CH 7: 1-3
SCHRODINGER EQ IN 3-D, 3-D INFINITE WELL AND
ENERGY QUANTIZATION AND SPECTRAL LINES IN H
TIME-DEPENDENT
SCHRODINGER EQ IN 3-D
• Time-dependent in 1-D:
ℏ2 𝜕𝜕 2 𝛹𝛹(𝑥𝑥, 𝑡𝑡)
𝜕𝜕𝛹𝛹(𝑥𝑥, 𝑡𝑡)
−
+ 𝑈𝑈(𝑥𝑥)𝛹𝛹(𝑥𝑥, 𝑡𝑡) = 𝑖𝑖𝑖
2
𝜕𝜕𝜕𝜕
2𝑚𝑚 𝜕𝜕𝜕𝜕
• Extend to 3-D
ℏ2 2
𝜕𝜕𝛹𝛹(𝒓𝒓, 𝑡𝑡)
−
∇ 𝛹𝛹(𝒓𝒓, 𝑡𝑡) + 𝑈𝑈(𝒓𝒓)𝛹𝛹(𝒓𝒓, 𝑡𝑡) = 𝑖𝑖𝑖
𝜕𝜕𝜕𝜕
2𝑚𝑚
Where in Cartesian coordinates:
𝜕𝜕2
2
∇ = 2
𝜕𝜕𝜕𝜕
+
𝜕𝜕2 𝜕𝜕2
+
𝜕𝜕𝑦𝑦2 𝜕𝜕𝑧𝑧2
TIME INDEPENDENT
SCHRODINGER EQ IN 3-D
• Assume as in the case of 1-D, the separation of
variables: 𝛹𝛹(𝒓𝒓, 𝑡𝑡) = 𝜓𝜓(𝒓𝒓)𝜙𝜙(𝑡𝑡)
• Then the time-independent Schrodinger EQ is
ℏ2 2
−
∇ 𝜓𝜓(𝒓𝒓) + 𝑈𝑈(𝒓𝒓)𝜓𝜓(𝒓𝒓) = 𝐸𝐸𝜓𝜓(𝒓𝒓)
2𝑚𝑚
PROBABILITY DENSITY AND
NORMALIZATION IN 3-D
• Probability density is probability per unit volume =
𝛹𝛹(𝒓𝒓, 𝑡𝑡) 2
• Normalization: The total probability of finding the
particle in 3-D space must be 1:
�
𝑎𝑎𝑙𝑙𝑙𝑙 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
𝛹𝛹(𝒓𝒓, 𝑡𝑡) 2 𝑑𝑑𝑑𝑑 = 1
3-D INFINITE WELL “QUANTUM DOT”
OR “DESIGNER ATOM”
• Inside the well the
potential energy is
zero.
• Outside the
potential energy is
infinite.
SOLVING SCHRODINGER EQ FOR 3-D
INFINITE WELL
• Assume the wave function can be written as the
product of three independent functions:
𝜓𝜓(𝒓𝒓) = 𝐹𝐹 𝑥𝑥 𝐺𝐺 𝑦𝑦 𝐻𝐻(𝑧𝑧)
• Sub into Schrodinger's eq and simplify:
•
1 𝜕𝜕2 𝐹𝐹(𝑥𝑥)
𝐹𝐹 𝜕𝜕𝜕𝜕 2
+
1 𝜕𝜕2 𝐺𝐺(𝑦𝑦)
𝐺𝐺 𝜕𝜕𝑦𝑦 2
1 𝜕𝜕2 𝐻𝐻(𝑧𝑧)
+
𝐻𝐻 𝜕𝜕𝑧𝑧 2
=−
2𝑚𝑚𝑚𝑚
ℏ2
• Each term must be a constant:
1 𝜕𝜕 2 𝐹𝐹(𝑥𝑥)
1 𝜕𝜕 2 𝐺𝐺(𝑦𝑦)
= 𝐶𝐶𝐶𝐶
= 𝐶𝐶𝐶𝐶
𝐹𝐹 𝜕𝜕𝜕𝜕 2
𝐺𝐺 𝜕𝜕𝑦𝑦 2
1 𝜕𝜕 2 𝐻𝐻(𝑧𝑧)
= 𝐶𝐶𝐶𝐶
𝐻𝐻 𝜕𝜕𝑧𝑧 2
• Same as the DiffEq we solved for 1-D infinite well.
• So F, G and H are sine functions
3-D INFINITE WELL SOLUTION
• Solution inside the well:
𝜓𝜓 𝒓𝒓 =
𝑛𝑛𝑥𝑥𝜋𝜋𝜋𝜋
𝑛𝑛𝑦𝑦𝜋𝜋𝑦𝑦
𝐴𝐴sin
sin
𝐿𝐿𝑥𝑥
𝐿𝐿𝑦𝑦
𝑛𝑛𝑧𝑧𝜋𝜋𝑧𝑧
sin
𝐿𝐿𝑧𝑧
• The energy is quantized, depending on three
quantum numbers:
𝑛𝑛𝑥𝑥2 𝑛𝑛𝑦𝑦2 𝑛𝑛𝑧𝑧2 𝜋𝜋𝜋 2
𝐸𝐸 = 2 + 2 + 2
𝐿𝐿𝑥𝑥 𝐿𝐿𝑦𝑦 𝐿𝐿𝑧𝑧 2𝑚𝑚
DEGENERACY IN CUBE
• If well is a cube, then each side has
length L.
• Then energy is:
𝜋𝜋𝜋 2
2
2
2
𝐸𝐸 = 𝑛𝑛𝑥𝑥 + 𝑛𝑛𝑦𝑦 + 𝑛𝑛𝑧𝑧
2𝑚𝑚𝑚𝑚2
• Different triplets of quantum
numbers can have the same
energy.
• Different wave functions with the
same energy is called degeneracy.
Nondegenerate
3-fold
degenerate
EXAMPLE 7.1
• For a cube: E 1,2,1 = E 2,1,1 = E 1,1,2
• But the probability densities are
not the same
SPLITTING OF DEGENERACY
• Degeneracy comes
from symmetry.
• If symmetry is lost so
is degeneracy.
• For example if two
sides equal L, but
the third is 0.9L
HOMEWORK BREAK
• Ch 7: 20
• Due Thursday
10DEC15
• More to come
BETTER MODEL OF HYDROGEN
𝑘𝑘𝑘𝑘2
−
𝑟𝑟
• Insert potential energy 𝑈𝑈 =
into 3-D Schrodinger
equation, apply boundary conditions and find the
energy is quantized:
𝑘𝑘2𝑚𝑚𝑚𝑚4 1
𝐸𝐸𝑛𝑛 = −
2ℏ2 𝑛𝑛2
• Same as Bohr model:
13.6 eV
𝐸𝐸𝑛𝑛 = −
𝑛𝑛2
• We’ll find other properties are also quantized.
(Remember in 3-D, we need to find 3 quantum
numbers)
HYDROGEN LINES
• Visible spectral lines were observed and empirical fit
1
1
1
7
−1
was found by Balmer: = 1.097 × 10 m
2 − 2
𝜆𝜆
2
𝑛𝑛
• We can find the lines using the quantized energy:
𝐸𝐸𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 = 𝐸𝐸𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑖𝑖 − 𝐸𝐸𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑓𝑓
𝑘𝑘2𝑚𝑚𝑚𝑚4 1
1
𝑘𝑘2𝑚𝑚𝑚𝑚4 1
1
ℎ𝑐𝑐
𝐸𝐸𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 = −
− 2 =+
− 2 =
2
2
2
2
2ℏ
2ℏ
𝜆𝜆
𝑛𝑛𝑖𝑖 𝑛𝑛𝑓𝑓
𝑛𝑛𝑓𝑓 𝑛𝑛𝑖𝑖
• Solve for λ , substitute values and in the case of
Balmer lines nf = 2.
HYDROGEN LINES
1
1
1
7
−1
= 1.097 × 10 m
− 2
2
𝜆𝜆
𝑛𝑛𝑓𝑓 𝑛𝑛𝑖𝑖
HYDROGEN LINES
1
1
1
7
−1
= 1.097 × 10 m
− 2
2
𝜆𝜆
𝑛𝑛𝑓𝑓 𝑛𝑛𝑖𝑖
HYDROGEN SPECTRUM FROM STARS
HOMEWORK BREAK
• Ch 7: 32
• Due Thursday
10DEC15
• More to come