General form for 2 x 2 unitary matrices

[jimmycricket]

I'm trying to show that any unitary matrix may be written in the form \begin{pmatrix}e^{i\alpha_1}\cos{\theta} & -e^{i\alpha_2}\sin{\theta}\\ e^{i\alpha_3}\sin{\theta} & e^{i\alpha_4}\cos{\theta}\end{pmatrix}

Writing the general form of a unitary matrix as
U=\begin{pmatrix} u_{11} & u_{12}\\ u_{21} & u_{22}\end{pmatrix}

gives

U^{\dagger}U=<br /> \begin{pmatrix}u_{11}^* &amp; u_{21}^*\\u_{12}^* &amp; u_{22}^*\end{pmatrix}\begin{pmatrix} u_{11} &amp; u_{12}\\ u_{21} &amp; u_{22}\end{pmatrix}=\begin{pmatrix}1 &amp; 0\\ 0 &amp; 1\end{pmatrix}

\Longrightarrow |u_{11}|^2+|u_{21}|^2=1 \:\:,\:\: |u_{12}|^2 + |u_{22}|^2=1\\<br /> <br /> \Longrightarrow |u_{11}|=\cos(\theta) \:\:,\:\: |u_{21}|=\sin(\theta) \:\:,\:\: |u_{12}|=\cos(\varphi) \:\:,\:\: |u_{22}|=\sin(\varphi)
for some \theta , \varphi

I'm not really sure where to go from here.

 

[vanhees71]

Don't be too quick :-). First write down all equations, and not only the diagonal ones.

 

[Orodruin]

I'm trying to show that any unitary matrix may be written in the form \begin{pmatrix}e^{i\alpha_1}\cos{\theta} &amp; -e^{i\alpha_2}\sin{\theta}\\ e^{i\alpha_3}\sin{\theta} &amp; e^{i\alpha_4}\cos{\theta}\end{pmatrix}

Writing the general form of a unitary matrix as
U=\begin{pmatrix} u_{11} &amp; u_{12}\\ u_{21} &amp; u_{22}\end{pmatrix}

gives

U^{\dagger}U=<br /> \begin{pmatrix}u_{11}^* &amp; u_{21}^*\\u_{12}^* &amp; u_{22}^*\end{pmatrix}\begin{pmatrix} u_{11} &amp; u_{12}\\ u_{21} &amp; u_{22}\end{pmatrix}=\begin{pmatrix}1 &amp; 0\\ 0 &amp; 1\end{pmatrix}

\Longrightarrow |u_{11}|^2+|u_{21}|^2=1 \:\:,\:\: |u_{12}|^2 + |u_{22}|^2=1\\<br /> <br /> \Longrightarrow |u_{11}|=\cos(\theta) \:\:,\:\: |u_{21}|=\sin(\theta) \:\:,\:\: |u_{12}|=\cos(\varphi) \:\:,\:\: |u_{22}|=\sin(\varphi)
for some \theta , \varphi

I'm not really sure where to go from here.

You have so far only used the conditions on the diagonal. You need to also fulfill the off-diagonal elements equal to zero.

Suggestion: Call $|u_{22}| =\cos\varphi$ instead of the 12 element. It makes no difference but will simplify your math slightly.

 

[jimmycricket]

Ok so for the orthogonal elements:
u_{11}^*u_{12}+u_{21}^*u_{22}=0
u_{12}^*u_{11}+u_{22}^*u_{21}=0

 

[Orodruin]

So what do you get when you insert into these the things you have already found? (Note that they are both equivalent by simple complex conjugation.)

 

[jimmycricket]

\cos(\theta)\sin(\varphi)+\sin(\theta)\cos(\varphi)=0
using your suggestion of u_{22}=\cos(\varphi)

 

[Orodruin]

\cos(\theta)\sin(\varphi)+\sin(\theta)\cos(\varphi)=0
using your suggestion of u_{22}=\cos(\varphi)

Well, almost, you forgot the phases. Once you have taken them into account, I suggest using some trigonometric identities.

 

[jimmycricket]

After using the product formula and some simplification I get

\frac{1}{2}\big{[}\sin(\theta+\varphi)(e^{i(\alpha_1+\alpha_2)}+e^{i(\alpha_3+\alpha_4)})+\sin(\theta-\varphi)(e^{i(\alpha_3+\alpha_4)}-e^{i(\alpha_1+\alpha_2)})\big{]}=0

 

[Orodruin]

A hint: Start by identifying the phase of both terms to be equal (note that this puts a condition on the phases! and that it is also a matter of choice of phases whether you put the phases equal or equal modulo π).

 

[Strilanc]

In addition to adding the must-equal-zero-off-diagonal constraints, I recommend breaking $|\mu_x|^2$ into $(a_x + b_x i)(a_x - b_x i)$.

Alternatively, you could take the approach of "Well, each column and row must have their 2-norm equal to 1. So if topleft has squared amplitude |a|=x then bottom-left and top-right must be |b|=|c|=1-x...".

 

[jimmycricket]

Heres everything I've written up now. Does this look ok?

U=\left(\!\begin{array}{cc}u_{11} &amp; u_{12}\\u_{21} &amp; u_{22}\end{array}\!\right)

By unitarity we have

U^{\dagger}U=\left(\!\begin{array}{cc}u_{11}^* &amp; u_{21}^*\\u_{12}^* &amp; u_{22}^*\end{array}\!\right)<br /> \left(\!\begin{array}{cc}u_{11} &amp; u_{12}\\u_{21} &amp; u_{22}\end{array}\!\right)=\left(\!\begin{array}{cc}1 &amp; 0\\0 &amp; 1\end{array}\!\right)

\Longrightarrow |u_{11}|^2+|u_{21}|^2=1 \:\:,\:\: |u_{12}|^2 + |u_{22}|^2=1

\Longrightarrow |u_{11}|=\cos(\theta) \:\:,\:\: |u_{21}|=\sin(\theta) \:\:,\:\: |u_{12}|=\sin(\varphi) \:\:,\:\: |u_{22}|=\cos(\varphi)<br />
UU^{\dagger}=<br /> \left(\!\begin{array}{cc}u_{11} &amp; u_{12}\\u_{21} &amp; u_{22}\end{array}\!\right)\left(\!\begin{array}{cc}u_{11}^* &amp; u_{21}^*\\u_{12}^* &amp; u_{22}^*\end{array}\!\right)=\left(\!\begin{array}{cc}1 &amp; 0\\0 &amp; 1\end{array}\!\right)
\Longrightarrow |u_{11}|^2+|u_{12}|^2=1 \:\:,\:\: |u_{21}|^2+|u_{22}|^2=1.
From these relations we get
\cos^2(\theta)+\sin^2(\varphi)=1 \:\:,\:\: \sin^2(\theta)+\cos^2(\varphi)=1

\Longrightarrow \cos^2(\theta)-\sin^2(\theta)=\cos^2(\varphi)-\sin^2(\varphi)=0
\Longleftrightarrow \frac{1}{2}\big{[}(1+\cos(2\theta))-(1-\cos(2\theta))\big{]}=\frac{1}{2}\big{[}(1+\cos(2\varphi))-(1-\cos(2\varphi))\big{]}<br />
\Longleftrightarrow \cos(2\theta)=\cos(2\varphi)\Longleftrightarrow \theta=\varphi\pm k\pi \:\:\:, k\in \mathbb{Z}
|u_{11}|=\cos(\theta)\Longrightarrow u_{11}=\pm e^{i\alpha_1}\cos(\theta).
Likewise, u_{12}=\pm e^{i\alpha_2}\sin(\theta) \:\:,\:\: u_{21}=\pm e^{i\alpha_3}\sin(\theta) \:\:,\:\: u_{22}=\pm e^{i\alpha_4}\cos(\theta)
So we may now write the general form for a unitary operation acting on 1 qubit as
\left(\!\begin{array}{cc}e^{i\alpha_1}\cos(\theta) &amp; -e^{i\alpha_2}\sin(\theta)\\e^{i\alpha_3}\sin(\theta) &amp; e^{i\alpha_4}\cos(\theta)\end{array}\!\right)

 

[Strilanc]

I think you mixed up $\theta$ and $\varphi$ halfway through.

 

[Orodruin]

You are missing one condition on the phases. The way you have written it down, the matrix is not necessarily unitary. A general unitary 2x2 matrix has 4 real parameters.

 

[jimmycricket]

You are missing one condition on the phases. The way you have written it down, the matrix is not necessarily unitary. A general unitary 2x2 matrix has 4 real parameters.

So do I need to eliminate one of the alphas?

 

[Orodruin]

Yes, but you cannot just do it randomly, you must do it in a way that ensures that the result is unitary.

 

[jimmycricket]

Soo check the unitarity condition again with my new U to get another set of equations?

 

[Orodruin]

That would be one way of doing it yes, but you can already get the condition from the contributions to the off diagonal elements in the product having the same phase.

Edit: Also, you should really not need to use the $UU^\dagger$ condition. Everything you need is already present from the $U^\dagger U = 1$ condition.

 

[jimmycricket]

Ok so using UU^{\dagger}=I I get for the orthogonal parts

e^{i(\alpha_1-\alpha_3)}\cos(\theta)\sin(\theta)-e^{i(\alpha_2-\alpha_4)}\sin(\theta)\cos(\theta)=0

Which gives

\alpha_1-\alpha_2-\alpha_3+\alpha_4=0

doing this for the other orthogonal row and collumn vector gives the same condition

 

[Orodruin]

Yes, so this is the final piece of the puzzle you need to reduce the number of real parameters of the general 2x2 unitary matrix to four.

 

[jimmycricket]

It seems to me that this should be the easy part but I am not really sure what to do here

 

[Orodruin]

Solve for one of the $\alpha$s and insert this into the form you have for the unitary matrix, using the others as free parameters.

 

[jimmycricket]

\left(\!\begin{array}{cc}e^{i\alpha_1}\cos(\theta) &amp; -e^{i\alpha_2}\sin(\theta)\\e^{i\alpha_3}\sin(\theta) &amp; e^{i(-\alpha_1+\alpha_2+\alpha_3)}\cos(\theta)\end{array}\!\right)

 

[Orodruin]

Something like that, yes. An alternative form is:
$$
e^{i\sigma}
\begin{pmatrix}
e^{i\delta} & 0 \\ 0 & 1
\end{pmatrix}
\begin{pmatrix}
\cos(\theta) & \sin(\theta) \\ -\sin(\theta) & \cos(\theta)
\end{pmatrix}
\begin{pmatrix}
e^{i\delta'} & 0 \\ 0 & 1
\end{pmatrix}
=
e^{i\sigma}
\begin{pmatrix}
\cos(\theta) e^{i(\delta+\delta')}& \sin(\theta)e^{i\delta} \\ -\sin(\theta)e^{i\delta'} & \cos(\theta)
\end{pmatrix}
$$

 

[jimmycricket]

The form you have there is exactly what I'm trying to achieve i.e. show that any U can be written as the product of a global phase factor and rotations in the Z and Y axes (regarding the bloch sphere representation). I'm trying to work backwards from my U matrix to the Z-Y decomposition. Is that possible with my parameters defined as they are or do I need to do some more work.

 

[Orodruin]

It is perfectly possible, just redefine the phase parameters in such a way that you can extract an overall phase.

 

[vanhees71]

Perhaps, everything becomes simpler by first using a bit of linear algebra and group theory?

I'd start with the determinant. From $U U^{\dagger}=1$, you get
$$\mathrm{det} (U U^{\dagger})=|\mathrm{det} U|^2=1 \; \Rightarrow \mathrm{det} U=\exp(\mathrm{i} \varphi), \quad \varphi \in [0,2 \pi[.$$
Thus you have
$$U=\exp(\mathrm{i} \varphi/2) S, \quad S \in \mathrm{SU}(2).$$
So we only need to find an SU(2) matrix $S$, i.e., a unitary matrix with $\mathrm{det} S=1$.

The Inverse of a matrix is given by Cramer's Rule, i.e.,
$$S=\begin{pmatrix}
a & b \\
c & d
\end{pmatrix} \; \Rightarrow \; S^{-1}=\begin{pmatrix} d & -b \\
-c & a
\end{pmatrix}.$$
On the other hand we have
$$S^{-1}=S^{\dagger}=\begin{pmatrix}
a^* & c^* \\
b^* & d^*
\end{pmatrix}.$$
Thus we have $d=a^*$, $c=-b^*$, i.e.,
$$S=\begin{pmatrix}
a & b \\
-b^* & a^*
\end{pmatrix}.$$
From $\mathrm{det} S=|a|^2+|b|^2$ one can write $|a|=\cos \theta$, $|b|=\sin \theta$ with $0 \leq \theta \leq \pi/2$. Thus we have
$$a=\cos \theta \exp(\mathrm{i} \chi), \quad b=\sin \theta \exp(\mathrm{i} \psi).$$
So the most general SU(2) matrix should be of the form
$$S=\begin{pmatrix}
\cos \theta \exp(\mathrm{i} \chi) & \sin \theta \exp(\mathrm{i} \psi) \\
-\sin \theta(-\mathrm{i} \psi) & \cos \theta \exp(-\mathrm{i} \chi )
\end{pmatrix}.$$
I hope I haven't overlooked anything.

An alternative way is to use the fact that the exponential mapping from su(2) to SU(2) is surjective (onto). Thus the most general SU(2) matrix is given by
$$S=\exp(\mathrm{i} \vec{T} \cdot \vec{n} \varphi)$$
with
$$\vec{T}=\frac{1}{2} \vec{\sigma},$$
where $\vec{\sigma}$ are the Pauli matrices, $\vec{n} \in \mathbb{R}^3$ is an arbitrary unit vector, and $\varphi \in [0,4 \pi[$. Then you have
$$S=\cos(\varphi/2) + \mathrm{i} \sin(\varphi/2) \vec{n} \cdot \vec{\sigma}.$$

 

[jimmycricket]

I'm using Nielsen and Chuang and would like to get an expression similar to theirs:

\left(\!\begin{array}{cc}e^{i(\alpha-\beta/2-\delta/2)}\cos(\theta) &amp; -e^{i(\alpha-\beta/2+\delta/2)}\sin(\theta)\\e^{i(\alpha+\beta/2-\delta/2)}\sin(\theta) &amp; e^{i(\alpha+\beta/2+\delta/2)}\cos(\theta)\end{array}\!\right)

which gives the Z-Y decomposition as

U=e^{i\alpha}R_z(\beta)R_y(\gamma)R_z(\delta)

Where the R's are the exponentiated pauli matrices.

 

[vanhees71]

That's of course another possibility. This you get immediately from the fact that SU(2) is the covering group of the rotation group SO(3), and you can decompose any rotation by the three rotations described by the Euler angles $\beta$, $\gamma$, and $\delta$. Finally, since you want not only SU(2) but the more general U(2) matrix, you have to multiply the general SU(2) matrix by an arbitrary phase factor. That this gives all U(2) matrices follows from the fact that SU(2) is the kernel of the group homomorphism, $\mathrm{det}:U(2) \rightarrow U(1)$, and thus U(2) factorizes into the invariant subgroup SU(2) and the coset U(1).

In terms of the Euler angles the SU(2) matrix reads
$$S=\exp(-\mathrm{i} \beta \sigma_3/2)\exp(-\mathrm{i} \gamma \sigma_2/2)\exp(-\mathrm{i} \delta \sigma_3/2)
=\begin{pmatrix}
\exp[-\mathrm{i} (\beta+\delta)/2] \cos (\gamma/2) & -\exp[-\mathrm{i} (\beta-\delta)/2] \sin (\gamma/2) \\
\exp[\mathrm{i} (\beta-\delta)/2] \cos (\gamma/2) & \exp[\mathrm{i} (\beta+\delta)/2] \cos (\gamma/2)
\end{pmatrix}.$$
This is of the same form as my solution above. You just have to map my angles to the Euler angles used here. The Euler angles are indeed a much nicer geometrical way to express the general SU(2) (and thus also U(2)) matrix.