Quantum Mechanics I Physics 5701

Quantum Mechanics I Physics 5701 Z. E. Meziani 02/10//2017

Outline 1 One Particle Wave Function Space F 2

One Particle Wave Function Space F One Particle Wave Function Space F The set of square-integrable functions is known as L 2 by mathematicians and it has a structure of a a Hilbert space. ψ ( r, t)ψ( r, t) d r = 1

One Particle Wave Function Space F One Particle Wave Function Space F The set of square-integrable functions is known as L 2 by mathematicians and it has a structure of a a Hilbert space. ψ ( r, t)ψ( r, t) d r = 1 This set is too wide in scope for treating only physics problems since we are only looking for square-integrable functions but functions that are

One Particle Wave Function Space F One Particle Wave Function Space F The set of square-integrable functions is known as L 2 by mathematicians and it has a structure of a a Hilbert space. ψ ( r, t)ψ( r, t) d r = 1 This set is too wide in scope for treating only physics problems since we are only looking for square-integrable functions but functions that are Everywhere defined

One Particle Wave Function Space F One Particle Wave Function Space F The set of square-integrable functions is known as L 2 by mathematicians and it has a structure of a a Hilbert space. ψ ( r, t)ψ( r, t) d r = 1 This set is too wide in scope for treating only physics problems since we are only looking for square-integrable functions but functions that are Everywhere defined Continuous

One Particle Wave Function Space F One Particle Wave Function Space F The set of square-integrable functions is known as L 2 by mathematicians and it has a structure of a a Hilbert space. ψ ( r, t)ψ( r, t) d r = 1 This set is too wide in scope for treating only physics problems since we are only looking for square-integrable functions but functions that are Everywhere defined Continuous Infinitely differentiable

One Particle Wave Function Space F One Particle Wave Function Space F The set of square-integrable functions is known as L 2 by mathematicians and it has a structure of a a Hilbert space. ψ ( r, t)ψ( r, t) d r = 1 This set is too wide in scope for treating only physics problems since we are only looking for square-integrable functions but functions that are Everywhere defined Continuous Infinitely differentiable We shall call F a subspace of L 2 containing a set of wave functions sufficiently regular for our description of physical systems

One Particle Wave Function Space F One Particle Wave Function Space F The set of square-integrable functions is known as L 2 by mathematicians and it has a structure of a a Hilbert space. ψ ( r, t)ψ( r, t) d r = 1 This set is too wide in scope for treating only physics problems since we are only looking for square-integrable functions but functions that are Everywhere defined Continuous Infinitely differentiable We shall call F a subspace of L 2 containing a set of wave functions sufficiently regular for our description of physical systems We now need to study the structure of this wave function space

F is a vector space that is if ψ( r) L 2 then where λ 1 and λ 2 are complex numbers ψ( r) = λ 1 ψ 1 ( r) + λ 2 ψ 2 ( r) ψ( r) 2 = λ 1 2 ψ 1 ( r) 2 + λ 2 2 ψ 2 ( r) 2 + λ 1 λ 2ψ1 ( r)ψ 2( r) }{{} + λ 1λ 2 ψ 1( r)ψ 2 }{{ ( r) }}{{} The modulus of these two quantities is the same and has an upper limit. The upper limit is λ 1 λ 2 [ ψ 1 2 + ψ 2] this implies that ψ( r) 2 is smaller than a function whose integral converges, thus it is a square integrable function. This property is an example of the fact that F is a vector space

Scalar Product For each pair of elements of F, φ( r) and ψ( r) taken in this order. We associate a complex number denoted (φ, ψ) equals to: (φ, ψ) = ψ ( r)ψ( r)d r (φ, ψ) is called scalar product of ψ( r) by φ( r) and the result is finite if φ, ψ F

Properties 1 The scalar product is linear with respect to the second function of the pair and anti-linear with respect to the first one (φ, ψ) = (ψ, φ) (φ, λ 1 ψ 1 + λ 2 ψ 2 ) = λ 1 (φ, ψ 1 ) + λ 2 (φ, ψ 2 ) (λ 1 φ 1 + λ 2 φ 2, ψ) = λ 1 (φ 1, ψ) + λ 2 (φ 2, ψ)

Properties 1 The scalar product is linear with respect to the second function of the pair and anti-linear with respect to the first one (φ, ψ) = (ψ, φ) (φ, λ 1 ψ 1 + λ 2 ψ 2 ) = λ 1 (φ, ψ 1 ) + λ 2 (φ, ψ 2 ) (λ 1 φ 1 + λ 2 φ 2, ψ) = λ 1 (φ 1, ψ) + λ 2 (φ 2, ψ) 2 If (φ, ψ) = 0 then φ and ψ are said to be orthogonal

Properties 1 The scalar product is linear with respect to the second function of the pair and anti-linear with respect to the first one (φ, ψ) = (ψ, φ) (φ, λ 1 ψ 1 + λ 2 ψ 2 ) = λ 1 (φ, ψ 1 ) + λ 2 (φ, ψ 2 ) (λ 1 φ 1 + λ 2 φ 2, ψ) = λ 1 (φ 1, ψ) + λ 2 (φ 2, ψ) 2 If (φ, ψ) = 0 then φ and ψ are said to be orthogonal 3 (ψ, ψ) = ψ( r) 2 d r is a real positive number and is zero only if ψ( r) = 0. (ψ, ψ) is called the norm of ψ( r)

Properties 1 The scalar product is linear with respect to the second function of the pair and anti-linear with respect to the first one (φ, ψ) = (ψ, φ) (φ, λ 1 ψ 1 + λ 2 ψ 2 ) = λ 1 (φ, ψ 1 ) + λ 2 (φ, ψ 2 ) (λ 1 φ 1 + λ 2 φ 2, ψ) = λ 1 (φ 1, ψ) + λ 2 (φ 2, ψ) 2 If (φ, ψ) = 0 then φ and ψ are said to be orthogonal 3 (ψ, ψ) = ψ( r) 2 d r is a real positive number and is zero only if ψ( r) = 0. (ψ, ψ) is called the norm of ψ( r) 4 Schwarz Inequality; (ψ 1, ψ 2 ) (ψ 1, ψ 1 ) (ψ 2, ψ 2 ) This becomes an equality when the two functions ψ 1 and ψ 2 are proportional.

Linear Operators A linear operator is a mathematical entity which associates with every function from ψ( r) F another function ψ ( r), the correspondence being linear. ψ ( r) = Âψ( r) [ ] Â [λ 1 ψ 1 ( r) + λ 2 ψ 2 ( r)] = λ 1 Âψ 1 ( r) + λ 2 Âψ 2 ( r)

Linear Operators A linear operator is a mathematical entity which associates with every function from ψ( r) F another function ψ ( r), the correspondence being linear. ψ ( r) = Âψ( r) [ ] Â [λ 1 ψ 1 ( r) + λ 2 ψ 2 ( r)] = λ 1 Âψ 1 ( r) + λ 2 Âψ 2 ( r) We have seen examples of linear operators: ˆP x = i x or ˆX

The parity operator is another example ˆΠψ(x, y, z) = ψ( x, y, z) ˆXψ(x, y, z) = xψ(x, y, z) ˆP ψ(x, y, z) = i ψ(x, y, z) x

The parity operator is another example ˆΠψ(x, y, z) = ψ( x, y, z) ˆXψ(x, y, z) = xψ(x, y, z) ˆP ψ(x, y, z) = i ψ(x, y, z) x Comment: When acting on ψ( r) in F these operators can transform it into a function which no longer belongs to F.

Product of Operators Assume  and ˆB two linear operators their product is defined by [ ] ( ˆB) ψ( r) =  ˆBψ( r) ˆB is first allowed to act on ψ( r) then it is  s turn to act on the resulting function. In general  ˆB ˆBÂ. as we have seen earlier with ˆX and ˆP x.

Product of Operators Assume  and ˆB two linear operators their product is defined by [ ] ( ˆB) ψ( r) =  ˆBψ( r) ˆB is first allowed to act on ψ( r) then it is  s turn to act on the resulting function. In general  ˆB ˆBÂ. as we have seen earlier with ˆX and ˆP x. The commutator of  and ˆB is defined by [Â, ˆB] =  ˆB ˆB Example: [ ˆX, ˆPx ] = i

Example of commutator calculation [ ] ˆX, ˆPx ψ( r) = ˆXi [ ] ψ( r) i ˆXψ( r) x x = i x ψ( r) i x x (xψ( r)) = i x ψ( r) ψ( r) i x x x i ψ( r) = i ψ( r) [ ˆX, ˆPx ] = i

Discrete orthonormal basis in F: { u i ( r) } Definition: Consider a countable set of functions of F labeled by a discrete index i (i = 1, 2, 3...n) : u 1 ( r), u 2 ( r),...u n( r), F This set is orthonormal if: (u i, u j ) = u i ( r)u j( r)d r = δ ij where δ ij is the kronecker symbol δ i,j = 0 for i j and δ i,j = 0 for i = j

Discrete orthonormal basis in F: { u i ( r) } Definition: Consider a countable set of functions of F labeled by a discrete index i (i = 1, 2, 3...n) : u 1 ( r), u 2 ( r),...u n( r), F This set is orthonormal if: (u i, u j ) = u i ( r)u j( r)d r = δ ij where δ ij is the kronecker symbol δ i,j = 0 for i j and δ i,j = 0 for i = j This set constitute a basis if every function ψ( r) F can be expanded in one and only one way in terms of u i ( r) ψ( r) = i c i u i ( r)

Components of a wave function in the { u i ( r) } basis u j ( r)ψ( r)d r = u j ( r) i c i u i ( r)d r One should make sure the i and d r are interchangeable (u j, ψ) = (u j, i c i u i ) = i c i (u j, u i ) = i c i δ ij = c j

Components of a wave function in the { u i ( r) } basis u j ( r)ψ( r)d r = u j ( r) i c i u i ( r)d r One should make sure the i and d r are interchangeable (u j, ψ) = (u j, i c i u i ) = i c i (u j, u i ) = i c i δ ij = c j We obtain: c j = (u j, ψ) = u j ( r)ψ( r)d r

Components of a wave function in the { u i ( r) } basis u j ( r)ψ( r)d r = u j ( r) i c i u i ( r)d r One should make sure the i and d r are interchangeable (u j, ψ) = (u j, i c i u i ) = i c i (u j, u i ) = i c i δ ij = c j We obtain: c j = (u j, ψ) = u j ( r)ψ( r)d r The components c j of ψ( r) on u j ( r) is equal to the scalar product of ψ( r) by u j ( r). The set of numbers c i is said to represent ψ( r) in the u i ( r) basis since it is equivalent to specify either ψ( r) or its set of components c i with respect to the basis u i ( r). The c i are in general complex numbers.

Expression of the scalar product in terms of the components Assume φ( r) = i b iu i ( r) and ψ( r) = j c ju j ( r) then (φ, ψ) = i b i u i, j c j u j = ij b i c j(u i, u j ) = ij b i c jδ ij = i b i c i (φ, ψ) = i b i c i

Expression of the scalar product in terms of the components Assume φ( r) = i b iu i ( r) and ψ( r) = j c ju j ( r) then (φ, ψ) = i b i u i, j c j u j = ij b i c j(u i, u j ) = ij b i c jδ ij = i b i c i (φ, ψ) = i b i c i The scalar product is expressed in terms of the components in the basis u i ( r). In particular (ψ, ψ) = c i c i = c i 2 i i

Expression of the scalar product in terms of the components Assume φ( r) = i b iu i ( r) and ψ( r) = j c ju j ( r) then (φ, ψ) = i b i u i, j c j u j = ij b i c j(u i, u j ) = ij b i c jδ ij = i b i c i (φ, ψ) = i b i c i The scalar product is expressed in terms of the components in the basis u i ( r). In particular (ψ, ψ) = c i c i = c i 2 i i The above is a generalization of the following in the 3D vector space. v, w R 3 v = i v i e i w = j w j e j v w = i v i w i

Closure relation Since {u i ( r)} is a basis of F thus for every function of F there is an expression such that ψ( r) = c i u i ( r) = (u i, ψ)u i ( r) = [ ] u i ( r )ψ( r ) u i ( r) i i i = d rψ( r ) u i ( r )u i ( r) i

Closure relation Since {u i ( r)} is a basis of F thus for every function of F there is an expression such that ψ( r) = c i u i ( r) = (u i, ψ)u i ( r) = [ ] u i ( r )ψ( r ) u i ( r) i i i = d rψ( r ) u i ( r )u i ( r) i The last result is possible if one can interchange i and d r. Now i u i ( r )u i ( r) = F ( r, r ) such that for every function we have ψ( r) = d r ψ( r )F ( r, r )

Closure relation Since {u i ( r)} is a basis of F thus for every function of F there is an expression such that ψ( r) = c i u i ( r) = (u i, ψ)u i ( r) = [ ] u i ( r )ψ( r ) u i ( r) i i i = d rψ( r ) u i ( r )u i ( r) i The last result is possible if one can interchange i and d r. Now i u i ( r )u i ( r) = F ( r, r ) such that for every function we have ψ( r) = d r ψ( r )F ( r, r ) This equation is characteristic of a δ( r r ) distribution function which takes the form of a Closure Relation u i ( r )u i ( r) = δ( r r ) i

Closure relation Since {u i ( r)} is a basis of F thus for every function of F there is an expression such that ψ( r) = c i u i ( r) = (u i, ψ)u i ( r) = [ ] u i ( r )ψ( r ) u i ( r) i i i = d rψ( r ) u i ( r )u i ( r) i The last result is possible if one can interchange i and d r. Now i u i ( r )u i ( r) = F ( r, r ) such that for every function we have ψ( r) = d r ψ( r )F ( r, r ) This equation is characteristic of a δ( r r ) distribution function which takes the form of a Closure Relation u i ( r )u i ( r) = δ( r r ) i Reciprocally, if an orthonormal set verify the closure relation it constitute a basis ψ( r) = d r ψ( r )δ( r r )

Introduction to bases" not belonging to F We will introduce bases which do not belong to F but are still useful since any function ψ( r) F can nevertheless be expanded in terms of these bases.

Introduction to bases" not belonging to F We will introduce bases which do not belong to F but are still useful since any function ψ( r) F can nevertheless be expanded in terms of these bases. Plane waves: One dimensional case We want to study the square integrable function ψ(x) which depend only on the x variable. We have already used the Fourier transform to treat ψ(x) = ψ(x) = 1 ψ(p)e ipx/ dp 2π 1 ψ(x)e ipx/ dx 2π if we consider the function v p(x) = 1 2π e ipx/ Then we can express ψ(x) F x in one and only way in terms of v p(x) ψ(x) = ψ(p)v p(x)dp ψ(x) = (v p, ψ) = vp (x)ψ(x)dx

Continued ψ(p) appears as the analog of the coefficients c i which represent the components of the function ψ(x) in a given bases. Using Parseval s theorem + (ψ, ψ) = dp ψ(p) 2

Continued ψ(p) appears as the analog of the coefficients c i which represent the components of the function ψ(x) in a given bases. Using Parseval s theorem + (ψ, ψ) = dp ψ(p) 2 ψ(p) play the same role as the c i but are in a continuous basis rather than a discrete basis. In fact ψ(p) fulfill the Closure Relation + dp v p(x)vp(x ) 2 = 1 2π dp e i p (x x ) = δ(x x )

Continued ψ(p) appears as the analog of the coefficients c i which represent the components of the function ψ(x) in a given bases. Using Parseval s theorem + (ψ, ψ) = dp ψ(p) 2 ψ(p) play the same role as the c i but are in a continuous basis rather than a discrete basis. In fact ψ(p) fulfill the Closure Relation + dp v p(x)vp(x ) 2 = 1 2π the scalar product (v p, v p ) is given by + dp v p(x)vp(x ) 2 = 1 2π dp e i p (x x ) = δ(x x ) dxe i x (px p x ) = δ(p x p x) v p(x) / F if we set p x = p x the δ function diverges. The three dimension generalization present no difficulties: (v p, v p ) = δ( p p )

Delta Function Basis ξ r0 ( r) = δ( r r 0 ) ψ( r) = ψ( r 0 ) = d r 0 ψ( r 0 )δ( r r 0 ) = d r 0 ψ( r 0 )ξ r0 ( r) d rψ( r)δ( r 0 r) = d rψ( r)ξ r 0 ( r) ψ ( r 0 ) is equivalent to c i, these complex numbers represent the component of ψ( r) in two different bases.

Delta Function Basis ξ r0 ( r) = δ( r r 0 ) ψ( r) = ψ( r 0 ) = d r 0 ψ( r 0 )δ( r r 0 ) = d r 0 ψ( r 0 )ξ r0 ( r) d rψ( r)δ( r 0 r) = d rψ( r)ξ r 0 ( r) ψ ( r 0 ) is equivalent to c i, these complex numbers represent the component of ψ( r) in two different bases. Properties:

Delta Function Basis ξ r0 ( r) = δ( r r 0 ) ψ( r) = ψ( r 0 ) = d r 0 ψ( r 0 )δ( r r 0 ) = d r 0 ψ( r 0 )ξ r0 ( r) d rψ( r)δ( r 0 r) = d rψ( r)ξ r 0 ( r) ψ ( r 0 ) is equivalent to c i, these complex numbers represent the component of ψ( r) in two different bases. Properties: (φ, ψ) = d r 0φ ( r 0)ψ( r 0)

Delta Function Basis ξ r0 ( r) = δ( r r 0 ) ψ( r) = ψ( r 0 ) = d r 0 ψ( r 0 )δ( r r 0 ) = d r 0 ψ( r 0 )ξ r0 ( r) d rψ( r)δ( r 0 r) = d rψ( r)ξ r 0 ( r) ψ ( r 0 ) is equivalent to c i, these complex numbers represent the component of ψ( r) in two different bases. Properties: (φ, ψ) = d r 0φ ( r 0)ψ( r 0) d r 0ξ r0 ( r)ξ r ( r ) = d r 0 0δ( r r 0)δ( r r 0) = δ( r r )

Delta Function Basis ξ r0 ( r) = δ( r r 0 ) ψ( r) = ψ( r 0 ) = d r 0 ψ( r 0 )δ( r r 0 ) = d r 0 ψ( r 0 )ξ r0 ( r) d rψ( r)δ( r 0 r) = d rψ( r)ξ r 0 ( r) ψ ( r 0 ) is equivalent to c i, these complex numbers represent the component of ψ( r) in two different bases. Properties: (φ, ψ) = d r 0φ ( r 0)ψ( r 0) d r 0ξ r0 ( r)ξ r ( r ) = d r 0 0δ( r r 0)δ( r r 0) = δ( r r ) (ξ r0, ξ r 0 ) = δ( r 0 r 0 )

Delta Function Basis ξ r0 ( r) = δ( r r 0 ) ψ( r) = ψ( r 0 ) = d r 0 ψ( r 0 )δ( r r 0 ) = d r 0 ψ( r 0 )ξ r0 ( r) d rψ( r)δ( r 0 r) = d rψ( r)ξ r 0 ( r) ψ ( r 0 ) is equivalent to c i, these complex numbers represent the component of ψ( r) in two different bases. Properties: (φ, ψ) = d r 0φ ( r 0)ψ( r 0) d r 0ξ r0 ( r)ξ r ( r ) = d r 0 0δ( r r 0)δ( r r 0) = δ( r r ) (ξ r0, ξ r 0 ) = δ( r 0 r 0 ) Compared to the discrete case we have changed i r 0 d r 0 i δ ij δ( r 0 r 0 )

A summary important formulas Discrete basis { u i ( r)} Continuous basis { w α( r)} Ortho-normalization relation (u i, u j ) = δ ij (w α, u α ) = δ(α α ) Closure Relation i u( r)u i ( r ) = δ( r r ) dα w α( r)w α( r ) = δ( r r ) Expansion of a wave function ψ( r) ψ( r) = i c iu i ( r) ψ( r) = dα c(α)w α( r) Expression of the components of ψ( r) c i = (u i, ψ) = d ru i ( r)ψ( r) c(α) = (wα, ψ) = d rw α ( r)ψ( r) Scalar product (φ, ψ) = i b i c i (φ, ψ) = dαb (α)c(α) Square of the Norm (ψ, ψ) = i c i 2 (ψ, ψ) = dα c(α) 2