Appendix A. Vector addition: - The sum of two vectors is another vector that also lie in the space:

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1 Tor Kjellsson Stockholm University Appendix A A.1 Q. Consider the ordinary vectors in 3 dimensions (a x î+a y ĵ+a zˆk), with complex components. Do the following subsets constitute a vector space? If so, what is its dimension? If not, why not? a) The subset of all vectors with a z = 0. Let us start by looking at what a Vector space is: A vector space consists of a set of vectors { α, β, γ...} together with a set of scalars {a, b, c...} is closed under the two operations: vector addition and scalar multiplication. Below we repeat some rules: Vector addition: - The sum of two vectors is another vector that also lie in the space: α + β = γ - It is commutative: - It is associative: α + β = β + α - There exists a null vector: α + ( β + γ ) = ( α + β ) + γ α + 0 = α - For every vector in the vector space there exists an inverse vector: α + α = 0 and another (more logical) way of writing α is α. Scalar multiplication: - The product of any scalar with any vector is within the space: a α = γ - It is distributive with respect to vector addition: a ( β + γ ) = a β + a γ 1

2 and scalar addition: (a + b) β = a β + b β - It is associative with respect to multiplication of scalars: a (b β ) = (ab) β - Multiplication with the scalars 0 and 1 is precisely what one would expect: 0 β = 0, 1 β = β now this was an awful lot of rules. Most of them are however quite intuitive and some can be grouped together. Let us sum them up: To check if a set is a vector space you check the following conditions: i) Any linear combination of vectors in the vector space also lie in the vector space. Especially check that the zero vector exists and that every vector has an inverse. ii) Vector addition is commutative and associative. iii) Scalar multiplication is distributive with respect to vector addition and scalar multiplication. It should also be associative with respect to multiplication of scalars. Most often the second and third criteria are a matter of course and you only need to briefly explain why they apply. Let us finally start with the problem! In this problem we want to check whether or not the set {(a x, a y, 0)} is a vector space. Now, for regular vectors the second and third criteria are a matter of course and we need not check them. The first criteria we however must check: The two vectors (a x, a y, 0), (b x, b y, 0) both lie in our subset. For any complex scalars λ, µ so do λ(a x, a y, 0), µ(b x, b y, 0) too. The sum of these: λ(a x, a y, 0) + µ(b x, b y, 0) = (λa x + µb x, λa y + µb y, 0) is also a vector in the same set. We have thus shown that the set is a vector space! (If you wonder about the null vector and the inverse vectors they are included in this statement. If you don t see why - pause and think about it - did we say anything about the unknowns we just used?) 2

3 b) The subset of vectors whose z-component is 1. This is not a vector space and there are at least two holes you can aim at (but you need only state one). First of all, there is no null vector in this set! Second of all, if you add the same vector to itself the z-component is now 2. However, this resulting vector does not meet the criteria of belonging to the set - thus the set cannot be a vector space. c) The subset of vectors whose components are all equal. Any vector in this set is of the type µ(1, 1, 1) for a complex scalar µ. Any linear combination of this is also of the same type: µ(1, 1, 1) + λ(1, 1, 1) = (µ + λ) (1, 1, 1) so the set is a vector space. Since all components are the same there is only one variable that is allowed to change and thus the dimension of the vector space is one. (Another way of seeing it is that all the vectors point in the same direction.) A.2 Q. Consider the collection of all polynomials (with complex coefficients) of degree less than N in x. For each of the following subsets, decide if it constitutes a vector space or not. If it does, suggest a convenient basis and give the dimension of the space. If not, explain why. a) The whole set. A polynomial in this set has the following appearance: p(x) = N 1 i=0 c i x i (1) and since different degrees of x are linearly independent we could represent them in the following way: p(x) = N 1 i=0 c i x i = (c 0, c 1, c 2,..., c N 1 ) (2) but this is the same representation as a vector! Given this representation it is not hard to convince oneself that the second and third criteria from the previous 3

4 problem (in the box) apply. We also easily see that any linear combination of two such elements also is a polynomial of the same type so we have a vector space! (as before - in the term any, the null element and inverse element are also included.) An appropriate basis is of course {x n } for integers n such that 0 n N 1. The dimension is the total number of values that n can take on - in this case N. b) The subset of even polynomials. The same reasoning as above applies but now with a basis {x 2n }. The dimension is once again given by the total number of values that n can take on. In the case of odd value of N this is for 0 n (N 1)/2 which amounts to (N + 1)/2. In the case of an even N we instead have 0 n (N 2)/2 (why? 1 ) and the dimension is thus N/2. c) The subset where the leading coefficient (the one in front of x n 1 ) is forced to be 1. This is not a vector space and two reasons are: i) The null polynomial is not in this set. ii) The sum of two such polynomials is not in the subset. d) The subset where we require that the polynomials have the value 0 at x = 1. This might seem tricky at first but it really is not. Given any two polynomials that meet this requirement any linear combination of them will still have the value 0 at x = 1. This is thus a vector space. An appropriate basis for this would be {(x 1) n } since the polynomials most easily can be written as linear combination of these. The dimension is in this case N 1 and not N as in problem a). Why? e) The subset where we require that the polynomials have the value 1 at x = 0. This is definitely not a vector space - the sum of two such polynomials has the value 2 at x = 0. 1 Look carefully at the problem statement. 4

5 A.3 Q. Prove that the components of a vector with respect to a given basis are unique. Assume that the vector α have two representations in the same basis: α = i c i e i (3) and α = i d i e i. (4) Now consider the subtraction of these: 0 = 0 = i c i e i i d i e i = i (c i d i ) e i. (5) Since the elements of the basis are linearly independent each term in the sum has to be 0 2. Thus c i d i = 0 for all i and the representation of a vector in a given basis must be unique. A.4 Q. Use the Gram-Schmidt procedure to orthonormalize the 3-space basis: e 1 = (1 + i, 1, i) T e 2 = (i, 3, 1) T e 3 = (0, 28, 0) T In the Gram-Schmidt orthonormalization procedure there are as many steps as the number of basis vectors. The procedure can be a bit involved but the idea is easy to grasp: Start by choosing one of the vectors as a basis and then normalize it: e 1 = e 1 e 1 = 1 2 (1 + i, 1, i)t. (6) 2 Think carefully about this, it is important that you understand it. 5

6 Now we proceed to construct the next basis vector. Recall that the scalar product of two vectors is a measure of the amount they share. In order to construct an orthogonal basis vector to e 1 we must thus subtract this shared amount : c 0 e 2 = e 2 e 1 e 2 e 1 = 1 2 ((1 + i), 1, i)t c (1 + i, 1, i)t. where the constant c 0 is the norm of the right hand side and c 1 is the following scalar product: and we thus have: c 1 = e 1 e 2 = 1 2 (1 i, 1, i) (i, 3, 1)T = 2 c 0 e 2 = (i, 3, 1) T (1 + i, 1, i) T = ( 1, 2, 1 i) T. (7) The norm of the right hand side is: so our basis vector is: ( i i) = 7 e 2 = 1 7 ( 1, 2, 1 i) T. (8) Now we proceed to find the last basis vector and we do this by repeating the process from above: d 0 e 3 = e 3 e 1 e 3 e 1 e 2 e 3 e 2 but now we subtract the projection along the second basis vector as well. As before, d 0 is just the norm of the right hand side that we later divide with to obtain our normalized basis vector e 3. Doing the algebra we obtain: d 0 e 3 = (1 7i, 5, 8 + i) T and the norm of the right hand side is 140 so d 0 = 140. This then gives: and our orthonormal basis is: e 3 = (1 7i, 5, 8 + i) T e 1 = 1 2 (1 + i, 1, i)t e 2 = 1 7 ( 1, 2, 1 i) T (9) e 3 = (1 7i, 5, 8 + i) T 6

7 A.5 Q. Prove the Schwartz inequiality: α β 2 α α β β (10) Hint: Define a vector γ = β ( α β / α α ) α Following the hint we define the vector γ = β ( α β / α α ) α. Now we consider the scalar product of this vector with itself (which must be real and non-negative): ( β β α α α α γ γ 0 ) ( β α β α α α ) 0 β β α β β α β α 2 β α α β + α α α α α α 2 α α 0 β β α β 2 α α β β β α 2 + β α 2 α α α α }{{} 0 α β 2 α α 0 β β α α α β 2. 0 Note also that α β = β α (although, we did not use it here). 7

8 A.14 Q. Using the standard basis (î, ĵ, ˆk) for vectors in three dimensions: a) Construct the matrix representing a rotation through angle θ (counter-clockwise, looking down the axis towards the origin) about the z-axis. When you are asked to give the matrix that represents a transformation, it is easiest to first find what the transformation does to the basis vectors. The components of any vector then transforms with the transpose of that matrix. For a rotation about the z-axis we see that the z-axis itself is left unaltered. The x-axis and y-axis however are rotated in the plane: y y θ θ x x We see that the new position are given, in the old basis, as: ˆx = cos(θ)ˆx + sin(θ)ŷ ŷ = cos(θ + π 2 )ˆx + sin(θ + π 2 )ŷ = sin(θ)ˆx + cos(θ)ŷ (11) so the matrix equation describing the transformation is (recall that ẑ = ẑ): ˆx cos(θ)ˆx + sin(θ)ŷ + 0ẑ cos(θ) sin(θ) 0 ˆx ˆx ŷ = sin(θ)ˆx + cos(θ)ŷ + 0ẑ = sin(θ) cos(θ) 0 ŷ = T ŷ ẑ 0ˆx + 0ŷ + ẑ ẑ ẑ and hence the components of any vector in this basis will transform with the following matrix: cos(θ) sin(θ) 0 T = sin(θ) cos(θ) 0 (12)

9 b) Construct the matrix representing a rotation through angle 120 (counter-clockwise, looking down the axis toward the origin) about an axis through the point (1, 1, 1). Below are the three basis vectors î, ĵ and ˆk together with the vector (1, 1, 1) in this basis: ĵ (1,1,1) î ˆk If we imagine looking down along the vector (1, 1, 1) we see that the three basis vectors are ordered symmetrically about this vector. This also means that the projection of the three basis vectors on the normal plane of the vector (1, 1, 1) must be situated symmetrically. In a plane, however, this means that they are 120 apart and since a rotation about the normal axis is the same as a rotation in this plane the projections of the basis vectors just change places with one turn. This means that î = ĵ, ĵ = ˆk and ˆk = î. The transformation matrix for this is: T = and the matrix we seek is: T = (13)

10 c) Construct the matrix representing reflection through the xy-plane. This is really easy, just think of what happens with the basis vectors: î = î, ĵ = ĵ and ˆk = ˆk. The matrix representing this transformation is: T = and the matrix we seek, the transpose of this, is the same matrix: T = (14) A.15 Q. In the usual basis (î, ĵ, ˆk), construct the matrix T x and T y representing a rotation through angle θ about the respective axis. There is nothing holy with the rotation about the z axis that we considered in problem A.14 a) - the principle is exactly the same. We just have to keep track of which vectors are rotated in what direction: About the x axis: Now we rotate the y axis towards the z axis, so our transformation of basis vectors is: ŷ = cos(θ)ŷ + sin(θ)ẑ ẑ = sin(θ)ŷ + cos(θ)ẑ (15) giving us the transformation T x = cos(θ) sin(θ) (16) 0 sin(θ) cos(θ) and T x = 0 cos(θ) sin(θ) (17) 0 sin(θ) cos(θ) 10

11 About the y axis: Now we rotate the z axis towards the x axis, so our transformation of basis vectors is: ẑ = cos(θ)ẑ + sin(θ)ˆx ˆx = sin(θ)ẑ + cos(θ)ˆx (18) giving us the transformation cos(θ) 0 sin(θ) T y = (19) sin(θ) 0 cos(θ) and cos(θ) 0 sin(θ) T y = (20) sin(θ) 0 cos(θ) Q. Suppose now we change bases to î = ĵ, ĵ = î, ˆk = ˆk. Construct the matrix S that effects this change of basis and check that ST x S 1 and ST y S 1 are what you would expect. Once again we first define the matrix that transforms the basis vectors: S = and then transponate this to get the matrix that transforms the coordinates: S = (21) Calculating the matrices T x, T y in the new basis we obtain: cos(θ) 0 sin(θ) ST x S 1 = = T y (θ) (22) sin(θ) 0 cos(θ) and ST y S 1 = 0 cos(θ) sin(θ) = T x ( θ). (23) 0 sin(θ) cos(θ) 11

12 A.18 Q. The 2 2 matrix representing a rotation of the xy plane is: ( ) cos(θ) sin(θ) T =. (24) sin(θ) cos(θ) Show that this matrix has no real eigenvalues except for certain values of θ and state which these are. This matrix has complex eigenvalues and eigenvectors - find them! Then construct a matrix S that diagonalizes T. Explicitly show that STS 1 is diagonal. The eigenvalues of a matrix can be found by solving the eigenvalue matrix equation: T α = λ α = (T λi) α = 0 under the assumption that α 0 and I is the identity matrix. Since α 0 the equation above is only satisfied if: which for our matrix becomes: The solutions are: det (T λi) = 0 (cos(θ) λ) 2 + sin(θ) 2 = 0 = λ 2 2 cos(θ)λ + 1 = 0. λ = cos(θ) ± cos(θ) 2 1 = cos(θ) ± i sin(θ) = e ±iθ. (25) Indeed these are in general not real numbers but for the angles θ = nπ for any integer n we have λ = 1 or λ = 1. To find the eigenvectors we now reinsert our eigenvalues into the matrix equation: T α = e ±iθ α ( ) a define α = and insert the matrix: b ( ) ( ) ( ) cos(θ) sin(θ) a = e ±iθ a sin(θ) cos(θ) b b to get the system of equations: a cos(θ) b sin(θ) = a (cos(θ) ± i sin(θ)) a sin(θ) + b cos(θ) = b (cos(θ) ± i sin(θ)). (26) and from this (it actually suffices with one of the equations - check this!) we get: b = ia. (27) 12

13 So the eigenvectors have the forms: ( ) ( ) 1 1 α 1 = a and α i 2 = a i which we of course want to normalize: α 1 = 1 ( ) 1 2 i and α 2 = 1 2 ( 1 i ). (28) To construct a matrix S that diagonalizes T we just need to use the eigenvectors of T as columns in S 1 : S 1 = 1 ( ) i i Leaving the algebraic details 3 to the dedicated students we get: STS 1 = 1 2 ( ) ( ) 2 cos(θ) + 2i sin(θ) 0 e iθ 0 = 0 2 cos(θ) 2i sin(θ) 0 e iθ. (29) Note that the entries in the diagonalized matrix are the eigenvalues of the original matrix. What do you think would have happened if you had put the eigenvectors in the opposite order in S 1? Answer is put in the following footnote 4. A.19 Q. Find the eigenvalues and the eigenvectors of the following matrix: Can this matrix be diagonalized? Seeking the eigenvalues: M = ( ) 1 1. (30) 0 1 ( ( ) ( 1 1 a a = λ 0 1) b b) we obtain the characteristic equation: (1 λ) 2 = 0 = λ = 1 (31) giving the eigenvectors: ( ) ( ) 1 1 a = 0 1 b ( ) a = a + b = a b b = b = b = 0. 3 You need to compute the inverse matrix S and then do the algebra that follows. You learn best if you do it yourselves! Hint: when computing the inverse, make sure that you include a factor so that the product SS 1 really becomes the identity matrix. 4 The eigenvalues switch places with each other. 13

14 ( 1 From the equations we only get one eigenvector: e 1 =. We need two 0) linearly independent eigenvectors to span the space so this matrix can not be diagonalized. A.22 a) Q. Show that if two matrices commute in one basis then they commute in any basis. We want to check that: [T e 1, T e 2] = 0 = [T f 1, Tf 2 ] = 0 and here we exploit the similarity equation 5 ST e S 1 = T f. Now assume that [T e 1, T e 2] = 0. Explore what happens to this commutator in another basis: [T f 1, Tf 2 ] = [STe 1S 1, ST e 2S 1 ] = ST e 1S 1 ST e }{{} 2S 1 ST e }{{} 2S 1 ST e }{{} 1S 1 = }{{} T f 1 T f 2 T f 2 T f 1 ST e 1T e 2S 1 ST e 2T e 1S 1 = S [T e 1, T e }{{ 2] S 1 = 0 } (32) 0 b) Q. Show that if two matrices are simultaneously diagonalizable they commute. If two matrices are simultaneously diagonalizable it means that: SM 1 S 1 = diag(a 1, b 1,...) and SM 2 S 1 = diag(a 2, b 2,...) (33) where the notation diag(a, b,...) means a diagonal matrix with entries a,b and so on. 5 Recall that a change of basis through a linear operator S gives the transformation between the bases of a given linear operator T through this equation. Two matrices connected in this way are said to be similar. 14

15 Now, from the previous problem we know that if two matrices commute in one basis they will commute in any basis. The two diagonal matrices above are the transformation of M 1 and M 2 to another basis. But diagonal matrices commute 6 so in this basis the commutator is 0. By the former problem then we have just concluded that two matrices that are simultaneously diagonalizable commute. A.25 Q. Let: a) Q. Verify that T is hermitian. T = ( ) 1 1 i. (34) 1 + i 0 T = ( (T) ) ( ) T T i = = 1 i 0 ( ) 1 1 i = T. (35) 1 + i 0 b) Q. Find its eigenvalues. gives the characteristic equation: ( ) ( ( 1 1 i a a = λ 1 + i 0 b) b) (1 λ)( λ) (1 i)(1 + i) = 0 = λ 1 = 2, λ 2 = 1. 6 If you don t see why: write out the product of two diagonal matrices. The product will be of the type diag(a 1 b 1, a 2 b 2,...). The product of M 2 and M 1 will be diag(b 1 a 1, b 2 a 2,...) which is the same as the former diagonal matrix. 15

16 c) Q. Find and normalize the eigenvectors (note that they are orthogonal). Seeking the eigenvectors: λ = 2 : ( ) ( 1 1 i a 1 + i 0 b) ( a = 2 b) = a + b(1 i) = 2a a(1 + i) = 2b = a = b(1 i) which gives us an eigenvector that we can normalize: e 1 = 1 ( ) 1 i. (36) 3 1 λ = 1 : ( ) ( ) 1 1 i a 1 + i 0 b = ( ) a b = a + b(1 i) = a a(1 + i) = b = b = a(1 + i) which gives us an eigenvector that we can normalize: e 2 = 1 ( ) i (37) and we note that this is orthogonal to the eigenvector. (as eigenvectors belonging to distinct eigenvalues of a hermitian matrix should be). d) Q. Construct the unitary diagonalizing matrix S and check explicitly that it diagonalizes T. A unitary matrix fulfils the following equation: U = U 1. Like in problem A.18 we can construct a matrix S that diagonalizes the matrix T by using the eigenvectors of T to first construct S 1 : S 1 = 1 ( ) 1 i 1 (38) i and then obtain (do the algebra!): S = 1 ( ) 1 + i i (39) and we indeed see that the matrix S is unitary. Once again we do not show the algebraic steps but indeed: ( ) STS = (40) 0 1 as it should. 16

17 e) Q. Check that det(t) and Tr(T) are the same for the regular and diagonalized form (they are invariant under linear transformations. Determinant: det(t) = 1 1 i 1 + i 0 = 2 and = 2. Trace: The trace of a matrix is defined as the sum of the diagonal elements. A quick glance at the two matrices shows that indeed the trace is the same: 1. A.27 Q. A unitary transformation is one for which Û Û = 1 : where 1 is the identity element, for instance the identity matrix if the operators are matrices. a) Q. Show that a unitary transformation preserve inner products, in the sense that Ûα Ûβ = α β for all vectors α β. Ûα Ûβ = Û Ûα β = α β (41) b) Q. Show that the eigenvalues of a unitary transformation have modulus 1. Let α be an element in Hilbert space. Let Û be a unitary transformation. 17

18 Now we consider the eigenvalue equation: Û α = λ α Ûα = λα If the above equation holds, so must the following: From problem a) we found that: Ûα Ûα = λα λα Û Ûα α = λ λ α α = λ 2 α α Û Ûα β = α β so combining this information we see that λ 2 = 1. c) Q. Show that the eigenvectors of a unitary transformation belonging to distinct eigenvalues are orthogonal. Let Û α = a α and expectation value: Û β = b β where a b. Consider the following α Ûβ = α bβ = b α β (42) Subtracting the lower equation from the upper: Û α β = a α β = a α β (43) 0 = (b a) α β (44) and since we made the assumption that a b we must have that α β = 0. 18