
2. Particle Confined
in a 3Dimensional Box or Square Well
One
of the more interesting applications is for a particle confined to a
3dimensional square ‘well’ or 3D box, as shown:
0 < x < a
0 < y < b
0 < z < c
The
potential energy inside the region is zero and outside is infinite so the
Schrodinger equation becomes:
T _{op} y = E_{op}_{
} y
Where
T _{op} is the kinetic energy operator which is written in 3D
as:
T
_{op }= p^ p^ / 2m = [p^ _{x} ^{2 } + p^ _{y} ^{2 } + p^
_{z} ^{2 }] / 2m
=  ħ^{2}/ 2m
[ ¶^{2} / ¶ x^{2} + ¶^{2} / ¶ y^{2} +
¶^{2} / ¶ z^{2} ) =  ħ^{2}/ 2m Ñ ^{2}
^{}
^{}
Where
Ñ
is the Laplacian operator.
The Schrodinger equation then becomes:
¶^{2} y/ ¶ x^{2} + ¶^{2} y/ ¶ y^{2} +
¶^{2} y/ ¶ z^{2} +
2mE/ ħ^{2 }y = 0
3. Solution
of the 3D Square Well  Box Schrodinger Equation:
This
is a partial differential equation easily solved by the separation of
variables:
y =
X(x) Y(y) Z(z)
Then:
X’
= ¶X/ ¶ x
Y’ = ¶Y/ ¶ y and Z’ =
¶Z/ ¶ z
This
leads to the equation:
X”YZ + XY”
Z + XYZ” + 2mE/ ħ^{2 } XYZ = 0
The
required normalization equation is then:
ò^{a}_{o}_{ } ò^{b}_{o}_{ }ò^{c}_{ o}_{ } ‖y‖ ^{2} dz dy dx = 1
Then,
dividing the Schrodinger equation by XYZ:
X”/
X + Y”/ Y + Z”/Z + 2mE / ħ^{2 } = 0
We
let:
X”/
X = 
a^{2}, Y”/ Y
= b^{2}, Z”/Z
= g^{2}
^{}
^{}
With a, b and g constants.
Then, we have:
x^{2} + a^{2 }x = 0,
y^{2} + b^{2 }y = 0, z^{2} + g^{2 }z
So the independent solutions will be:
x= Ö(2/a) sin ( n _{x}px/a)]
y= Ö(2/b) sin ( n _{y }px/b)]
z= Ö(2/c) sin ( n _{z}px/c)]
where: n _{x } = 1, 2, 3, 3 etc.
Then:
y = (8/ abc)^{1/2}
sin ( n _{x}px/a)
sin ( n _{y }px/b) sin ( n _{z}px/c)
And further, to obtain the quantized energy, E:
n^{2} _{x
}p^{2} /a^{2}  n^{2}
_{y }p^{2 }/b^{2}  n^{2}
_{z }p^{2} /c^{2} + 2mE/ ħ^{2 } = 0
And:
E = p^{2 }ħ^{2 }/ 2m
[n^{2}_{x } /a^{2}
+ n^{2} _{y } /b^{2} + n^{2} _{z } /c^{2} )
If the box is a cube, so a = b = c:
E = p^{2 }ħ^{2 }/ 2m a^{2} (n^{2}_{x } + n^{2} _{y } + n^{2} _{z })
(It should be noted
that in this system, and for states higher in energy than the ground state, many
levelsdistinct states have the same energy. This property is called the
degeneracy of states.)
4. The Hydrogen Atom
The
Hamiltonian for the hydrogen atom Schrodinger equation can also be written in
concise form as:
H _{op} y = E_{op}_{
} y
For
which we only need to know the operators in order to expand it to its proper
representation. In this case:
H
_{op}
= p^r
^{2}/ 2m + ℓ^^{2}/ 2mr^{2} + V(r)
And: p^r =  i ħ (1/r ¶^{2} / ¶ r^{2}
)
The operators are radically different because we
are dealing with an atom and treating it along the lines of radial symmetry –
so the operator must be a function of r. (Spherical coordinate system)
We also note the wave function: y = y (r,
q, j)
Again, this
is to preserve spherical symmetry. The
full Schrodinger equation for the hydrogen atom can then be written as:
[p^r ^{2}/ 2m + ℓ^{2}/
2mr^{2} + V(r)]
y (r, q, f) = Ey (r,
q, j)
This
equation can now be used, after further expansion of the assorted operators, to
obtain the properties of the hydrogen atom, including its energy levels.
Problems (Continued from earlier set):
3) Find the expectation value for the momentum of a particle with a wave function:
U(x)
= A exp [i(a  a^{2 }ħ t/ 2m]
4)Using
the operators, p^r and ℓ^^{2 }
^{}
^{}
=  ħ^{2}/ sin^{2}
q [ sin q ¶ / ¶ q (sin
q ¶ / ¶ q) + ¶^{2} / ¶ j^{2 }]
Write
out the full form of the Schrodinger equation for the hydrogen atom in
spherical coordinates. Thence, obtain
the final form such that H _{op} =
 [E – V] y (r, q,
j)
Hint:
Replace m by the reduced mass,
m
= mM/ m+M
5)
In the form shown, the variables r, q, j may be separated by letting:
y (r, q,
j) =
R(r) Q (q ) F( j)
And performing suitable manipulations. Do this in a similar
way to the approach used to solve for the 3D box and then
show the solutions which result are:
i) F( j)
= exp (i m _{ℓ }j)
ii) Q
(q ) = Q _{ℓ}_{ m} =
sin ^{m
ℓ} q F_{ ℓ}  m _{ℓ}  (cos q)
iii) R _{nℓ}_{ }= exp(Zr/na_{o}) [Zr/ a_{o}]^{
ℓ} G _{ℓ} (Zr/ a_{o})
Explain how your solutions help to account for the Table
entries below:
6)
In the case of the first solution (i) we demand that the function be
singlevalued and continuous so that:
F( j + 2p) = F( j)
So that:
where m _{ℓ} is the magnetic quantum number. Dividing by
exp
(i m _{ℓ }j) we obtain:
exp [ i m _{ℓ }(2p)] = 1
exp [ i m _{ℓ }(2p)] = 1
Indicate the condition on m _{ℓ} for which this is satisfied.
7) The solutions Q _{ℓ}_{ m} are associated Legendre polynomials and require two quantum numbers, the magnetic and the azimuthal, ℓ. If ℓ is quantized and can only be a
positive integer, show its acceptable values based on the values of m.
8) The solutions R _{nℓ}_{ }are Laguerre functions which also require two
quantum numbers, ℓ and n to identify acceptable solutions. (Note: n is
the principal quantum number)
The eigenfunctions y (r, q, j) are formed by taking the product of the 3
types of functions so that:
y_{ nℓ }_{m ℓ} (r, q,
j) = R _{nℓ}_{ }(r) Q _{ℓ}_{
m} (q)F_{m ℓ} ( j)
By
inspection of the table (Prob. 4), identify the part or whole of the expression for the
wave function y_{ 210 } that comes closest to the form for R _{nℓ}_{ }(r).
8) Complete the normalization condition for the Hatom:
ò ^{¥}_{o}_{ } ò^{p}_{o}_{ }ò^{2}^{p}_{ o}_{ } y*_{ nℓ }_{m
ℓ} (r, q,
j) y_{ nℓ }_{m ℓ} (r, q,
j) dV = ? = 1
dV = ( ) dr dq dj
9)
Let the ground state wave function of the hydrogen atom be given by: U
= exp (ra)
Assume
there is no relative motion of the nucleus so that q = 0 and j = 0
a)
Show
that the relevant Schrodinger equation would now be:
 ħ^{2 }/ 2m
[1/ r^{2} ¶/ ¶ r ( r^{2} (¶ / ¶ r)]U = [E – V(r))]U
b)
If V(r) =  e^{2} / r for the hydrogen atom,
show that the quantized energy for this system is:
E
=  ħ^{2 }/ 2m
[a^{2}  2a/ r]  e^{2} /
r
c) Find the specific energy in the ground state if:
r
= ħ^{2 }/ m e^{2}
and: a
= m e^{2} / ħ^{2}
10) The Hamiltonian operator for the quantum harmonic oscillator is:
H^ = p^^{2}/
2m + mw^{2} x^{2}/ 2.
Also the momentum operator: p^ = i h (¶ /¶x). Use these to write out the applicable Schrodinger
equation then indicate the form of the expected solutions – with a diagram for
the potential.
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