I'm trying to reconcile (i) the statement that swapping two subsystems constitutes a rotation by $2\pi$ and (ii) the angle that is implied by the Hermitian generator of a SWAP gate.

I haven't tracked down an explicit statement of (i) but I think the idea is, using Dirac's "belt trick" one can show that the ends of a belt must be swapped twice to remove $4\pi$ worth of twist (rotations) applied to the belt buckle. So for Fermions (antisymmetric with respect to particle exchange) the SWAP of two subsystems results in a $2\pi$ rotation and some relative phase between the systems.

The statement (ii) is just that $$ \text{SWAP} = \exp (i \frac{\pi}{4} (XX + YY + ZZ)) $$

where $X,Y,Z$ are pauli operators. By analogy with the relationship between $SU(2)$ and $SO(3)$ this seems to imply a rotation of some kind, but I'm not confident in this analogy. So my questions are

  1. Does the analogy for understanding single-qubit rotation gates in terms of $SO(3)$ extend to understanding multi-qubit gates in terms of some other rotation group?

  2. If not, how should I understand the $\text{SWAP}$ gate in terms of rotations in the context of quantum computing (since there does seem to be a physics explanation connecting the two)?


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.

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.

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

$$ R_{\hat n}(\alpha) = \exp\left(-i\frac{\alpha}{2}(n_x X + n_y Y + n_z Z)\right) = I\cos\frac{\alpha}{2} -i (n_x X + n_y Y + n_z Z)\sin\frac{\alpha}{2} $$

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

$$ \exp\left(i\frac{\beta}{2} A\right) = I \cos\frac{\beta}{2} + i A \sin\frac{\beta}{2}\tag1 $$

where $\beta\in\mathbb{R}$ 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

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

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 $\exp\left(i\frac{\theta}{4}\right)$ to obtain

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

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

$$ \text{SWAP}(\theta) = II \cos\frac{\theta}{2} + i \frac{II + XX + YY + ZZ}{2} \sin\frac{\theta}{2}. $$

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 $2\pi$ yields $-I$ and a rotation by $4\pi$ returns to $I$, just as expected from the "belt trick". Note however that the full SWAP takes place for $\theta=\pi$ and $\theta=3\pi$, not $2\pi$. This is analogous to how the full bit-flip is effected by $R_X(\alpha)$ for $\alpha=\pi$ and $3\pi$.

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

$$ \text{SWAP}(\theta) = \begin{pmatrix} e^{i\frac{\theta}{2}} & & & \\ & \cos\frac{\theta}{2} & i\sin\frac{\theta}{2} & \\ & i\sin\frac{\theta}{2} & \cos\frac{\theta}{2} & \\ & & & e^{i\frac{\theta}{2}} \\ \end{pmatrix} $$

and we recognize the middle $2\times 2$ block as $R_X(-\theta) \in SU(2)$. In other words, if we relabel the North Pole of the Bloch sphere to $|10\rangle$ and the South Pole to $|01\rangle$ then the action of $\text{SWAP}(\theta)$ on the $\mathrm{span}(|01\rangle, |10\rangle)$ 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 $\mathrm{span}(|01\rangle, |10\rangle)$ and $\mathrm{span}(|00\rangle, |11\rangle)$.


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