2.5 Distinguishing quantum states

Alice is watching a running competition for robot donkeys, and she notes down whether her favourite donkey wins: a $1$ when this happens, and a $0$ otherwise. She could also encode this information into a qubit: in the most general case, she prepares the state $\left|\psi(\theta_{0})\right\rangle$ in situation $0$ (no win), or a state $\left|\psi(\theta_{1})\right\rangle$ in the case $1$ (her favourite donkey wins). Alice can make these states by simply applying $U(\theta_{0})$ or $U(\theta_{1})$ to $\left|0\right\rangle$ , as in Eq. 2.27. Now, assume Alice gives this qubit to Bob. Can Bob guess what bit value ( $0$ or $1$ ) was encoded, based purely on this one qubit? Would Bob’s odds improve if he can first perform a rotation or reflection? You can practice this idea in the following exercise.

Exercise 2.6 (Plus and minus).

Imagine that you are given a qubit that is one of the following two states:

\left|+\right\rangle=\frac{\left|0\right\rangle+\left|1\right\rangle}{\sqrt{2}% },\qquad\left|-\right\rangle=\frac{\left|0\right\rangle-\left|1\right\rangle}{% \sqrt{2}}.

You want to guess which state it is. You can apply some rotation and then measure. What rotation should you apply and with what probability can you guess the correct state?

Solution.

Apply

U(-\pi/4)

and measure. You can guess the state with certainty!

If you want to represent different bit values by different quantum states, you should be careful to not use $\left|\psi(\theta)\right\rangle$ and $\left|\psi(\theta+\pi)\right\rangle$ since these states cannot be distinguished.

Exercise 2.7 (Indistinguishable states).

Show that the two states $\left|\psi(\theta)\right\rangle$ and $\left|\psi(\theta+\pi)\right\rangle=-\left|\psi(\theta)\right\rangle$ cannot be distinguished in any way. That is, no matter what qubit operation you apply and then measure the resulting state, the measurement outcomes in both cases will always have the same probabilities.

Solution.

We saw above that any combination

M

of rotations and reflections is linear. Thus if

M\left|\psi(\theta)\right\rangle=\left|\psi(\theta^{\prime})\right\rangle=% \bigl{(}\begin{smallmatrix}\cos\theta^{\prime}\\ \sin\theta^{\prime}\end{smallmatrix}\bigr{)}

, then

M\big{\lparen}-\left|\psi(\theta)\right\rangle\big{\rparen}=-\left|\psi(\theta% ^{\prime})\right\rangle=\bigl{(}\begin{smallmatrix}-\cos\theta^{\prime}\\ -\sin\theta^{\prime}\end{smallmatrix}\bigr{)}

. In view of Eq. 2.6, the probabilities

p_{0}

and

p_{1}

of measurement outcomes are the same for both states.

It is quite interesting to compare 2.6 and 2.7. When two states differ by an overall minus sign, they are completely indistinguishable, as in 2.7. For all practical purposes, the two vectors $\pm\left|\psi(\theta)\right\rangle$ describe the same state. In contrast, ‘relative’ minus signs as in 2.6 are important and can even lead to states that can be perfectly distinguished!

In the following homework problem you can figure out the optimal way of distinguishing two arbitrary quantum states.

Homework 2.5 (Telling two states apart).

Let $\theta,\theta^{\prime}$ be two angles. For simplicity, assume that $-\frac{\pi}{2}\leq\theta\leq\theta^{\prime}\leq\frac{\pi}{2}$ . Suppose that Eve hands you a single qubit, either in state $\left|\psi(\theta)\right\rangle$ or in state $\left|\psi(\theta^{\prime})\right\rangle$ , with 50% probability each. (For example, she could toss a fair coin to decide which of the two states to give to you.) Your task is to identify which of the two states you received. In a few steps you will find an optimal procedure:

1.

First apply a rotation $U(\phi)$ by some angle $\phi$ . Which two possible states will you obtain?
2.

Next, measure the qubit and interpret the outcome as follows: If the outcome is 0 then your guess is that you were handed the state $\left|\psi(\theta)\right\rangle$ , otherwise your guess is $\left|\psi(\theta^{\prime})\right\rangle$ . What is the probability of correctly identifying the state that was given to you? Write down a formula in terms of $\theta$ , $\theta^{\prime}$ and $\phi$ .

Hint: First compute the success probability assuming you are given the first state, then the success probability assuming you are given the second state, and then remember that in reality you get one of the two states with 50% probability each.
3.

You still have the freedom of choosing the rotation angle $\phi$ in a clever way. What is the success probability as a function of $\theta$ and $\theta^{\prime}$ if you choose $\phi$ optimally?

Hint: Try using the trigonometric identities from Eq. 2.30. In particular, from these you can show that

$\sin^{2}\alpha=\frac{1}{2}\big{\lparen}1-\cos(2\alpha)\big{\rparen},\qquad\cos% ^{2}\alpha=\frac{1}{2}\big{\lparen}1+\cos(2\alpha)\big{\rparen}.$ (2.36)

If you are stuck, you can also use Wolfram Alpha.

Hack.

1.

If we started out in the state $\left|\psi(\theta)\right\rangle$ then we obtain the following state after rotating:

$U(\phi)\left|\psi(\theta)\right\rangle=\left|\psi(\theta+\phi)\right\rangle=% \begin{pmatrix}\cos(\theta+\phi)\\ \sin(\theta+\phi)\end{pmatrix}.$

If instead the state $\left|\psi(\theta^{\prime})\right\rangle$ was handed to us then we obtain the following state:

$U(\phi)\left|\psi(\theta^{\prime})\right\rangle=\left|\psi(\theta^{\prime}+% \phi)\right\rangle=\begin{pmatrix}\cos(\theta^{\prime}+\phi)\\ \sin(\theta^{\prime}+\phi)\end{pmatrix}.$
2.

First of all, we calculate the probability that we guess correctly if we are handed the state $\left|\psi(\theta)\right\rangle$ . In this case, the probability of indeed measuring $0$ is given by:

$\cos^{2}(\theta+\phi).$

Similarly, the probability of guessing $1$ when starting out with state $\left|\psi(\theta^{\prime})\right\rangle$ is given by:

$\sin^{2}(\theta^{\prime}+\phi).$

As both initial states are given with equal probability, the total success probability is the average of these two success probabilities:

$\frac{1}{2}\left(\cos^{2}(\theta+\phi)+\sin^{2}(\theta^{\prime}+\phi)\right).$

We want to find the value of $\phi$ that maximizes the above expression. To that end, we rewrite it using some trigonometric identities:

	$\displaystyle\frac{1}{2}\left(\cos^{2}(\theta+\phi)+\sin^{2}(\theta^{\prime}+% \phi)\right)$	$\displaystyle=\frac{1}{4}+\frac{1}{4}\cos(2(\theta+\phi))+\frac{1}{4}-\frac{1}% {4}\cos(2(\theta^{\prime}+\phi))$
		$\displaystyle=\frac{1}{2}+\frac{1}{2}\sin(\theta+\theta^{\prime}+2\phi)\sin(% \theta^{\prime}-\theta),$

where the second line follows from $\frac{1}{2}(\cos(2\alpha)-\cos(2\beta))=\sin(\beta+\alpha)\sin(\beta-\alpha)$ . Note that $\sin(\theta^{\prime}-\theta)\geq 0$ , since $\theta^{\prime}>\theta$ and both angles are in the interval $[-\pi/2,\pi/2]$ . Hence, the maximum is attained when $\theta+\theta^{\prime}+2\phi=\pi/2$ , which yields:

\phi=\frac{\pi}{4}-\frac{\theta+\theta^{\prime}}{2}.

The success probability is therefore

\frac{1}{2}+\frac{1}{2}\sin(\pi/2)\sin(\theta^{\prime}-\theta)=\frac{1}{2}+% \frac{1}{2}\sin(\theta^{\prime}-\theta).

If you are interested, here is some intuitive explanation of how this procedure works: The rotation $M=U(\pi/4-(\theta+\theta^{\prime})/2)$ rotates the state $\left|\psi(\frac{\theta+\theta^{\prime}}{2})\right\rangle$ whose angle is the average of the two angles to the state $\left|\psi(\pi/4)\right\rangle$ , since

U\left\lparen\frac{\pi}{4}-\frac{\theta+\theta^{\prime}}{2}\right\rparen\left|% \psi\left\lparen\frac{\theta+\theta^{\prime}}{2}\right\rparen\right\rangle=% \left|\psi\left\lparen\frac{\theta+\theta^{\prime}}{2}+\frac{\pi}{4}-\frac{% \theta+\theta^{\prime}}{2}\right\rparen\right\rangle=\left|\psi(\pi/4)\right\rangle.

As a result, the state $\psi(\theta^{\prime})$ is rotated as close as possible to $\left|1\right\rangle$ while $\psi(\theta)$ is simultaneously rotated as close as possible to $\left|0\right\rangle$ , see the illustration below:

Now let us examine the probability that this procedure succeeds, which was $\frac{1}{2}+\frac{1}{2}\sin(\theta^{\prime}-\theta)$ . In the extreme scenario that $\theta^{\prime}-\theta=0$ , the success probability is $1/2$ . This makes sense, since the two states are equal and hence we cannot do better than guessing randomly. Actually, if $\theta^{\prime}-\theta=\pi$ , then the success probability is also $1/2$ . This can be understood since then $\left|\psi(\theta^{\prime})\right\rangle=-\left|\psi(\theta)\right\rangle$ . Since this global minus sign is undetectable by any measurement, we can also not do better than guessing. On the other hand, if $\theta^{\prime}-\theta=\pi/2$ , then the success probability is $1$ , which also makes sense. We simply rotate the states until they coincide with $\left|0\right\rangle$ and $\left|1\right\rangle$ , and then we measure. We see that for all other values of $\theta^{\prime}-\theta$ , the probability of making a correct guess increases as $\theta-\theta^{\prime}$ gets closer to $\pi/2$ . Hence, the closer the states $\left|\psi(\theta)\right\rangle$ and $\left|\psi(\theta^{\prime})\right\rangle$ are to being orthogonal, the higher the probability with which you can distinguish them! (Can you see this from the above figure?)

Exercise 2.8 (Broken leg and arm (challenging)).

Problem: Alice and Bob like to explore the wilderness surrounding their town. For this purpose they have built two large gorilla robots that can navigate through rough terrain while carrying them comfortably on their backs. This is not a good day for Bob since his robot accidentally falls off a cliff! Luckily, Bob survives the fall with only a few bruises, but his robot gets damaged pretty badly: one arm, one leg, and its communications device are all broken. Bob does not have any spare parts for legs and arms, but at least he manages to fix his communications device for a brief moment. Unfortunately, it can send only one bit or one qubit before it stops working. Bob would like to communicate to Alice which leg (left or right) and which arm (also, left or right) of his robot is broken so that she could take off the corresponding one from her robot and send it down to him. Alice can send him just one limb (either leg or arm) because both robots need to be able to walk back home (which they can still do on three limbs). The situation is made more complicated by the fact that Alice does not have all the required tools for removing an arbitrary limb from her robot. Bob recalls that Alice took with her either the tools for legs or for arms (and not both), but he cannot remember which.

There are four possible combinations of which leg and which arm of Bob’s robot broke – you can assume that each of them happened with probability $1/4$ . Similarly, there are two types of limbs Alice can remove from her robot (she has either tools for removing legs or arms) and you can assume that she took the right tools for each with probability $1/2$ .

Questions:

1.

If Bob can send only one bit to Alice, how should he decide on its value depending on which of the four ways his robot could be broken? How should Alice interpret his message and decide on whether to send the left or the right limb? (Recall that Alice can send only legs or only arms, and Bob does not know whether it is legs or arms.) If they both use an optimal strategy, with what probability will Alice interpret Bob’s message correctly and send the right limb for his robot?
2.

What if Bob can instead send one quantum bit? Depending on his situation, he can choose one of four states and, depending on her situation, Alice can apply one of two rotations before measuring it. What is their optimal joint strategy and with what probability does it succeed?

You can assume that Alice and Bob know how to interpret each other’s messages, since they have discussed beforehand what to do if this particular emergency situation ever happens.

Solution.

Ideally, Bob would like to send 2 bits indicating which leg and which arm is broken. However, Alice only cares about one of these bits since she has only one type of tools with her.

1.

Let us denote the two possible values of each bit by $L$ (left) and $R$ (right). One strategy Bob can use is to send the “majority” of his two bits. Namely, he can use the following encoding: $LL\mapsto L$ , $RR\mapsto R$ . The remaining two cases he can encode arbitrarily, say, $LR\mapsto L$ , $RL\mapsto R$ . Alice’s strategy is simply to send the limb corresponding to Bob’s message (the left limb if she received $L$ and the right limb if she received $R$ ). This works with probability

$\frac{1}{4}\left\lparen 1+1+\frac{1}{2}+\frac{1}{2}\right\rparen=\frac{3}{4}=0% .75,$ (2.37)

where the four terms inside the brackets are the probabilities that Alice makes the right decision for each of the four Bob’s situations.

Bob can send the following qubit state, depending on what limbs are broken (i.e., LL means left leg and left arm, LR means left leg and right arm, etc.)

	$\displaystyle\left\|LL\right\rangle$	$\displaystyle=\cos(\pi/8)\left\|0\right\rangle-\sin(\pi/8)\left\|1\right\rangle$
	$\displaystyle\left\|LR\right\rangle$	$\displaystyle=\cos(\pi/8)\left\|0\right\rangle+\sin(\pi/8)\left\|1\right\rangle$
	$\displaystyle\left\|RR\right\rangle$	$\displaystyle=\cos(3\pi/8)\left\|0\right\rangle+\sin(3\pi/8)\left\|1\right\rangle$
	$\displaystyle\left\|RL\right\rangle$	$\displaystyle=\cos(3\pi/8)\left\|0\right\rangle-\sin(3\pi/8)\left\|1\right\rangle$

To recover the leg bit, Alice just measures. To recover the arm bit, Alice applies $U(\pi/4)$ and then measures. (Note that she cannot recover both bits since the original state is no longer around after the measurement.) They will succeed with probability $\cos^{2}(\pi/8)=\frac{1}{2}+\frac{1}{2\sqrt{2}}\approx 0.85$ . This is better than in the first scenario!

	$\displaystyle\left\|LL\right\rangle$	$\displaystyle=\cos(\pi/8)\left\|0\right\rangle-\sin(\pi/8)\left\|1\right\rangle$
	$\displaystyle\left\|LR\right\rangle$	$\displaystyle=\cos(\pi/8)\left\|0\right\rangle+\sin(\pi/8)\left\|1\right\rangle$
	$\displaystyle\left\|RR\right\rangle$	$\displaystyle=\cos(3\pi/8)\left\|0\right\rangle+\sin(3\pi/8)\left\|1\right\rangle$
	$\displaystyle\left\|RL\right\rangle$	$\displaystyle=\cos(3\pi/8)\left\|0\right\rangle-\sin(3\pi/8)\left\|1\right\rangle$