Mathematics of quantum mechanics bridge course calicut university

1. Mathematical Foundations of Quantum Mechanics Dr.Randeep N C Department of Physics Government Arts & Science College, Kozhikode May 17-24, 2021 Dr.Randeep N C Mathematical foundations of QM

2. Matrix Representations Lecture-4 Dr.Randeep N C Mathematical foundations of QM

3. Matrix Representations Consider two operator Â and X̂ in a vector space V with eigenstates {|ai i} of Â is complete, then X̂ = 1X̂1 = N X j=1 N X k=1 |aj i haj | X̂ |ak i hak | X̂ = N X j=1 N X k=1 (haj | X̂ |ak i) |aj i hak | Here there are altogether N2 number of the form haj | X̂ |ak i, where N is the dimensionality of the ket space. We can arrange them as N × N square matrix such that the column and row indices appears as follows haj | row X̂ |ak i column = X̂jk is called matrix elements of the operator X̂. Dr.Randeep N C Mathematical foundations of QM

4. Matrix Representations Therefore, the operator X̂ is represented by X̂ . =       ha1| X̂ |a1i ha1| X̂ |a2i . . . ha2| X̂ |a1i ha2| X̂ |a2i . . . . . . . . . . . . . haN| X̂ |a1i haN| X̂ |a2i . . .       For a general operator X̂, haj | X̂ |ak i = hak | X̂† |aj i ∗ If operator X̂ is Hermitian haj | X̂ |ak i = hak | X̂ |aj i ∗ Representation of matrix multiplication Ẑ = X̂Ŷ in terms of matrix elements as haj | Ẑ |ak i = haj | X̂Ŷ |ak i Dr.Randeep N C Mathematical foundations of QM

7. Matrix Representations Suppose |αi = P i ci |ai i and |βi = P i di |ai i, then the inner product hβ|αi = d∗ 1 d∗ 2 . . .       c1 c2 . . .       = d∗ 1 c1 + d∗ 2 c2 + ... and the outer product |αi hβ| =       c1 c2 . . .       d∗ 1 d∗ 2 . . . =       c1d∗ 1 c1d∗ 2 . . . c2d∗ 1 c2d∗ 2 . . . . . . . . . . . . . . . . . .       Dr.Randeep N C Mathematical foundations of QM

9. Problems Problem 6: Consider a three dimensional vector space spanned by an orthonormal basis |1i , |2i , |3i. Kets |αi and |βi are given by |αi = i |1i − 2 |2i − i |3i , |βi = i |1i + 2 |3i . Find all nine matrix elements of the operator Â = |αihβ|, in this basis, and construct the matrix A. Is it Hermitian. Answer: Â = |αihβ| = (i |1i − 2 |2i − i |3i)(−i h1| + 2 h3|) = |1i h1| + 2i |1i h3| + 2i |2i h1| − 4 |2i h3| − |3i h1| − 2i |3i h3| Matrix elements of an operator Â are h1|Â|1i = 1, h1|Â|2i = 0, h1|Â|3i = 2i, h2|Â|1i = 2i, h2|Â|2i = 0, h2|Â|3i = −4, h3|Â|1i = −1, h3|Â|2i = 0, h3|Â|3i = −2i. Dr.Randeep N C Mathematical foundations of QM

10. Problems Therefore, A matrix is A =   1 0 2i 2i 0 −4 −1 0 2i   then A† =   1 −2i −1 0 0 0 −2i −4 −2i   Here A† 6= A, therefore matrix A is not Hermitian. Problem 7: For any operator A, i(A† − A) is (a) Hermitian, (b) Anti-Hermitan, (c) Unitary, (d) Orthonormal. Answer: For a Hermitan operator X† = X, (i(A† − A))† = −i((A† )† − A† ) = −i(A − A† ) = i(A† − A) =⇒ i(A† − A) is Hermitian. Dr.Randeep N C Mathematical foundations of QM

11. Problems Problem 8: If a Hamiltonian H is given as H = (|0i h0| − |1i h1|) +i(|0i h1| − |1i h0|), where |0i and |1i are orthonormal states. The eigenvalues of H are [JEST- 2015] (a) ± 1, (b) ± i, (c) ± √ 2, (d) ± i √ 2 Answer: Hamiltonian in matrix notation is H = 1 i −i −1 To find eigenvalues, solve the determinant |H − λ1| =

15. 1 − λ i −i −1 − λ

19. = 0 −(1 − λ)(1 + λ) + i2 = 0 −(1 − λ2 ) − 1 = 0 λ2 = 2 λ = ± √ 2 Dr.Randeep N C Mathematical foundations of QM

20. Problems Problem 9: The Hamiltonian operator for a two state system is given by H = α(|1i h1| − |2i h2| + |1i h2| + |2i h1|) where α is a positive number with the dimension of energy. The energy eigenstates corresponding to the larger and smaller eigenvalue respectively are [JEST-2014] (a) |1i − ( √ 2 + 1) |2i , |1i + ( √ 2 − 1) |2i , (b) |1i + ( √ 2 − 1) |2i , |1i − ( √ 2 + 1) |2i , (c) |1i + ( √ 2 − 1) |2i , ( √ 2 + 1) |1i − |2i , (d) |1i − ( √ 2 + 1) |2i , ( √ 2 − 1) |1i + |2i . Answer: H = α α α −α |H − λ1| =

24. α − λ α α −α − λ

28. = 0 −(α − λ)(α + λ) − α2 = 0 λ2 = 2α2 =⇒ λ = ± √ 2α That is eigenvalues of H matrix are λ1 = + √ 2α and λ2 = − √ 2α Dr.Randeep N C Mathematical foundations of QM

29. Problems To find eigenstate |λ1i corresponding to eigenvalue λ1 = + √ 2α, α − √ 2α α α −α − √ 2α x y = 0 (α − √ 2α)x + αy = 0 (1 − √ 2)x + y = 0 =⇒ x = 1, y = √ 2 − 1 Therefore, |λ1i = |1i + ( √ 2 − 1) |2i To find eigenstate |λ2i corresponding to eigenvalue λ1 = − √ 2α, α + √ 2α α α −α + √ 2α x y = 0 (α + √ 2α)x + αy = 0 (1 + √ 2)x + y = 0 =⇒ x = 1, y = −( √ 2 + 1) Therefore, |λ2i = |1i − ( √ 2 + 1) |2i Dr.Randeep N C Mathematical foundations of QM

30. Spin 1 2 system Spin 1/2 system is a two level quantum system, one of the simplest quantum system we can consider. The component of spin (S) in different directions are denoted by the operators Ŝx , Ŝy and Ŝz . The eigenvalue equations of Ŝz operator are Ŝz |0i = + ~ 2 |0i , Ŝz |1i = − ~ 2 |1i Spin is an observable, then the operator Ŝz is Hermitian and its eienstates span the 2 dimensional Hilbert space, that is |0i h0| + |1i h1| = 1 =⇒ Completeness reation An operator Ŝz in its eigenbasis can be written as Ŝz = ~ 2 [|0i h0| − |1i h1|] Dr.Randeep N C Mathematical foundations of QM

31. Spin 1 2 system Consider two non Hermitian operators S+ = ~(|0i h1|), S− = ~(|1i h0|) Act these operators in the eigenstates of Ŝz operator S+ |0i = ~(|0i h1|) |0i = 0, S+ |1i = ~(|0i h1|) |1i = ~ |0i S− |0i = ~(|1i h0|) |0i = ~ |1i , S− |1i = ~(|1i h0|) |1i = 0 Because of this results, S+ is called raising operator and S− is called lowering operator. This Ŝ± can be written as Ŝ± = Ŝx ± iŜy . Let |0i = 1 0 , |1i = 0 1 , then Ŝz = ~ 2 [|0i h0| − |1i h1|] = ~ 2 1 0 0 −1 , Dr.Randeep N C Mathematical foundations of QM

32. Spin 1 2 system S+ = ~(|0i h1|) = ~ 0 1 0 0 , S− = ~(|1i h0|) = ~ 0 0 1 0 Therefore, Ŝx = S+ + S− 2 = ~ 2 0 1 1 0 , Ŝy = S+ − S− 2i = ~ 2 0 −i i 0 That is spin component can be written as Ŝz = ~ 2 σ̂z , Ŝx = ~ 2 σ̂x , Ŝy = ~ 2 σ̂y where σ̂x , σ̂y , σ̂z are Pauli spin matrices. Therefore, eigenvectors of an operator Ŝz are |0i and |1i with eigenvalues ~ 2 and −~ 2 respectively. Also eigenvectors of an operator Ŝx are |+i and |−i with eigenvalues ~ 2 and −~ 2 respectively. Dr.Randeep N C Mathematical foundations of QM

33. Spin 1 2 system Eigenvalue equations of Ŝz operator can be written in another form, Ŝz |Sz ; +i = ~ 2 |Sz ; +i, Ŝz |Sz ; −i = − ~ 2 |Sz ; −i where, |Sz ; +i = |0i = 1 0 is an eigenvector with eigenvalue +~ 2 . |Sz ; −i = |1i = 0 1 is an eigenvector with eigenvalue −~ 2 . Eigenvalue equations of Ŝx operator are Ŝx |Sx ; +i = ~ 2 |Sx ; +i, Ŝx |Sx ; −i = − ~ 2 |Sx ; −i where, |Sx ; +i = |+i = 1 √ 2 1 1 is an eigenvector with eigenvalue +~ 2 . |Sx ; −i = |−i = 1 √ 2 1 −1 is an eigenvector with eigenvalue −~ 2 . Dr.Randeep N C Mathematical foundations of QM

35. Thank You Dr.Randeep N C Mathematical foundations of QM

36. Mathematical Foundations of Quantum Mechanics Dr.Randeep N C Department of Physics Government Arts Science College, Kozhikode May 17-24, 2021 Dr.Randeep N C Mathematical foundations of QM

37. Linear Vector Space Lecture-1 Dr.Randeep N C Mathematical foundations of QM

38. Question Question: How do we specify the state of the system in quantum mechanics ? Dr.Randeep N C Mathematical foundations of QM

39. Question Question: How do we specify the state of the system in quantum mechanics ? State of the system in classical mechanics and thermodynamics. X P Phase space (X, P ) V P PV diagram (P, V ) Dr.Randeep N C Mathematical foundations of QM

40. Question Question: How do we specify the state of the system in quantum mechanics ? Answer:- Linear Vector Spaces Inner Product Spaces Hilbert Spaces |ψi Dr.Randeep N C Mathematical foundations of QM

41. Linear Vector Space A linear vector space V is a collection of objects (|αi , |βi , |γi , ....) called vectors, for which there exists A definite rule for forming the vector sum, |αi + |βi A definite rule for multiplication by scalars (a, b, c, ....), a |αi with following features 1 Closure: |αi + |βi ∈ V 2 Vector addition is commutative: |αi + |βi = |βi + |αi 3 Vector addition is associative: |αi + (|βi + |γi) = (|αi + |βi) + |γi 4 There exists a null vector |0i obeying |αi + |0i = |αi 5 For every vector |αi there exist an inverse under addition |−αi such that |αi + |−αi = |0i 6 Scalar multiplication is distribute in the vectors: a(|αi + |βi) = a |αi + a |βi 7 Scalar multiplication is distributive in scalars: (a + b) |αi = a |αi + b |αi 8 Scalar multiplication is associative: a(b |αi) = ab |αi. Dr.Randeep N C Mathematical foundations of QM

42. Linear Vector Space This scalars a, b, c, .... are either real or complex numbers If scalars are real numbers =⇒ Real vector space If scalars are complex numbers =⇒ Complex vector space Examples:- Set of real numbers is a vector space over real scalars. 1 2 + 3 = 5 ∈ V 2 2 + 3 = 3 + 2 3 2 + (3 + 4) = (2 + 3) + 4 4 2 + 0 = 2 5 2 + (−2) = 0 6 2(3 + 4) = 2 × 3 + 2 × 4 7 (2 + 3)4 = 2 × 4 + 3 × 4 8 2(3 × 4) = (2 × 3)4 Set of complex numbers over real scalars. Set of complex numbers over complex scalars. Set of real N × N matrices over real scalars. Set of real 2 × 1 matrices over real scalars. Dr.Randeep N C Mathematical foundations of QM

43. Example: Set of real 2 × 1 matrices 1 1 0 + 0 1 = 1 1 ∈ V 2 1 0 + 0 1 = 0 1 + 1 0 3 1 0 + 0 1 + 1 −1 ! = 0 1 + 1 0 ! + 1 −1 4 1 0 + 0 0 = 1 0 5 1 0 + −1 0 = 0 0 Dr.Randeep N C Mathematical foundations of QM

44. Example: Set of real 2 × 1 matrices 6 2 1 0 + 0 1 ! = 2 1 0 + 2 0 1 7 2 + 3 1 0 = 2 1 0 + 3 1 0 8 2 3 1 0 ! = 2 × 3 1 0 Let us denote |0i = 1 0 and |1i = 0 1 Dr.Randeep N C Mathematical foundations of QM

45. Linear independence Linear independence: A set of vectors {|α1i , |α2i , ..., |αni} is said to be linearly independent if c1 |α1i + c2 |α2i + .... + cn |αni = 0 =⇒ c1, c2, ...., cn = 0. Consider the set {|0i , |1i} c1 |0i + c2 |1i = 0 c1 1 0 + c2 0 1 = c1 c2 = 0 0 =⇒ c1, c2 = 0. Therefore, the set of vectors {|0i , |1i} are linearly independent and the vector |0i cannot be written in terms of |1i. Linear dependence: A set of vectors {|α1i , |α2i , ..., |αni} is said to be linearly dependent if there exist a set of complex numbers c1, c2, ..., cn with ci 6= 0 for atleast one value of i, such that c1 |α1i + c2 |α2i + .... + cn |αni = 0. Dr.Randeep N C Mathematical foundations of QM

46. Basis and Dimension Span: A collection of vectors is said to span the space if every vector can be written as a linear combination of the members of this set. Any arbitrary vector of 2 × 1 matrices can be written as a b = a 1 0 + b 0 1 = a |0i + b |1i That is the set of vectors {|0i , |1i} span the space of 2 × 1 real matrices. Basis: A set of linearly independent vectors that span the space is called a basis. The set of vectors {|0i , |1i} are linearly independent and span the space of 2 × 1 real matrices. Therefore, they are basis for the space. Dr.Randeep N C Mathematical foundations of QM

47. Basis and Dimension Standard Basis: A standard basis (also called natural basis) is a set of vectors in which each basis vector has a single nonzero entry with value 1. For 2D, standard basis set contain the vectors, 1 0 , 0 1 For 3D,   1 0 0   ,   0 1 0   ,   0 0 1   The set of vectors {|e1i , |e2i , ...., |eni} represents the basis for N dimension space. Dimension: The number of vectors in any basis is called the dimension of the vector space. Dr.Randeep N C Mathematical foundations of QM

48. Dual Space Dual Space: The dual space of V, is denoted by V∗ , is a space of all linear functionals on V. A linear functional is a linear map from V to F, T : V → F Suppose {|α1i , |α2i , ...} ∈ V and a ∈ F, then T(|αi i) → F this linear functional satisfy the conditions T(|α1i + |α2i) = T(|α1i) + T(|α2i) and T(a |α1i) = aT(|α1i) The elements of dual space are denoted by the set of vectors {hα1| , hα2| , ....}, calling ‘ hα| 0 as bra vector. Dr.Randeep N C Mathematical foundations of QM

49. Dual Space Example: Vector space V is a space of 2 × 1 real matrices with a1 a2 , b1 b2 , ... ∈ V Then the corresponding dual space V∗ is a space of 1 × 2 real matrices with elements, a1 a2 , b1 b2 , ....., ∈ V∗ These two sets of vectors satisfy the condition a1 a2 b1 b2 = a1b1 + a2b2 ∈ F We can show that dual space V∗ is also a vector space with dim(V∗ )=dim(V) The basis for the dual space is represented by the set of bra vectors {he1| , he2| , ....}. Dr.Randeep N C Mathematical foundations of QM

50. Dual Correspondence In quantum mechanics, we are calling the space V as ket space and its dual space V∗ as bra space. The elements of ket space are ket vectors and the elements of bra space are bra vectors. There exist a one to one correspondence between a ket space and bra space calling dual correspondence (DC):. |αi DC ← → hα| |αi + |βi DC ← → hα| + hβ| c |αi DC ← → c∗ hα| |e1i , |e2i , ...., |eni DC ← → he1| , he2| , ...., hen| Dr.Randeep N C Mathematical foundations of QM

51. Problems Problem:1 Consider a set of three vectors {(1, 1, 1), (1, −1, 1), (1, 1, −1)}, check whether they are linearly independent or not ? Dr.Randeep N C Mathematical foundations of QM

52. Problems Problem:1 Consider a set of three vectors {(1, 1, 1), (1, −1, 1), (1, 1, −1)}, check whether they are linearly independent or not ? Answer: Let |1i =   1 1 1   , |2i =   1 −1 1   , |3i =   1 1 −1   c1 |1i + c2 |2i + c3 |3i = c1   1 1 1   + c2   1 −1 1   + c3   1 1 −1   = 0   c1 + c2 + c3 c1 − c2 + c3 c1 + c2 − c3   =   0 0 0   (c1 − c2 + c3) + (c1 + c2 − c3) = 0 =⇒ 2c1 = 0, c1 = 0 (c1 + c2 + c3) + (c1 + c2 − c3) = 0 =⇒ 2c1 + 2c2 = 0, c2 = 0 c1 = 0, c2 = 0, c1 + c2 + c3 =⇒ c3 = 0 Here c1 = c2 = c3 = 0. Therefore the set of vectors {|1i , |2i , |3i} are linearly independent. Dr.Randeep N C Mathematical foundations of QM

53. Problems Another method: c1 + c2 + c3 = 0 c1 − c2 + c3 = 0 c1 + c2 − c3 = 0   1 1 1 1 −1 1 1 1 −1     c1 c2 c3   = 0 This equation of the form AX = 0. In this equation, 1 For nontrivial solutions (X 6= 0) exist, |A| = 0 =⇒ the set of vectors are linearly dependent. 2 Only trivial solution (x = 0) exist, |A| 6= 0 =⇒ the set of vectors are linearly independent. |A| =

59. 1 1 1 1 −1 1 1 1 −1

65. = 1(1 − 1) − 1(−1 − 1) + 1(1 + 1) = 4 Here |A| 6= 0, then the set vectors are linearly independent. Dr.Randeep N C Mathematical foundations of QM

66. Problems Problem:2 Consider the three vectors ~ V1 = 2î + 3k̂, ~ V2 = î + 2ĵ + 2k̂ and ~ V3 = 5î + ĵ + αk̂, where î, ĵ and k̂ are the standard unit vectors in a three dimensional Euclidean space. These vectors will be linearly dependent if the value of α is (a) 31/4 (b) 23/4 (c) 27/4 (d) 0 [ NET-June-2018 ] Dr.Randeep N C Mathematical foundations of QM

67. Problems Problem:2 Consider the three vectors ~ V1 = 2î + 3k̂, ~ V2 = î + 2ĵ + 2k̂ and ~ V3 = 5î + ĵ + αk̂, where î, ĵ and k̂ are the standard unit vectors in a three dimensional Euclidean space. These vectors will be linearly dependent if the value of α is (a) 31/4 (b) 23/4 (c) 27/4 (d) 0 [ NET-June-2018 ] Answers: ~ V1 =   2 0 3   , ~ V2 =   1 2 2   , ~ V3 =   5 1 α   For a linearly dependent set of vectors |A| = 0, |A| =

73. 2 1 5 0 2 1 3 2 α

79. = 0 |A| = 2(2α − 2) + 3(1 − 10) = 4α − 4 − 27 = 4α − 31 = 0 =⇒ α = 31/4 Dr.Randeep N C Mathematical foundations of QM

80. Problems Problem:3 Check whether the following set of vectors are linearly independent or not ? (a) {sin x, cos x, ex } (b) {sinh x, cosh x, ex } Dr.Randeep N C Mathematical foundations of QM

81. Problems Problem:3 Check whether the following set of vectors are linearly independent or not ? (a) {sin x, cos x, ex } (b) {sinh x, cosh x, ex } Answer (a): c1 sin x + c2 cos x + c3ex = 0 c1 cos x − c2 sin x + c3ex = 0 −c1 sin x − c2 cos x + c3ex = 0 |A| =

87. sin x cos x ex cos x − sin x ex − sin x − cos x ex

93. |A| = sin x(−ex sin x + ex cos x) − cos x(ex cos x + ex sin x) + ex (− cos2 x − sin2 x) = −ex − ex = −2ex Here |A| 6= 0 =⇒ the set of vectors are linearly independent. Answer (b): In this problem |A| = 0 =⇒ the set of vectors are linearly dependent. Dr.Randeep N C Mathematical foundations of QM

96. State of the System Lecture-2 Dr.Randeep N C Mathematical foundations of QM

99. Inner product space Norm of the vector: The inner product of any vector with itself is a non negative real number. Its square root is called norm of the vector. Norm of the vector |αi is ||α|| = p hα|αi Normalized ket: A ket |αi with its norm ||α|| = 1 is called a normalized ket. If |αi is not normalized, corresponding normalized ket |α̃i is obtained by |α̃i = |αi p hα|αi Orthogonal vectors: Two vectors |αi, |βi whose inner product is zero are called orthogonal vectors. hα|βi = 0 =⇒ |αi and |βi are orthogonal vectors. Dr.Randeep N C Mathematical foundations of QM

100. Complete Orthonormal set of vectors: A collection {|α1i , |α2i , ....} of mutually orthogonal normalized vectors is called orthonormal set of vectors. hαi |αj i = δij =⇒ {|αi i} are othonormal set of vectors. Complete: A vector space is said to be complete if every Cauchy sequence of vectors in the space converges to a limit vector which is also in the space. The set of rational number Q is a linear vector space which is not complete. Since Cauchy sequence like (3, 3.1, 3.14, 3.141, 3.1415, ...) converges to a limit point π, which is not there in Q. Every finite dimensional inner product space is complete. But an infinite dimensional inner product space is not necessarily complete. Hilbert space: A complete inner product space is called Hilbert space. Set of all 2 × 1 real matrices is a complete inner product space, and therefore a Hilbert space. Dr.Randeep N C Mathematical foundations of QM

101. Problems Problem 4: Consider a three dimensional vector space spanned by an orthonormal basis |1i , |2i , |3i. Kets |αi and |βi are given by |αi = i |1i − 2 |2i − i |3i , |βi = i |1i + 2 |3i (a) Construct hα| and hβ| (in terms of the dual basis h1| , h2| , h3|). (b) Find hα|βi and hβ|αi, and confirm that hβ|αi = hα|βi∗ . (c) Normalize ket vectors |αi and |βi. Dr.Randeep N C Mathematical foundations of QM

102. Problems Problem 4: Consider a three dimensional vector space spanned by an orthonormal basis |1i , |2i , |3i. Kets |αi and |βi are given by |αi = i |1i − 2 |2i − i |3i , |βi = i |1i + 2 |3i (a) Construct hα| and hβ| (in terms of the dual basis h1| , h2| , h3|). (b) Find hα|βi and hβ|αi, and confirm that hβ|αi = hα|βi∗ . (c) Normalize ket vectors |αi and |βi. Answer: (a) hα| = −i h1| − 2 h2| + i h3| , hβ| = −i h1| + 2 h3| (b) hα|βi = (−i h1| − 2 h2| + i h3|)(i |1i + 2 |3i) hα|βi = h1|1i − 2ih1|3i − 2ih2|1i − 4h2|3i − h3|1i + 2ih3|3i hα|βi = 1 + 2i hβ|αi = (−i h1| + 2 h3|)(i |1i − 2 |2i − i |3i) hβ|αi = h1|1i − 2ih1|2i − h1|3i + 2ih3|1i − 4h3|2i − 2ih3|3i hβ|αi = 1 − 2i = hα|βi∗ Dr.Randeep N C Mathematical foundations of QM

103. Problems (c) To normalize |αi, take innerproduct hα|αi, hα|αi = (−i h1| − 2 h2| + i h3|)(i |1i − 2 |2i − i |3i) hα|αi = 1 + 4 + 1 = 6 Therefore, normalized ket corresponding to |αi is |α̃i = |αi p hα|αi = i √ 6 |1i − 2 √ 6 |2i − i √ 6 |3i To normalize |βi, take innerproduct hβ|βi, hβ|βi = (−i h1| + 2 h3|)(i |1i + 2 |3i) hβ|βi = 1 + 4 = 5 Therefore, normalized ket corresponding to |βi is

105. β̃ = |βi p hβ|βi = i √ 5 |1i + 2 √ 5 |3i Dr.Randeep N C Mathematical foundations of QM

106. Problems Problem 5: Two vectors a 0 and b c are orthonormal if (a) a = ±1, b = ± 1 √ 2 , c = ± 1 √ 2 , (b) a = ±1, b = ±1, c = 0, (c) a = ±1, b = 0, c = ±1, (d) a = ±1, b = ± 1 √ 2 , c = 1 2 [NET-JUNE 2017] Dr.Randeep N C Mathematical foundations of QM

107. Problems Problem 5: Two vectors a 0 and b c are orthonormal if (a) a = ±1, b = ± 1 √ 2 , c = ± 1 √ 2 , (b) a = ±1, b = ±1, c = 0, (c) a = ±1, b = 0, c = ±1, (d) a = ±1, b = ± 1 √ 2 , c = 1 2 [NET-JUNE 2017] Answer: For orthonormal vectors hai |aj i = δij , therefore, we get a 0 b c = ab = 0 a 0 a 0 = a2 = 1 =⇒ a = ±1 a = ±1, ab = 0 =⇒ b = 0 b c b c = b2 + c2 = 1 b2 + c2 = 1, b = 0 =⇒ c = ±1 Therefore, the solution is a = ±1, b = 0, c = ±1. Dr.Randeep N C Mathematical foundations of QM

108. State of the System Postulate 1: State of the system The state of a system at any time t is represented by a vector |ψ(t)i in the Hilbert space. The state ket is postulated to contain complete information about the physical state, everything we are allowed to ask about the state is contained in the ket. In quantum mechanics, the state kets |αi and c |αi, with c 6= 0 represent the same physical state. Superposition principle: If |αi and |βi are possible states of a system, then their linear combination |γi = c1 |αi + c2 |βi is also a possible state of the system. Dr.Randeep N C Mathematical foundations of QM

109. Operator Operator: An operator T̂ is a mapping or a function that acts on an element in one space V to produce an element in another space V 0 . T̂ : V −→ V 0 Linear Operator: A linear operator T̂ on a vector space V is a mapping from V to V with the properties: 1 T̂(|αi + |βi) = T̂(|αi) + T̂(|βi), ∀ |αi , |βi ∈ V 2 T̂(c |αi) = cT̂(|αi), ∀ c ∈ F and |αi ∈ V Combining above equations, the condition for linear operator as T̂(c1 |αi+c2 |βi) = c1T̂(|αi)+c2T̂(|βi) ∀ c1, c2 ∈ F and |αi , |βi ∈ V An operator acting on a ket vector |αi will give another ket |βi. T̂ |αi = |βi An operator acting on a bra vector hα| will give another bra hβ|, hα| T̂ = hβ| Dr.Randeep N C Mathematical foundations of QM

110. Eigenvalue equation An operator acting on a ket vector from left to right, while it act on a bra vector from right to left. Dual correspondence T̂ |αi DC ← → hα| T̂† where T̂† is called Hermitian adjoint of the operator T̂. Eigenvalue equation: An equation of the form Â |ai i = ai |ai i is called eigenvalue equation. Where ai is the eigenvalue of an operator Â with corresponding eigen vector |ai i. For an operator Â in n dimension with a set of eigenvalue equations, Â |a1i = a1 |a1i , Â |a2i = a2 |a2i , ...., Â |ani = an |ani , then {|a1i , |a2i , ..., |ani} = {|ai i} is the set of eigen vectors and {a1, a2, ..., an} is the set of eigenvalues of the operator Â. Dr.Randeep N C Mathematical foundations of QM

111. Eigenstates and eigenvalues of an operator ˆ σz Consider a Pauli spin matrix ˆ σz , ˆ σz = 1 0 0 −1 eigenvalues of ˆ σz can be obtained using the determinant, | ˆ σz − λ| = 0

115. 1 − λ 0 0 −1 − λ

119. = 0, (1 − λ)(−1 − λ) = 0 Therefore eigenvalues of ˆ σz are λ1 = +1 and λ2 = −1. Let |0i be the eigenvector corresponding to eigenvalue λ1 = +1 1 − 1 0 0 −1 − 1 x y = 0 From this, we get −2y = 0, =⇒ y = 0, x = a constant = 1 Dr.Randeep N C Mathematical foundations of QM

120. Eigenstates and eigenvalues of an operator ˆ σz therefore, |0i = 1 0 Let |1i be the eigenvector corresponding to eigenvalue λ1 = −1 1 + 1 0 0 −1 + 1 x y = 0 From this, we get 2x = 0, x = 0, y = a constant = 1 therefore, |1i = 0 1 That is eigenstates of ˆ σz operator are |0i = 1 0 and |1i = 0 1 . Dr.Randeep N C Mathematical foundations of QM

121. Eigenstates and eigenvalues of an operator ˆ σx Consider another Pauli spin matrix ˆ σx , ˆ σx = 0 1 1 0 eigenvalues of ˆ σx can be obtained using | ˆ σx − λ| = 0

125. −λ 1 1 −λ

129. = 0, (λ2 − 1) = 0 Therefore eigenvalues of ˆ σx are λ3 = +1 and λ4 = −1. Let |3i be the eigenvector corresponding to eigenvalue λ3 = +1 −1 1 1 −1 x y = 0 From this, we get −x + y = 0, y = 1, x = 1 Then |3i = 1 1 Dr.Randeep N C Mathematical foundations of QM

130. Eigenstates and eigenvalues of an operator ˆ σx Let |4i be the eigenvector corresponding to eigenvalue λ4 = −1 1 1 1 1 x y = 0 From this, we get x + y = 0, x = 1, y = −1 Then |4i = 1 −1 Consider the possible inner products h3|4i = 1 1 1 −1 = 0, Orthogonal vectors h3|3i = 1 1 1 1 = 2, Not normalized h4|4i = 1 −1 1 −1 = 2, Not normalized Dr.Randeep N C Mathematical foundations of QM

131. Normalization These two vectors |3i and |4i are orthogonal vectors, but not normalized. This can be normalized by defining |3i = c3 |3i = c3 1 1 and |4i = c4 |4i = c4 1 −1 Then, h+|+i = c∗ 3 c3 1 1 1 1 = 1 = 2|c3|2 =⇒ c3 = 1 √ 2 and h−|−i = c∗ 4 c4 1 −1 1 −1 = 1 = 2|c4|2 =⇒ c4 = 1 √ 2 Therefore, normalized eigenvectors of the operator ˆ σx are |+i = 1 √ 2 1 1 , |−i = 1 √ 2 1 −1 Dr.Randeep N C Mathematical foundations of QM

134. Hermitian Operator Lecture-3 Dr.Randeep N C Mathematical foundations of QM

136. Hermitian Operator Hermitain adjoint of X̂ = |αi hβ| is X̂† = |βi hα| Hermitian Operator: Operators that are their own Hermitian adjoint are called Hermitian operators. That is Â† = Â For any arbitrary operator X̂, hα|X̂|βi = hα|(X̂|βi) = (hα|X̂)|βi = hβ|X̂† |αi∗ For a Hermitian operator Â, hα|Â|βi = hβ|Â|αi∗ Postulates 2: Observable and Operator To every physically measurable quantity A, called observables, there corresponds a Hermitian operator Â. Dr.Randeep N C Mathematical foundations of QM

138. Properties of Hermitian operator Then for Hermitian operator ha1|Â|a2i = ha2|Â|a1i∗ a2ha1|a2i = a∗ 1 ha2|a1i∗ a2ha1|a2i = a1ha1|a2i (a2 − a1)ha1|a2i = 0 Here a2 − a1 6= 0 =⇒ ha1|a2i = 0. Therefore, eigenvector of a Hermitian transformation belonging to distinct eigenvalues are orthogonal. Property 3: The set of eigenvectors of a Hermitian operator span the space. Proof: A matrix Â is said to be normal if ÂÂ† = Â† Â Every normal matrix is diagonalizable, therefore its eigenvectors span the space. Since Hermitian matrix is normal matrix, therefore its eigenvectors also span the space. Dr.Randeep N C Mathematical foundations of QM

143. Completeness Relation Pai = |ai i hai | Completeness relation is obtained from projection operator as X i |ai i hai | = X i Pai = 1 Consider eigenstates |0i and |1i of ˆ σz operator and take following outer products, |0i h0| = 1 0 1 0 = 1 0 0 0 , |1i h1| = 0 1 0 1 = 0 0 0 1 1 X i=0 |ii hi| = |0i h0| + |1i h1| = 1 0 0 1 = 1 That is eigenstates of ˆ σz operator satisfy completeness relation, therefore its eigenbasis is complete. Dr.Randeep N C Mathematical foundations of QM

144. Completeness Relation Consider eigenstates |+i and |−i of ˆ σx operator and take following outer products, |+i h+| = 1 √ 2 1 1 1 √ 2 1 1 = 1 2 1 1 1 1 , |−i h−| = 1 √ 2 1 −1 1 √ 2 1 −1 = 1 2 1 −1 −1 1 X i |ii hi| = |+i h+| + |−i h−| = 1 0 0 1 = 1 That is eigenstates of ˆ σx operator also satisfy completeness relation, therefore its eigenbasis is complete. Dr.Randeep N C Mathematical foundations of QM

147. Measurements Lecture-5 Dr.Randeep N C Mathematical foundations of QM

148. Measurements Postulate 3: Measurements and Eigenvalues The only possible result of a measurement is the one of the eigenvalues of the operator. That is, if we are measuring the ẑ component of the spin Ŝz in an arbitrary state, we will get either the result +~ 2 or −~ 2 . A quantum mechanical measurement always causes the system to jump into an eigenstate of the dynamical variable that is being measured. That is before a measurement of observable A is made, the system is assumed to represent by linear combination of eigenstates of the operator Â: |αi = X i |ai ihai |αi = X i ci |ai i Dr.Randeep N C Mathematical foundations of QM

149. Measurements When measurement is performed in this state, the system is thrown into one of the eigenstate |ai i of the operator Â |αi A measurement − − − − − − − − − → |ai i and the corresponding eigenvalue ai is obtained. Thus a measurement always changes the state of the system in quantum mechanics. If the state of the system is already in one of the eigenstate of the observable, then |ai i A measurement − − − − − − − − − → |ai i , state will not change. That is repeated measurement of same observable in succession yield the same result. During measurement we do not know in advance into which of the various |ai i’s the system will thrown as a result of measurement. Dr.Randeep N C Mathematical foundations of QM

150. Measurements Postulate 4: Probabilistic Outcome of a Measurements In the state |αi = P i |ai ihai |αi = P i ci |ai i, the probability for jumping into a particular eigenstate |ai i is given by P(ai ) = |hai |αi|2 = |ci |2 , provided |αi is normalized. If |αi is not normalized, then P(ai ) = |hai |αi|2 hα|αi = |hai |αi|2 P i |hai |αi|2 = |ci |2 P i |ci |2 Expectation value: Expectation value is the average of all possible outcome of a measurement. Expectation value of an operator Â taken with respect to a state |αi is defined as hÂi = hα|Â|αi, if |αi is normalized Dr.Randeep N C Mathematical foundations of QM

151. Measurements hÂi = hα|Â|αi hÂi = X ij hα|ai ihai |Â|aj ihaj |αi = X ij aj hα|ai ihai |aj ihaj |αi hÂi = X ij aj hα|ai iδij haj |αi = X i ai hα|ai ihai |αi hÂi = X i ai |hai |αi|2 = X i ai |ci |2 If |αi is not normalized, then hÂi = hα|Â|αi hα|αi hÂi = P i ai |hai |αi|2 P i |hai |αi|2 = P i ai |ci |2 P i |ci |2 Dr.Randeep N C Mathematical foundations of QM

152. Problems Problem 10: The state of the system is given by |ψi = |φ1i + 2 |φ2i +3 |φ3i, where |φ1i , |φ2i and |φ3i form an orthonormal set. The probability of finding the system in the state |φ2i is (a)1/7 (b)1/3 (c)2/7 (d)3/7 Answer: P(|φ2i) = 22 12 + 22 + 32 = 4 14 = 2 7 Problem 11: An electron is in a spin state |φi = √ 3/2 1/2 in the Sz representation. What is the probability of finding the z component of its spin along the −ẑ direction [GATE-2002] (a).75 (b).5 (c).35 (d).25 Answer: |φi = √ 3/2 1/2 = √ 3/2 1 0 + 1/2 0 1 P(−ẑ) = (1/2)2 ( √ 3/2)2 + (1/2)2 = 1/4 3/4 + 1/4 = 1/4 = .25 Dr.Randeep N C Mathematical foundations of QM

153. Problems Problem 12: The wavefunction of a particle is given by ψ = 1/ √ 2φ0 + iφ1, where φ0 and φ1 are normalized eigenfunctions with energies E0 and E1 corresponding to the ground state and first excited state respectively. The expectation value of the Hamiltonian in the state ψ is [NET-JUNE 2011] (a)E0 2 + E1, (b)E0 2 − E1, (c)E0−2E1 3 , (d)E0+2E1 3 Answer: |ψi = 1 √ 2 |φ0i + i |φ1i hÂi = hα|Â|αi hα|αi = P i ai |hai |αi|2 P i |hai |αi|2 = P i ai |ci |2 P i |ci |2 hĤi = E0 × 1/2 + E1 × 1 1/2 + 1 hĤi = E0 + 2E1 3 Dr.Randeep N C Mathematical foundations of QM

154. Problems Problem 13: A spin 1/2 particle is in a linear superposition 0.8 |0i + 0.6 |1i of spin up and spin down states. If |0i and |1i are eigenstates of σz then what is the expectation value of the operator 10σz + 5σx . (a)7.6, (b)10, (c)8.8, (d)1.2 Answer:Let |ψi = 0.8 |0i + 0.6 |1i , Â = 10σz + 5σx |ψi = .8 1 0 + .6 0 1 = .8 .6 Â = 10 1 0 0 −1 + 5 0 1 1 0 = 10 5 5 −10 hÂi = hψ|Â|ψi = .8 .6 10 5 5 −10 .8 .6 hÂi = 11 −2 .8 .6 = 8.8 − 1.2 = 7.6 Dr.Randeep N C Mathematical foundations of QM

155. Commutator and Anticommutator Commutator [ , ] is defined by [Â, B̂] = ÂB̂ − B̂Â Anticommutator { , } is defined by {Â, B̂} = ÂB̂ + B̂Â Properties of Commutator: 1 [Â, Â] = 0 [Â, Â] = ÂÂ − ÂÂ = 0 2 [Â, B̂] = −[B̂, Â] [Â, B̂] = ÂB̂ − B̂Â = −(B̂Â − ÂB̂) = −[B̂, Â] 3 [Â, c] = 0, c is a constant [Â, c] = Âc − cÂ = 0 Dr.Randeep N C Mathematical foundations of QM

156. Commutator and Anticommutator 4 [Â + B̂, Ĉ] = [Â, Ĉ] + [B̂, Ĉ] [Â + B̂, Ĉ] = (Â + B̂)Ĉ − Ĉ(Â + B̂) = ÂĈ + B̂Ĉ − ĈÂ − ĈB̂ = ÂĈ − ĈÂ + B̂Ĉ − ĈB̂ = [Â, Ĉ] + [B̂, Ĉ] 5 [Â, B̂Ĉ] = [Â, B̂]Ĉ + B̂[Â, Ĉ] [Â, B̂]Ĉ + B̂[Â, Ĉ] = (ÂB̂ − B̂Â)Ĉ + B̂(ÂĈ − ĈÂ) = ÂB̂Ĉ − B̂ÂĈ + B̂ÂĈ − B̂ĈÂ = ÂB̂Ĉ − B̂ĈÂ = [Â, B̂Ĉ] 6 [Â, [B̂, Ĉ]] + [B̂, [Ĉ, Â]] + [Ĉ, [Â, B̂]] = 0 [Â, [B̂, Ĉ]] + [B̂, [Ĉ, Â]] + [Ĉ, [Â, B̂]] = ÂB̂Ĉ − ÂĈB̂ − B̂ĈÂ + ĈB̂Â+ B̂ĈÂ − B̂ÂĈ − ĈÂB̂ + ÂĈB̂ + ĈÂB̂ − ĈB̂Â − ÂB̂Ĉ + B̂ÂĈ = 0 Dr.Randeep N C Mathematical foundations of QM

157. Compatible and Incompatible Observables A and B are defined to be compatible when the corresponding operators commute, [Â, B̂] = 0 =⇒ A, B are compatible observable Example: (x, y), (px , py ), (x, py ), (S2 , Sx ), .. Observables A and B are incompatible when the corresponding operators do not commute, [Â, B̂] 6= 0 =⇒ A, B are incompatible observable Example: (x, Px ), (Sx , Sy ), (Jx , Jy ), ... Theorem: Suppose that A and B are compatible observables, and eigenvalues of Â are non degenerate. Then the matrix elements hai |B̂|aj i are all diagonal. Proof: If A and B are compatible observable, then [Â, B̂] = 0 Dr.Randeep N C Mathematical foundations of QM

159. Compatible and Incompatible Therefore, |ai i is also an eigenvector of B̂ with eigenvalue bi . Then the simultaneous eigenstates of Â and B̂ represented by |ai i = |bi i = |ai , bi i In the simultaneous eigenbasis, operators Â and B̂ are diagonal matrices, Â =       a1 0 0 . . 0 a2 0 . . 0 0 a3 . . . . . . . . . . . .       , B̂ =       b1 0 0 . . 0 b2 0 . . 0 0 b3 . . . . . . . . . . . .       Degenerate case: Suppose eigenvalues of an operator Â are degenerate, then Â in its eigenbasis is of the form Â =       a1 0 0 . . 0 a1 0 . . 0 0 a2 . . . . . . . . . . . .       Dr.Randeep N C Mathematical foundations of QM

160. Compatible and Incompatible Here eigenvalue ai of an operator Â are mi fold degenerate, then Â |ai , δi = ai |ai , δi , δ = 1, 2, ..mi where |ai , δi are mi mutually orthogonal eigenket of Â with same eigenvalue ai . Also {|ai , δi} span the subspace Vmi i in which eigenvalues of Â are ai ’s. ÂB̂ |ai , δi = B̂Â |ai , δi , δ = 1, 2, 3..., mi ÂB̂ |ai , δi = ai B̂ |ai , δi Which doesn’t means that B̂ |ai , δi = bi |ai , δi, we can say only that B̂ |ai , δi ∈ Vmi i . That is B̂ |ai , δi will not go out of the space Vmi i . Therefore, the B̂ matrix will be a block diagonal matrix, in which Dr.Randeep N C Mathematical foundations of QM

161. Compatible and Incompatible hai , ζ|aj , δi = δij In each block, we can diagonalize B̂ matrix independently and obtain simultaneous eigenstates of Â and B̂ as Â |ai , bi i = ai |ai , bi i , B̂ |ai , bi i = bi |ai , bi i Consider two commuting operators Â and B̂ with eigenvalues of Â are non degenerate, |αi A measurement − − − − − − − − − → |ai , bi i B measurement − − − − − − − − − → |ai , bi i A measurement − − − − − − − − − → |ai , bi i If eigenvalues of Â are degenerate, then |αi A measurement − − − − − − − − − → mi X j=1 |ai , bj i B measurement − − − − − − − − − → |ai , bi i A measurement − − − − − − − − − → |ai , bi i That is A measurements and B measurement do not interfere, that is why we are calling it as compatible observables. Dr.Randeep N C Mathematical foundations of QM

165. The Uncertainty Relation Lecture-6 Dr.Randeep N C Mathematical foundations of QM

166. The Uncertainty Relation In a quantum state |αi, when we measure an observable A, there exist a probability distribution of all position outcome of that measurement. Consider a state |αi = 6 P i=1 ci |ai i, the probability distribution of all possible outcomes of an observable A is as shown figure, The probability distribution is characterized by mean or expectation value and is given by hÂi = hα|Â|αi Dr.Randeep N C Mathematical foundations of QM

167. The Uncertainty Relation Another quantity associated with the probability distribution is standard deviation, which is a measure the average fluctuation around the mean, and is defined by (∆A) = [h(Â − hAi)2 i]1/2 It is the measure of uncertainty of an operator Â in the state |αi. It is also called root mean square deviation. The expectation value h(Â − hÂi)2 i is called dispersion or variance or mean square deviation. The variance, h(Â − hÂi)2 i = h(Â2 − 2hÂiÂ + hÂi2 )i = hÂ2 i − h2hÂiÂi + hhÂi2 i = hÂ2 i − 2hÂi2 + hÂi2 h(Â − hÂi)2 i = hÂ2 i − hÂi2 Dr.Randeep N C Mathematical foundations of QM

168. The Uncertainty Relation Therefore, the uncertainty in Â is (∆A) = q hÂ2i − hÂi2 That is the uncertainty of Â in the state |αi is the standard deviation of expansion coefficients ci ’s of the state |αi in the eigenbasis of Â. Suppose state in question is an eigenstate of an operator, then only one non zero ci exist and standard deviation (uncertainty) vanishes. Dr.Randeep N C Mathematical foundations of QM

169. The Uncertainty Relation Let A and B be two observable, then for any state we must have the following inequality, (∆Â)(∆B̂) ≥ 1 2 |h[Â, B̂]i| is called general uncertainty relation If A and B are two compatible observables, then [Â, B̂] = 0 and (∆Â)(∆B̂) = 0 that is product of uncertainty vanishes. Dr.Randeep N C Mathematical foundations of QM

170. Problem Problem 14: Â and B̂ are two quantum mechanical operators. If [Â, B̂] stands for the commutator of Â and B̂, then [[Â, B̂], [B̂, Â]] is equal to [GATE-2002] (a)ÂB̂ÂB̂ − B̂ÂB̂Â, (b)Â(ÂB̂ − B̂Â), (c)0, (d)([Â, B̂])2 Answer: [[Â, B̂], [B̂, Â]] = −[[Â, B̂], [Â, B̂]] = 0 Problem 15: For the operators P and Q, the commutator [P, Q−1 ] is [JEST-2016] (a)Q−1 [P, Q]Q−1 , (b) − Q−1 [P, Q]Q−1 , (c)Q−1 [P, Q]Q, (d) − Q[P, Q]Q−1 Answer: [P, Q−1 ] = PQ−1 − Q−1 P Q−1 [P, Q]Q−1 = Q−1 (PQ − QP)Q−1 = Q−1 PQQ−1 − Q−1 QPQ−1 Q−1 [P, Q]Q−1 = Q−1 P − PQ−1 = −[P, Q−1 ] that is [P, Q−1 ] = −Q−1 [P, Q]Q−1 Dr.Randeep N C Mathematical foundations of QM

171. Problem Problem 16: The matrices A =   0 −1 0 1 0 0 0 0 0   , B =   0 0 1 0 0 0 0 0 0   and C =   0 0 0 0 0 1 0 0 0   satisfy the commutation relations [NET-JUNE-2014] (a) [A, B] = B + C, [B, C] = 0, [C, A] = B + C (b) [A, B] = C, [B, C] = A, [C, A] = B (c) [A, B] = B, [B, C] = 0, [C, A] = A (d) [A, B] = C, [B, C] = 0, [C, A] = B Answer: AB =   0 0 0 0 0 1 0 0 0   = C, BA =   0 0 0 0 0 0 0 0 0   = 0, =⇒ [A, B] = C Dr.Randeep N C Mathematical foundations of QM

172. Problem BC =   0 0 0 0 0 0 0 0 0   = 0, CB =   0 0 0 0 0 0 0 0 0   = 0, =⇒ [B, C] = 0 CA =   0 0 0 0 0 0 0 0 0   = 0, AC =   0 0 −1 0 0 0 0 0 0   = −B, =⇒ [C, A] = B Problem 17: The operators A and B share all the eigenstates. Then the least possible value of the product of uncertainties (∆A)(∆B) is (a) ~, (b) 0, (c) ~/2, (d) |AB| [JEST-2014] Answer: General uncertainty relation is (∆A)(∆B) ≥ 1 2 |h[A, B]i| If operators A and B share all eigenstates, then [A, B] = 0, then (∆A)(∆B) ≥ 0 That is least possible product of uncertainties is 0. Dr.Randeep N C Mathematical foundations of QM

173. Problem Problem 18: Given that ψ1 and ψ2 are eigenstates of a Hamiltonian with eigenvalues E1 and E2 respectively. What is the uncertainty in the state ψ1 + ψ2 [JEST-2015] (a) − √ E1E2, (b) 1 2 |E1 − E2|, (c)1 2 (E1 + E2), (d) 1 √ 2 |E2 − E1| Answer: Given |ψi = 1 √ 2 (|ψ1i + |ψ1i), Ĥ |ψ1i = E1 |ψ1i , Ĥ |ψ2i = E2 |ψ2i hĤi = 1 2 × E1 + 1 2 × E2 = E1 + E2 2 hĤ2 i = 1 2 × E2 1 + 1 2 × E2 2 = E2 1 + E2 2 2 (∆Ĥ) = q hĤ2i − hĤi2 = r E2 1 + E2 2 2 − E1 + E2 2 2 (∆Ĥ) = r E2 1 + E2 2 − 2E1E2 4 = |E1 − E2| 2 Dr.Randeep N C Mathematical foundations of QM

174. Change of Basis Suppose we have two incompatible observables A and B, then the ket space in question can be spanned either by the set {|ai i} or by the set {|bi i}. Changing the base ket from one set of basis to another set of basis is called change of basis or change of representation. The basis in which the basekets are given by {|ai i} is called A-representation. Theorem: Given two sets of basekets, both satisfying orthonormality and completeness, there exist an unitary operator Û (ÛÛ† = Û† Û = 1) such that |b1i = Û |a1i , |b2i = Û |a2i , ........, |bNi = Û |aNi . Proof: Consider the operator Û = X k |bk i hak | Û |aì = X k |bk i hak | aì = X k |bk i δk` = |bì Dr.Randeep N C Mathematical foundations of QM

175. Change of Basis To check Û = P k |bk i hak |, a unitary operator, Û† Û = X k X ` |aì hb`| bk i hak | = X k` |aì hak | δk` = X k |ak i hak | = 1 That is Û = X k |bk i hak | is a unitary operator transform |aì → |bì. The matrix element of the operator Û in the old basis {|ai i} is haj | Û |ak i = haj |bk i That is the matrix elements of the Û operator is the innerproduct of old base bras and new base kets. Then the Û matrix in old basis is Dr.Randeep N C Mathematical foundations of QM

176. Change of Basis Û =       ha1|b1i ha1|b2i . . . ha2|b1i ha2|b2i . . . . . . . . . . . . . . . . . .       which is a transformation matrix from old {|ai i} basis to new {|bi i} basis. Consider an arbitrary ket |αi = X k |ak ihak |αi here hak |αi’s are the expansion coefficients in the old basis. The expansion coefficient in the new basis is obtained by using hbj |αi, hbj |αi = X k hbj |ak ihak |αi = X k haj |Û† |ak ihak |αi Dr.Randeep N C Mathematical foundations of QM

181. The Stern-Gerlach Experiment Lecture-7 Dr.Randeep N C Mathematical foundations of QM

182. Change of basis: Eigenbasis of Ŝx and Ŝz Consider Ŝz operator with eigenstates {|0i , |1i} and Ŝx with eigenstates {|+i , |−i}. These sets of eigenstates can be transform from one to another by a unitary transformation Û: {|0i , |1i} Û − → {|+i , |−i} where Û = h0 | +i h0 | −i h1 | +i h1 | −i , h0|+i = 1 0 1 √ 2 1 1 = 1 √ 2 , h0|−i = 1 0 1 √ 2 1 −1 = 1 √ 2 h1|+i = 0 1 1 √ 2 1 1 = 1 √ 2 , h1|−i = 0 1 1 √ 2 1 −1 = − 1 √ 2 Dr.Randeep N C Mathematical foundations of QM

183. Change of basis: Eigenbasis of Ŝx and Ŝz Therefore, Û matrix is Û = h0 | +i h0 | −i h1 | +i h1 | −i = 1 √ 2 1 1 1 −1 Then the state |+i can be expressed in terms |0i , |1i as |+i = 1|+i = (|0i h0|) |+i + (|1i h1|) |+i = |0i × 1 √ 2 + |1i × 1 √ 2 that is |+i = 1 √ 2 [|0i + |1i] The state |−i can be expressed in terms |0i , |1i as |−i = 1|−i = (|0i h0|) |−i + (|1i h1|) |−i = |0i × 1 √ 2 + |1i × − 1 √ 2 |−i = 1 √ 2 [|0i − |1i] Dr.Randeep N C Mathematical foundations of QM

184. Change of basis: Eigenbasis of Ŝx and Ŝz The state |0i can be expressed in terms |+i , |−i as |0i = 1|0i = (|+i h+|) |0i + (|−i h−|) |0i = |+i × 1 √ 2 + |−i × 1 √ 2 |0i = 1 √ 2 [|+i + |−i] The state |1i can be expressed in terms |+i , |−i as |1i = 1|1i = (|+i h+|) |1i + (|−i h−|) |1i = |+i × 1 √ 2 + |−i × − 1 √ 2 |1i = 1 √ 2 [|+i − |−i] Dr.Randeep N C Mathematical foundations of QM

185. Measurement of ˆ Sz in the eigenstates of ˆ Sx In the state |+i = 1 √ 2 [|0i + |1i] When we measure ˆ Sz , Probability for getting the eigenvalue +~ 2 = |h0|+i|2 = | 1 √ 2 |2 = 1 2 Probability for getting the eigenvalue −~ 2 = |h1|+i|2 = | 1 √ 2 |2 = 1 2 . Also in the state |−i = 1 √ 2 [|0i − |1i] When we measure ˆ Sz , Probability for getting the eigenvalue +~ 2 = |h0|−i|2 = | 1 √ 2 |2 = 1 2 Probability for getting the eigenvalue −~ 2 = |h1|−i|2 = | − 1 √ 2 |2 = 1 2 . Dr.Randeep N C Mathematical foundations of QM

186. Measurement of ˆ Sx in the eigenstates of ˆ Sz In the state |0i = 1 √ 2 [|+i + |−i] When we measure ˆ Sx , Probability for getting the eigenvalue +~ 2 = |h+|0i|2 = | 1 √ 2 |2 = 1 2 Probability for getting the eigenvalue −~ 2 = |h−|0i|2 = | 1 √ 2 |2 = 1 2 . Also in the state |1i = 1 √ 2 [|+i − |−i] When we measure ˆ Sx , Probability for getting the eigenvalue +~ 2 = |h+|1i|2 = | 1 √ 2 |2 = 1 2 Probability for getting the eigenvalue −~ 2 = |h−|1i|2 = | − 1 √ 2 |2 = 1 2 . Dr.Randeep N C Mathematical foundations of QM

187. Problems Problem 19: Two different set of orthronormal basis vectors n 1 0 , 0 1 o and n 1 √ 2 1 1 , 1 √ 2 1 −1 o are given for a two dimensional real vector space. The matrix representation of a linear operator Â in these basis are related by a unitary transformation. The unitary matrix may be chosen to be [NET-JUNE-2015] (a) 0 −1 1 0 , (b) 0 1 1 0 , (c) 1 √ 2 1 1 1 −1 , (d) 1 √ 2 1 0 1 1 Answer: |a1i = 1 0 , |a2i = 0 1 , |b1i = 1 √ 2 1 1 , |b2i = 1 √ 2 1 −1 {|ai i} Û − → {|bi i}, Û = ha1|b1i ha1|b2i ha2|b1i ha2|b2i ha1|b1i = 1 √ 2 × 1 = 1 √ 2 , ha1|b2i = 1 √ 2 × 1 = 1 √ 2 , ha2|b1i = 1 √ 2 × 1 = 1 √ 2 , ha2|b2i = 1 √ 2 × −1 = − 1 √ 2 Dr.Randeep N C Mathematical foundations of QM

188. Problems Therefore, Û matrix is Û = 1 √ 2 1 1 1 −1 Problem 20: Two different set of orthronormal basis vectors 1 √ 2 1 1 , 1 √ 2 1 −1 and 1 √ 2 1 i , 1 √ 2 1 −i are given for a two dimensional real vector space. The matrix representation of a linear operator Â in these basis are related by a unitary transformation. The unitary matrix may be chosen to be (a) 1 √ 2 1 1 1 −1 , (b) 1 √ 2 1 i 1 −i , (c)1 2 1 + i 1 + i 1 − i 1 + i , (d)1 2 1 + i 1 − i 1 − i 1 + i Answer: |a1i = 1 √ 2 1 1 , |a2i = 1 √ 2 1 −1 |b1i = 1 √ 2 1 i , |b2i = 1 √ 2 1 −i Dr.Randeep N C Mathematical foundations of QM

189. Problems {|ai i} Û → {|bi i} , Û = ha1 | b1i ha1 | b2i ha2 | b1i ha2 | b2i ha1 | b1i = 1 √ 2 1 1 1 √ 2 1 i = 1 2 (1 + i) ha1 | b2i = 1 √ 2 1 1 1 √ 2 1 −i = 1 2 (1 − i) ha2 | b1i = 1 √ 2 1 −1 1 √ 2 1 i = 1 2 (1 − i) ha2 | b1i = 1 √ 2 1 −1 1 √ 2 1 −i = 1 2 (1 + i) Then U matrix is U = 1 2 1 + i 1 − i 1 − i 1 + 1 Dr.Randeep N C Mathematical foundations of QM

190. Problems Problem 21: If the eigenvalue of a symmetric 3 × 3 matrix A are 0, 1, 3 and the corresponding eigenvector can be written as   1 1 1   ,   1 0 −1   ,   1 −2 1   respectively, then the matrix A4 is [TIFR-2016] (a)   41 −81 40 −81 0 −81 40 −81 41   , (b)   −82 −81 79 −81 81 −81 79 −81 83   (c)   14 −27 13 −27 54 −27 13 −27 14  , (d)   14 −13 27 −13 54 −13 27 −13 14   Answer: The normalized eigenvectors of operator A are |a1i = 1 √ 3   1 1 1   , |a2i = 1 √ 2   1 0 −1   , |a3i = 1 √ 6   1 −2 1   Dr.Randeep N C Mathematical foundations of QM

191. Problems Operator A in the basis {|a1i , |a2i , |a3i} can be written as A =   0 0 0 0 1 0 0 0 3   Then the standard basis in three dimensions is |b1i =   1 0 0   , |b2i =   0 1 0   , |b3i =   0 0 1   Then U matrix is U =   ha1 | b1i ha1 | b2i ha1 | b3i ha2 | b1i ha2 | b2i ha2| b3i ha3 | b1i ha3 | b2i ha3 | b3i   =   1/ √ 3 1/ √ 3 1/ √ 3 1/ √ 2 0 −1/ √ 2 1/ √ 6 −2/ √ 6 1/ √ 6   Operator A4 in the basis {|a1i , |a2i , |a3i} is of the form A4 =   0 0 0 0 1 0 0 0 81   Dr.Randeep N C Mathematical foundations of QM

192. Problems A4 in the basis {|b1i , |b2i , |b3i} can be written as (A4 ) 0 = U† (A4 )U (A4 ) 0 =   1/ √ 3 1/ √ 2 1/ √ 6 1/ √ 3 0 −2/ √ 6 1/ √ 3 −1/ √ 1 1/ √ 6     0 0 0 0 1 0 0 0 81     1/ √ 3 1/ √ 3 1/ √ 3 1/ √ 2 0 −1/ √ 2 1/ √ 6 −2/ √ 6 1/ √ 6   (A4 ) 0 =   1/ √ 3 1/ √ 2 1/ √ 6 1/ √ 3 0 −2/ √ 6 1/ √ 3 −1/ √ 1 1/ √ 6     0 0 0 1/ √ 2 0 −1/ √ 2 81/ √ 6 −162/ √ 6 81/ √ 6   (A4 ) 0 =   14 −27 13 −27 54 −27 13 −27 81   Dr.Randeep N C Mathematical foundations of QM

193. Problems Problem 22: A two state quantum system has has two observables A and B. It is known that the observable A has eigenstates |α1i and |α2i with eigenvalues a1 and a2 respectively, while B has eigenstates |β1i and |β2i with eigenvalues b1 and b2 respectively, and that these eigenstates are related by |β1i = 3 5 |α1i − 4 5 |α2i , |β2i = 4 5 |α1i + 3 5 |α2i Suppose a measurement is made of the observable A and a value a1 is obtained. If the observable B is now measured, the probability of obtaining the value b1 will be [TIFR-2015] (a)0.80, (b)0.64, (c)0.60, (d)0.36 Answer: Given A |α1i = a1 |α1i , A |α2i = a2 |α2i , B |β1i = b1 |β1i , B |β2i = b2|β2i {|αi i} û → {|βi i} , Û = hα1 | β1i hα1 | β2i hα2 | β1i hα2 | β2i Dr.Randeep N C Mathematical foundations of QM

194. Problems |β1i = 1 |β1i = 2 X i=1 |αi i hαi | ! |β1i = hα1 | β1i |α1i + hα2 | β1i |α2i |β2i = hα1 | β2i |α1i + hα2 | β2i |α2i Similarly, |α1i = hβ1 | α1i |β1i + hβ2 | α1i |β2i |α2i = hβ1 | α2i |β1i + hβ2 | α2i |β2i Given |β1i = 3 5 |α1i − 4 5 |α2i , |β2i = 4 5 |α1i + 3 5 |α2i, then |α1i = 3 5 |β1i + 4 5 |β2i , |α2i = − 4 5 |β1i + 3 5 |β2i Suppose we get a1, after measuring the observable A, then the system is collapse into state |α1i. Then probability for getting the eigenvalue b1 in this state |α1i = 3 5 |β1i + 4 5 |β2i is P (b1) = (3/5)2 (3/5)2 + (4/5)2 = 9 25 = .36 Dr.Randeep N C Mathematical foundations of QM

195. Stern-Gerlach Experiment It is an experiment to demonstrate that the spin of an electron is quantized. In which silver (Ag) atoms are heated in an oven. The oven has a small hole through which some of the silver atoms escape. As in figure, the beam goes through a collimator and then subjected to a inhomogeneous magnetic field and finally fall on a detector. 1 1https : //cronodon.com/Atomic/quantum angular momentum.html Dr.Randeep N C Mathematical foundations of QM

196. Stern-Gerlach Experiment The silver atoms have 47 electrons, in which 46 electrons can be visualized as forming a spherically symmetrical electron cloud with no net angular momentum. Therefore, atoms as a whole have an angular momentum, which is due to spin of the 47th (5s) electron. Then the magnetic moment of the entire atom is due to spin magnetic moment of the 47th electron. The magnetic moment µ of the atom is proportional to the electron spin S, µ ∝ S The interaction energy of the magnetic moment with the magnetic field B is U = −µ.B Then the corresponding z component of force on the atom is Fz = − ∂U ∂z = ∂ ∂z (µ.B) ≈ µz ∂Bz ∂z Dr.Randeep N C Mathematical foundations of QM

197. Stern-Gerlach Experiment If electron is like a classical spinning object, then there is no preferred direction for the orientation of µ and we get values of µz between |µ| and −|µ|. Instead, what we experimentally observed is a two beams on the screen. In other words, the SG apparatus splits the original silver beam from the oven into two distinct components. From this results we can conclude that, this splitting is due to spin S of electron and it has two components in z direction, we call them Sz + and Sz −. Block diagram of Stern-Gerlach experiment Oven SG(ẑ) Sz(+) Sz(−) Dr.Randeep N C Mathematical foundations of QM

198. Sequential Stern-Gerlach Experiment Experiment:1 Oven SG(ẑ) Sz(+) Sz(−) SG(ẑ) Sz(+) × Experiment:2 Oven SG(ẑ) Sz(+) Sz(−) SG(x̂) Sx(+) Sx(−) Experiment:3 Oven SG(ẑ) Sz(+) Sz(−) SG(x̂) Sx(+) Sx(−) SG(ẑ) Sz(+) Sz(−) Dr.Randeep N C Mathematical foundations of QM

199. Sequential Stern-Gerlach Experiment Experiment:1 Oven SG(ẑ) |0i |1i SG(ẑ) |0i × Experiment:2 Oven SG(ẑ) |0i = 1 √ 2 [|+i + |−i] |1i SG(x̂) |+i |−i Experiment:3 Oven SG(ẑ) |0i = 1 √ 2 [|+i + |−i] |1i SG(x̂) |+i = 1 √ 2 [|0i + |1i] |−i SG(ẑ) |0i |1i Dr.Randeep N C Mathematical foundations of QM

202. Position and Momentum Representations Lecture-8 Dr.Randeep N C Mathematical foundations of QM

203. Continuous basis (Example: Position representation) Consider a position operator x̂ in one dimension, then the eigenvalue equation of position operator is given by x̂ |x0 i = x0 |x0 i Orthonormality condition hx0 | x00 i = δ (x0 − x00 ) Completeness relation Z ∞ −∞ |x0 i hx0 | dx0 = 1 Any arbitrary ket, |αi can be expanded in terms of {|x0 i} as |αi = Z ∞ −∞ dx0 |x0 i hx0 | αi Dr.Randeep N C Mathematical foundations of QM

204. Continuous basis (Example: Position representation) Measurement of an observable x |αi = Z ∞ −∞ dx0 |x0 i hx0 | αi x-measurement −→ Z x0 + ∆ 2 x0− ∆ 2 dx00 |x00 i hx00 | αi that is after measurement, particle is located between the narrow range (x0 − ∆/2, x0 + ∆/2) In |αi = R ∞ −∞ dx0 |x0 i hx0 | αi, the expansion coefficient hx0 | αi is the probability amplitude of finding the particle at x0 . This inner product hx0 | αi is what we called the wave function Ψα (x0 ) for the state |αi, Ψα (x0 ) = hx0 | αi That is the wave function Ψα (x0 ) is the expansion coefficient of a state |αi in the position representation. Dr.Randeep N C Mathematical foundations of QM

207. Continuous basis (Example: Position representation) Suppose Â = x̂2 , then x0

209. x̂2

211. x00 = (x00 ) 2 δ (x0 − x00 ) β

213. x̂2

215. α = Z ∞ −∞ dx0 Z ∞ −∞ dx00 Ψ∗ β (x0 ) (x00 ) 2 δ (x0 − x00 ) Ψα (x00 ) β

217. x̂2

219. α = Z ∞ −∞ dx0 Ψ∗ β (x0 ) (x0 ) 2 Ψα (x0 ) In general for a function of x, ˆ f (x) hβ|ˆ f (x)|αi = Z ∞ −∞ dx0 Ψ∗ β (x0 ) f (x0 ) Ψα (x0 ) Then the expectation value of x̂ in the state |αi is given by hα |x̂| αi = Z ∞ −∞ dx0 Ψ∗ α (x0 ) x0 Ψα (x0 ) Dr.Randeep N C Mathematical foundations of QM

220. Continuous basis (Example: Position representation) In three dimension, |X0 i = |x0 , y0 , z0 i is the position eigenket. Any arbitrary ket |αi can be expanded in terms of {|X0 i} as |αi = Z d3 x0 |X0 i hX0 | αi This |X0 i = |x0 , y0 , z0 i is the simultaneous eigenket of the observable x, y and z: x̂ |X0 i = x0 |X0 i , ŷ |X0 i = y0 |X0 i , ẑ |X0 i = z0 |X0 i That is the three components of the position vector can be measured simultaneously to arbitrary degree of accuracy. Therefore, [x̂i , x̂j ] = 0, i, j = 1, 2, 3 That is [x̂, ŷ] = [x̂, ẑ] = [ŷ, ẑ] = 0 and [x̂, x̂] = [ŷ, ŷ] = [ẑ, ẑ] = 0 Dr.Randeep N C Mathematical foundations of QM

221. Continuous basis (Example: Momentum representation) The eigenvalue equation of momentum operator p̂x is given by p̂x |p0 x i = p0 x |p0 x i Orthonormality condition hp0 x | p00 x i = δ (p0 x − p00 x ) Completeness relation Z ∞ −∞ |p0 x i hp0 x | dp0 x = 1 Any arbitrary ket, |αi can be expanded in terms of {|p0 x i} as |αi = Z ∞ −∞ dp0 x |p0 x i hp0 x | αi Here hp0 x | αi = Φα (p0 x ) is called momentum space wave function. Dr.Randeep N C Mathematical foundations of QM

222. Continuous basis (Example: Momentum representation) Normalization hα | αi = Z ∞ −∞ dp0 x hα | p0 x i hp0 x | αi = Z ∞ −∞ dp0 x Φ∗ α (p0 x ) Φα (p0 x ) = 1 The momentum eigenstates in three dimensions are represented by |P 0 i = |p 0 x , p 0 y , p 0 z i Which is a simultaneous eigenket of the observable px , py and pz : p̂x |P0 i = p0 x |P0 i , p̂y |P0 i = p0 y |P0 i , p̂z |P0 i = p0 z |P0 i That is the three components of the momentum vector can be measured simultaneously to arbitrary degree of accuracy. Therefore, [p̂i , p̂j ] = 0, i, j = 1, 2, 3 That is [p̂x , p̂y ] = [p̂x , p̂z ] = [p̂y , p̂z ] = 0 and [p̂x , p̂x ] = [p̂y , p̂y ] = [p̂z , p̂z ] = 0 Dr.Randeep N C Mathematical foundations of QM

224. Translation This property is imposed for demanding the probability conservation. That is |αi and translated ket T̂(dx 0 ) |αi satisfy, D α

227. T̂† (dx0 ) T̂ (dx0 )

230. α E = hα | αi 2 Successive translation satisfy, T̂ (dx0 ) T̂ (dx00 ) = T̂ (dx0 + dx00 ) 3 Translation in opposite direction is same as the inverse of original translation T̂ (−dx0 ) = T̂−1 (dx0 ) 4 In the limit dx 0 → 0, the translation operator reduces to the identity operator lim dx0→0 T̂ (dx0 ) = 1 Dr.Randeep N C Mathematical foundations of QM

231. Translation We can take infinitesimal translation operator as T̂ (dx0 ) = 1 − i p̂x ~ dx0 where p̂x is the generator of translation called x component of momentum. To check above four properties, 1 T̂† (dx0 ) T̂ (dx0 ) = 1 + i p̂† x ~ dx0 1 − i p̂x ~ dx0 = 1 + i ~ p̂† x − p̂x dx0 + O dx02 = 1 2 T̂ (dx0 ) T̂ (dx00 ) = 1 − i p̂x ~ dx0 1 − i p̂x ~ dx00 = 1 − i ~ p̂x (dx0 + dx00 ) = T̂ (dx0 + dx00 ) 3 T̂ (−dx0 ) = 1 + i p̂x ~ dx0 = T̂† (dx0 ) = T̂−1 (dx0 ) 4 limdx0→0 T̂ (dx0 ) = limdx0→0 1 − i p̂x ~ dx0 = 1 that is all four properties of translation operators are satisfied. Dr.Randeep N C Mathematical foundations of QM

234. Time Evolution Operator Lecture-9 Dr.Randeep N C Mathematical foundations of QM

235. Translation Now let us find the commutation relation of x̂ and T̂(dx 0 ), x̂T̂ (dx0 ) |x0 i = x̂ |x0 + dx0 i = (x0 + dx0 ) |x0 + dx0 i T̂ (dx0 ) x̂ |x0 i = x0 T̂ (dx0 ) |x0 i = x0 |x0 + dx0 i From above two equations, we get x̂T̂ (dx0 ) − T̂ (dx0 ) x̂ |x0 i = dx0 |x0 + dx0 i h x̂, T̂ (dx0 ) i |x0 i = dx0 |x0 + dx0 i h x̂, T̂ (dx0 ) i |x0 i ≈ dx0 |x0 i That is the commutation relation between x̂ 0 and T̂(dx 0 ) is h x̂, T̂ (dx0 ) i = dx0 x̂T̂ (dx0 ) − T̂ (dx0 ) x̂ = dx0 x̂ 1 − i p̂x ~ dx0 − 1 − i p̂x ~ dx0 x̂ = dx0 Dr.Randeep N C Mathematical foundations of QM

236. Translation − i ~ x̂p̂x + i ~ p̂x x̂ dx0 = dx0 − i ~ (x̂p̂x − p̂x x̂) = 1 [x̂, p̂x ] = i~ That is x̂ and p̂x are incompatible observable. Therefore it is impossible to find simultaneous eigenket of x̂ and p̂x . Therefore, position and momentum uncertainty relation is (∆x) (∆px ) ≥ 1 2 |h[x̂, p̂x ]i| (∆x) (∆px ) ≥ ~ 2 Dr.Randeep N C Mathematical foundations of QM

237. Finite Translation Similarly for other position and momentum components, [ŷ, p̂y ] = i~ =⇒ (∆y) (∆py ) ≥ ~ 2 [ẑ, p̂z ] = i~ =⇒ (∆z) (∆pz ) ≥ ~ 2 Consider a finite translation in the x direction by an amount ∆x 0 : T̂ (∆x0 ) |x0 i = |x0 + ∆x0 i By compounding N infinitesimal translations, each of which is characterized by a spatial displacement ∆x 0 /N in the x direction, and letting N → ∞, we obtain T̂ (∆x0 ) = lim N→∞ 1 − ip̂x ∆x0 N~ N T̂ (∆x0 ) = exp − ip̂x ∆x0 ~ which is the finite translation operator. Dr.Randeep N C Mathematical foundations of QM

241. Momentum Operator in the Position Basis Therefore, the expectation value of an operator p̂x in the state |αi in terms of wavefunction is hp̂x i = hα|p̂x |αi = Z ∞ −∞ dx0 Ψ∗ α(x 0 ) − i~ ∂ ∂x0 Ψα(x 0 ) . Which is the expectation value of momentum in position representation. The expectation value of p̂2 x in the position representation is obtained by hα|p̂2 x |αi = Z ∞ −∞ dx0 Z ∞ −∞ dx00 hα | x0 i x0 |p̂2 x |x00 hx00 | αi hp̂2 x i = hα|p̂2 x |αi = Z ∞ −∞ dx0 Ψ∗ α(x 0 ) − ~2 ∂2 ∂x02 Ψα(x 0 ) . Dr.Randeep N C Mathematical foundations of QM

242. Connection between the x-representation and the p-representation The transformation matrix from the old basis {|ai i} to the new basis {|bi i} is given by hai |Û|aj i = hai |bj i Likewise, hx 0 |p 0 x i is the transformation function from x-representation to the p-representation. we know that hx 0 |p̂ 0 x |αi = −i~ ∂ ∂x0 hx 0 |αi Let us take |αi = |p 0 x i, then hx 0 |p̂x |p 0 x i = −i~ ∂ ∂x0 hx 0 |p 0 x i p 0 x hx 0 |p 0 x i = −i~ ∂ ∂x0 hx 0 |p 0 x i i ~ p 0 x dx 0 = dhx 0 |p 0 x i hx0 |p0 x i Dr.Randeep N C Mathematical foundations of QM

243. Connection between the x-representation and the p-representation After integrating above equation, we get i ~ p 0 x x 0 = ln hx 0 |p 0 x i + ln C exp i ~ p 0 x x 0 = Chx 0 |p 0 x i hx 0 |p 0 x i = N exp i ~ p 0 x x 0 Which is the wavefunction of the momentum eigenstate |p 0 x i calling momentum eigenfunction. To get the normalization constant N, let us consider hx 0 |x 00 i = Z dp 0 x hx 0 |p 0 x ihp 0 x |x 00 i δ(x 0 − x 00 ) = Z dp 0 x N exp i ~ p 0 x x 0 N∗ exp − i ~ p 0 x x 00 Dr.Randeep N C Mathematical foundations of QM

244. Connection between the x-representation and the p-representation δ(x 0 − x 00 ) = |N|2 Z dp 0 x exp i ~ p 0 x (x 0 − x 00 ) Using the following Fourier transform result, 1 2π~ Z exp i ~ p 0 x (x 0 − x 00 ) dpx = δ(x 0 − x 00 ) we get 1 2π~ = |N|2 =⇒ N = 1 √ 2π~ Therefore, momentum eigenfunction is hx 0 |p 0 x i = 1 √ 2π~ exp ip 0 x x 0 ~ Dr.Randeep N C Mathematical foundations of QM

245. Relation between position space wavefunction and momentum space wavefunction The position wavefunction is given by Ψα(x 0 ) = hx 0 |αi = Z dp 0 x hx 0 |p 0 x ihp 0 x |αi Substituting the expression of hx 0 |p 0 x i , we get position eigenfunction in terms of momentum eigenfunction. That is Ψα(x 0 ) = 1 √ 2π~ Z dp 0 x exp i ~ p 0 x x 0 Φα(p 0 x ) The momentum wavefunction is given by Φα(p 0 x ) = hp 0 x |αi = Z dx 0 hp 0 x |x 0 ihx 0 |αi Substituting the expression of hp 0 x |x 0 i, we get momentum eigenfunction in terms of position eigenfunction. That is Φα(p 0 x ) = 1 √ 2π~ Z dx 0 exp − i ~ p 0 x x 0 Ψα(x 0 ) Dr.Randeep N C Mathematical foundations of QM

247. Properties: Time evolution operator 1 Time evolution operators are unitary operators Û† (t, t0)Û(t, t0) = 1 Like translation operator, this property imposed for probability conservation: hα, t0|α, t0i = hα, t0; t|α, t0; ti = 1 Suppose that at time t0 the state ket is expanded in terms of the eigenkets of some observable A as |α, t0i = X i ci (t0)|ai i Likewise, at later time |α, t0; ti = X i ci (t)|ai i In general, the modulus of individual expansion coefficient varying with time |ci (t)| 6= |ci (t0)| Dr.Randeep N C Mathematical foundations of QM

248. Properties: Time evolution operator But because of probability conservation, we have X i |ci (t)|2 = X i |ci (t0)|2 2 Composition property Û(t2, t0) = Û(t2, t1)Û(t1, t0), t2 t1 t0 An infinitesimal time evolution operator is defined as |α, t0; t0 + dti = Û(t0 + dt, t0)|α, t0i 3 The infinitesimal time evolution operator reduces to identity as dt → 0, lim dt→0 Û(t0 + dt, t0) = 1 As in the case of translation, the infinite time evolution operator can be written as Û(t0 + dt, t0) = 1 − iĤdt ~ , Ĥ is the Hamiltonian Dr.Randeep N C Mathematical foundations of QM

249. Properties: Time evolution operator Using composition property of time evolution operator by letting t1 → t, t2 → t + dt, we get Û(t + dt, t0) = Û(t + dt, t)Û(t, t0) = 1 − iĤdt ~ Û(t, t0) Û(t + dt, t0) − Û(t, t0) = − iĤdt ~ Û(t, t0) i~ Û(t + dt, t0) − Û(t, t0) dt = ĤÛ(t, t0) =⇒ i~ ∂ ∂t Û(t, t0) = ĤÛ(t, t0) (1) is the Schrodinger equation for the time evolution operator. Multiplying both side of the above equation by |α, t0i on the right, we get i~ ∂ ∂t Û(t, t0)|α, t0i = ĤÛ(t, t0)|α, t0i Dr.Randeep N C Mathematical foundations of QM

250. Schrodinger equation Therefore, the Schrodinger equation for the state ket is i~ ∂ ∂t |α, t0; ti = Ĥ|α, t0; ti (2) If Ĥ is time independent, then from the equation (5), i~ d dt Û(t, t0) = ĤÛ(t, t0) dÛ(t, t0) Û(t, t0) = − i ~ Ĥdt ln Û(t, t0) + lnC = − i ~ Ĥt =⇒ Û(t, t0) = C exp − i ~ Ĥt When t → t0, Û(t, t0) → 1, then 1 = C exp − i ~ Ĥt0 =⇒ C = exp i ~ Ĥt0 Dr.Randeep N C Mathematical foundations of QM

251. Schrodinger equation Therefore, Û(t, t0) = exp − i ~ Ĥ(t − t0) is the time evolution operator when H is time independent. To evaluate the effect of time evolution operator on a general initial ket |αi, expand Û(t, t0) in the eigenbasis of energy. For simplicity consider [Â, Ĥ] = 0, then eigenket of Â is also an eigenket of Ĥ: Ĥ|ai i = Ei |ai i Set initial time t0 = 0 in the expression Û(t, t0), then Û(t, 0) = exp − iĤt ~ = X i,j |ai ihai | exp − iĤt ~ |aj ihaj | = X i,j |ai i exp − iEj t ~ hai |aj ihaj | Dr.Randeep N C Mathematical foundations of QM

253. Schrodinger equation Therefore, after a time t, the expansion coefficient changes with time as ci (t = 0) = ci → ci (t) = c(t0 = 0) exp − iEi t ~ with its modulus unchanged. Act hx 0 | on equation (7), we get hx 0 |α, t0 = 0; ti = X i hx 0 |ai ici exp − iEi t ~ Hence, we get time dependent wavefunction as Ψ(x 0 , t) = X i ψi (x)ci exp − iEi t ~ where ψi (x) = hx 0 |ai i is the eigenfunction of energy or observable A. Dr.Randeep N C Mathematical foundations of QM

256. The time dependent Schrodinger equation That is i~ ∂ ∂t Ψ(x 0 , t) = hx 0 |Ĥ|α, t0; ti Replace RHS of above expression by equation (8), we get i~ ∂ ∂t Ψ(x 0 , t) = hx 0 | p̂2 x 2m |α, t0; ti + hx 0 |V̂ (x)|α, t0; ti i~ ∂ ∂t Ψ(x 0 , t) = Z ∞ −∞ dx 00 hx 0 | p̂2 x 2m |x 00 ihx 00 |α, t0; ti+ hx 0 |V̂ (x)|x 00 ihx 00 |α, t0; ti We know that, hx 0 | p̂2 x 2m |x 00 i = (−i~)2 2m ∂2 ∂x002 δ(x 0 − x 00 ) hx 0 |V̂ (x)|x 00 i = V (x 00 )δ(x 0 − x 00 ) Dr.Randeep N C Mathematical foundations of QM

257. The time dependent Schrodinger equation Therefore, above equation become i~ ∂ ∂t Ψ(x 0 , t) = Z ∞ −∞ dx 00 (−i~)2 2m ∂2 ∂x002 δ(x 0 − x 00 )Ψ(x 00 , t)+ V (x 00 )δ(x 0 − x 00 )Ψ(x 0 , t) That is i~ ∂ ∂t Ψ(x 0 , t) = −~2 2m ∂2 ∂x02 Ψ(x 0 , t) + V (x 0 )Ψ(x 0 , t) (5) Which is the time dependent Schrodinger equation in one dimension. Dr.Randeep N C Mathematical foundations of QM

258. Time independent Schrodinger equation Choose simultaneous eigenstates of Â and Ĥ as initial state, that is |α, t0i = |ai i Then the corresponding time dependent state is |α, t0; ti = |ai i exp − iEi t ~ Act hx 0 | on the left, we get hx 0 |α, t0; ti = hx 0 |ai i exp − iEi t ~ ψ(x 0 , t) = ψi (x 0 ) exp − iEi t ~ Substitute this in time dependent Schrodinger wave equation (9), i~ ∂ ∂t ψi (x 0 ) exp − iEi t ~ = −~2 2m ∂2 ∂x02 ψi (x 0 ) exp − iEi t ~ + V (x 0 )ψi (x 0 ) exp − iEi t ~ Dr.Randeep N C Mathematical foundations of QM

259. Time independent Schrodinger equation i~ψi (x 0 ) × − iEi ~ exp − −iEi t ~ = −~2 2m ∂2 ψi (x 0 ) ∂x02 exp − iEi t ~ + V (x 0 )ψi (x 0 ) exp − iEi t ~ −~2 2m ∂2 ψi (x 0 ) ∂x02 + V (x 0 )ψi (x 0 ) = Ei ψi (x 0 ) is the time independent Schrodinger equation. This equation is of the form Hψi = Ei ψi . Therefore which is an eigenvalue equation. By using appropriate boundary conditions, we can solve time independent Schrodinger equation for different potential problems. Dr.Randeep N C Mathematical foundations of QM

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