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If I have one of the following pair of qubits: $$ \frac{1}{\sqrt{2}}(|0\rangle+i|1\rangle)\text{ , }\frac{1}{\sqrt{2}}(i|0\rangle+|1\rangle) $$ And I measure it, I'll not be able to distinguish one from the other since they'll have the same probability on the computational basis. But I know that there is a unitary matrix $U$ that I can apply to the one that I have before measuring it, and the result from measuring one after that application of $U$ is different than measuring other, so that way I'll be able to distinguish which one I have.

How can I find what is that unitary matrix $U$? I've tried with Pauli matrices and some others but still can't figure it out, and I think that is a simple problem... I just can't visualize it yet!

(Sorry for any mistake, I'm new to quantum mechanics and quantum information theory)

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  • $\begingroup$ To answer this question I would like to see more effort from you first. Let's start with: which Pauli matrix are your states eigenvectors of? What about the computational basis states? What axis does one rotate about to get the one to the other? And if you start with that direction's Pauli matrix as a Hamiltonian, what unitary operators does it create as time goes on? $\endgroup$ – CR Drost Sep 5 '17 at 3:42
  • $\begingroup$ Draw these states on the Bloch sphere. $\endgroup$ – DanielSank Sep 5 '17 at 4:04
  • $\begingroup$ @BrunoReis please feel free to edit your question to include some extra working-through? $\endgroup$ – CR Drost Sep 5 '17 at 4:12
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Lets drop the braket notation to keep this consistent. Let the left and right components in the sum correspond to the top and bottom components of a vector respectively. Then We want a unitary matrix

$$M = \left[ \begin{matrix} M_{11} \ M_{12} \\ M_{21} \ M_{22} \end{matrix} \right]$$

Such that absolute value of $A_1$ in

$$ M \left[ \begin{matrix}1 \\ i \end{matrix} \right] = \left[ \begin{matrix} A_1 \\ B_1\end{matrix} \right] $$

Is different than the absolute value of $A_2$ in:

$$ M \left[ \begin{matrix}i \\ 1 \end{matrix} \right] = \left[ \begin{matrix} A_2 \\ B_2\end{matrix} \right] $$

Why do we want this? Because those vectors on the left side are the original qubits you wanted, up to a square root of 2 factor. And on the right hand side the top components $A_1, A_2$ correspond to the probability of seeing a 0.

So now we look at the left hand sides and expand them out, to make our problem more concrete:

We want a unitary $$M = \left[ \begin{matrix} M_{11} \ M_{12} \\ M_{21} \ M_{22} \end{matrix} \right]$$

Such that

$$ |M_{11} + M_{12}i | \ne |i M_{11} + M_{12} | $$

So we could consider a system of equations for unitarity, but assuming you're more interested in the physics than the math just citing: https://en.wikipedia.org/wiki/Unitary_matrix#Elementary_constructions

We have that

$$M = \left[ \begin{matrix} M_{11} & M_{12} \\ -e^{i\phi}(M_{12})^* & -e^{i\phi}(M_{11})^* \end{matrix} \right]$$

Is unitary if $$ |M_{11}|^2 + |M_{12}|^2 = 1$$

Now that top equation can always be satisfied by picking any $M_{11}$ and $M_{12}$ to STARt, and then dividing by a common number of your choice later.

So lets pick 2 numbers, say $M_{11} = (1+i)$, $M_{12} = (1-i)$. Observe that

$$ |M_{11} + M_{12}i| = |(1+i) + i(1-i)| = |2+i|$$

and

$$ |iM_{11} + M_{12}i| = | i(1+i) + (1-i) = |0|=0$$

So clearly these 2 aren't the same!. Now we normalize these, to $$ \frac{1+i}{\sqrt{2}}$$ and $$\frac{1-i}{\sqrt{2}} $$

And now using that matrix formula from earlier we cook up our operator as:

$$M = \left[ \begin{matrix} \frac{1+i}{\sqrt{2}} & \frac{1-i}{\sqrt{2}} \\ -e^{i\phi}\frac{1+i}{\sqrt{2}} & -e^{i\phi}\frac{1-i}{\sqrt{2}} \end{matrix} \right]$$

And now you're free to pick a phase $\phi$ of your desires.

Now this operator is hardly a "natural" one, or a simple one, or a famous one. But the process of construction here is very brainless and direct. So that should be the main take away here.

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Since you've already got a "hard way" answer let me give the "easy way" answer.

Your two states are the $\pm 1$ eigenstates of the $\sigma_y$ Pauli matrix, and your two target states are the $\pm 1$ eigenstates of the $\sigma_z$ Pauli matrix. These must be related therefore by a 90 degree rotation about the $x$ axis.

We know that if we have a time-independent Hamiltonian $H$ it causes unitary evolution like $\exp(-i H t/\hbar).$ Similarly we can see that if $\sigma_x$ were a Hamiltonian it would leave invariant the $x$-eigenstates but would not be trivial -- it would rotate about the $x$-axis. So what we need is $\exp(-i\theta\sigma_x)$ for some $\theta.$

Now $\sigma_x^2 = I,$ the identity matrix, and therefore if we do the Taylor expansion out, this unitary matrix is just:$$\exp(-i\theta\sigma_x) = \cos(\theta) I - i \sin(\theta) \sigma_x = \begin{bmatrix}\cos\theta & -i\sin\theta\\-i\sin\theta&\cos\theta\end{bmatrix}.$$

Operating on the state $\begin{bmatrix}1\\i\end{bmatrix}$ we find $\begin{bmatrix}\cos\theta + \sin\theta\\-i\sin\theta + i \cos\theta\end{bmatrix}.$

The lower entry looks like $\cos\theta - \sin\theta$ which equals zero at $\theta=\pi/4,$ so this parameter $\theta$ actually rotates about the $x$-axis by $2\theta$. Committing to that we get that $\cos\theta = \sin\theta = \sqrt{1/2}$ and then $$U = \sqrt{\frac12}\begin{bmatrix}1&-i\\-i&1\end{bmatrix}.$$

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