lec14.ppt

NPTEL – Physics – Mathematical Physics - 1
Lecture 14
Matrix representing of Operators
While representing operators in terms of matrices, we use the basis kets to compute the
matrix elements of the operator as shown below
< 𝛷1|𝑥|𝛷1 >< 𝛷1|𝑥|𝛷2 > … … … … … … < 𝛷1|𝑥|𝛷𝑛 >
𝑋 =
(< 𝛷𝑛|𝑥|𝛷1 > … … … … … … … … … … … … . < 𝛷𝑛|𝑥|𝛷𝑛 >)
Where |𝛷𝑛 >’s are the eigenkets of X
The expectation value of X is
< 𝑋 > =< 𝛷|𝑋|𝛹 >
= ∑𝑛,𝑚 < 𝛹|𝛷𝑛 >< 𝛷𝑛|𝑥|𝛷𝑛 >< 𝛷𝑚|𝛹 >
If two operators A, B commute then they have same set of eigenkets.
Change of basis
Suppose we have t observables A and B. The ket space is spanned by {|𝛷
>} and
.
.
.
.
.
.
.
.
{|𝑥 >}. e.g. for a spin 1
system, we can have a description in terms of |𝑠 ±> or |𝑠 ±> or
2 𝑧 𝑥
|𝑠𝑦±> where 𝑠𝑖, (𝑖 ∈ 𝑥, 𝑦, 𝑧) are the components of the spin 2
matrices. Either of
these
different set of base kets span the same space. We are interested in finding out how these
two descriptions are related.
Changing the set of base kets is referred to as a change of basis or a change of
representation. Thus given two sets of basis kets, both satisfying orthonormality
and completeness, there exists an unitary operator U such that
{|𝑥 >} = 𝑈{|𝛷 >}
Where U is an unitary operator satisfying,
𝑈𝑈† = 1
and 𝑈†𝑈 = 1
Eigenvectors corresponding to different eigenvalues are linearly independent. Let 𝑢, 𝑣 are
eigenvectors of the matrix A corresponding to 𝜆1and 𝜆2,
then 𝑎𝑢 + 𝑏𝑣 = 0 if they are not independent.
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1
Opearating by A. 𝑎, 𝑏 ≠ 0.
𝑎𝜆1𝑢 + 𝑏𝜆2𝑢 = 0

(1) and (2) have unique solutions if
Det|
𝑎 𝑏
| = 0 ⇒ 𝑎𝑏(𝜆 − 𝜆 ).
𝑎𝜆1 𝑏𝜆2
Thus 𝑎, 𝑏 = 0
2 1
Hence they are linearly independent.
Summarizing, a change of basis kets is an unitary operation that establishes the
link between the two kets.
Linear Algebra in Quantum Mechanics
We have seen in quantum mechanics that the state of an electron subjected
to some potential is given by a wave function Ψ(𝑟⃗, 𝑡) which is obtained from the
solution of a linear second order differential equation, called the Schrodinger
equation. This implies that if Ψ1 and Ψ2 are the solutions for the Schrodinger
equation, then 𝐶1Ψ1 + 𝐶2Ψ2 is also a solution of the equation with 𝐶1 and 𝐶2 as
arbitrary complex numbers.
It is actually the linearity of Schrodinger equation that allows for the superposed
solution as a valid solution. The commutativity, associativity for Ψ are trivially
satisfied.
Another close link between the content of this chapter to quantum mechanics can
be mentioned as follows. Since all the physical observables correspond to
real values, they must correspond to Hermitian matrices. As an example, the
students can be asked
that operator like the momentum (𝑃̂𝑥 = −𝑖ℎ ) or the Pauli matrix 𝜎𝑦 (
𝜕
𝜕𝑥
contain 𝑖(= √−1),
however are still unitary
and yield real
eigenvalues.
0 − 𝑖
𝑖 0
) which
Also for another example, one can show that eigenvectors of 𝜎𝑦 corresponding to
the (real) eigenvalues (±1) are orthogonal.
Consider a basis set {𝑢𝑖} for (𝑖 = 1, 2 … … . 𝑛) in a 𝑅𝑛 space. The set {𝑢𝑖} is
orthonormal if < 𝑢𝑖 |𝑢𝑗 > = ∫ 𝛼3𝑟 𝑢𝑖
∗ (𝑟⃗)𝑢𝑗 (𝑟⃗) = 𝛿𝑖𝑗
Thus every function Ψ(𝑟⃗) ∈ 𝑅𝑛 can be expanded in one (and only one) way in
terms of
𝑢𝑖(𝑟⃗)
Ψ(𝑟⃗) = ∑𝑖 𝑐𝑖𝑢𝑖 (𝑟⃗) which means that the set 𝑐𝑖 is unique and can be found by,
𝑐𝑖 = < 𝑢𝑖 |Ψ > = ∫ 𝑑3𝑟 𝑢𝑖
∗(𝑟⃗)Ψ(𝑟⃗).
Let 𝛷(𝑟⃗) and Ψ(𝑟⃗) be two functions which are expand using the basis {𝑢𝑖}
𝛷(𝑟⃗) = ∑𝑖 𝑏𝑖 𝑢𝑖(𝑟⃗) Ψ(𝑟⃗) = ∑𝑖 𝑐𝑗 𝑢𝑗 (𝑟⃗)
The inner product of these two functions is expressed as,
< 𝛷|Ψ > = ∑𝑖 𝑏𝑖
∗
𝑐𝑖
and < Ψ|Ψ > = ∑𝑖|𝑐𝑖|2.

Change of basis
For a plane wave,
Ψ(𝑥) =
Ψ̃ (p)
=
1
√2𝜋ℎ −∞
∞ 𝑖𝑝𝑥
∫ 𝑑𝑝
Ψ̃ (𝑝)𝑒
⁄ℎ
1
√2πℎ −∞
∞ −𝑖𝑝𝑥
∫ 𝑑𝑥
Ψ(𝑥)𝑒
⁄ℎ
< Ψ(𝑥)|Ψ(𝑥) > = < Ψ̃ (𝑝)|Ψ̃ (𝑝) >
This is known as Perseval’s identity.
Projection Operators
A projection operator p is a self adjoint operator satisfying the condition,
𝑝2 = 𝑝
Let 𝑝1 and 𝑝2 be two projection operators, then the question is whether 𝑝1 + 𝑝2 is also a
projection operator.
(𝑝1 + 𝑝2)2 = 𝑝1
2 + 𝑝2
2 + 𝑝1𝑝2 + 𝑝2𝑝1
𝑝1 + 𝑝2 is also a projection operator if
𝑝1𝑝2 + 𝑝2𝑝1 = 0 (1)
That is if they anticommute. The relation between 𝑝1 and 𝑝2 is more clearly expressed by
left multiplying (1) by 𝑝1,
𝑝2 + 𝑝1𝑝2𝑝1 = 0
Again right multiplying by 𝑝1
𝑝1𝑝2𝑝1 + 𝑝2 = 0
Subtracting (2) and (3)
𝑝1𝑝2 − 𝑝2𝑝1 = 0
Thus (4) and (1) are simultaneously satisfied if 𝑝1𝑝2 = 0
= 𝑝2𝑝1.
(2)
(3)
(4)
Thus the necessary and sufficient condition are that the two operators have to be mutually
orthogonal.
Some important properties of Hermitian Matrices
1.Two Hermilian matrices can be simultaneously diagonalised if and only if they
commute. The converse is also true.
2.The angle of rotation 𝜃 for the rotation described by a proper orthogonal matrix
𝑅𝜃 is given by,
2𝑐𝑜𝑠𝜃 = 𝑡𝑟[𝑅𝜃] − 1
3.For a matrix:
a) All eigenvalues 𝜆 are distinct. There are three mutually orthogonal
eigenvectors.
b) Two of the eigenvalues are equal 𝜆1 = 𝜆2. The corresponding eigenvector is
unique. But one can always generate an arbitrary orthogonal eigenvector.

Thus these two along with the eigenvector corresponding to 𝜆3 constitutes the three
eigenvectors.
c) 𝜆1 = 𝜆2 = 𝜆3 any three mutually perpendicular eigenvectors will do.
To summarize this chapter, a little more can be said about the Hermitian matrices and
their relevance to quantum mechanics. To prove that the eigenvalues as real, one may
proceed in the following manner.
𝐴|𝛼 >= 𝑎|𝛼 >
Inner product with 𝛼 > yields
< 𝛼|𝐴|𝛼 >= 𝑎 < 𝛼|𝛼 >
The adjoint equation can be denoted as,
< 𝛼|𝐴 = 𝑎∗ < 𝛼| (remembering A = 𝐴†)
Again taking an inner product with |𝛼 > yields the result 𝑎 = 𝑎∗ , or in other words,
the eigenvalues of Hermilian matrices are real.
Likewise, the eigenkets corresponding to different eigenvalues are orthogonal.
𝐴|𝛼 >= 𝑎|𝛼 >
𝐴|𝛽 >= 𝛽|𝛽 >
< 𝛼|𝐴|𝛽 >= 𝛽 < 𝛼|𝛽 >
< 𝛽|𝐴|𝛼 >= 𝛼 < 𝛽|𝛼 >
Because of the Hermilian nature of the eigenvalues,
< 𝛼|𝐴|𝛽 >= 𝛽|𝐴|𝛼 >
Hence (𝛼 − 𝛽) < 𝛼|𝛽 > = 0
Thus if 𝛼 ≠ 𝛽 < 𝛼|𝛽 > = 0
So they are orthogonal. Further eigenvectors of a Hermitian matrix span the space.

Tutorials
1.Prove each of the following,
a) P is orthogonal if and only if 𝑃𝑇 is orthogonal
b) If P is orthogonal, then 𝑃−1is orthogonal.
c) If P and Q are orthogonal, then PQ is orthogonal.
Ans: a) (𝑃𝑇)𝑇 = 𝑃.
Thus P is orthogonal if and only if 𝑃𝑃𝑇 = 𝐼 if and only if (𝑃𝑇)𝑇𝑃𝑇 = 𝐼, that is, if and
only if 𝑃𝑇 is orthogonal.
b) We have 𝑃𝑇 = 𝑃−1 since P is orthogonal. Thus by part (a) 𝑃−1 is orthogonal. c) We
have 𝑃𝑇 = 𝑃−1, 𝑄𝑇 = 𝑄−1
2. Prove that momentum operator in quantum mechanics is a Hermitian Operator.
< 𝑝𝑥 > = ∫ ∂𝑥
Ψ∗ (− ) Ψdx
∞ 𝑖ℎ ∂
−∞
∞ 𝜕Ψ
= - 𝑖ℎ ∫
Ψ∗
𝑑𝑥
−∞ 𝜕𝑥
= -𝑖ℏ [Ψ∗
Ψ ∫ − ∫ Ψ(𝑥) 𝑑𝑥]
dΨ ∗
dx
∞ ∞
−∞ −∞
0
= iℏ ∫ Ψ(𝑥) 𝑑𝑥
𝑑Ψ ∗
𝑑𝑥
∞
−∞
= < 𝑝𝑥 >∗
3. A particle of mass m is in a one-dimensional box of width a. At
𝑡 = 0, the particle is
in the state Ψ(𝑥, 0) =
3𝛷 +4𝛷
2 9
√25
, The 𝛷 function are orthogonal eigenstates of the
𝑛
Hamitian 𝐻𝛷𝑛 = √2⁄a 𝑠𝑖𝑛 (𝜋𝑥⁄𝑎) what will be the measurement of energy, E yield at
𝑧 = 0. What is the probability of finding this value ?
Ans: First check that 𝛹 is properly normalized.
< 𝛹|𝛹 > = 1
Ψ = ∑𝑛 𝑐𝑛|𝛷𝑛 >, 𝑐2 = , 𝑐9 = , 𝑐𝑛 = 0
3 4
√25 √25
for 𝑛 ≠ 2 or 9.
P(𝐸2) = 25
and 𝑃(𝐸9) = 25
.
9 16

In an ensemble of 2500 identical one-dimensional boxes each containing
an identical particle in the same state Ψ(𝑥, 0). Measurement E at 𝑡 = 0 , finds
900 particle of each 𝐸2 = 4𝐸, and 1600 particles 𝐸9 = 81𝐸.
4. It can be seen that 𝑎⃗⃗⃗⃗1⃗, ⃗𝑎⃗⃗⃗2⃗ and 𝑎⃗⃗⃗⃗3⃗ are all linearly independent as they do
not lie in a plane. Thus, a unit vector in the direction of ⃗𝑎⃗⃗⃗1⃗ is
𝑣⃗⃗⃗ = = [2, 3, 0] = [0.554, 0.831,
0].
1
𝑎⃗⃗⃗⃗⃗1⃗
1
|𝑎⃗⃗⃗⃗⃗⃗| √3
1
We subtract from 𝑎⃗⃗⃗⃗2⃗, the component of 𝑎⃗⃗⃗⃗2⃗ in the
direction of ⃗𝑣⃗⃗1⃗.
𝑣⃗⃗⃗⃗2⃗ = ⃗𝑎⃗⃗⃗2⃗−< ⃗𝑣⃗⃗1⃗|𝑎⃗⃗⃗⃗2⃗ > ⃗𝑢⃗⃗⃗1⃗
< ⃗𝑣⃗1⃗|𝑎⃗⃗⃗⃗2⃗ > = 4.16 ⇒< ⃗𝑣⃗⃗1⃗|𝑎⃗⃗⃗⃗2⃗ > ⃗𝑣⃗⃗1⃗ = [2.30,
3.45, 0]
Hence, 𝑣⃗⃗⃗⃗2⃗ = [3.7, −2.45, 0].
Hence calculate the unit vector in the direction of ⃗𝑢⃗⃗⃗2⃗
𝑢
⃗⃗⃗⃗⃗
=
2
𝑣
⃗⃗⃗⃗
2⃗
|𝑣
⃗⃗⃗⃗⃗
|
2
Finally calculate 𝑣⃗⃗⃗⃗3⃗ = ⃗𝑎⃗⃗⃗3⃗− < ⃗𝑣⃗⃗1⃗|𝑎⃗⃗⃗⃗3⃗ > ⃗𝑣
⃗⃗1⃗
The components of ⃗𝑎⃗⃗⃗3⃗ in the direction of
⃗𝑣⃗1⃗ and ⃗𝑣⃗⃗⃗2⃗ are taken out. That is substract − < ⃗𝑣
⃗⃗⃗2⃗|𝑎⃗⃗⃗⃗3⃗ > ⃗𝑣⃗⃗⃗2⃗
So the corresponding orthonormal set is given by,
𝑒̂
=
𝑖
𝑣
⃗
⃗𝑖
|⃗𝑣
⃗|
𝑖
5. Consider a basis {𝑒⃗⃗⃗1 , ⃗𝑒⃗⃗2 } = {( ) , (
)} to represent
1 −2
2 3
a vector 𝑣 = ( ) in a two
3
5
dimensional space such that,
𝑣 = 𝑐1⃗𝑒⃗⃗1 + 𝑐2⃗𝑒⃗⃗2
Find a 2x2 matrix [A], such that
𝑣 = [𝐴].
𝑢⃗
where 𝑢⃗ = (𝑐
2
)
𝑐1
Solution: First let us find 𝑢
⃗
𝑐1 ( ) + 𝑐2 ( ) = ( )
1 −2
2 3
3
5

𝑐1 − 2𝑐2 = 3
2𝑐1 + 3𝑐2 = 5} solving for 𝑐1 and 𝑐2
2𝑐1 − 4𝑐2 = 6
−2𝑐1 + 3𝑐2 = −5
7𝑐2 = + 1
𝑐 1
19
2 = − 7
; 𝑐1 = 7
Thus 𝑢⃗ = ( 7
)
19
− 1
7
Using [𝐴]. 𝑢⃗
= 𝑣
[𝐴] = (
2
3
)
1 − 2

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