2.5.1 Another mysterious operation

We still have not discussed the yellow box in Quirky. Unlike last week’s mystery operation, which operated on bits, this week’s mystery box operates on quantum bits. Let us call this mysterious quantum operation $M$ . How can we figure what is going on inside the box? As a first step, let us consider the problem of determining $M\left|0\right\rangle$ . In Quirky, we can create this state by the following setup:

The problem of determining an unknown state is called quantum state tomography, since we want to reconstruct an unknown quantum state ‘from the outside’, by performing various measurements. It is a fundamental task that experimentalists are faced with every day, when they want to make sure that quantum state that they prepared in the laboratory is the state that they intended to create!

We can already get substantial information by performing a measurement operation on the unknown state. To see this, let us write

\displaystyle M\left|0\right\rangle=\begin{pmatrix}\psi_{0}\\ \psi_{1}\end{pmatrix}.

If we perform a measurement then by Eq. 2.6 we obtain outcome $1$ with probability $\psi_{1}^{2}$ . What this means is that if we repeat the above experiment a large number of times then we would expect that the fraction of times that we obtain outcome $1$ is roughly $\psi_{1}^{2}$ . This is completely analogously to how one can estimate a coin by tossing it many times and counting the number of heads and tails, as we discussed last week in Section 1.3. This provides us with a procedure for estimating $\psi_{1}^{2}$ . In Quirky, we can simply use the probability display after the measurement to determine the probability of outcome $1$ :

Thus, we find that $\psi_{1}^{2}\approx 11.7\%$ . Since $M\left|0\right\rangle$ is a unit vector, we can also infer that $\psi_{0}^{2}=1-\psi_{1}^{2}\approx 88.3\%$ . However, amplitudes can be negative, so this only determines $\psi_{0}$ and $\psi_{1}$ up to signs! Now, remember from 2.7 that $\left|\psi\right\rangle$ and $-\left|\psi\right\rangle$ are indistinguishable, so we can only hope to determine $\left|\psi\right\rangle=M\left|0\right\rangle$ up to an overall sign. Thus, we are left with two possibilities:

\displaystyle\pm\begin{pmatrix}\sqrt{88.3\%}\\ \sqrt{11.7\%}\end{pmatrix},\quad\pm\begin{pmatrix}\sqrt{88.3\%}\\ -\sqrt{11.7\%}\end{pmatrix}

Note that this situation is very similar to 2.6, where we had to decide between $\left|+\right\rangle$ and $\left|-\right\rangle$ . In the last homework problem, you will clarify the situation and reveal the inner workings of the mystery box.

Homework 2.6 (Mystery time).

1.

How can you decide which of the two options is the case? Use Quirky to determine the quantum state $M\left|0\right\rangle$ up to sign.
2.

Similarly, determine the quantum state $M\left|1\right\rangle$ up to sign.
3.

Bonus question: Do steps 1 and 2 specify the quantum operation $M$ completely? If yes, write down a formula for $M$ . If not, how can you learn $M$ ?

Hack.

1.

We must determine which of the two forms the state $M\left|0\right\rangle$ is:

$M\left|0\right\rangle=\pm\begin{pmatrix}\sqrt{0.883}\\ \sqrt{0.117}\end{pmatrix}=\pm\left|M_{0}\right\rangle\qquad\text{or}\qquad M% \left|0\right\rangle=\pm\begin{pmatrix}\sqrt{0.883}\\ -\sqrt{0.117}\end{pmatrix}=\pm\left|M_{1}\right\rangle.$

We can picture the two states $\left|M_{0}\right\rangle$ and $\left|M_{1}\right\rangle$ as follows: Calculating the angle $\theta$ can be done as follows:

$\theta=\arctan\left(\frac{\sqrt{0.117}}{\sqrt{0.883}}\right)\approx 0.349.$

If we are dealing with $\left|M_{0}\right\rangle$ , then rotating the state by an angle of $-\theta$ gives the state $\left|0\right\rangle$ , and hence a measurement would have a $0\%$ chance of yielding the value $1$ . On the other hand, if we are dealing with $\left|M_{1}\right\rangle$ , then rotating by an angle of $-\theta$ will yield neither $\left|0\right\rangle$ nor $\left|1\right\rangle$ , hence the probability of the measurement outcome will be neither $0\%$ nor $100\%$ . So, let’s build this circuit and see what probability we get!

The probability of getting outcome $1$ is exactly $0\%$ , meaning that we are dealing with the state $\left|M_{0}\right\rangle$ . So, we conclude that:

$M\left|0\right\rangle=\pm\begin{pmatrix}\sqrt{0.883}\\ \sqrt{0.117}\end{pmatrix}.$
2.

We insert a NOT operation in front of the mystery operation in the circuit shown above. This will change the state from $\left|0\right\rangle$ to $\left|1\right\rangle$ , and hence after the application of the mystery operation, we have the state $M\left|1\right\rangle$ . We measure this state and check the probability of getting $1$ :

Hence, we find that $M\left|1\right\rangle$ is of the following form:

$M\left|1\right\rangle=\pm\begin{pmatrix}\sqrt{0.117}\\ \sqrt{0.883}\end{pmatrix}=\pm\left|N_{0}\right\rangle\qquad\text{or}\qquad M% \left|1\right\rangle=\pm\begin{pmatrix}\sqrt{0.117}\\ -\sqrt{0.883}\end{pmatrix}=\pm\left|N_{1}\right\rangle$

Now we can make a similar drawing:

And again we can calculate $\theta$ :

$\theta=\arctan\left(\frac{\sqrt{0.883}}{\sqrt{0.117}}\right)\approx 1.221.$

So, we can do a similar trick as before:

We see that the probability of getting outcome $1$ is not $0$ , hence we must be dealing with the state $\left|N_{1}\right\rangle$ , and so we find:

$M\left|1\right\rangle=\pm\begin{pmatrix}\sqrt{0.117}\\ -\sqrt{0.883}\end{pmatrix}.$
3.

The somewhat surprising answer is no, since the sign of $M\left|0\right\rangle$ relative to $M\left|1\right\rangle$ is important and still undetermined. However, even though we cannot determine the signs of $M\left|0\right\rangle$ and $M\left|1\right\rangle$ individually, we can find out whether they are the same or not. This can be done by applying $M$ to some intermediate state such as $\left|+\right\rangle$ . If the signs are the same, we obtain:

$M\left|+\right\rangle=\pm\frac{1}{\sqrt{2}}\begin{pmatrix}\sqrt{0.883}+\sqrt{0% .117}\\ \sqrt{0.117}-\sqrt{0.883}\end{pmatrix}\approx\pm\begin{pmatrix}0.906\\ -0.442\end{pmatrix}.$

On the other hand, if the signs are opposite, we have:

$M\left|+\right\rangle=\pm\frac{1}{\sqrt{2}}\begin{pmatrix}\sqrt{0.883}-\sqrt{0% .117}\\ \sqrt{0.117}+\sqrt{0.883}\end{pmatrix}\approx\pm\begin{pmatrix}0.442\\ -0.906\end{pmatrix}.$

We can have a look at what Quirky tells us:

Apparently, the probability of getting $1$ is $82.1\%$ . Since $0.821=(-0.906)^{2}$ , we must be in the second case. So, the signs of $M\left|0\right\rangle$ and $M\left|1\right\rangle$ must be different. This tells us everything we can possibly figure out about $M$ .