I have one critical problem with quantum computing. When I have one quantum state on bloch sphere, I do some transformation on that state for example PauliX transformation. As far as I know, we can present quantum state as |y> = cos (theta/2) |0> + e^(i*phi) sin (theta/2) |0> So if we have theta=PI/4 and phi=3/2PI cos PI/8 = 0.924; sin PI/8=0.383 and e^(i*(3/2PI))= -i When I want to visualize it on Bloch sphere, the coordinates I get from: x = sin(theta)*cos(phi) y = sin(theta)*sin(phi) z = cos(theta) For these parameters quantum state looks like that: |y> = 0.924 |0> + (-0.383i) |1> I can write it as a matrix [ 0.924 ] [ -0.383i ] But problem for me appears when I do an PauliX transformation [ 0 1 ] [ 1 0 ] and I want to get phi and theta angles from new state. After PauliX new state is: [ -0.383i ] [ 0.924 ] I thought that when matrix is again: [ cos(theta/2) ] [ e^(i*phi)*sin(theta/2) ] then I can take theta from arccos of upper -0.383i (which is 0.75PI) so theta is calculated for me (0.75PI) Then I thought I can divide lower part of matrix by sin(theta/2) to get only e^(i*phi), but when I divide 0.924 by sin(0.75/2 * PI)=0.924 I get "1". As we know e^(i*phi)=cos(phi)*isin(phi). but arccos(1) = 0 Phi cannot be 0, because when I enter these data (theta=45 and phi=270) in some applet on http://www.pha.jhu.edu/~javalab/qubit/qubit.html [Broken] result after PauliX = theta'=135 (which is in fact 0.75PI) and phi'=90 (PI/2). I really don't know how to calculate phi angle. I'm confused. No one can help me. I know that it can be really simple solution to my problem. Could You please help me and answer me how can I get these angles?