# fixed input qubit state to an arbirary pure state using two variable rotations and one fixed rotation

It is a theorem that any arbitrary unitary transformation in SU(2) can be factored into the following form:

$O = U_X(\theta) U_Y(\phi) U_X(\delta)$

Where $U_X$ is a Bloch sphere rotation. I believe it is possible to fix one of these angles, leaving only two angles as degrees of freedom. So, I am saying that, given that X and Y might be different, and the fixed angle is some number like $\frac{\pi}{2}$ any unitary can be written in a form something like:

$O = U_X(\theta) U_Y(\frac{\pi}{2}) U_X(\delta)$

Does this sound right? Perhaps I am working on the assumption that, if a pure input state is fixed, then you only need two degrees of freedom to map that input state to any other pure output state. This thinking comes from the fact that the pure states are all on the surface of the Bloch sphere.

Here is a quote from Wikipedia:

Since polarization states are defined by two degrees of freedom, for example azimuth angle and ellipticity angle of the polarization state, such a polarization controller needs two degrees of freedom. The same holds for the task of transforming an arbitrary polarization into a fixed, known one.

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Isn't this just equivalent to the issue fixing one of the three Euler angles for real rotations? Also you missed a word after "arbitrary unitary" in the first line. ("rotation"? "operator"?) – David Z Mar 23 '13 at 1:13
David, thank you, yes I think the problem is the same. I just need some help in writing down a form for the operation (unitary transformation or Euler rotation) that highlights the order of operations. I, myself, can't derive the form by hand. – Ben Sprott Mar 25 '13 at 13:21
@BenSprott I'm really curious as to what your application was: whether it was indeed the preparation of an arbitrary quantum state as I surmised in my answer below? If so, do you have any published research on such preparations? – WetSavannaAnimal aka Rod Vance Oct 28 '13 at 0:01

Another way to see that your result does not hold is to imagine the effect of your general transformation on the sphere.

Let us map $SU(2)$ into $SU(2) / \mathbb{Z}_2 \cong SO(3)$ by the wonted homomorphism condensing the coset $\chi(\zeta) = \{\zeta,\,-\zeta\};\,\zeta \in SU(2)$ into a unique element in $\chi \in SO(3)$ and consider the action of $\chi$ on the unit sphere $\mathbb{S}^2$.

Now for any two distinct points $P_1, P_2 \in \mathbb{S}^2$ on the sphere there is a rotation $\chi(P_1, P_2) \in SO(3)$ mapping $P_1$ onto $P_2$, namely the rotation in the plane containing the great circle segment joining $P_1, P_2$ through an angle given by the length of the great circle segment.

So, if it were true that all $\chi \in SU(2)$ could be represented as a product of the stated form with two adjustable parameters $\theta$ and $\delta$, then we could find a corresponding $\chi \in SO(3)$ to map any point $P_1\in \mathbb{S}^2$ to any other $P_2\in \mathbb{S}^2$.

However, by considering certain points on the sphere and the set of all their possible images under the rotations corresponding to the elements of the form $U_X(\theta)\,U_Y(\alpha)\,U_X(\delta)\in SU(2)$ ($\alpha$ fixed for a given discussion), we can show that this set of all possible images of certain points is not the whole sphere, and therefore your proposed form cannot realize all elements of $SU(2)$.

To this end, consider the point $A$. Its image under $U_X(\delta)\in SU(2)$ for any $\delta$ is simply $A$ again. Then we rotate by the fixed angle $\alpha$ about the $y$-axis: the image is the point $C$ (or $B$, if $\alpha=\pi/2$). The set of images of this point under $U_X(\theta)\in SU(2)$ for all possible $\theta \in\mathbb{R}$ is the circle $\Gamma$. If, as in your example $\alpha = \pi/2$, then $\Gamma$ is the great circle $\{(x,\,y,\,z):\,y^2 + z^2 = 1,\,x=0\}$, more generally it is the great circle $\{(x,\,y,\,z):\,y^2 + z^2 = (\sin\alpha)^2,\, x = \cos\alpha\}$. In any case, it is not the whole of $\mathbb{S}^2$, so therefore your sequence of $SU(2)$ operations cannot realize the whole of $SU(2)$.

## Further discussion on realizing a general operator for preparing a quantum state

If you are trying to build a laboratory setup to prepare a general quantum state and, for some reason, you can realise only $U_Y(\pi/2)$ but arbitrary $U_X(\theta)$ then you can think along the following lines: consider a Lie group $\mathfrak{G}$ and suppose you have apparatus that can realise the one-parameter groups say $\mathfrak{G}_j = \{\exp(H_j \theta):\,\theta \in \mathbb{R}\}$ (the $\theta_j$ are the "dials" you can twiddle in your preparation aparatus and here you only have one $H_j$, the matrix $i\,\sigma_x$) and you also have a few discrete members of your Lie group $\gamma_1,\,\gamma_2,\,\cdots$ (not needfully distinct - here you have only $U_Y(\pi/2)$, say). Suppose further that, for some reason, the $H_j$ do not span the Lie algebra $\mathfrak{g}=\operatorname{Lie}(\mathfrak{G})$. Then a general preparation operator realised when your dials are set to $\theta_1, \, \theta_2,\,\cdots$ is:

$$\begin{array}{lcll}\chi &=& \gamma_1\,\exp\left(H_1\,\theta_1\right)\,\gamma_2\,\exp\left(H_2\,\theta_2\right)\,\gamma_3\,\exp\left(H_3\,\theta_3\right)\cdots\\ &=& \gamma_1\,\exp\left(H_1\,\theta_1\right)\,\gamma_1^{-1}\gamma_1 \gamma_2\,\exp\left(H_2\,\theta_2\right)\,\gamma_2^{-1}\gamma_1^{-1} \gamma_1\gamma_2\gamma_3\,\exp\left(H_3\,\theta_3\right)\cdots &\\ &=&\exp\left( \gamma_1\,H_1\,\gamma_1^{-1}\,\theta_1\right)\,\exp\left(\gamma_1 \gamma_2\,H_2\,(\gamma_1 \gamma_2)^{-1}\theta_2\right)\,\exp\left(\gamma_1\gamma_2\gamma_3\,H_3\,(\gamma_1\gamma_2\gamma_3)^{-1}\,\theta_3\right)\cdots&\\&&&(1)\end{array}$$

(This is simply using $\gamma \exp(Z) \gamma ^{-1} = \exp(\operatorname{Ad}_\gamma Z)$ for $\gamma\in\mathfrak{G}$ and $Z\in\mathfrak{g}$ in the adjoint representation of $\mathfrak{G}$). So, to check whether you can realize the whole of $\mathfrak{G}$ simply by adjusting the $\theta_j$ dials, what you do is form the array of Lie algebra members

$$\begin{array}{ll}\left\{\gamma_1\,H_1\,\gamma_1^{-1},\,\,\gamma_1 \gamma_2\,H_2\,(\gamma_1 \gamma_2)^{-1},\,\,\gamma_1\gamma_2\gamma_3\,H_3\,(\gamma_1\gamma_2\gamma_3)^{-1},\,\,\cdots\right\}&\\=\left\{\operatorname{Ad}_{\gamma_1} H_1,\,\operatorname{Ad}_{\gamma_1\gamma_2} H_2,\,\operatorname{Ad}_{\gamma_1\gamma_2\gamma_3} H_3\,\cdots\right\}&\\&(2)\end{array}$$

and stop when you find an array that spans the whole Lie algebra. By the existence theorem for canonical co-ordinates of the second kind (see, for example, Rossmann, "Lie Groups: An introduction through linear groups" chapters 1 and 2), a product of the form in (1), as the $\theta_j$ range through all of $\mathbb{R}$, can then realise any member of some open neighbourhood $\mathbf{N}\subset\mathfrak{G}$ of the identity in $\mathfrak{G}$ right translated by $\Pi \gamma_j\in\mathfrak{G}$, (i.e. the neighbourhood $\mathbf{N}\,\Pi \gamma_j\subset\mathfrak{G}$) so some finite product of products of the form in (1) can realise the neighbourhood $\mathbf{N}^M(\Pi \gamma_j)^M$ for some integer $M$. If $\mathfrak{G}$ is compact (as here for $\mathfrak{G} = SU(2)$), this means that the whole identity connected component in $\mathfrak{G}$ can be realized by such a product. Clearly some arrays will be "more" linearly dependent than others (i.e. have better condition numbers as a basis for the vector space $\mathfrak{g}$) and will thus realize the Lie group in shorter concatenations, so you need some ingenuity to get the best combination of $H_j$ and $\gamma_j$.

In $SU(2)$, things are simpler than this general case because it can be shown that only one such product of the form in (1) is needed if the array in (2) both spans the Lie algebra $su(2)$ and is orthogonal with respect to the Killing form $\left<A,\,B\right>=\operatorname{trace}(A\,B)$ for $A,\,B\in\mathfrak{g}$. So, for example, let's look at a product of the form:

$$\begin{array}{ll}U_X(\theta_1)\,U_Y(\pi/2)\,U_X(\theta_2)\,U_Y(\pi/2)\,U_X(\theta_3)&\\ = \exp\left(\frac{i}{2}\,\sigma_x\,\theta_1\right) \exp\left(\frac{i}{2}\operatorname{Ad}_{U_Y(\pi/2)}\, \sigma_x \,\theta_2\right) \,\exp\left(\frac{i}{2}\operatorname{Ad}_{U_Y(\pi)}\, \sigma_x\, \theta_3\right)\, U_Y(-\pi)&\\&(3)\end{array}$$

We have:

$$\begin{array}{lcll}U_X(\theta) &=& \exp\left(\frac{\theta}{2} \left(\begin{array}{cc}0&i\\i&0\end{array}\right)\right)&\\ U_Y(\frac{\pi}{2}) &=& \exp\left(\frac{\pi}{4} \left(\begin{array}{cc}0&-1\\1&0\end{array}\right)\right) = \frac{1}{\sqrt{2}}\left(\begin{array}{cc}1&-1\\1&1\end{array}\right)&\\ U_Y(\pi) &=& \exp\left(\frac{\pi}{2} \left(\begin{array}{cc}0&-1\\1&0\end{array}\right)\right) = \left(\begin{array}{cc}0&-1\\1&0\end{array}\right)&\\ &&&(4)\end{array}$$

so the candidate Lie algebra members are:

$$\begin{array}{lcll}i\,\operatorname{Ad}_I \sigma_x &=& \left(\begin{array}{cc}0&i\\i&0\end{array}\right) = i\,\sigma_x&\\ i\,\operatorname{Ad}_{U_Y(\pi/2)} \sigma_x &=& \frac{1}{2}\left(\begin{array}{cc}1&-1\\1&1\end{array}\right)\left(\begin{array}{cc}0&i\\i&0\end{array}\right)\left(\begin{array}{cc}1&1\\-1&1\end{array}\right) = \left(\begin{array}{cc}-i&0\\0&i\end{array}\right) = -i\,\sigma_z&\\ i\,\operatorname{Ad}_{U_Y(\pi)} \sigma_x &=& \left(\begin{array}{cc}0&-1\\1&0\end{array}\right)\left(\begin{array}{cc}0&i\\i&0\end{array}\right)\left(\begin{array}{cc}0&1\\-1&0\end{array}\right) = \left(\begin{array}{cc}0&-i\\-i&0\end{array}\right)=-i\,\sigma_x&\\ &&&(5)\end{array}$$

which are linearly dependent, so there is no point in having third term in (4); the product will not realise a neighbourhood in $SU(2)$. It is not too hard to show that there is no point in using any two matrices $U_Y(\alpha),\,U_Y(\beta)$ for any two arbitrary, fixed angles $\alpha$ and $\beta$, the array in (5) will still not span the Lie algebra $su(2)$.

What you're going to need in the above array is a Lie algebra member of the form $i\,\operatorname{Ad}_{U_Z(\pi /2)} \sigma_x = -i\,\sigma_y$ and we shall not only span the Lie algebra, but moreover do it with algebra members that are orthogonal with respect to the Killing form. Therefore our third term on the first line of (3) needs to be:

$$V=U_Y(-\pi /2) U_Z(\pi /2) = \frac{1}{\sqrt{2}}\left(\begin{array}{cc}e^{i\frac{\pi}{4}}&e^{-i\frac{\pi}{4}}\\-e^{i\frac{\pi}{4}}&e^{-i\frac{\pi}{4}}\end{array}\right)\quad\quad\quad(6)$$

and then we can realise any $SU(2)$ matrix as a product of the form:

$$U_X(\theta_1)\,U_Y(\pi/2)\,U_X(\theta_2)\,U_Y(-\pi /2) U_Z(\pi /2)\,U_X(\theta_3)= U_X(\theta_1)\,U_Y(\pi/2)\,U_X(\theta_2)\,V\,U_X(\theta_3)\quad\quad\quad(7)$$

then the last matrix in the test array (5) is now:

$$i\,\operatorname{Ad}_{U_Y(\pi/2) U_Y(-\pi/2) U_Z(\pi/2)} \sigma_x = i\,\operatorname{Ad}_{U_Z(\pi/2)} \sigma_x = \left(\begin{array}{cc}e^{i\frac{\pi}{4}}&0\\0&e^{-i\frac{\pi}{4}}\end{array}\right)\left(\begin{array}{cc}0&i\\i&0\end{array}\right)\left(\begin{array}{cc}e^{-i\frac{\pi}{4}}&0\\0&e^{i\frac{\pi}{4}}\end{array}\right) = \left(\begin{array}{cc}0&-1\\1&0\end{array}\right)=-i\,\sigma_y\quad\quad\quad(8)$$

and so our test array (5) now both spans the whole of $su(2)$ and is orthonormal with respect to the Killing form so that we can indeed realise any $SU(2)$ matrix with three "dials" $\theta_1,\,\theta_2,\,\theta_3 \in [0,\,4\pi)$ and the discrete Lie group members in (7).

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