Download Lectures 3-4: Quantum mechanics of one

Lectures 3-4: Quantum mechanics of one-electron atoms
o Schrödinger equation for one-electron atom.
The Schrödinger equation
o
One-electron atom is simplest bound system in nature.
o
Consists of electron moving in 3D Coulomb potential of nucleus :
o
Z =1 for hydrogen, Z =2 for helium, etc.
o
As mass of nucleus (M) >> mass of electron, electron moves moves relative to
nucleus as if nucleus was fixed, and the mass (m) of were slightly reduced to µ.
o
Classically, the total energy of system is:
o Solving the Schrödinger equation.
o Wavefunctions and eigenvalues.
o Atomic orbitals.
" = "(x, y,z,t)
!
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The Schrödinger equation
o
The Schrödinger equation
Using the Equivalence Principle, the classical dynamical quantities can be replaced with their
associated differential operators:
o
Substituting, we obtain the operator equation:
o
Assuming electron can be described by a wavefunction of form
o
Since V(x,y,z) does not depend on time, "(x, y,z,t) = # (x, y,z)e$iEt / h is a solution to the
Schrödinger equation.
o
The time-independent Schrödinger
! equation can therefore be written:
o
As V = V(r), convenient to use spherical polar coordinates.
(1)
can write
where
"2 =
or
o
o
This is the Schrodinger equation for the system, where,
Laplacian operator.
is the
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1 # 2#
1
#
#
1
#2
(r
)+ 2
(sin$
)+ 2 2
2
r #r
#r r sin$ #$
#$
r sin $ #% 2
Can now use separation of variables to split the partial differential
equation
into a set of ordinary differential equations.
!
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Separation of the Schrödinger equation
Separation of the Schrödinger equation
o As the LHS of Eqn 3 does nor depend on r or ! and RHS does not depend on " their
common value cannot depend on any of these variables.
o
Assuming that the eigenfunction is separable:
o
Substituting (2) into the time-independent Schrodinger equation (1) and using the Laplacian:
(2)
o Therefore set the LHS of Eqn 3 to a constant:
(4)
and
o
Carrying out the differentiations,
o
Note total derivatives now used, as R is a function of r alone, etc.
o
Now multiply through by "2µr 2 sin 2 # /R$%h 2 and taking transpose,
o As LHS only depends on r and RHS on !, both sides must equal a constant, which we
choose as l(l+1):
(5)
(6)
o We have now separated the time-independent Schrödinger equation into three
differential equations, which each only depend on one of # (4), $ (5) and R(6).
(3)
!
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Azimuthal solutions (#("))
o
A particular solution of (4) is
o
As the einegfunctions must be single valued, i.e., #(0)=#(2%) =>
Polar solutions (!
(!(!))
o
(7)
o
or using Euler’s formula, 1 = cos ml 2" + isin ml 2"
o
This is only satisfied if ml = 0, ±1, ±2, ...
o
Therefore, acceptable solutions to (4) only exist
when ml can only have certain integer values,
i.e. it is a quantum number.
Solutions to Eqn. 7 are of the form
where Fl|m | (cos" ) are associated Legendre polynomial functions.
!
o
By making a change of variables using z = rcos", Eqn. 5 is transformed into a
differential equation called the associated Legendre equation:
l
o
ml is called the magnetic quantum number as plays role when atom in magnetic
field.
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The requirement that ! remains finite leads to integer values of l and certain
! restrictions on ml as follows:
l = 0, 1, 2, 3, ...
ml = -l, -l+1, .., 0, .., l-1, l
o
l is the orbital or angular momentum quantum number.
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Polar solutions (!
(!(!)) and spherical harmonics
o
We can write the associated Legendre functions
Spherical harmonic solutions
o
Y 00= 1
Y 01= cos"
Y ±11= (1-cos2")1/2 e±i%
Y 02= 1-3cos2"
Y ±12= (1-cos2")1/2cos" e±i%
with the quantum number subscripts as:
#
#
o
!00 = 1
!10 = cos"
!20 = 1-3cos2"
!2±2 = 1-cos2"
The first few spherical harmonics are:
!1±1 = (1-cos2")1/2
!2±1 = (1-cos2")1/2cos"
Customary to multiply the $(") and !(!) functions to form the so-called spherical
harmonic functions which can be written as:
i.e., product of trigonometric and polynomial functions.
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Radial solutions (R( r ))
Radial solutions (R( r ))
o
What is the ground state of hydrogen (Z=1)? Assuming that the ground state has n
= 1, l = 0, the radial wave equation (Eqn. 6) can be written:
o
In general, the radial wave equation has many solutions, one for each positive
integer of n. Can therefore write
o
Taking the derivative
o
More generally (Appendix N of Eisberg & Resnick) solutions are of form
(7)
o
Try a solution of the form R = Ae"r / a , where A is a constant and a0 is a constant
with the dimension of length. Sub into Eqn. 7:
o
To satisfy this Eqn. for any r, both expressions in brackets must equal zero. Setting
the second expression to zero =>
where a0 is the Bohr radius. Bound-state solutions are only acceptable if
0
!
o
Setting first term to zero =>
where n is the principle quantum number, defined by n = l +1, l +2, l +3, …
eV
Same as Bohr’s results
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o
En only depends on n: all l states for a given n are degenerate (i.e. have the same
energy).
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Radial solutions (R( r ))
Summary of separation of variables
o
Gnl(Zr/a0) are called associated Laguerre polynomials, which depend on n and l.
o
Express electron wavefunction as product of three functions: " (r,#, $ ) = R(r)%(# )&($ )
o
Several resultant radial wavefunctions (Rnl( r )) for the hydrogen atom are given
below
o
As V ! V(t), attempt to solve time-independent Schrodinger equation.
o
Separate into three ordinary differential equations for R(r),"(# ) and "(# ).
o
Eqn. 4 for "(# ) only has acceptable solutions for certain value of ml.
o
Using these values for ml in Eqn. 5, !(!) only has acceptable values for certain
!
values of l.
o
With these values for l in Eqn. 6, R(r) only has acceptable solutions for certain
values of En.
o
=> three quantum numbers!
!
!
!
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Born interpretation of the wavefunction
Born interpretation of the wavefunction
o
Principle of QM: the wavefunction contains all the dynamical information about the
system it describes.
o
In H-atom, the ground state orbital has the same sign everywhere => sign of orbital
must be all positive or all negative.
o
Born interpretation of the wavefunction: The probability (P(x,t)) of finding a
particle at a position between x and x+dx is proportional to |&(x,t)|2dx:
o
Other orbitals vary in sign. Where orbital changes sign, & = 0 (called a node) =>
probability of finding electron is zero.
o
Consider first excited state of hydrogen: sign of
wavefunction is insignificant (P = &2 = (-&)2).
o
P(x,t) is the probability density.
o
Immediately implies that sign of wavefunction has no
direct physical significance.
&(x,t)
P(x,t)
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Born interpretation of the wavefunction
o
o
Radial solutions (R( r ))
Next excited state of H-atom is asymmetric about origin. Wavefunction has
opposite sign on opposite sides of nucleus.
o
The radial probability function Pnl(r ), is the probability that the electron is found
between r and r + dr:
o
Some radial probability functions are given at right:
o
Some points to note:
o The r2 factor makes the radial probability density
vanish at the origin, even for l = 0 states.
The square of the wavefunction is identical on
opposite sides, representing equal distribution
of electron density on both side of nucleus.
o For each state (given n and l), there are n - l - 1
nodes in the distribution.
o The distribution for states with l = 0, have n maxima,
which increase in amplitude with distance from origin.
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Radial solutions (R( r ))
Radial probability density
o
Radial probability distributions for an electron in several of the low energy orbitals
of hydrogen.
o
The abscissa is the radius
in units of a0.
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o
The radial distribution function, P(r ) gives the
probability that the electron will be found in a shell
of radius r.
o
For a 1s electron in hydrogen, P(r ) is a maximum
when r = a0/Z.
o
For hydrogen, Z = 1 => rmax= a0
o
Most likely to find the electron at the Bohr radius.
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Hydrogen eigenfunctions
s orbitals
0.2
o
The eigenfunctions for the state described by the quantum numbers (n, l, ml) are of
the form:
o
Named from “sharp” spectroscopic lines.
o
l = 0, ml = 0
Depend on quantum numbers:
o
&n,0,m = Rn,0 (r ) Y0,m (!, ")
n = 1, 2, 3, …
l = 0, 1, 2, …, n-1
ml = -l, -l+1, …, 0, …, l-1, l
o
Angular solution:
o
Value of Y0,0 is constant over sphere.
Energy of state on dependent on n:
o
For n = 0, l = 0, ml = 0 => 1s orbital:
o
The probability density is
0
-0.2
0.2
0
-0.2
o
o
En = "
o
13.6Z 2
n2
Usually more than one state has
same energy, i.e., are degenerate.
-0.2
0
0.2
!
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p orbitals
d orbitals
o
Named from “principal” spectroscopic lines.
o
Named from “diffuse” spectroscopic lines.
o
l = 1, ml = -1, 0, +1 (n must therefore be >1)
o
l = 2, ml = -2, -1, 0, +1, +2 (n must
therefore be >2)
2
1
o
&n,1,m = Rn1 (r ) Y1,m (!, ")
o
Angular solution:
o
A node passes through the nucleus and separates the two
lobes of each orbital.
o
Dark/light areas denote opposite sign of the
wavefunction.
o
0
-1
-2
-1
Three p-orbitals denoted px, py , pz
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o
&n,2,m = Rn1 (r ) Y2,m (!, ")
o
Angular solution:
o
There are five d-orbitals, denoted
o
m = 0 is z2. Two orbitals of m = -1 and +1
are xz and yz. Two orbitals with m = -2 and
+2 are designated xy and x2-y2.
-0.5
0
-1
-0.5
0
0.5
0.5
1
1
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Quantum numbers
o
Principal quantum number:
o n = 1 (K shell)
o n = 2 (L shell)
o n = 3 (M shell)
o …
o
Quantum numbers
Orbital quantum number:
o l = 0 (s subshell)
o l = 1 (p subshell)
o l = 2 (d subshell)
o l = 3 (f subshell)
o …
o
o
If n = 1 and l = 0 = > the state is designated 1s. n = 3, l = 2 => 3d state.
o
The eigenvalues of the one-electron atom depend only on n, but the eigenfunctions
depend on n, l and ml, since they are the product of Rnl(r ), $lml (') and !ml(!).
o
For given n, there are generally several values of l and ml => degenerate
eigenfunctions.
o
Possible values for l and ml for n = 1, 2, 3
n
1
2
3
l
0
0|1
0, 1, 2
ml
0
0 | -1, 0, +1
0 | -1, 0, +1 | -2, -1, 0, 1, 2
Number of
degenerate
eigenfunctions for
each l
1
1|3
1|3|5
Number of
degenerate
eigenfunctions for
each n
1
4
9
See Table 7.1 of Eisberg & Resnick.
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Atomic orbitals
o
Orbital transitions for hydrogen
Quantum mechanical equivalent of orbits in Bohr model.
o
Transition between different energy
levels of the hydrogenic atom must
follow the following selection rules:
(l = ±1
(m = 0, ±1
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o
A Grotrian diagram or a term diagram
shows the allowed transitions.
o
The thicker the line at right, the more
probable and hence more intense the
transitions.
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