Find the Inverse of the Rotation Matrix

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IMPORTANT NOTE: These are the answers. In the Exercise 1.2-30 exam you need to add justifications Find the polynomial of degree 3 whose graph Example (3.3-27) passes through the points (0,1),(1,0),(-1,0), (2,-15) Determine whether the following vectors form a basis of R4 (1,1,1,1), (1,-1,1,-1),(1,2,4,8),(1,-2,4,-8) Answer: Yes, (the matrix with these vectors as columns is invertible) Find the inverse of the rotation Let T be a clockwise rotation in R2 matrix. by π/2 followed by an orthogonal projection onto the y axis. 1. Find the matrix of T. cos(a) -sin(a) Answer= cos(a) sin(a) 2. Determine whether T is invertible 3. Find im(T) and ker(T) sin(a) cos(a) -sin(a) cos(a) Answer: It was given in class. For the matrix A below, find Find the inverse of the matrix and 1 2 a b all the 2x2 matrices X that check your answer. Interpret your satisfy the equation A.X=I2. 3 5 result geometrically. b -a Answer: The matrix is a reflection about a line L followed by a scaling by (a2 + b2 )1/2. The inverse is the same reflexion, followed -5 2 Answer: by a scaling (a2 + b2 )-1/2 3 -1 a b (a2 + b2 )-1/2 b -a 1 3.2-46 Find a basis of the kernel and image of the matrix. (2.4-31)For which values of the Determine the dimensions of the constants a, b and c is the kernel and image. following matrix invertible? Determine the rank. Justify your answers. There are 0 a b no values of a, b, c 1 2 0 3 5 -a 0 c that make the matrix invertible 0 0 1 4 6 -b -c 0 Done in class Note: The problem was first Example (3.3-31) stated that T was from R to Give an example of a 5 x 4 matrix R4 such. In this case, the Let V be the subspace of R4 answer is there are no A with dim(ker A)=3. defined by the equation x1- transformations (why? Prove Compute dim(im A). x2 + 2x3 + 4 x4 = 0 this!). Find a linear transformation T One answer for a transf. 3 4 1 0 0 0 from R to R such that from R3 to R4 such is The image has ker(T)={0}, im(T)=V. 0 1 0 0 dimension 2. 0 0 0 0 Describe T by its matrix. 1 0 0 0 0 0 0 1 2 0 0 0 0 0 0 -1 2 0 0 -1 Example (3.3-29) Find a basis of the subspace of R3 defined by the equation 2x1+3x2 + x3 =0 Answer: (3,-2,0), (0,1,-3) 2.

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Lecture Notes: Qubit Representations and Rotations
Phys 711 Topics in Particles & Fields | Spring 2013 | Lecture 1 | v0.3 Lecture notes: Qubit representations and rotations Jeffrey Yepez Department of Physics and Astronomy University of Hawaiì at Manoa Watanabe Hall, 2505 Correa Road Honolulu, Hawaiì 96822 E-mail: [email protected] www.phys.hawaii.edu/∼yepez (Dated: January 9, 2013) Contents mathematical object (an abstraction of a two-state quan- tum object) with a \one" state and a \zero" state: I. What is a qubit? 1 1 0 II. Time-dependent qubits states 2 jqi = αj0i + βj1i = α + β ; (1) 0 1 III. Qubit representations 2 A. Hilbert space representation 2 where α and β are complex numbers. These complex B. SU(2) and O(3) representations 2 numbers are called amplitudes. The basis states are or- IV. Rotation by similarity transformation 3 thonormal V. Rotation transformation in exponential form 5 h0j0i = h1j1i = 1 (2a) VI. Composition of qubit rotations 7 h0j1i = h1j0i = 0: (2b) A. Special case of equal angles 7 In general, the qubit jqi in (1) is said to be in a superpo- VII. Example composite rotation 7 sition state of the two logical basis states j0i and j1i. If References 9 α and β are complex, it would seem that a qubit should have four free real-valued parameters (two magnitudes and two phases): I. WHAT IS A QUBIT? iθ0 α φ0 e jqi = = iθ1 : (3) Let us begin by introducing some notation: β φ1 e 1 state (called \minus" on the Bloch sphere) Yet, for a qubit to contain only one classical bit of infor- 0 mation, the qubit need only be unimodular (normalized j1i = the alternate symbol is |−i 1 to unity) α∗α + β∗β = 1: (4) 0 state (called \plus" on the Bloch sphere) 1 Hence it lives on the complex unit circle, depicted on the j0i = the alternate symbol is j+i: 0 top of Figure 1.
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