 0 x sin 2 ' x (dx , change variable # = ' x (
L0
)L *
)L *
L
!
2
1
2
2
+ <x>=
#
sin
#
,
use
sin
#
=
(1 - cos 2# )
2 0
L! 0
2
!
!
1
2L ,
+ <x>= 2 . 0 # d# - 0 # cos2# d# 2 use 0 udv=uv-0 vdu
2! / 0
0
3
L %!2 & L
+ <x>= 2 ' ( =
(same result as from graphing " 2 ( x))
! ) 2 * 2
!
L2
L2
Similarly <x >= 0 x sin ( x)dx = - 2
L
3 2!
0
L
2
2
2
L2
L2
L2
and 4X= <x > - < x > =
- 2= 0.18 L
3 2!
4
4X= 20% of L, Particle not sharply confined in Box
2
2
Expectation Values & Operators: More Formally
• Observable: Any particle property that can be measured
– X,P, KE, E or some combination of them,e,g: x2
– How to calculate the probable value of these quantities for a QM state ?
• Operator: Associates an operator with each observable
– Using these Operators, one calculates the average value of that Observable
– The Operator acts on the Wavefunction (Operand) & extracts info about the
Observable in a straightforward way gets Expectation value for that
observable
< Q >=
+!
$
#* ( x, t ) [Qˆ ] #* ( x, t )dx
"!
Q is the observable, [Qˆ ] is the operator
& < Q > is the Expectation value
Examples:
[X] = x ,
[P]2 -! 2 % 2
[K] =
=
2m 2m %x 2
! d
i dx
%
[E] = i!
%t
[P] =
Operators  Information Extractors
! d
Momentum Operator
i dx
gives the value of average mometum in the following way:
[p] or p̂ =
+"
<p> = # ! (x)[ p]! (x)dx =
*
Plug & play form
-"
+"
$ ! ' d!
*
!
(x)
#
&% i )( dx dx
-"
Similarly :
!2 d 2
[K] or K̂ = gives the value of average KE
2m dx 2
+"
+"
$ ! 2 d 2! (x) '
*
*
<K> = # ! (x)[K]! (x)dx = # ! (x) & *
dx
)
2
% 2m dx (
-"
-"
Similarly
+"
<U> = # ! * (x)[U (x)]! (x)dx : plug in the U(x) fn for that case
-"
$ ! 2 d 2! (x)
'
and <E> = # ! (x)[K + U (x)]! (x)dx = # ! (x) & *
+ U (x)) dx
2
2m
dx
%
(
-"
-"
+"
+"
*
*
Hamiltonian Operator [H] = [K] +[U]
The Energy Operator [E] = i!
+
informs you of the average energy
+t
[H] & [E] Operators
• [H] is a function of x
• [E] is a function of t …….they are really different operators
• But they produce identical results when applied to any solution of
the time-dependent Schrodinger Eq.
•
[H]Ψ(x,t) = [E] Ψ(x,t)
! !2 # 2
"
! #"
& $ 2m #x 2 + U ( x, t ) ' % ( x, t ) = &i! #t ' % ( x, t )
(
)
(
)
• Think of S. Eq as an expression for Energy conservation for a
Quantum system
Example : Average Momentum of particle in box
! n (x) =
2
n"
sin( x)
L
L
+(
&
! *n (x) =
2
n"
sin( x)
L
L
(
* #! d %
#
%
!
p
!
dx
=
!
) $ &
) *$ i dx +&! dx
'(
'(
L
! 2 n"
n"
n"
< p >=
(
x)cos(
x)dx
i L L )0
L
L
< p >=
*
Since ) sinax cosax dx =
!#
n"
,< p >= *sin 2 (
iL $
L
1
n"
sin 2 ax ...here a =
2a
L
x= L
%
x + = 0 since Sin 2 (0) = Sin 2 (n" ) = 0
& x =0
We knew THAT before doing any math !
Quiz 1: What is the <p> for the Quantum Oscillator in its symmetric ground state
Quiz 2: What is the <p> for the Quantum Oscillator in its asymmetric first excited state
But what about the <KE> of the Particle in Box ?
p2
< p >= 0 so what about expectation value of K=
?
2m
< K >= 0 because < p >= 0; clearly not, since we showed E=KE ! 0
Why ? What gives ?
n" !
Because p n = ± 2mEn = ±
; " ± " is the key!
L
AVERAGE p =0 , particle is moving back & forth
p2
<p > 2
<KE> = <
> ! 0 not
2m
2m
!
Be careful when being "lazy"
Quiz: what about <KE> of a quantum Oscillator?
Does similar logic apply??
Schrodinger Eqn: Stationary State Form
d
1 df (t )
Since [ln f (t ) ] =
dt
f (t ) dt
"! (t )
1 "! (t ) E
iE
In i!
= E! (t ) , rewrite as
= =#
"t
! (t ) "t
i!
!
and integrate both sides w.r.t. time
t =t
1 "! (t )
iE
1 d! (t )
iE
&t=0 ! (t ) "t dt = &0 # ! dt % &0 ! (t ) dt dt = # !
t
t
iE
$ ln ! (t ) # ln ! (0) = # t , now exponentiate both sides
!
% ! (t ) = ! (0)e
% ! (t ) = e
#
iE
t
!
#
iE
t
!
; ! (0) = constant= initial condition = 1 (e.g)
& Thus ( (x,t)=' (x)e
#
iE
t
!
where E = energy of system
Schrodinger Eqn: Stationary State Form
+
iE
t
!
"
iE
t
!
iE iE
t" t
!
!
P ( x, t ) = # * # = ! * ( x ) e ! ( x ) e
= ! * ( x)! ( x)e
=| ! ( x) |2
In such cases, P(x,t) is INDEPENDENT of time.
These are called "stationary" states because Prob is independent of time
Examples : Particle in a box (why?)
: Quantum Oscillator (why?)
Total energy of the system depends on the spatial orientation
of the system : charteristic of the potential situation !