We need the diffusion operator $D = 2 \vert \psi\rangle \langle \psi\vert - I$ to get the ball rolling. You can see how it works geometrically in the next exercise. --- ***Exercise G.2.1.*** Show that $D$ flips a vector representing the state $\vert\phi\rangle$ of the system *around* the axis of $\vert \psi\rangle$, represented by the purple arrow in the image below:

Hint.

Use geometry! To start with, suppose the vector representing the state of the system $\vert\phi\rangle$ makes an angle $\pi/2 - \varphi$ to $\vert \psi\rangle$. You can assume that $-D = I - 2\vert \psi\rangle\langle\psi\vert $ reflects parallel to $\vert \psi\rangle$, and go from there.

Solution.

Applying $-D$ reflects across the dotted red line orthogonal to $\vert \psi\rangle$. This gives a total rotation by $2\varphi$, as shown below middle:

We now apply $-I$, so that the total effect is $-D \cdot -I = D$. This reflects the vector to the other side of the circle. But it's easy to see that the angle from the $\vert \psi\rangle$ axis is now $$ \pi - 2\varphi - \left(\frac{\pi}{2} - \varphi\right) = \frac{\pi}{2} - \varphi, $$ so that the final vector is indeed flipped around the $\vert \psi\rangle$ axis, i.e., makes the same angle as the initial vector but on the other side. ▢

--- ***Exercise 5.2.2. (Bonus)*** If you prefer algebra, show that the overlap of the state of the system $\vert\phi\rangle$ with $\vert \psi\rangle$ is unchanged after applying $D$, but reflected in the orthogonal subspace.

Solution.

Algebraically, let's write our arbitrary state $\vert \phi\rangle = \alpha \vert \psi\rangle + \beta \vert \phi_\perp\rangle$, where $\vert \phi_\perp\rangle$ is the projection onto the subspace orthogonal to $\vert \psi\rangle$. Then $$ D\vert \phi\rangle = \alpha (2 \vert \psi\rangle\langle \psi\vert - I)\vert \psi\rangle + \beta (2 \vert \psi\rangle\langle \psi\vert - I)\vert \phi_\perp\rangle = \alpha \vert \psi\rangle - \beta \vert \phi_\perp\rangle. $$ Hence, $\langle\psi\vert D\vert \phi\rangle = \alpha = \langle\psi\vert \phi\rangle$, while $\langle\phi_\perp\vert D\vert \phi\rangle = -\beta = -\langle\phi_\perp\vert \phi\rangle$. So the overlap is unchanged, and state is reflected in the orthogonal subspace, as required. ▢

Codercise G.2.1 - Rotating the state using the Grover operator

The oracle as a reflection

The geometry of the diffusion operator

The combined geometric picture