|
이곳은 종잇조각의 사용자 연습장으로, 사용자 문서 아래에 위치합니다. 이곳은 위키 문법 연습 및 문서 발전을 위한 공간입니다.
|
12.5.3[편집]
(1)[편집]
Since , , and are Hermitian,
, , and , all of their eigenvalues are real, and there is a set of eigenvectors that forms a complete basis of the Hilbert space.
For any Hermitian operator, say , let its eigenbasis, . Then for any ,
(2)[편집]
By symmetry, .
(3)[편집]
and
.
Since and ,
and . So
(4)[편집]
When , . Then
12.5.10[편집]
For ,
Let , and .
diverges if for any , there is such that .
Since and cannot both be zero at the same time, one of or diverges unless the series terminates.
If is even and , is even. If is odd and , is odd. For other values of or the series diverges and lacks any physical meaning.
For , for all .
For , for all .
For , for all and .
They are equal to if we ignore the scaling factors.
12.6.9[편집]
The eigenvalue equation on is
Since ,
Let , , and
.
The general solutions of each differential equations are and .
Boundary conditions:
(1)
(2)
(3) continuity
(4) continuity of the derivative
Since , if we take and , then and .
The minimum value of on the first quadrant of the graph is .
- If there are no bound states.
12.6.11[편집]
(1)[편집]
If , then for every , , which leads to a nonphysical zero wave function.
For , .
Unless the series terminates,
, which goes to infinity as r increases.
This cannot be a physical solution, so the series must terminate at certain
(2)[편집]
For each , the allowed values are .
For each , there are -fold degeneracy, so the total degeneracy is .
If , .
If , .
- has -fold degeneracy
Also are odd if is odd and even if is even.
- parity of is
The results are equal to the results obtained from the Cartesian solutions.
(3)[편집]
Then because for all .
To normalize , .
Then because for all .
To normalize , .
Cartesian solutions for are:
By the same logic,
So and
13.1.5[편집]
13.2.1[편집]
(1)[편집]
(2)[편집]
By energy conservation, .
Let the angle between and be . i.e.
When the particle arrives at the maximum or minimum distance from the origin, .
So and .
If , , and are all pointing to the same direction, .
This does not hold for general cases, so .
, so and are antiparallel to each other.
For circular orbits, .
(1)[편집]
,
(2)[편집]
(3)[편집]
Take
, , and .
Then and , because , , , and are all hermitian operators.
Take and . Then
, so . Also,
and
.
Other commutation relations are:
,
,
, and
.
By the same logic with the harmonic oscillators, and . Since and commute, an eigenstate of one operator is an eigenstate of another operator.
- where
(1)[편집]
Since the spherical harmonics are orthonormal to each other, the wave function only has the component.
(2)[편집]
Since the spherical harmonics are orthonormal to each other, the probability of observing , , and are same and others are impossible.
(3)[편집]
and
The spherical harmonics are orthonormal to each other, so
and .
Since is a energy eigenstate, it satisfies the T.I.S.E.
is independent of and .
- and
(1)[편집]
For ground state, . Then the only possible .
So the corresponding eigenstate is .
To normalize ,
By the same logic, .
.
- cf.)
(2)[편집]
Reduced mass of is so its bohr radius is .
Reduced mass of is so its bohr radius is .
From the result of (1), .
The potential is rotationally invariant. The Hamiltonian operator is only the function of and .
So the Hamiltonian and Angular momentum operators mutually commute, which means that are eigenstates.
Dividing each side with we get,
Let .
Take and .
As the equation becomes .
The solution to this equation is . Take and .
Take .
This equation is same as (13.1.8). .
Let .
- where
- , so always exists and is real.