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## Main Question or Discussion Point

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?

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?

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