mathmatical physics

1. NPTEL – Physics – Mathematical Physics - 1
Lecture 19
Example 2
Eigenvalues and eigenfunctions of a spin,
1
2
particle
Let us consider a spin-𝑆⃗ pointing along an arbitrary direction in space, along the
unit vector 𝑛̂ . The component of the vector along 𝑛̂ is 𝑆𝑛 = 𝑛̂. 𝑆⃗. In order to find
the eigenvalues and eigenvectors of 𝑆𝑛, we have to solve the eigenvalue equation,
𝑆𝑛𝜆 = 𝜈ℎ𝜆 (1)
where the eigenvalue 𝜈ℎ (ℎ has the dimension of angular momentum) is written for
a later convenience. The Cartesian components of 𝑛̂ are (𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝛷, 𝑠𝑖𝑛𝜃𝑠𝑖𝑛𝛷, 𝑐𝑜𝑠𝜃).
So we can write 𝑆𝑛 as,
𝑆𝑛 = 𝑆𝑥𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝛷 + 𝑆𝑦𝑠𝑖𝑛𝜃𝑠𝑖𝑛𝛷 + 𝑆𝑧𝑐𝑜𝑠𝜃
=
ℎ
2
( 𝑐𝑜𝑠𝜃 𝑠𝑖𝑛𝜃𝑒−𝑖𝛷
)
𝑠𝑖𝑛𝜃𝑒 − 𝑐𝑜𝑠𝜃
−𝑖𝛷 (2)
where 𝑆𝑥,𝑦,𝑧 are pauli
matrices.
𝑆𝑥 = ( ) , 𝑆𝑦 = (
0 1 0 − 𝑖
1 0 𝑖 0
) , 𝑆𝑧 = (
1 0
0 − 1
)
Writing the spin eigenfunction 𝜆 = (𝑏)
𝑎
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2. NPTEL – Physics – Mathematical Physics - 1
𝑆𝑛𝜆 = 2
(
𝑎𝑠𝑖𝑛𝜃𝑒𝑖𝛷 − 𝑏𝑐𝑜𝑠𝜃
)
ℎ 𝑎𝑐𝑜𝑠𝜃 + 𝑏𝑠𝑖𝑛𝜃𝑒−𝑖𝛷
(3)
The set of equations (4) and (5) have non-trivial solutions only if,
|
𝑐𝑜𝑠𝜃 − 2𝜈 𝑠𝑖𝑛𝜃𝑒−𝑖𝛷
𝑠𝑖𝑛𝜃𝑒𝑖𝛷 − (𝑐𝑜𝑠𝜃 + 2ν)
| = 0
The eigenvalues are obtained as,
−𝑖𝜙
| 𝑐𝑜𝑠𝜃 − 2𝜈 − 𝜆 𝑠𝑖𝑛𝜃𝑒
𝑠𝑖𝑛𝜃𝑒𝑖𝛷 – (𝑐𝑜𝑠𝜃 + 2𝜈) − 𝜆
| = 0
𝜆1 = + 2
, 𝜆2 = − 2
1 1
For 𝜆1 = 2 𝑎
= 𝑡𝑎𝑛 2
𝑒
1 𝑏 𝜃 𝑖𝛷
We shall not make a mistake, apart from an arbitrary phase factor, if we write the
eigenvector corresponding to this eigenvalues as,
𝜆1 = (
𝑐𝑜𝑠 𝜃⁄2
𝑠𝑖𝑛 𝜃⁄2 𝑒𝑖𝛷)
Similarly for 𝜆2 = − 2
1
𝑏
𝑎
= −𝑐𝑜𝑡
𝜃
2
𝑒𝑖𝛷
With a similar reasoning as above,
𝜆2 = (
𝑠𝑖𝑛 𝜃⁄2
−𝑐𝑜𝑠 𝜃⁄2 𝑒𝑖𝜃)
Thus the original eigenvalue equation (Eq.(1)) reduces to a set of two linear,
homogeneous equations,
(𝑐𝑜𝑠𝜃 − 2𝜈)𝑎 + 𝑠𝑖𝑛𝜃𝑒−𝑖𝛷𝑏 = 0
(4)
𝑠𝑖𝑛𝜃𝑒𝑖𝛷𝑎 − (𝑐𝑜𝑠𝜃 + 2𝜈)𝑏 = 0 (5)
One can easily check that for the spin oriented along + 𝑧 axis, i.e. 𝜃 = 0 and 𝜃 = 𝜋 will
reduce the eigenvectors to
𝜆1 = (1
)
0
𝜆2 = (0)
1
As are usually expected for a 𝑠 = 1
particle.
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2

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NPTEL – Physics – Mathematical Physics - 1
Numerical Evaluation of Eigenvalue problems
Gauss elimination method
The centre issue in the eigenvalue problem is the solution of the system of m
linear equations containing n unknowns denoted by 𝐴𝑥 = 𝑏. In expanded form it is
written as,
( 𝐴𝑚1 … … … … … 𝐴𝑚𝑛 ) ( 𝑥𝑚 )
𝐴11𝐴12 … … . 𝐴1𝑛
.
.
.
.
.
.
𝑥1
.
.
.
.
= .
.
.
( 𝑏𝑚 )
.
𝑏1
.
(1)
There are various ways of solving the above equation. Gauss Elimination method
or forward elimination method is one among them. It proceeds by reducing A to an upper
diagonal form.
𝐶11
0 𝐶22 … … .
. 𝐶2𝑛
.
0 … … …
… .
.
.
.
.
.
.
[ 0
𝐶12 … … . 𝐶1𝑛
.
.
.
.
.
.
0 𝐶𝑚𝑛 ]
We present an example here.
Example 1. Solve the system of equations using Gaussian elimination
2𝑥 + 𝑦 + 3𝑧 = 1
2𝑥 + 6𝑦 + 8𝑧 = 3
6𝑥 + 8𝑦 + 18𝑧 = 5
We shall try to achieve the upper diagonal nature via the following steps. Apply
𝑅2 → −𝑅1 + 𝑅2 (where 𝑅𝑖’s are each of the equations)
And 𝑅3 → −3𝑅1 + 𝑅3
This will transform the system of equation to
2𝑥 + 𝑦 + 3𝑧 = 1
5𝑦 + 5𝑧 = 2
5𝑦 + 9𝑧 = 2
Proceeding 𝑅3 → −𝑅2 + 𝑅3, the system of equation become,
2𝑥 + 𝑦 + 3𝑧 = 1

4. NPTEL – Physics – Mathematical Physics - 1
5𝑦 + 5𝑧 = 2
4𝑧 = 0
𝑅3 → 4
𝑅3
Next
1
Now,
1
1
2𝑥 + 𝑦 + 3𝑧 = 1
5𝑦 + 5𝑧 = 2
𝑧 = 0
𝑅1 → 2
𝑅1
𝑅2 → 5
𝑅2
𝑥 + 𝑦 + 𝑧 =
1 3 1
2 2 2
𝑦 + 𝑧 =
𝑧 = 0
2
5
It is very clear that 𝑦 = 2
and
5
𝑧 = 0 anyway. Thus 𝑥 = 3
10
(𝑥, 𝑦, 𝑧) = ( , , 0)
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3 2
10 5