# Quantum computing - bloch sphere coordinates

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?

Last edited by a moderator:

## Answers and Replies

DrClaude
Mentor
I thought that when matrix is again:
[ cos(theta/2) ]
[ e^(i*phi)*sin(theta/2) ]
In that case, you are imposing that the coefficient in front of ##| 0 \rangle## is real. Since the result you got is imaginary, you need to take out a complex phase from such that the coefficient will be real, i.e.,
$$c_0 |0 \rangle + c_1 |1 \rangle = e^{i \alpha} \left( \tilde{c}_0 |0 \rangle + \tilde{c}_1 |1 \rangle \right)$$
where ##\alpha## is chosen such that ##\tilde{c}_0## is real. The angle ##\phi## can then be found from ##\tilde{c}_1##, not ##c_1##.