## Single-qubit rotations on a subspace within two-qubit unitary

4

I would like to implement the operation $$U(a,b) = \exp\left(i \frac{a}{2} (XX + YY) + i \frac{b}{2} (XY - YX) \right)$$

($$a,b \in \mathbb{R}$$) without using Baker-Campbell-Hausdorf expansion, which at first seems necessary since $$[(XY - YX), (XX + YY)] \neq 0$$. My intuition is that this can be done in the same way that $$\exp(i(aX + bY))$$ does not require a BCH expansion to implement. The above operation is generated by these two matrices:

\begin{align} i \frac{a}{2} (XX + YY)\rightarrow i a \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \\ i \frac{b}{2} (XY - YX)\rightarrow i b\begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & \text{-}i & 0 \\ 0 & i & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}. \end{align}

Since sum of these matrices is proportional to the operator $$(aX - bY)$$ in the $$(|01\rangle,|10\rangle)$$ subspace it seems possible that the operation can be done with a general single-qubit rotation $$\text{R}_\hat{n}$$ in that subspace. Taking the (unnormalized) unit vector $$\hat{n} = a\hat{x} - b\hat{y}$$ this rotation is given by

$$\text{R}_\hat{n} (\theta) = \cos\frac{\theta}{2} + i \sin \frac{\theta}{2} (a X - b Y)$$

so that the operation can be implemented as

$$U(a, b) = \text{CNOT}^{2\rightarrow 1} \text{CR}_{\hat{n}}(\theta)^{1\rightarrow 2} \text{CNOT}^{2\rightarrow 1}$$

where $$\text{CR}_{\hat{n}}(\theta)$$ is a controlled version of $$\text{R}_\hat{n}$$ and $$i\rightarrow j$$ indicates an operation on qubit $$j$$ controlled by qubit $$i$$. My main concern is that since neither $$(XY - YX)$$ nor $$(XX + YY)$$ has support in the $$|00\rangle, |11\rangle$$ subspace that there's something missing or wrong in this process.

My question is, is this a valid decomposition for $$U(a, b)$$ or is there something wrong in the above reasoning?

You can just multiply the 4 by 4 matrices to check your answer. – AHusain – 2021-01-15T00:55:31.787

I don't know the matrix representation of $U(a,b)$ – forky40 – 2021-01-15T00:56:57.530

You can find a matrix representation of a 2-qubit gate using a math software package, e.g. scipy enables you to compute matrix exponentials using scipy.linalg.expm.

– Adam Zalcman – 2021-01-15T17:37:46.810

4

Your approach is correct. In particular, sandwiching a controlled rotation between two CNOT gates is a common technique for implementing rotations on the $$|01\rangle, |10\rangle$$ subspace on hardware that does not implement it natively.

We can justify your approach using the fact that if $$A$$ has eigendecomposition

$$A = \sum_i \lambda_i|i\rangle\langle i|$$

then $$e^A$$ has eigendecomposition

$$e^A = \sum_i e^{\lambda_i}|i\rangle\langle i|.$$

Consequently, if $$A$$ is block diagonal

$$A = \begin{pmatrix} A_1 & & & \\ & A_2 & & \\ & & \ddots & \\ & & & A_k \\ \end{pmatrix}$$

then so is $$e^A$$

$$e^A = \begin{pmatrix} e^{A_1} & & & \\ & e^{A_2} & & \\ & & \ddots & \\ & & & e^{A_k} \\ \end{pmatrix}.$$

In the present case

$$\frac{a}{2}(XX + YY) + \frac{b}{2}(XY - YX) = \begin{pmatrix} 0 & & \\ & aX - bY & \\ & & 0 \end{pmatrix}$$

so

$$\exp\left(i\frac{a}{2}(XX + YY) + i\frac{b}{2}(XY - YX)\right) = \begin{pmatrix} e^0 & & \\ & e^{iaX-ibY} & \\ & & e^0 \end{pmatrix} = \begin{pmatrix} 1 & & \\ & R_{\hat n}(2t) & \\ & & 1 \end{pmatrix}$$

where $$n$$ is the normalized real 3-vector $$(-a, b, 0) / t = (-\alpha, \beta, 0)$$, $$\hat n = \beta Y-\alpha X$$ and $$t = \|n\|_2 = \sqrt{a^2 + b^2}$$.

In particular, we do not need to be concerned about the subspace $$|00\rangle, |11\rangle$$ because it is the eigenspace of the operator in the exponent associated with eigenvalue zero which means that $$U(a, b)$$ acts on it as identity.

Note that in practice it is sometimes possible to avoid using Baker-Campbell-Hausdorff by bringing the terms of the exponent into a form in which they commute, e.g. by regrouping the terms

\begin{align} U(a,b) &= \exp\left(i \frac{a}{2} (XX + YY) + i \frac{b}{2} (XY - YX) \right) \\ &= \exp\left(\frac{i}{2} X\otimes(aX + bY) + \frac{i}{2} Y\otimes(aY - bX) \right) \\ \end{align}

where $$X\otimes(aX+bY)$$ and $$Y\otimes(aY-bX)$$ commute. Therefore,

\begin{align} U(a, b) &= \exp\left(\frac{i}{2} X\otimes(aX + bY)\right) \exp\left(\frac{i}{2} Y \otimes(aY - bX)\right) \\ &= \exp\left(\frac{it}{2} X\otimes(\alpha X + \beta Y)\right) \exp\left(\frac{it}{2} Y\otimes(\alpha Y - \beta X)\right) \end{align}

where $$t = \sqrt{a^2 + b^2}, \alpha = \frac{a}{t},\beta = \frac{b}{t}$$ as before. Notice that $$[X\otimes(\alpha X + \beta Y)]^2 = I$$ and $$[Y\otimes(\alpha Y - \beta X)]^2 = I$$ so

$$\exp\left(\frac{it}{2} X\otimes(\alpha X + \beta Y)\right) = I \cos\frac{t}{2} + i X\otimes(\alpha X + \beta Y) \sin\frac{t}{2} \\ \exp\left(\frac{it}{2} Y\otimes(\alpha Y - \beta X)\right) = I \cos\frac{t}{2} + i Y\otimes(\alpha Y - \beta X) \sin\frac{t}{2}$$

(c.f. equation $$(4.7)$$ on p.175 in Nielsen & Chuang). Thus,

$$U(a, b) = II \cos^2\frac{t}{2} + ZZ\sin^2\frac{t}{2} + i\left(\frac{\alpha}{2}(XX + YY) + \frac{\beta}{2}(XY - YX)\right)\sin t.$$

In matrix representation

$$U(a, b) = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & \cos t & (-\beta + i\alpha)\sin t & 0 \\ 0 & (\beta + i\alpha)\sin t &\cos t & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}$$

where we recognize the middle $$2\times 2$$ block as $$R_{\hat n}(2t)$$ with $$\hat n = \beta Y-\alpha X$$ as before.