# Does the general form of a unitary operator define strict signs for the second column?

As per IBM's documentation for quantum circuits, the general unitary operator is defined as:

$$\hat{U}=\begin{bmatrix}\cos(\frac{\theta}{2})&-e^{i\lambda}\sin(\frac{\theta}{2})\\e^{i\phi}\sin(\frac{\theta}{2})&e^{i\phi+i\lambda}\cos(\frac{\theta}{2})\end{bmatrix}$$

For a better understanding I was working out the algebra with the given constraints ($$\hat{U}^{\dagger}\hat{U}$$, $$0\leq\theta\leq\pi$$, and $$0\leq\phi\leq2\pi$$), but I didn't find any step that required a constraint on the sign of $$a$$ and $$b$$ (rows 1 and 2 of the second column, respectively) such that $$a$$ is negative and $$b$$ is positive. That said, is it not true that the true general form is:

$$\hat{U}=\begin{bmatrix}\cos(\frac{\theta}{2})&\pm e^{i\lambda}\sin(\frac{\theta}{2})\\e^{i\phi}\sin(\frac{\theta}{2})&\mp e^{i\phi+i\lambda}\cos(\frac{\theta}{2})\end{bmatrix}$$

or am I missing some constraint?

• The page you're referencing shows as "not found"; nbviewer.jupyter.org/github/Qiskit/qiskit-tutorials/blob/master/… seems to be the closest one? That page has a different expression – Mariia Mykhailova Aug 17 '19 at 16:03
• Hmm that's odd. But yes that link is what I was looking for. I wrote it slightly incorrectly in my question, it has been updated however – Shivashriganesh Mahato Aug 18 '19 at 3:53
• What is the range on $\lambda$ you are imposing? Can you replace $\lambda \to \lambda + \pi$? If you can, that gets rid of the signs. – AHusain Aug 18 '19 at 15:21

See this other answer of mine for a full derivation of the general form of a unitary $$2\times2$$ matrix.

As shown there, unitary matrices (notice I'm not imposing a constraint on the determinant here) can be written as $$U=\begin{pmatrix}e^{i\alpha_{11}}\cos\theta& e^{i\alpha_{12}}\sin\theta\\ e^{i\alpha_{21}}\sin\theta & e^{i\alpha_{22}}\cos\theta \end{pmatrix},\tag A$$ with the parameters $$\alpha_{ij}$$ satisfying the condition $$\alpha_{11}-\alpha_{12}=\alpha_{21}-\alpha_{22}+\pi.\tag B$$

There are then many different ways to choose how to write these coefficients. This freedom mostly arises from the fact that a global phase change of the matrix doesn't change anything in the physics, and thus $$U\simeq e^{i\phi}U$$ for any $$\phi\in\mathbb R$$.

One relatively standard way to fix a notation is to fix the determinant. In the above notation, this determinant reads $$\det U=e^{i(\alpha_{11}+\alpha_{22})}$$, where I've used (B) and standard trigonometry to simplify the coefficients. Imposing $$\det U=1$$ thus corresponds to the constraint $$\alpha_{11}+\alpha_{22}=0$$ (or $$2k\pi$$ for $$k$$ integer of course, but nothing changes if we pick $$k>0$$ here). Using this again in (B) then also implies $$\alpha_{21}+\alpha_{12}=-\pi$$.

With this added constraint, the $$U$$ now reads $$U=\begin{pmatrix}e^{i\alpha_{11}}\cos\theta& e^{i\alpha_{12}}\sin\theta\\ -e^{-i\alpha_{12}}\sin\theta & e^{-i\alpha_{11}}\cos\theta \end{pmatrix}.$$ Equivalently, I can say that a general SU(2) matrix has the form $$U=\begin{pmatrix}x& y\\ -\bar y & \bar x\end{pmatrix},\tag C$$ with $$x,y\in\mathbb C$$ satisfying $$|x|^2+|y|^2=1$$.

Now, if I want to describe a general unitary matrix, and not just one with unit determinant, I can simply write $$U=e^{i\phi}(e^{-i\phi}U)$$ where $$\det U=e^{2i\phi}$$ (remember that the determinant of a unitary matrix is always a phase). Then, $$\tilde U\equiv e^{-i\phi}U$$ has unit determinant (remember that $$\det(\lambda U)=\lambda^2 \det U$$), and can thus be written as per (C).

Putting the above results together, we conclude that a generic unitary matrix has the form $$e^{i\varphi}\begin{pmatrix}x & y\\ -\bar y& \bar x\end{pmatrix},\tag D$$ with the additional constraint $$|x|^2+|y|^2=1$$, and arbitrary phase $$\varphi\in\mathbb R$$.

What is nice about (D) is that it gives us a pretty straightforward recipe to figure out how the minuses and other phases should be placed: simply put out phases from the matrix in order to make it into a unit-determinant one, and then verify that what is left looks like (C).

Applying this to your example, we get $$\begin{pmatrix}\cos(\frac{\theta}{2})&-e^{i\lambda}\sin(\frac{\theta}{2})\\e^{i\phi}\sin(\frac{\theta}{2})&e^{i\phi+i\lambda}\cos(\frac{\theta}{2})\end{pmatrix}= e^{i(\phi+\lambda)/2} \begin{pmatrix}e^{-i(\phi+\lambda)/2}\cos(\frac{\theta}{2})&-e^{i(-\phi+\lambda)/2}\sin(\frac{\theta}{2})\\e^{i(\phi-\lambda)/2}\sin(\frac{\theta}{2})&e^{i(\phi+\lambda)/2}\cos(\frac{\theta}{2})\end{pmatrix},$$ which therefore looks like it should. It is also worth noting here that this is not a general parametrisation of a $$2\times 2$$ unitary matrix. Global phase aside, a general parametrisation involves three parameters, not two. It is, however, a general parametrisation of unit determinant unitaries, identifying $$U\sim e^{i\alpha}U$$.