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

2

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?

Shivashriganesh Mahato

Posted 2019-08-17T06:46:21.323

Reputation: 21

The page you're referencing shows as "not found"; https://nbviewer.jupyter.org/github/Qiskit/qiskit-tutorials/blob/master/qiskit/advanced/terra/summary_of_quantum_operations.ipynb seems to be the closest one? That page has a different expression

– Mariia Mykhailova – 2019-08-17T16:03:21.130

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 – 2019-08-18T03:53:09.437

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 – 2019-08-18T15:21:13.213

Answers

1

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 $U\sim e^{i\alpha}U$.

glS

Posted 2019-08-17T06:46:21.323

Reputation: 12 247