schrodinger eqn
DESCRIPTION
quantum mechanicsTRANSCRIPT
Introduction to Quantum MechanicsSchrodinger’s Equation
Instructor: Gautam Dutta
DA-IICT
winter 2012
Schrodinger’s Equation
Schrodinger’s Equation
Particles are associated with waves.
Waves of what? – Electric field, Magnetic field, mass density...
Schrodinger: Probability waves
P (x) = e2πih (px−Et) = e
i~(px−Et) (1)
Here ~ = h2π.
Can probability be negative? Experimentally these probabilities must
form an interference pattern.
Electromagnetic waves have amplitude ~E which can be both positive
[1]
Schrodinger’s Equation
and negative but intensity that we measure is |~E|2
We denote the wave amplitude by ψ(x) given by Eq.1 and the
non-negative probability P (x) will be given as
P (x) = |ψ(x)|2 = ψ∗(x)ψ(x)
ψ(x) can be +ve or -ve. We allow it to be complex in general. We
need such a quantity to produce the interference effect.
ψ(x) is called the probability amplitude and P (x) is the probability
density for the particle to be found at x.
[2]
Schrodinger’s Equation
Since it is certain to find the particle somewhere between −∞ and
+∞, it is obvious to expect the following normalization for ψ∫ ∞
−∞ψ∗ψdx = 1
What is the Equation that governs the evolution of ψ Consider
ψ = ei~(px−Et). Then
−~2∂2ψ
∂x2= p2ψ
Also
i~∂ψ
∂t= Eψ
[3]
Schrodinger’s Equation
Classically E = p2/2m for a free particle. So ψ = ei~(px−Et) satisfies
the following equation
i~∂ψ
∂t= − ~2
2m∂2ψ
∂x2(2)
If the particle is under the influence of a pottential V (x) then
E =p2
2m+ V
An appropriate equation for the wavefunction of such a particle would
be
i~∂ψ
∂t= − ~2
2m∂2ψ
∂x2+ V (x)ψ (3)
[4]
Schrodinger’s Equation
Obviously ψ = ei~(px−Et) with a single value of E and p will not
satisfy this equation. A superposition
ψ(x, t) =∫f(p)e
i~(px−Et)dp
might satisfy.
The above Equation was suggested by Erwin Schrodinger for the
evolution of the wavefunction of a particle of mass m in a pottential
V .
It is like the Newton’s Equation in Classical Mechanics. Once we
solve Schrodinger’s Equation and find ψ evrything about the particle
is known that is possible to be known in Quantum Mechanics.
ψ(x, t) gives the probability amplitude of finding the particle at the
[5]
Schrodinger’s Equation
point x at the time t. The probability of finding the particle between
x and x + dx is ψ∗ψdx. So the probability of finding the particle
between x1 and x2 is
P (x1 ≤ x ≤ x2) =∫ x2
x1
ψ∗ψdx
[6]
Schrodinger’s Equation
Operators: How do we get the values of physical quantities like
Energy, Momentum, Position Angular Momentum etc from ψ ?
Here we don’t have a well defined position and momentum of the
particle. So we can’t assign values of physical quantities to the
particle, like energy, angular momentum etc, which are functions of
its position and momentum.
But here we have states of particle defined by the wavefunctions
satisfying the Schrodinger’s equation. The collection of all possible
states of the particle constitute a vector space since the Schrodinger’s
equation is a linear differential equation. We can define linear
operators on this vector space. Certain states may be eigenstates of
[7]
Schrodinger’s Equation
these operators. It can be shown that the eigenstates of an operator
are linearly independent and they constitute a basis of the vector
space. We take the example of a free particle described by the
wavefunction ψ = ei~(px−Et). Then
−i~ ∂∂xψ = pψ
So the free particle states are eigenstates of the operator p = −i~ ∂∂x.
Similarly as we have seen earlier these states are eigenstates of the
energy operator E = i~ ∂∂t.
[8]
Schrodinger’s Equation
How do we construct the position operator?
Consider the fourier transform of the spatial wave function ψ(x)
f(p) =∫ψ(x)e−
i~pxdx; ψ(x) =
∫f(p)e
i~pxdp
Then
pψ(x) = −i~ ∂∂xψ(x) =
∫pf(p)e
i~pxdp
and
i~∂f(p)∂p
=∫xψ(x)e−
i~pxdx
[9]
Schrodinger’s Equation
We have the following pairs of fourier transforms:
f(p) ↔ ψ(x)
pf(p) ↔ −i~ ∂∂xψ(x)
i~∂
∂pf(p) ↔ xψ(x)
In the position representation, −i~ ∂∂x is the momentum operator.
By analogy i~ ∂∂p is the position operator in the momentum
representation.
The third fourier transform pair above suggests that in the position
representation the position operator is x. So in the position
representation the position operator is a multiplicative operator i.e
[10]
Schrodinger’s Equation
xψ(x) = xψ(x).
Every measurable physical quantity is associated with an operator.
The form of the operator has to be known in a particular
representation. An Operator can be a differential form or a
multiplicative form depending upon the representation we are working
with. For example in the position representation the momentum
operator is given by the differential form
p = −i~ ∂∂x
while the position operator is given by the multiplicative form
x = x
If the wavefunction of the particle is an eigenfunction of an operator
[11]
Schrodinger’s Equation
then it is easy to find the physical quantity corresponding to that
operator.
Oψ = oψ
ψ describes the state of a particle in which the value of the physical
quantity corresponding to the operator O is o. E.g
pei~(px−Et) = pe
i~(px−Et)
Eei~(px−Et) = Ee
i~(px−Et)
However in general ψ may not be in an eigenstate of momentum and
energy. If ψ(x, t) =∫f(p)e
i~(px−Et)dp then
pψ =∫f(p)pe
i~(px−Et)dp
[12]
Schrodinger’s Equation
Probability density of finding the particle with a momentum p is given
by |f(p)|2 = f(p)∗f(p). The average momentum or the Expectation
value of momentum is
p =∫|f(p)|2pdp
Note that this is equal to∫∞−∞ψ∗pψdx.
Try this exercise.
[13]
Schrodinger’s Equation
Steady state or Stationary statesIf the total energy of a particle is constant in a state then its wave
function Ψ(x, t) can be written as
Ψ(x, t) = e−i~ Etψ(x)
Substituting into Schrodinger’s Equation
− i~∂Ψ∂t
=−12m
~2∂2Ψ∂x2
+ VΨ
∴ Eei~Etψ(x) = e
i~Et
[−~2
2m∂2ψ(x)∂x2
+ V ψ(x)]
[14]
Schrodinger’s Equation
∴
[−~2
2m∂2
∂x2+ V
]ψ(x) = Eψ(x)
Hψ(x) = Eψ(x) (4)
Where the operator
H =−~2
2m∂2
∂x2+ V
is called the Hamiltonian operator and equation4 is the steady state
form of the Schrodinger Equation.
It is an eigenvalue equation and it cannot in general be solved for any
value of E and it depends upon the boundary condition. Only specific
values En are allowed. These values are called the eigenvalues of the
of H. The corresponding solution to the equation are labelled by n.
[15]
Schrodinger’s Equation
Thus
Hψn(x) = Enψn(x)
Quantization comes naturally in Schrodinger’s formalism.
[16]
Schrodinger’s Equation
Application of Schrodinger EquationParticle in a rigid box Consider a particle of mass m confined within
a rigid box of side L. In one dimension the potential in which the
particle is kept can be given as
V = 0 for 0 < x < 1
V = ∞ for x ≤ 0 and x ≥ L
In the region outside the box the particle cannot exist since the
potential is infinite. Within the box the Schrodinger Eqn is
− ~2
2m∂2ψ
∂x2= Eψ
∴∂2ψ
∂x2=
2mE~2
ψ
[17]
Schrodinger’s Equation
ψ(x) = A sin√
2mE~
x+B cos√
2mE~
x
Boundary conditions: ψ = 0 at x = 0=⇒ B = 0
ψ(x) = A sin√
2mE~
x
ψ = 0 at x = L.
ψ(L) = A sin√
2mE~
L = 0
=⇒√
2mE~
L = nπ
∴ E =n2π2~2
2mL2
Normalisation:
[18]
Schrodinger’s Equation∫ ∞
−infty
|ψ(x)|2dx = 1
∴ A2
∫ L
0
sin2 nπx
Ldx = 1
∴ A2L
2= 1 =⇒ A =
√2L
Find the average value of momentum p, p2, position x, energy E
Find (∆p)2 = p2 − p2, (∆x)2 and ∆p∆x
[19]
Schrodinger’s Equation
Physical Interpretation of Energy EigenfunctionIf Ψ(x, t) = e
i~Entψn(x), the particle has a fixed energy En. If we
have several identical systems and we measure the energy of the
particle independently in each of these systems then the energy will
turn out to be En.
ψn
En
ψn
En
ψn
En
ψn
En
[20]
Schrodinger’s Equation
In general the particle may not be in any eigenstate. In such
a case a measurement in identical systems will give different values
but each measurement will allways give only one of the eigenvalues
E1, E2, E3, .....En, ..... The same applies for any other observable
O that we measure.
ψ
E k
ψ
E E
ψ
E
ψ
4 1 2
[21]
Schrodinger’s Equation
Let
Ψ(x, t) = a1Ψ1(x, t) + a2Ψ2(x, t) + ....+ anΨn(x, t)
Since energy is a physical observable with real values, the energy
eigenfunctions are real.
The average value of the energy in the linearly superposed state
Ψ(x, t) must also be real.
〈E〉 =∫ ∞
−∞Ψ∗(x, t)
i~∂∂t
Ψ(x, t)dx
=n∑
i=1
|ai|2Ei +∑i 6=j
a∗iajEje− i
~(Ei−Ej)t
∫ ∞
−∞ψ∗i (x)ψj(x)dx
[22]
Schrodinger’s Equation
Since 〈E〉 must be a real number in any state, we must have∫ ∞
−∞ψ∗i (x)ψj(x)dx = 0 for Ei 6= Ej
∫∞−∞ψ∗(x)φ(x)dx is defined as the scalar product in the space of the
states of the quantum system.
So we conclude that eigenstates corresponding to different eigenvalues
of a physical observable are orthogonal. We will see that the linear
operators corresponding to physical observable are hermitian.
∫ ∞
−∞Ψ∗Ψ = 1 =
∑i
|ai|2
[23]
Schrodinger’s Equation
Also
〈E〉 =∑
i
|ai|2Ei
These implies the following interpretations:
We measure the energy of the system in this state. We will always
get the values E1, E2...En with the following probabilities
P (E1) = |a1|2
P (E2) = |a2|2
..... = ....
P (En) = |an|2
Mesurement in Quantum Mechanics
[24]
Schrodinger’s Equation
Let ψ(x) be an eigenstate corresponding to a potential V (x) with
energy E. The measuring system will be some potential V ′(x), say
the infinite well potential.
Then the process of measurement spontaneously throws(captures)
the particle in one of the eigenstates of V ′(x) ψ1, ψ2,... ψn
with probability |a1|2, |a2|2, ....|an|2. The process of measurement
inherrently have to disturb the system and we cannot measure the
exact value E in any measurement.
[25]
Schrodinger’s Equation
V V’
ψ
ψ
ψ
ψ
1
2
3E
E 1
E
E 1
2
3
Given system Measuring system
[26]
Schrodinger’s Equation
The Harmonic Oscillator
V(x)
V (x) =12kx2
[27]
Schrodinger’s Equation
H =p2
2m+ V (x)
= − ~2
2m∂2
∂x2+
12kx2 (5)
Time independent Schrodinger’s Eqn:
− ~2
2m∂2ψ(x)∂x2
+12kx2ψ(x) = Eψ(x)
ψ(x) = e−αx2with α =
√mk
2~is a solution with energy
E =~2
√k
m=
12~ω
[28]
Schrodinger’s Equation
In general the solutions which are ”well defined” are given by
ψn(x) = AnHn(√
2αx)e−αx2
where
H0(y) = 1
H1(y) = 2y
H2(y) = 4y2 − 2
H3(y) = 8y3 − 12y
H4(y) = 16y4 − 48y2 + 12
The energy in the state ψn(x) is En = (n+ 1/2)~ω
[29]
Schrodinger’s Equation
V(x)
E
E
E
E
0
1
2
3
The wavefunction exist even in the classically forbidden region.
[30]
Schrodinger’s Equation
V(x)
E 4
ψ4
−A A
The wavefunction exist even in the classically forbidden region.
[31]
Schrodinger’s Equation
Particle in a non-rigid boxFinite potential well.
Asin(kx) +B Cox(kx)
V
L
e k(L−x)kxe
Properties of wavefunctions1) ψ should be continuous
2) lim|x|→∞ψ(x) = 0
[32]
Schrodinger’s Equation
3)∫∞−∞ |ψ|
2dx = 14) The first spatial derivative of ψ must be continuous wherever V
is finite. Let us divide the whole space into three distinct regions.
Region 1: −∞ < x < 0; V (x) = V0
Region 2: 0 < x < L; V (x) = 0Region 3: L < x <∞; V (x) = V0
We solve the Schrodinger’s Equation in the three regions and
match Ψ(x) and its derivative at the boundaries of the regions.
In region 2 we have
− ~2
2m∂2Ψdx2
= EΨ
[33]
Schrodinger’s Equation
Here the soln is
Ψ2(x) = A2ek2x +B2e
−k2x k2 =
√−2mE
~2
where A2 and B2 are arbitrary constants to be determined from the
boundary conditions.
In region 1 the time independent Eqn is
− ~2
2m∂2Ψdx2
+ V0Ψ = EΨ
Here the soln is
Ψ1(x) = A1ek1x +B1e
−k1x k1 =
√−2m(E − V0)
~2
[34]
Schrodinger’s Equation
Similarly in region 3 we have
Ψ3(x) = A3ek3x +B3e
−k3x k3 =
√−2m(E − V0)
~2
Note that k2 is imaginary, while k1 and k3 are real.
As x → −∞, B1e−k1x → ∞. So we put B1 = 0 As x →
∞, A3ek3x →∞. So we put A3 = 0 So we have
Ψ1(x) = A1ek1x
Ψ2(x) = A2ek2x +B2e
−k2x
Ψ3(x) = B3e−k3x
[35]
Schrodinger’s Equation
At x = 0 we have Ψ1(0) = Ψ2(0). This gives
A1 = A2 +B2 (6)
At x = L we have Ψ2(L) = Ψ3(L). This gives
A2ek2L +B2e
−k2L = B3e−k3L (7)
We match the first derivative of Ψ at the two boundaries of the well,
Ψ′1(0) = Ψ′
2(0) and Ψ′2(L) = Ψ′
3(L) This gives
A1k1 = (A2 −B2)k2 (8)
−B3k3e−k3L = (A2e
k2L −B2e−k2L)k2 (9)
[36]
Schrodinger’s Equation
From Eq.6 and Eq.8 we get
A2 =12
(1 +
k1
k2
)A1 B2 =
12
(1− k1
k2
)A1
From 7 and 9 we have
B3 =k2 + k1
k2 − k3A1e
(k2+k3)L
So we have
Ψ1(x) = A1ek1x
Ψ2(x) =A1
2
(1 +
k1
k2
)ek2x +
A1
2
(1− k1
k2
)e−k2x
Ψ3(x) = A1k2 + k1
k2 − k3ek2Lek3(L−x)
[37]
Schrodinger’s Equation
A1 can be obtained by normalizing Ψ(x).
k1
k2=√V0 − E
i√E
= −i√V0
E− 1
∴ 1 +k1
k2=
√V0
Ee−iα where tanα =
√V0
E− 1
1− k1
k2=
√V0
Eeiα
k2 + k1
k2 − k3= e−2iα
[38]
Schrodinger’s Equation
Now Ψ2 can be written as
Ψ2(x) = A1
√V0
Ecos
(√2mE~
x− α
)
At x = L, Ψ2(L) = Ψ3(L). This gives√V0
Ecos
(√2mE~
L− α
)= ei(−2α+
√2mE~ L) (10)
The imaginary part of the R.H.S must equal to 0. So we have
√2mE~
L− 2α = nπ
[39]
Schrodinger’s Equation
So Eqn.(10) becomes
cos
(√2mE2~
L+nπ
2
)= (−1)n
√E
V0
The energy values satisfying the above transcendental equation gives
the discrete energies that the particle can have within the box.
Though n can be any integer, distinct values of energy are obtained
only for n = 0, 1, 2and3. Beyond this we get the cosine curves are
the same and hence we get the same energies. We can plot the
two graphs (−1)n√
EV0
and cos(√
2mE2~ L+ nπ
2
)as a function of
√E
and look for their intersection. All the solutions will lie for E < V0.
Beyond this the absolute value of R.H.S is greater than one and
hence it can’t intersect the sine or cosine curve on the L.H.S. The
[40]
Schrodinger’s Equation
number of discrete energy solutions we have depends upon the height
of the well V0 and the width L of the well. For every period of cosine
cycle we have 2 possible energies. So for n = 0, 1, 2, 3) we will have
2 × 4 = 8 distinct energy eigenvalues. The cosine cycle completes
when√E changes by 4π~√
2mL. So the number of discrete eigenvalues
less than V0 is given by
8×√V0
4π~√2mL
=2√
2mV0L
π~
When E > V0, the system is not bounded and we can have a valid
wavefunction for the particle for any continuous value of energy.
[41]
Schrodinger’s Equation
Figure 1: Eigen values of energy are given by the Eqn. cos�√
2mE2~ L + nπ
2
�= (−1)n
qEV0
. We
plot the two functions on the r.h.s and the l.h.s as a function of√
E. The solid curves are for even
n while the dashed curves are for odd n. The intersections give the values of E. There will be no
bound states beyond E = V0 .
[42]
Schrodinger’s Equation
Examples
1. Schrodinger Equation is a linear differential Equation. Show that
if Ψ1 and Ψ2 are two solutions of the Schrodinger Equation then
Ψ = a1Ψ1 + a2Ψ2
is also a solution of Schrodinger Equation.
2. The angular momentum along z direction is given by Lz = xpy −ypx. Treating px and py as operators viz..
px = −i~ ∂∂x
andpy = −i~ ∂∂y
[43]
Schrodinger’s Equation
show that
Lz = −i~ ∂
∂Φ
3. Solve the Schrodinger Equation for a free particle which is in a
stationary state with energy E. Show that
ψ(x) = Aei~√
2mEx +Be−i~√
2mEx
is a solution to the steady state Schrodinger Equation. Find the
average momentum in this state.
[44]
Schrodinger’s Equation
Two and three dimensionsSteady state Schrodinger Eqn:
p2
2mψ(x, y, z) + V (x, y, z)ψ(x, y, z) = Eψ(x, y, z)
p2 = px2 + py
2 + pz2
∴ Eψ(x, y, z) = − ~2
2m
(∂2
∂x2+
∂2
∂y2+
∂2
∂z2
)ψ(x, y, z)
+V (x, y, z)ψ(x, y, z)
= − ~2
2m∇2ψ(x, y, z) + V (x, y, z)ψ(x, y, z)
[45]
Schrodinger’s Equation
The momentum operator is a vector operator and is given by
~p = −i~~∇ = −i~(i∂
∂x+ j
∂
∂y+ k
∂
∂z
)
[46]
Schrodinger’s Equation
Two Dimensional box
V= 8
V= 8
V= 8
V= 8
V=0
a
b
x
y
V (x, y) = 0 0 ≤ x ≤ a and 0 ≤ y ≤ b
= ∞ elsewhere
[47]
Schrodinger’s Equation
− ~2
2m
(∂2
∂x2+
∂2
∂y2
)ψ(x, y) = Eψ(x, y)
Assume variable separable form
ψ(x, y) = ψ(x)ψ(y)
ψ(x, y) = A sin(nxπx
a) sin(
nyπy
b)
E =
(n2
x
a2+n2
y
b2
)~2π2
2m
[48]
Schrodinger’s Equation
For 3-dim box of size (a,b,c)
ψ(x, y, z) = A sin(nxπx
a) sin(
nyπy
b) sin(
nzπz
c)
E =
(n2
x
a2+n2
y
b2+n2
z
c2
)~2π2
2m
[49]
Schrodinger’s Equation
V=0
V= 8
a
V=0
a
b
V=0
a
b
c
[50]
Schrodinger’s Equation
Spherically symmetric potential
V (r, θ, φ) = V (r)
E.g Hydrogen atom
V (r) = − Ze2
4πε0r2Harmonic Oscillator
V (r) = kr2
Particle in spherically symmetric box
V (r) = 0, when 0 ≤ r ≤ a
= ∞. when r > a
[51]
Schrodinger’s Equation
a−a
V=0
V= 8Quantum Dot
[52]
Schrodinger’s Equation
Schrodinger’s Equation in Sphericle polar co-ordinate:
Eψ(r, θ, φ) = − ~2
2m∇2ψ(r, θ, φ) + V (r)ψ(r, θ, φ)
= − ~2
2m
[1r2∂
∂r
(r2∂
∂r
)+
1r2 sin θ
∂
∂θ
(sin θ
∂
∂θ
)+
1r2 sin2 θ
∂2
∂φ2
]ψ(r, θ, φ)
+V (r)ψ(r, θ, φ)
Comparing with the classical Equation
E =p2
r
2m+
L2
2mr2+ V (r)
[53]
Schrodinger’s Equation
L2 = −~2
[1
sin θ∂
∂θ
(sin θ
∂
∂θ
)+
1sin2 θ
∂2
∂φ2
]Exercise: Starting with ~L = ~r× ~p show that L
2is given by the above
operator.
[54]
Schrodinger’s Equation
Separate r and the angular variables
ψ(r, θ, φ) = R(r)Y (θ, φ)
After rearrangement the Sch Eqn gets decoupled into r and θ, φ
− ~2
2m1R
∂
∂r
(r2∂R
∂r
)+ r2V (r)− r2E
− ~2
2m1Y
[1
sin θ∂
∂θ
(sin θ
∂
∂θ
)+
1sin2 θ
∂2
∂φ2
]Y = 0
[55]
Schrodinger’s Equation
Multiplying this by 2m2 gives
− ~2 1R
∂
∂r
(r2∂R
∂r
)+ r22m(V (r)− E)
−~2 1Y
[1
sin θ∂
∂θ
(sin θ
∂
∂θ
)+
1sin2 θ
∂2
∂φ2
]Y = 0
The angular equation can have acceptable solutions only if it is equal
to a constant given by ~2l(l + 1) where l is an integer.[1
sin θ∂
∂θ
(sin θ
∂
∂θ
)+
1sin2 θ
∂2
∂φ2
]Y = −l(l + 1)Y
Y (θ, φ) are called spherical harmonics. Again by separation of
[56]
Schrodinger’s Equation
variables θ and φ it can be shown that
Y (θ, φ) = Θ(θ)Φ(φ) = Pml (θ)e±imφ
e±imφ is the solution to the φ part of the above equation after
separation into θ and φ
∂2Φ∂φ2
= −m2Φ
The periodic boundary condition along φ requires m to be an integer.
The functions Pml (θ) are known as Associated Lagendre polynomials
given as
Pml (θ) = (−1)m(1− cos2 θ)m/2 dm
d cosm θPl(cos θ)
[57]
Schrodinger’s Equation
Here Pl(cos θ) is the Legendre Polynomial of degree l.
So 0 ≤ m ≤ l.
Operating Lz on ψ(r, θ, φ) gives (check this)
Lzψ = −i~∂ψ∂φ
= ~mψ
Also
L2ψ = ~2l(l + 1)ψ
So the energy eigen states of any spherically symmetrical pottential
are also eigenstates of Lz and L2
[58]
Schrodinger’s Equation
−h v/ l(l+1) −h
z
m
L and L2z
[59]
Schrodinger’s Equation
/l(l+1)v h−
h−
h−
h−
h−
h−
z
m=2
1
0
−1
−2
l=2
Eg. Show that the wave function given by R(r)P 11 (cos θ)eiφ,
R(r)P1(cos θ) and R(r)P 11 (cos θ)e−iφ are eigenstates of L
2and Lz
operators and find the corresponding eigenvalues. (P1(cos θ) = cos θ)
[60]
Schrodinger’s Equation
Summary
• Given any system specified by a potential V (r) we can find
energy eigen states of the system which are also called stationary
states. These states are physicaly acceptable solutions to the time
independent Schrodinger Equation.
HΨ(~r) = EΨ(~r)
The probability of finding the particle at the position ~r will be given
as |Ψ(~r)|2 ∫allspace
|Ψ(~r)|2dV = 1
[61]
Schrodinger’s Equation
• Physical Quantities like momentum, angular momentum etc. are
replaced by their corresponding operators. A given stationary state
Ψ may or may not be also an eigenstate of these operators. If they
are then the physical quantity corresponding operator has a well
defined value in the state Ψ given by the eigenvalue. Otherwise we
can only talk about the average value of that Physical quantity.
• Any general state Ψ of a particle can be expressed
as a linear combination(superposition) of the statinary
states Ψ1,Ψ2,Ψ3,.....,Ψn with energy eigenvalues E1, E2, .....En
respectively.
Ψ = a1Ψ1 + a2Ψ2 + .....+ anΨn
If we measure its energy the probability that the particle will be
[62]
Schrodinger’s Equation
found in state Ψi with energy Ei is given by |ai|2.The average energy of the particle will be given by∫
V
Ψ∗HΨdV =∑
i
|ai|2Ei
[63]