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?

forky40

Posted 2021-01-15T00:19:53.463

Reputation: 1 518

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

Answers

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.

Adam Zalcman

Posted 2021-01-15T00:19:53.463

Reputation: 4 307