# Question

Consider two single qubit states $$\left\{|\alpha_0\rangle,|\alpha_1\rangle\right\}$$ which are not orthogonal or parallel, i.e. $$\left|\langle\alpha_0|\alpha_1\rangle\right|\ne0,1$$. Additionally, consider the unitary operation: $$U|\alpha_i\rangle|0\rangle=|c_{ii}\rangle\forall i\in\left\{0,1\right\}$$; where we want $$|c_{ii}\rangle$$ to approximate $$|\alpha_i\rangle|\alpha_i\rangle$$, i.e. approximate cloning - note that $$|c_{ii}\rangle\ne|\alpha_i\rangle|\alpha_i\rangle\forall i$$ due to the no-cloning theorem. We can define the fidelity $$F_i\equiv\left|\langle\alpha_i|\langle\alpha_i|c_{ii}\rangle\right|^2$$ as a measure of the quality of the approximation such that the best approximation is given by maximising $$F_i$$ with respect to $$|c_{ii}\rangle$$ subject to the following constraints:

1. $$|c_{ii}\rangle\in\operatorname{span}\left\{|\alpha_j\rangle|\alpha_j\rangle\right\}$$
2. $$|\alpha_i\rangle|0\rangle\to|c_{ii}\rangle\forall i$$ is unitary $$\iff\langle\alpha_i|\alpha_j\rangle=\langle c_{ii}|c_{jj}\rangle\forall i,j$$
3. $$F=F_0=F_1$$

Contraint 1 allows us to express:

$$|c_{ii}\rangle\equiv\sum_ja_{ji}|\alpha_j\rangle|\alpha_j\rangle$$

Now if we let $$A_{ij}=\langle\alpha_i|\alpha_j\rangle$$ then constraint 2 can be expressed as:

$$\boldsymbol A=\boldsymbol a^\dagger\boldsymbol A^{\circ 2}\boldsymbol a$$

where $$\boldsymbol A^{\circ 2}$$ is the Hadamard power i.e. $$\left[\boldsymbol A^{\circ 2}\right]_{ij}\equiv A_{ij}^2$$. In this notation we can also express $$F_i=\left|\left[\boldsymbol A\boldsymbol a\right]_{ii}\right|^2$$.

We have already applied constraint 1 in the notation used leaving both constraints 2 and 3 to be applied.

As far as I see the problem of finding the values of $$|c_{ii}\rangle$$ that maximise $$F$$ from here can then be solved in one of three ways - each of which I have tried with no luck. I would be very grateful if someone was able to outline a solution for complex $$\boldsymbol a\in\mathbb C^{2\times2}$$. However, if no elegant solution for $$\boldsymbol a\in\mathbb C^{2\times2}$$ then a solution for $$\boldsymbol a\in\mathbb R^{2\times2}$$ where $$|\alpha_i\rangle$$ can be expressed in terms of $$|0\rangle$$ and $$|1\rangle$$ with only real coefficients would suffice to be accepted as the correct answer. I know there should be an elegant solution for $$\boldsymbol a\in\mathbb R^{2\times2}$$ as this was what was asked on a problem sheet I have been using for some self-learning, but I am interested in the more general solution $$\boldsymbol a\in\mathbb C^{2\times2}$$ if possible.

Finally, while I am an avid StackExchange user, this is my first post on Quantum Computing StackExchange and so I apologise if I have broken any site-specific rules and please do let me know. Additionally, if the moderators feel this is more appropriate for either Physics or Maths SE do migrate.

# Attempts

Below I will outline three of my attempted methods in case these are of any help.

$$\boldsymbol a\in\mathbb C^{2\times2}$$ has 8 degrees of freedom, but as $$F_i$$ is independent of an overall phase factor the degrees of freedom reduce to 7 and we can express:

$$\boldsymbol a\equiv a\begin{bmatrix}1&be^{i\beta}\\ce^{i\gamma}&de^{i\delta}\end{bmatrix}$$

Using the method of Lagrange multipliers with a matrix constraint we must maximise:

$$L_i\equiv\left|\left[\boldsymbol A\boldsymbol a\right]_{ii}\right|^2+\operatorname{Tr}\left(\boldsymbol \mu\left[\boldsymbol A-\boldsymbol a^\dagger\boldsymbol A^{\circ 2}\boldsymbol a\right]\right)+\lambda\left(\left|\left[\boldsymbol A\boldsymbol a\right]_{ii}\right|^2-\left|\left[\boldsymbol A\boldsymbol a\right]_{\overline{ii}}\right|^2\right)\quad\forall i$$

where $$\boldsymbol\mu\in\mathbb C^{2\times2}$$ and $$\lambda\in\mathbb C$$ are the lagrange multipliers; and $$\overline i\equiv\begin{cases}0,&i=1\\1,&i=0\end{cases}$$. Thus,

$$\text{d}L_i=\left(1+\lambda\right)\left(\vec\pi_i^\dagger\boldsymbol a^\dagger\boldsymbol A\vec\pi_i\vec\pi_i^\dagger\boldsymbol{A\text{d}a}\vec\pi_i+\vec\pi_i^\dagger\text{d}\boldsymbol a^\dagger\boldsymbol A\vec\pi_i\vec\pi_i^\dagger\boldsymbol{Aa}\vec\pi_i\right)-\operatorname{Tr}\left(\boldsymbol{\mu a}^\dagger\boldsymbol A^{\circ2}\text{d}\boldsymbol a\right)-\operatorname{Tr}\left(\text{d}\boldsymbol a^\dagger\boldsymbol A^{\circ2}\boldsymbol{a\mu}\right)-\lambda\left(\vec\pi_\overline{i}^\dagger\boldsymbol a^\dagger\boldsymbol A\vec\pi_\overline{i}\vec\pi_\overline{i}^\dagger\boldsymbol{A\text{d}a}\vec\pi_\overline{i}+\vec\pi_\overline{i}^\dagger\text{d}\boldsymbol a^\dagger\boldsymbol A\vec\pi_\overline{i}\vec\pi_\overline{i}^\dagger\boldsymbol{Aa}\vec\pi_\overline{i}\right)$$

As $$\left\{\text{d}\boldsymbol a\vec\pi_i,\vec\pi_i^\dagger\text{d}\boldsymbol a^\dagger\middle|\forall i\right\}$$ are independent variables and $$\operatorname{Tr}\left(\boldsymbol M\right)\equiv\sum_i\vec\pi_i^\dagger\boldsymbol M\vec\pi_i$$ then for $$\text{d}L_i=0$$ the simultaneous equations are:

$$\begin{cases}\left(\delta_{ij}\left(1+2\lambda\right)-\lambda\right)\vec\pi_i^\dagger\boldsymbol a^\dagger\boldsymbol A\vec\pi_i\vec\pi_i^\dagger\boldsymbol A-\pi_i^\dagger\boldsymbol{\mu a}^\dagger\boldsymbol A^{\circ2}=0&\forall i,j\\\left(\delta_{ij}\left(1+2\lambda\right)-\lambda\right)\boldsymbol A\vec\pi_i\vec\pi_i^\dagger\boldsymbol{Aa}\vec\pi_i-\boldsymbol A^{\circ2}\boldsymbol{a\mu}\vec\pi_i=0&\forall i,j\\\vec\pi_i^\dagger\boldsymbol a^\dagger\boldsymbol A\vec\pi_i\vec\pi_i^\dagger\boldsymbol{A\text{d}a}\vec\pi_i=F&\forall i\\\boldsymbol A=\boldsymbol a^\dagger\boldsymbol A^{\circ 2}\boldsymbol a\end{cases}$$

This gives $$\lambda=-\frac{1}{2}$$ and so:

$$\begin{cases}\frac{1}{2}\vec\pi_i^\dagger\boldsymbol a^\dagger\boldsymbol A\vec\pi_i\vec\pi_i^\dagger\boldsymbol A-\pi_i^\dagger\boldsymbol{\mu a}^\dagger\boldsymbol A^{\circ2}=0&\forall i\\\frac{1}{2}\boldsymbol A\vec\pi_i\vec\pi_i^\dagger\boldsymbol{Aa}\vec\pi_i-\boldsymbol A^{\circ2}\boldsymbol{a\mu}\vec\pi_i=0&\forall i\\\vec\pi_i^\dagger\boldsymbol a^\dagger\boldsymbol A\vec\pi_i\vec\pi_i^\dagger\boldsymbol{A\text{d}a}\vec\pi_i=F&\forall i\\\boldsymbol A=\boldsymbol a^\dagger\boldsymbol A^{\circ 2}\boldsymbol a\end{cases}$$

Then we can easily see that $$F=\vec\pi_i^\dagger\boldsymbol{A\mu}\vec\pi_i=\vec\pi_i^\dagger\boldsymbol{\mu A}\vec\pi_i$$ for all $$i$$. However, I do not know how I would proceed further than this.

Another approach would be to use constraint 3 to reduce the degrees of freedom to 6 by showing:

$$de^{i\delta}=-\alpha^*be^{i\beta}+\left|1+\alpha ce^{i\gamma}\right|e^{i\epsilon}$$

where $$\alpha\equiv\langle\alpha_0|\alpha_1\rangle$$. Now our remaining degrees of freedom are $$\left\{a,b,c,\beta,\gamma,\epsilon\right\}$$ and we need only maximise:

$$L_i'\equiv\left|\left[\boldsymbol A\boldsymbol a\right]_{ii}\right|^2+\operatorname{Tr}\left(\boldsymbol \mu\left[\boldsymbol A-\boldsymbol a^\dagger\boldsymbol A^{\circ 2}\boldsymbol a\right]\right)\quad\forall i$$

I won't write out the resulting simultaneous equations again as this time they are more involved as the reduction in degrees of freedom means $$\left\{\text{d}\boldsymbol a\vec\pi_i,\vec\pi_i^\dagger\text{d}\boldsymbol a^\dagger\middle|\forall i\right\}$$ are not all independent degrees of freedom.

Or finally, we could apply both constraints 2 and 3 before maximising; however, I have not managed to navigate my way through the swamp of simultaneous equations for this but in this case, we now only need to maximise $$F$$ with respect to the remaining degrees of freedom.

Now I have in all three cases managed to find the conditions under which these are maximised; however, I then have been struggling to solve the resulting simultaneous equations.

Without loss of generality, you can assume $$|\alpha_0\rangle=|0\rangle,\qquad |\alpha_1\rangle=\cos\theta|0\rangle+\sin\theta|1\rangle$$ and clearly $$U|\alpha_0\rangle|0\rangle=a_0|00\rangle+a_1|\alpha_1\alpha_1\rangle,\qquad U|\alpha_1\rangle|0\rangle=a_1|00\rangle+a_0|\alpha_1\alpha_1\rangle.$$ There are two constraints. Firstly, normalisation: $$|a_0|^2+|a_1|^2+2\text{Re}(a_0a_1^*)\cos^2\theta=1.$$ Secondly, unitarity (preservation of inner product): $$2\text{Re}(a_0a_1^*)+(|a_0|^2+|a_1|^2)\cos^2\theta=\cos\theta.$$ For fixed $$\theta$$, we can therefore solve for $$|a_0|^2+|a_1|^2$$ and $$\text{Re}(a_0a_1^*)$$.
The fidelity of the transformation is $$F=|a_0+a_1\cos^2\theta|^2=|a_0|^2+|a_1|^2\cos^4\theta+2\text{Re}(a_0a^*)\cos^2\theta=1+|a_1|^2(\cos^4\theta-1).$$ So, you want to minimise $$|a_1|^2$$ subject to the given constraints (we can take $$a_0$$ to be real). Let $$a_1=re^{i\phi}$$. So, we now have that $$a_0^2+r^2=\frac{1-\cos^3\theta}{1-\cos^4\theta},\qquad 2a_0r\cos\phi=\frac{\cos\theta(1-\cos\theta)}{1-\cos^4\theta}.$$ Next, let $$a_0=\sqrt{\frac{1-\cos^3\theta}{1-\cos^4\theta}}\cos\gamma, r=\sqrt{\frac{1-\cos^3\theta}{1-\cos^4\theta}}\sin\gamma.$$ The first condition is now automatically satisfied, and we have $$\frac{1-\cos^3\theta}{1-\cos^4\theta}\sin(2\gamma)\cos\phi=\frac{\cos\theta(1-\cos\theta)}{1-\cos^4\theta}$$ for the other. We are trying to minimise $$\sin\gamma$$ for fixed $$\theta$$, so we pick $$\cos\phi=1$$, and have $$\sin(2\gamma)=\frac{\cos\theta}{1+\cos\theta+\cos^2\theta}.$$ Finally, this yields a fidelity $$F=1+\frac{\cos^3\theta-1}{2}\left(1-\sqrt{1-\frac{\cos^2\theta}{(1+\cos\theta+\cos^2\theta)^2}}\right).$$
We might make a couple of quick checks: if $$\cos\theta=0$$ or 1, we know that $$F=1$$. These both hold.
• Thank you, thid method is significantly easier! My two questions about it though are why can we assume that the values of $a_0$ and $a_1$ are the same in $U|\alpha_0\rangle|0\rangle=a_0|00\rangle+a_1|\alpha_1\alpha_1\rangle$ and $U|\alpha_1\rangle|0\rangle=a_1|00\rangle+a_0|\alpha_1\alpha_1\rangle$? Intuitively it makes sense to me but I wondered if there is a mathematical reasoning behind it? And secondly, it appears you don't think these are appropriate constraints and so am intrigued as to why? Aug 16 at 8:48
• (for example, you might include $(|0\rangle|\alpha_1\rangle+|\alpha_1\rangle|0\rangle)$ in the span) Aug 16 at 8:59
• I suppose I don't have a formal mathematical reasoning behind my assumption. Probably you can derive it from the requirement that the two fidelities are equal. But I was mostly thinking about symmetry. I should be able to exchange my definition of $|\alpha_0\rangle$ and $|\alpha_1\rangle$, and it should all work just the same. Aug 16 at 9:01