# How to find the required rotation on the Bloch sphere, knowing the start and end

I'm trying to figure out the following situation. Say we have a Bloch sphere with $|g\rangle$ on the positive z-axis and $|e\rangle$ on the negative z-axis. The state is initially in $|g\rangle$, but undergoes a $\pi/2$ rotation about some axis, and ends up in $\frac{|g\rangle+|e\rangle}{\sqrt{2}}$. Now, I want to find the operator $\hat{R}$ that performs this rotation.

I'm not exactly sure how, but I started thinking along the lines of a general pure qubit state. In this case it is given by $\cos({\theta/2})|g\rangle + e^{i \phi}\sin({\theta/2})|e\rangle$. So we begin with $\theta = 0$, and after the rotation we have $\theta = \pi/2$, $\phi = 0$. As we start parallel to the z-axis, I suppose the rotation is about a combination of the x and the y axis. But I don't really see how to work that out. I also know that $R(\theta)_{x,y} = \cos({\theta/2})*1 - i \sin({\theta/2}) X,Y$ where the 1 is the identity matrix and $X,Y$ the pauli matrices. So its some combination of these two operators, that lead to the state I want. I just can't figure out how to construct it, and I was wondering if you could help me with this.

What you need is spherical linear interpolation, which is usually done with quaternions. Well, quaternions can be represented by the pauli matrices. If you don't mind the hand-waviness, a rotation operator about the axis $\vec{s}$ is given by $$R = e^{i \frac{\theta}{2} \vec{s} \cdot \vec{\sigma}}$$.

The vector $\vec{s}$ is perpendicular to the great circle that passes between the two points. You could then find the required angle with a bit of spherical trigonometry.

Edit:

Let $R_e$ be the rotation that takes $\left|0\right\rangle$ to $\left|e\right\rangle$, and similarly $R_g$ takes $\left|0\right\rangle$ to $\left|g\right\rangle$.Then $R_e R_g^\dagger$ takes $\left|g\right\rangle$ to $\left|e\right\rangle$. Suppose I write a qubit as

$$\left|\psi\right\rangle = e^{-i \frac{\phi}{2}} \cos\left(\frac{\theta}{2}\right) \left|0\right\rangle + e^{i \frac{\phi}{2}} \sin\left(\frac{\theta}{2}\right) \left|1\right\rangle$$ I can also write this as $$\left|\psi\right\rangle = \frac{1}{2}\begin{pmatrix} e^{ -i \frac{\phi}{2} } & 0 \\ 0 & e^{ +i \frac{\phi}{2} }\end{pmatrix} \begin{pmatrix} e^{ -i \frac{\theta}{2} } & -e^{ -i \frac{\theta}{2} } \\ e^{ +i \frac{\theta}{2} } & e^{ +i \frac{\theta}{2} }\end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix}$$.

Edit 2: To answer your actual question :) in your case, $$\left|\psi\right\rangle = \frac{1}{2}\begin{pmatrix} e^{ -i \frac{\theta}{2} } & -e^{ -i \frac{\theta}{2} } \\ e^{ +i \frac{\theta}{2} } & e^{ +i \frac{\theta}{2} }\end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix}$$ which you can easily write in terms of Pauli matrices.

• Hm, I see. It does make sense, but I'll have to think about for a bit to figure out how to actually construct it. Commented Nov 23, 2014 at 16:36
• Do let me know if it ends up being bunk :) Commented Nov 23, 2014 at 19:15
• I suppose I should have been more clear in the main post, but g is actually what is normally 0 and e is what is normally 1, on the Bloch sphere. I'm working with a two level emitter, hence the unusual notation. Commented Nov 23, 2014 at 20:14

This is more of a comment, but my rep is too low... You say,

The state is initially in $|g\rangle$, but undergoes a $\pi/2$ rotation about some axis...

and you end up with $\theta=\pi/2$, $\phi=0$, ie you are on the $x$-axis and your rotation was simply about $y$, surely?

• Well I'd agree, if it wasn't for the fact that this rotation doesn't give me the desired result Commented Nov 23, 2014 at 20:12