Is there a way to write a generic low dimensional Clifford matrix?

Question

Suppose I want to write a general $2\times2$ special unitary matrix in a given basis, I can write it as such: $$\begin{pmatrix} \alpha & -\overline\beta\\ \beta & \overline \alpha\end{pmatrix}$$ with $|\alpha|^2+|\beta|^2=1$. However I do not know if such a form exists for Clifford matrices/gates. Is there such a way to represent $2\times2$ or $4\times 4$ or even higher dimensions Clifford in terms of conditions between its matrix elements?

Answer 1

The unitary group is an infinite group, which is why unitary matrices can be parameterized by continuous parameters. Your representation is incomplete. A $2\times 2$ unitary matrix has four parameters, and one represenation is $$e^{i\phi/2} \begin{pmatrix} e^{i\phi_1/2}\cos\theta & e^{i\phi_2/2}\sin\theta \\ -e^{-i\phi_2/2}\sin\theta & e^{-i\phi_1/2}\cos\theta \end{pmatrix} $$

The Clifford group is a finite, i.e. The $n$-qubit Clifford group has a finite number of elements. The first few such sizes are $$ |C(1)| = 192, \quad |C(2)| = 92160, \quad |C(3)| = 743178240. $$

So, it's not possible to parameterize them with continuous parameters.

If instead, you are interested in generating all possible Clifford gates (for small n), there are algorithms that systematically construct them all from the generators.

Answer 2

For an n-qubit gate, the stabilizer tableau of a clifford operation $C$ maps each Pauli generator $P \in \{X_1,Z_1,X_2,Z_3,\dots,X_n,Z_n\}$ to the conjugated Pauli string $C^{-1}P C$. I like to lay these out into a table where each input generator corresponds to a column and you read down the column to get the output:

$$ \begin{array}{c|cc|cc|cc|c|cc|} &X_1&Z_1&X_2&Z_2&X_3&Z_3&\dots&X_n&Z_n\\ \hline \text{sign}&\pm&\pm&\pm&\pm&\pm&\pm&\dots&\pm&\pm\\ \hline q_1&C_{X_{1,1}}&C_{Z_{1,1}}&C_{X_{1,2}}&C_{Z_{1,2}}&C_{X_{1,3}}&C_{Z_{1,3}}&\dots&C_{X_{1,n}}&C_{Z_{1,n}}\\ q_2&C_{X_{2,1}}&C_{Z_{2,1}}&C_{X_{2,2}}&C_{Z_{2,2}}&C_{X_{2,3}}&C_{Z_{2,3}}&\dots&C_{X_{2,n}}&C_{Z_{2,n}}\\ q_3&C_{X_{3,1}}&C_{Z_{3,1}}&C_{X_{3,2}}&C_{Z_{3,2}}&C_{X_{3,3}}&C_{Z_{3,3}}&\dots&C_{X_{3,n}}&C_{Z_{3,n}}\\ \vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ q_n&C_{X_{n,1}}&C_{Z_{n,1}}&C_{X_{n,2}}&C_{Z_{n,2}}&C_{X_{n,3}}&C_{Z_{n,3}}&\dots&C_{X_{n,n}}&C_{Z_{n,n}}\\ \hline \end{array} $$

A tableau is valid if, for each $k$, the column $C_{X_{\ast,k}}$ anticommutes with the column $C_{Z_{\ast,k}}$, and all other column pairs commute. Note that, for example, this means the signs are totally irrelevant to whether a tableau is valid.

Let's consider single qubit tableaus:

$$ \begin{array}{c|cc|} &X_1&Z_1\\ \hline \text{sign}&\pm&\pm\\ \hline q_1&A&B\\ \hline \end{array} $$

The only condition here is that $A$ anticommutes with $B$. If we define

$$\overline{P} = \text{next}(P) = \begin{cases} X &\rightarrow Y\\ Y &\rightarrow Z\\ Z &\rightarrow X\\ \end{cases}$$

then we can summarize this as requiring that the tableau be of this form:

$$ \begin{array}{c|cc|} &X_1&Z_1\\ \hline \text{sign}&(-1)^a&(-1)^b\\ \hline q_1&A&\overline{A} \oplus A^c\\ \hline \end{array} $$

where $\oplus$ means to multiply while discarding scalar factors, and $a,b,c$ are each set to 0 or to 1.

We can move the $c$ bit dependence into the choice of whether to use $Z_1$ or $Y_1$ as the generator:

$$ \begin{array}{c|cc|} &X_1&Z_1 \oplus X_1^c\\ \hline \text{sign}&(-1)^a&(-1)^b\\ \hline q_1&A&\overline{A}\\ \hline \end{array} $$

If we set all the bits to 0, we get

$$ \begin{array}{c|cc|} &X_1&Z_1\\ \hline \text{sign}&+&+\\ \hline q_1&A&\overline{A}\\ \hline \end{array} $$

which is kinda reminiscent of

$$\begin{bmatrix} a & -\overline{b}\\ b & \overline{a} \end{bmatrix}$$

but is perhaps even more reminiscent of the real and imaginary parts of the single complex entry in a 1x1 unitary matrix having to form a unit vector.

I'm not aware of a "nice" way to phrase the two qubit tableaus in this way, but probably there is one. It may also be the case that breaking the Paulis down into pairs of bits reveals nicer patterns.

Answer 3

I don't know if this is what you would consider a parameterization but here goes:

As you note above every element of $ SU_2 $ can be written $$\begin{pmatrix} \alpha & -\overline\beta\\ \beta & \overline \alpha\end{pmatrix}$$ with $|\alpha|^2+|\beta|^2=1$, in other words all $ (\alpha,\beta) $ on the unit sphere $ S^3 $ in $ \mathbb{C}^2 $. This is the same thing as all $$ a_1I+a_2iZ+b_1XZ+b_2iX $$ with $ (a_1,a_2,b_1,b_2) $ on the unit sphere $ S^3 $ in $ \mathbb{R}^4 $. There are $ 48 $ Clifford gates in $ SU_2 $. We can write them down by the values of $ (a_1,a_2,b_1,b_2)\in S^3 \subset \mathbb{R}^4 $ they correspond to.

$ (1,0,0,0) $ is the identity matrix. This has order $ 1 $ and projective order $ 1 $.

$ (-1,0,0,0) $ is the identity matrix. This has order $ 2 $ and projective order $ 1 $

The $ 6 $ signed permutations of $ (0,1,0,0) $ correspond to the $ 6 $ other determinant $ 1 $ Pauli matrices $ \pm iZ, \pm iX, \pm XZ $. These have order $ 4 $ and projective order $ 2 $.

The $ 12 $ signed permutations of $ (\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0,0) $ for which the first entry is nonzero correspond to the square roots of the Pauli matrices. These have order $ 8 $ and projective order $ 4 $.

The $ 12 $ signed permutations of $ (0,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0) $ for which the first entry is zero correspond to the Hadamard type Cliffords. These have order $ 4 $ and projective order $ 2 $.

The $ 16 $ signed permutations of $ (\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2}) $ correspond to the Cliffords of order divisible by $ 3 $. These all have order $ 3 $ or $ 6 $ and have projective order $ 3 $.

The $ 48 $ determinant $1$ single qubit Cliffords all correspond to $ \mathbb{Z}[\frac{1}{\sqrt{2}}] $ points on the unit sphere $ S^3 $. It is tempting to say that the implication reverses but there are $ \mathbb{Z}[\frac{1}{\sqrt{2}}] $ points on the unit sphere $ S^3 $, like $ (\frac{1}{2},\frac{1}{2},\frac{1}{\sqrt{2}},0) $, which correspond to non-Clifford gate.