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

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 $$\mathrm{span}(|00\rangle, |11\rangle)$$.