Can we understand multi-qubit gates in terms of rotation groups?

Extending the relationship between $SU(2)$ and $SO(3)$ to higher dimensions

The analogy for understanding single-qubit gates in terms of $SO(3)$ is provided by an accidental isomorphism $Spin(3)cong SU(2)$ where the spin group $Spin(n)$ is the double cover of $SO(n)$. As the name suggests, the isomorphism is not part of a recurring pattern, so the answer to the first question is negative in general.

However, $Spin(3)cong SU(2)$ isn't the only accidental isomorphism. Other interesting examples include $Spin(4) cong SU(2) times SU(2)$ which is related to the fact any 4D real rotation can be described by two quaternions (one acting by left-multiplication and the other by right-multiplication) and $Spin(6) cong SU(4)$ (see this answer for details of this isomorphism). Nevertheless, real rotations in 4D and 6D are not as intuitive as those in 2D and 3D. A better approach to gaining intuitive understanding of multi-qubit gates is implicit in the second question: identify an interesting type of gate which is easier to describe.

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General remark about understanding unitary gates as rotations

A key realization that aids in understanding quantum gates in terms of rotations is that all operators in $SU(n)$ can be diagonalized in $mathbb{C}^n$, but a generic element of $SO(n)$ cannot be diagonalized in $mathbb{R}^n$. A geometric consequence of this fact is that a generic element of $SO(n)$ changes direction of some basis vectors. By contrast, in the appropriate basis the action of a unitary operator is the multiplication of all vector components by various scalar phase factors. In a sense, the action of the operator is decomposed into 2D rotations in the field of complex numbers.

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Origin of the $pi/4$ angle in the formula for SWAP

Perhaps the most familiar example of a unitary gate understood via its counterpart in $SO(3)$ is the single-qubit rotation

$$ R_{hat n}(alpha) = expleft(-ifrac{alpha}{2}(n_x X + n_y Y + n_z Z)right) = Icosfrac{alpha}{2} -i (n_x X + n_y Y + n_z Z)sinfrac{alpha}{2} $$

where $hat n = (n_x, n_y, n_z)$ is a real 3-vector of unit length. This can be generalized as

$$ expleft(ifrac{beta}{2} Aright) = I cosfrac{beta}{2} + i A sinfrac{beta}{2}tag1 $$

where $betainmathbb{R}$ and $A$ is a matrix such that $A^2 = I$ (see exercise 4.2. on p.175 in section 4.2 of Nielsen & Chuang). This formula has the benefit of making it clear which values of the angle $beta$ correspond to the identity and to the unitary $A$.

Now, before we write the SWAP gate in the form of $(1)$ let us first generalize it to a single-parameter group

$$ text{SWAP}(theta) = expleft(i frac{theta}{4} (XX + YY + ZZ)right). $$

Unfortunately, the naive attempt of writing the gate as $(1)$ by setting $A = XX + YY + ZZ$ fails because as we see from the multiplication table

$$ begin{array}{c|ccc} & XX & YY & ZZ hline XX & II & -ZZ & -YY YY & -ZZ & II & -XX ZZ & -YY & -XX & II end{array} $$

the non-identity terms in $(XX + YY + ZZ)^2$ do not cancel. However, since the global phase has no physical meaning, we can multiply any unitary matrix by a phase factor without changing the corresponding quantum gate. In our case, we multiply by $expleft(ifrac{theta}{4}right)$ to obtain

$$ text{SWAP}(theta) = expleft(i frac{theta}{4} (II + XX + YY + ZZ)right) $$

with exponent that squares to a multiple of identity as can be seen from the new multiplication table

$$ begin{array}{c|cccc} & II & XX & YY & ZZ hline II & II & XX & YY & ZZ XX & XX & II & -ZZ & -YY YY & YY & -ZZ & II & -XX ZZ & ZZ & -YY & -XX & II end{array} $$

The table shows that $(II + XX + YY + ZZ)^2 = 4II$ and so

$$ text{SWAP}(theta) = II cosfrac{theta}{2} + i frac{II + XX + YY + ZZ}{2} sinfrac{theta}{2}. $$

This explains why $pi$ is divided by $4$ in the exponential formula given in the question. One factor of $2$ normalizes $II + XX + YY + ZZ$ and the other factor of $2$ comes from formula $(1)$. Consequently, $text{SWAP}(theta)$ behaves analogously to $R_{hat n}(alpha)$: rotation by $2pi$ yields $-I$ and a rotation by $4pi$ returns to $I$, just as expected from the "belt trick". Note however that the full SWAP takes place for $theta=pi$ and $theta=3pi$, not $2pi$. This is analogous to how the full bit-flip is effected by $R_X(alpha)$ for $alpha=pi$ and $3pi$.

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Limited application of the Bloch sphere to multi-qubit gates

Another way of understanding gates in terms of rotations is available when the gate in question acts non-trivially on a subspace of dimension two. In this case, we can use the Bloch sphere to get a limited intuitive picture of the action of the gate. For example, $text{SWAP}(theta)$ can be written as

$$ text{SWAP}(theta) = begin{pmatrix} e^{ifrac{theta}{2}} & & & & cosfrac{theta}{2} & isinfrac{theta}{2} & & isinfrac{theta}{2} & cosfrac{theta}{2} & & & & e^{ifrac{theta}{2}} end{pmatrix} $$

and we recognize the middle $2times 2$ block as $R_X(-theta) in SU(2)$. In other words, if we relabel the North Pole of the Bloch sphere to $|10rangle$ and the South Pole to $|01rangle$ then the action of $text{SWAP}(theta)$ on the $mathrm{span}(|01rangle, |10rangle)$ subspace is the same as the action of $R_X(-theta)$ on a qubit. Note however, that this interpretation ignores the change in relative phase imparted by the gate between $mathrm{span}(|01rangle, |10rangle)$ and $mathrm{span}(|00rangle, |11rangle)$.