Linear Vector Space | Modern Physics PDF Download

A set of vectors Ψ, ϕ, X, ... and set of scalars a, b, c defined over vector space which will follow a rule for vector addition and rule for scalar multiplication.
(i) Addition Rule
If Ψ and ϕ are vectors of elements of a space, their sum Ψ + ϕ is also vector of the same space.

• Law of Commutativity: Ψ + ϕ = ϕ + Ψ
• Law of Associativity: (Ψ + ϕ) + X = Ψ + (ϕ + X)
• Law of Existence of a null vector and inverse vector: Ψ + (-Ψ) = (-Ψ) + Ψ = 0

(ii) Multiplication rule 

• The product of a scalar with a vector gives another vector. If Ψ and ϕ are two vectors of the space, any linear combination aΨ + bϕ is also a vector of the same space, where a and b being scalars. 
• Distributive with respect to addition:
  a(ϕ + Ψ) = aϕ + aΨ , (a + b)Ψ = aΨ + bΨ 
• Associativity with respect to multiplication of scalars: a (bΨ) = (ab)Ψ 
• For each element Ψ, there must exist a unitary element I and a non-zero scalar O such that: I ·Ψ = Ψ·I = Ψ, O·Ψ = Ψ·O = O

Scalar Product 

The scalar product of two functions ϕ(x) and Ψ(x) is given by (Ψ ϕ) = ∫Ψ*(x)ϕ(x)dx. where ϕ(x) and Ψ(x) are two com plex function of variable x , ϕ* (x) and Ψ* (x) are complex conjugate of ϕ(x) and Ψ(x) respectively.
The scalar product of two function ϕ (x, y, z) and Ψ(x, y, z) in 3-dimension is defined as (Ψ,ϕ) = ∫Ψ*ϕdxdydz

Hilbert Space

The Hilbert space H consists of a set of vectors Ψ, ϕ, X and set of scalar a,b,c which satisfies the following four properties:
(i) H is a linear space
(ii) H is a linear space that defines the scalar product which is strictly positive.

• (Ψ, ϕ) = (ϕ, Ψ)*
• (Ψ, aϕ₁ + bϕ₂) = a (Ψ, ϕ₁) + b(Ψ, ϕ₂)
• (Ψ, Ψ) = |Ψ|² ≥ 0  

(iii) H is separable i.e.,

• ║Ψ - Ψ_n║ ≤ 0

(iv) H is complete ║Ψ - Ψ_m║ = 0 , when m →∞, n →∞ 

Dimension and Basis of a Vectors.

Linear independency:  
A set of N vectors ϕ₁ ,ϕ₂, ϕ₃ ......ϕ_n,  is said to be linearly independent if and only if the solution of the equation  is a₁ = a₂ = a₃ = a₄ = 0.... a_n = 0 , otherwise ϕ₁ ,ϕ₂, ϕ₃ ......ϕ_n is said to be linear dependent.
The dimension of a space vector is given by the maximum number of linearly     independent vectors that a space can have.
The maximum number of linearly independent vectors of a space is N i.e., ϕ₁ ,ϕ₂, ϕ₃ ......ϕ_N , this space is said to be N dimensional. In this case any vector Ψ of the vector space can be expressed as linear combination,

Orthonormal Basis 
Two vectors ϕ_i ,ϕ_j is said to be orthonormal, if their scalar product (ϕ_i, ϕ_j) = δ_{i, j}, where δ_{i, j} is kronekar delta that means δ_{i, j} = 0 , when i ≠ j and δ_{i, j} = 1 , if i = j. 

Square Integrable Function

If scalar product  where α is a positive finite number, then Ψ (x) is said to be square integrable.
The square intergable function can be treated as probability distribution function, if α = 1 and Ψ is said to be normalized.

Dirac Notation

Dirac introduced what was to become an invaluable notation in quantum mechanics, state vector Ψ which is square integrable function  to what he called a ket vector |Ψ〉 and its conjugate Ψ by a bra 〈Ψ| and scalar product (ϕ, Ψ) bra-ket 〈ϕ|Ψ〉 (In summary Ψ → |Ψ〉, Ψ* → 〈Ψ| and (ϕ, Ψ) = 〈ϕ|Ψ〉), where 〈ϕ, Ψ〉 = ∫ϕ*(r, t)Ψ(r, t)d³r.
Properties of kets, bras and bra-kets. 

• |Ψ〉 is normalized, if 〈Ψ|Ψ〉 = 1 
• Schwarz inequality
  |〈Ψ| ϕ〉|² ≤ 〈Ψ|Ψ〉 〈ϕ|ϕ〉, which is analogically derived from
• Triangular inequality,
• Orthogonal states, 〈Ψ| ϕ〉 = 0
• Orthonormal state, 〈Ψ| ϕ〉 = 0, 〈Ψ| Ψ〉 = 1, 〈ϕ|ϕ〉 = 1
• Forbidden quantities: If |Ψ〉 and |ϕ〉 belong to same vector space, then product of the type |Ψ〉|ϕ〉 and 〈ϕ|〈Ψ| are forbidden. They are nonsensical.
• If |Ψ〉 and |ϕ〉 however belong to two different vector space, then |ϕ〉 ⊗|Ψ〉 represent tensor product of |Ψ〉 and |ϕ〉

Operator 

An operator A is the mathematical rule that when applied to a ket |ϕ〉 will transform it into another ket |Ψ〉 of the same space and when it acts on any bra 〈X| , it transforms it into another bra 〈ϕ| , that means A|ϕ〉 = |Ψ〉 and 〈X| A = 〈ϕ|
Example of Operator-
Identity operator, I | Ψ〉 = |Ψ〉
Pairity operator, π|Ψ(r)〉 = |Ψ(-r)〉
Gradiant operator, ∇ Ψ (r) and Linear momentum operator,

Linear Operator

A is linear operator if,  

• A(λ₁ | Ψ₁〉 + λ₂ | Ψ₂〉) = λ₁ A| Ψ₁〉 + λ₂A|Ψ₂〉
• Product of two linear operator A and B is written as AB , which is defined as, (AB) | Ψ〉 = A(B|Ψ〉)

Matrix Representation of Operator

If |Ψ〉 is in orthonormal basis of |u_i〉 is defined as
|Ψ〉 c₁|u₁〉 + c₂ |u₂〉 + ... |Ψ〉 = Σc_i|u_i〉 , where c_i =〈u_i|Ψ〉  
Then, the ket |Ψ〉 is defined as  vector. 

• The corresponding bra 〈Ψ| is defined as (〈u₁|Ψ〉*〈u₂ |Ψ〉*...) or (c*₁,c*₂ ...c*_j), which is a row matrix. 
• Operator A is represented as Matrix A whose Matrix element A_ij is defined as 〈u_i|A|u_j〉 in basis of |u_i〉.
• Transpose of operator A is represented as Matrix A^T, whose matrix element is defined as A^T_ij = 〈u_j |A| u_i〉
• Hermitian adjoint of Matrix A is represented as A^†, whose matrix element is defined as A^† = 〈u_j |A| u_i〉
  Hermitian conjugate A^† of a matrix A can be found in two steps:

Step I: Find transpose of A i.e., convert row into column i.e., A^T 
Step II: Then take complex conjugate of each element of A^T.
Properties of Hermitian Adjoint A^† 

• (A^†)^† = A
• (λA)^† = λ*A^†
• (A + B)^† = A^† + B^†
• (AB)^† = B^† A^†

Eigen Value of Operator

If an operator A is defined as, A |ψ〉 = λ|ψ〉, then
λ is said to be eigen value and |ψ〉 is said to be eigenvector corresponding to operator.

Correspondence between Ket and Bra 

If A |ϕ〉 = |ψ〉, then 〈ϕ|A^† 〈ψ| , where A^† is Hermitian adjoint of matrix or operator A. 

Hermitian Operator

An operator A is said to be Hermitian if,  
A^† = A i.e., Matrix element 〈u_i| A|u_j〉 = (〈u_j |A|u_i〉)*

• The eigen values of Hermitian matrix is real 
• The eigen vectors corresponding to different eigen values are orthogonal. 

Commutator

If A and B are two operators, then the commentator [A, B] is defined as AB - BA.
If [A,B] = 0 , then it is said that operators A and B commute to each other. 

Properties of commutator: 

• Antisymmetry: [A, B] = -[B, A] 
• Linearity: [A, B + C + D] =  [A, B] + [A, C] + [A, D] 
• Distributive: [AB, C] + [A, C] B + A[B, C]
• Jacobi Identity: [A, [B, C]] + [B,[C, A]] + C[A, B]] = 0

Set of Commuting Observables

• If two operators A and B commute and if |ψ〉 is eigen vector of A , then B |ψ〉 is also an eigen vector of A , with the same eigenvalue. 
• If two operators A and B commute and if |ψ₁〉 and |ψ₂〉 are two eigenvectors of A with different eigenvalues, then Matrix element 〈ψ₁ | B | ψ₂〉 is zero. 
• If two operators A and B commute, then one can construct an orthonormal basis of state with eigenvectors common to A and B.

Projection Operator

• The operator P is said to be projection operator, if P^† = P and P² = P
• The product of two commutating projection operators P₁ and P₂ is also a projection operator. 
• The sum of two projection operators is generally not a projection operator. 
• The sum of projection operator, P₁ + P₂ + P₃ + .... is projection operator, if P_i ,P_j are mutually orthogonal.

Example 1: Prove that f(x) = x, g (x) = x² , h(x) = x³ are linearly independent.

For linear independency 12 3
a₁f(x) + a₂g(x) + a₃h(x) = 0 ⇒ a₁x + a₂x² + a₃x³ = 0
Equating the coefficient of x, x² and x³ on both sides, one can get.
a₁ = 0, a₂ = 0, a₃ = 0, so f(x), g(x) and h(x) are linearly independent.

Example 2: Prove that vector  are linearly dependent.

So,  are linearly dependent. 

Example 3:  then prove that f (x) and g (x) are orthogonal as well as linearly independent.

For linear independency:

So, f (x) and g (x) are linearly independent.
For orthogonality: (f(x)g (x)) =  ∫f*(x)g(x)dx
 Scalar product of f (x) and g (x) is zero i.e., orthogonal. 

Example 4:

(b) Find the value of A such that |ψ〉 is normalized.

(b) For normalization condition-

Example 5:
It is given that 〈ϕ_i | ϕ_j〉 = δ_ij then,
(a) Find the condition for |ψ₁〉 and |ψ₂〉 to be normalized.
(b) Find the condition for |ψ₁〉 and |ψ₂〉 to be orthogonal.

(a) If |ψ₁〉 is normalized, then |ψ₁ | ψ₁〉 = 1

It is given that

So, |a₁|² + |a₂|² = 1
Similarly, for |ψ₂〉 to be normalized-

⇒ |b₁|² + |b₂|² = 1
(b) For |ψ₁〉 and |ψ₂〉 to be orthogonal,

⇒ a*₁b*₁ + a*₂b₂ = 0
Similarly, from 〈ψ₂ | ψ₁〉 = 0 ⇒ b*₁a₁ + b*₂a₂ = 0

Example 6: If S operator is defined as

and 〈u_i | u_j〉 = δ_ij ;  i,j = 1, 2, 3
(a) Construct S matrix
(b) Prove that S is hermitian matrix

The Matrix

where matrix element S_ij = 〈u_i |S| u_j〉

∵ S = S^† i.e., S matrix is Hermitian.

Example 7: If D_x is defined as ∂/∂x and ψ(x) = A sin nπx/a
(a) Operate D_x on ψ (x)
(b) Operate D²_x on ψ(x)
(c) Which one of the above gives eigen value problem?

(c) When D²_x operate on

So, operation of  D²_x(x) on ψ(x) =A sin nπx/a give eigen value problem with eigen value

Example 8: If operator A is given by,  then
(a) find eigen value and eigen vector of A.
(b) normalized the eigen vector.
(c) prove both eigen vector are orthogonal.

(a)
for eigen value
| A - λI | = 0

The eigen vector corresponding to λ =  1,
A |u₁〉 = λ|u₁〉

so, eigen vector corresponding to

eigen vector corresponds to λ = -1

(b) For normalised eigen vector.

Similarly,



(c) for orthogonality, 〈u₁ | u₂〉 = 〈u₂ | u₁〉 = 0

Example 9: If momentum operator P_x is defined as  and position operator X is defined as XΨ(x) = xΨ(x)
(a) Find the value of commutator [X, Px]
(b) Find the value of [X² , Px]
(c) Find the value of [X , P²_x]

(a): [X, P_x]  (X P_x - P_xX)
Operate on both side ψ

(b) [X², P_x] = [X·X, P_x] = X [X, P_x]+ [X, P_x]X = Xiℏ + iℏX= 2iℏX
(c) [X, P²_x] = [X,P_x ·P_x]

Example 10: (a) Prove that P_Ψ = |Ψ〉〈Ψ| is projection operator
(b) Operate P_Ψ on |ϕ〉
(c) Operate P_Ψ on 〈ϕ|
(d) Operate P_Ψ on |Ψ〉 and 〈Ψ|
(e) Find the eigen value of any projection operator.

(a)

So, P_Ψ is projection operator.

⇒ P_Ψ = 0, P_Ψ = I, so eigen value of P_Ψ = 0 or 1

Example 11:
(a) Find the value of [A, B]
(b) Write down eigen vector of B in the basis of eigen vector of A.

(a): [A, B] = AB - BA

{As [A, B] = 0 , so A and B will commute}
(b) Eigen vector of A is  for eigen value λ₁ = a
 for eigen value λ₂ = a
Eigen vector of B is  for eigen value λ₁ = b
 for eigen value λ_2∂ = -b

Example 12:
(a) find A^† 
(b) find B^† 
(c) which one of A and B have real eigen value?

(a)

A^† = A , so A is Hermitian.

i.e., B^† = -B
So, it is not Hermitian rather it is Anti-Hermitian.
(c) The eigen value of A matrix is real because A is Hermitian.