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

Question

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.

Answer 1

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.

Answer 2

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?