Proof of the 4.11 exercise in the Nielsen & Chuang book

Question

In question 4.11 in Nielsen and Chuang's book, it states that there is a formula to describe any unitary matrix $U$ with two vectors $\vec{n}$ and $\vec{m}$ in the following way:

$$U=\exp(i \alpha) R_n(\beta_1)R_m(\gamma_1)R_n(\beta_2)R_m(\gamma_2) ...$$

(see this question)

How can we prove this formula? I would say writing $\vec{n}$ and $\vec{m}$ as a decomposition in the $\{X, Y, Z\}$ basis.

Or also, maybe we can think that two unitary vectors that are not parallel are enough to describe any rotation in the Hilbert space (but I don't find this explanation rigorous enough).

Or maybe, can we use something more theoretical like the Lie Group ?

Answer 1

One way that I like to do this is as follows: $$ R_n(\pi)R_m(\gamma)R_n(\pi)R_m(-\gamma) $$ defines a rotation about some axis $n^\perp$ that is orthogonal to $n$. So long as we pick $\gamma$ such that the rotation angle is an irrational multiple of $\pi$, repeated application can give us any arbitrary angle of rotation about that axis.

So, now we have rotations by arbitrary angles about two orthogonal axes, $n$ and $n^\perp$, which we know can decompose an arbitrary single-qubit unitary. That means, overall, we have the required decomposition.

Proof of the first claim: it must be that $R_n(\pi)R_m(\gamma)R_n(\pi)R_m(-\gamma)=R_q(\theta)$, so I just want to prove that $n\cdot q=0$. This is equivalent to proving that $$ \text{Tr}(R_q(\theta) n\cdot\sigma)=0. $$ But this I can do with some manipulation of trace: \begin{align*} \text{Tr}(R_q(\theta) n\cdot\sigma)&=\text{Tr}(R_n(\pi)R_m(\gamma)R_n(\pi)R_m(-\gamma) n\cdot\sigma) \\ &=-\text{Tr}(n\cdot\sigma R_m(\gamma)n\cdot\sigma R_m(-\gamma) n\cdot\sigma) \\ &=-\text{Tr}(R_m(\gamma)n\cdot\sigma R_m(-\gamma)) \\ &=-\text{Tr}(n\cdot\sigma) \\ &=0. \end{align*}

Proof of second claim: I'm relying on the fact that we already know (e.g. Nielsen & Chuang) that any single-qubit $U$ can be written as $$ U=R_z(\alpha)R_X(\beta)R_z(\gamma) $$ up to a global phase. Now, let's introduce a $V$ such that $$ V(n\cdot\sigma)V^\dagger=Z,\qquad V(n^\perp\cdot\sigma)V^\dagger=X. $$ This means that $$ V^\dagger R_z(\alpha)R_X(\beta)R_z(\gamma)V=R_n(\alpha)R_{n^\perp}(\beta)R_n(\gamma). $$ So, if I want to design a particular unitary $U'$, I define $U=VU'V^\dagger$, solve for the angles $\alpha,\beta,\gamma$ in the X/Z decomposition, and the same angles will create $U'$ with the $n^\perp/n$ decomposition.

Answer 2

For an arbitrary unitary operator $U$ one has (theorem 4.1 in Nielsen and Chuang):

$$U=e^{i\alpha}R_z(\beta)R_y(\gamma)R_z(\delta)$$

We are asked to write this as:

$$U=e^{i\beta}R_{\hat n}(\beta_1)R_{\hat m}(\gamma_1)R_{\hat n}(\beta_2)R_{\hat m}(\gamma_2)...$$

I will prove:

$$U=e^{i\beta}R_{z}(\beta_1)R_{\hat m}(\gamma_1)R_{z}(\beta_2)R_{\hat m}(\gamma_2)...$$

Using the following lemma:

Given $\hat m=(m_x, m_y, m_z)$ such that $\hat m \cdot z \ne 0$ and $0 \lt \theta \lt 2\arcsin(\sqrt{m_x^2+m_y^2})$, there exists $\beta$, $\delta$, and $\gamma$ such that:

$$R_z(\beta)R_{y}(\gamma)R_z(\delta)=R_{\hat m}(\theta)$$

This is proven as follows:

One has:

$$R_z(\beta)R_{y}(\gamma)R_z(\delta)=\left[\cos(\beta/2)I-i\sin(\beta/2)Z\right]\times\left[\cos(\gamma/2)I-i\sin(\gamma/2)Y\right]\times\left[\cos(\delta/2)I-i\sin(\delta/2)Z\right]$$

and $$R_{\hat m}(\theta)=\cos(\theta/2)I-i\sin(\theta/2)(m_xX+m_yY+m_zZ)$$

Both expressions can be expanded out into terms proportional to $I$, $X$, $Y$, $Z$ and like terms equated, giving:

I terms: $\cos(\gamma/2)\cos(\beta/2+\delta/2)=\cos(\theta/2)$

Z terms: $\cos(\gamma/2)\sin(\beta/2+\delta/2)=\cos(\theta/2)m_z$

X terms: $-\sin(\gamma/2)\sin(\beta/2-\delta/2)=\sin(\theta/2)m_x$

Y terms: $\sin(\gamma/2)\cos(\beta/2-\delta/2)=\sin(\theta/2)m_y$

Squaring and adding the last two expressions gives:

$$\sin^2(\gamma/2)=\frac{\sin^2(\theta/2)}{(m_x^2+m_y^2)}$$

And thus:

$$\gamma/2=\arcsin\left(\frac{\sin(\theta/2)}{\sqrt{m_x^2+m_y^2}}\right)$$

Given the provided limits on $\theta$, $\gamma$ will have a solution. The lemma is completed by using the second and third equations (Z terms and X terms) to solve for $\beta/2+\delta/2$ and $\beta/2-\delta/2$ which are then easily solved to find $\beta$ and $\delta$ as desired.

To complete the proof, one notes that for a given $\hat m$ one can partition $R_y(\gamma)$ as follows:

$$R_y(\gamma) = R_y(\gamma/N)^N = (R_z(-\beta)R_{\hat m}(\theta(N))R_z(-\delta))^N$$

Where N is chosen to be large enough that the second equality holds.

Thus on has:

$$U=e^{i\alpha}R_z(\beta_1)R_y(\gamma)R_z(\delta_{N+1})=e^{i\alpha}R_z(\beta_1)R_y(\gamma/N)^NR_z(\delta_{N+1})=e^{i\alpha}R_z(\beta_1)(R_z(-\beta)R_{\hat m}(\theta(N))R_z(-\delta))^NR_z(\delta_{N+1})$$

Which has the desired form. I will only make a hand-waving argument that a global redefinition of the coordinate frame allows one to replace $R_z$ with an arbitrary axis $R_{\hat n}$

Answer 3

Here's another proof that does not rely too much on geometric intuition but in algebraic properties of SU(2). Any U(2) matrix can be written, ignoring a not measurable global phase, as \begin{equation} U = R_z(\beta)R_y(\gamma)R_z(\delta). \end{equation} One wants to write this $U$ as a finite set of rotations under two non-orthogonal axis $\hat{n}$ and $\hat{m}$. This is easy if one uses the group composition law of SU(2): \begin{equation} R_\hat{k}(c)= R_\hat{n}(a)R_\hat{m}(b), \end{equation} where in the link one can see the substitution of $\hat{k}$ and $c$ in terms of $a,b,\hat{n},\hat{m}$. One can invert this relations to play the other game around, e.g. choose $\hat{k}$, $c$ and find $a,b$ ($\hat{n},\hat{m}$ are known already in section 4.5.3 of Nielsen). Hence, it is straightforward to write \begin{equation} U =R_\hat{n}(\beta_1)R_\hat{m}(\gamma_1)R_\hat{n}(\beta_2)R_\hat{m}(\gamma_12)R_\hat{n}(\beta_3)R_\hat{m}(\gamma_3), \end{equation} as requested.