So far we only talked about the description of a probabilistic and a quantum system. We now look into observing/measuring those systems.
Observing a probabilistic system
Observing a probabilistic system is the process of learning the outcome from a probability distribution. If our probability distribution for example represents a coin flip, observing this distribution is equivalent to actually flipping the coin. In the probabilistic case, an observation is just about updating our knowledge or beliefs. This will be different in the quantum case.
Definition 4.1 (Observing a probabilistic system) Given a probability distribution \(d \in \mathbb{R}^n\), we will get the outcome \(i\) with a probability \(d_i\). The new distribution is then \[
e_i = \begin{pmatrix} 0 \\ \vdots \\ 0 \\ 1 \\ 0 \\ \vdots \\ 0 \end{pmatrix} \leftarrow \text{1 at the $i$-th position}.
\]
The intuition for the new distribution is that we know after observing that \(i\) is the deterministic possibility for sure.
When observing a probabilistic system, the observation is just a passive process with no impact on the system. This means that there is no difference to the end result, whether we observe during the process or not. We take a look at an example to further understand this.
We use a random 1-bit number example similar to the random 2-bit example from Chapter 2. We have a distribution \(d_{\text{1-bit}} = \begin{pmatrix} \frac{1}{2} \\ \frac{1}{2} \end{pmatrix}\) which represents the probability distribution of generating a 1-bit number with equal probability. We also have a process \(A_{\text{flip}} = \begin{pmatrix} \frac{2}{3} & \frac{1}{3} \\[2pt] \frac{1}{3} & \frac{2}{3} \end{pmatrix}\) which flips the bit with a probability of \(\frac{1}{3}\).
We look at two different cases: For the first case, we observe only the final distribution and for the second case we observe after the generation of the 1-bit number and we also observe the final distribution.
Observing the final distribution
From Section 2.3 we know that the final distribution \(d\) is \[
d = A_{\text{flip}} \cdot d_{\text{1-bit}} = \begin{pmatrix} \frac{2}{3} & \frac{1}{3} \\[2pt] \frac{1}{3} & \frac{2}{3} \end{pmatrix} \begin{pmatrix} \frac{1}{2} \\[2pt] \frac{1}{2} \end{pmatrix} = \begin{pmatrix} \frac{1}{2} \\[2pt] \frac{1}{2} \end{pmatrix}.
\] We observe this distribution and will get outcome \(0\) and the new distribution \(d = e_0 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}\) with a probability of \(\Pr[0] = d_0 = \frac{1}{2}\). We get the outcome \(1\) and the new distribution \(d = e_1 = \begin{pmatrix} 0 \\ 1 \end{pmatrix}\) with a probability of \(\Pr[1] = d_1 = \frac{1}{2}\).
Observing after generation and the final distribution
We now observe the system after the generation of the 1-bit number and also observe the final distribution. After the generation, we will get outcome \(0\) and the new distribution \(d = e_0 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}\) with a probability of \(\Pr[0] = d_0 = \frac{1}{2}\). We get the outcome \(1\) and the new distribution \(d = e_1 = \begin{pmatrix} 0 \\ 1 \end{pmatrix}\) with a probability of \(\Pr[1] = d_1 = \frac{1}{2}\).
We now apply in each case the matrix \(A_\text{flip}\). This will give us the outcome \(A_\text{flip} \cdot \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} \frac{2}{3} \\ \frac{1}{3} \end{pmatrix}\) for the case of the outcome \(0\) and the outcome \(A_\text{flip} \cdot \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} \frac{1}{3} \\ \frac{2}{3} \end{pmatrix}\) for the case of the outcome \(1\). If we observe the distribution \(\begin{pmatrix} \frac{2}{3} \\ \frac{1}{3} \end{pmatrix}\), we will get the outcome \(0\) and the new distribution \(d = e_0 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}\) with a probability of \(\Pr[0] = \frac{2}{3}\) and the outcome \(1\) and the new distribution \(d = e_1 = \begin{pmatrix} 0 \\ 1 \end{pmatrix}\) with a probability of \(\Pr[1] =\frac{1}{3}\). If we observe the distribution \(\begin{pmatrix} \frac{1}{3} \\ \frac{2}{3} \end{pmatrix}\), we will get the outcome \(0\) and the new distribution \(d = e_0 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}\) with a probability of \(\Pr[0] = \frac{1}{3}\) and the outcome \(1\) and the new distribution \(d = e_1 = \begin{pmatrix} 0 \\ 1 \end{pmatrix}\) with a probability of \(\Pr[1] =\frac{2}{3}\).
Combining these probabilities, we get the total probability \(\Pr[0]=\frac{1}{2}\frac{2}{3} + \frac{1}{2}\frac{1}{3} = \frac{1}{2}\) for the outcome \(0\) and the probability \(\Pr[1]=\frac{1}{2}\frac{1}{3} + \frac{1}{2}\frac{2}{3} = \frac{1}{2}\) for the outcome \(1\). This is the same as observing the final distribution.
Measuring a quantum system
Unlike in the probabilistic system, the “observation” of a quantum system is called measuring. The definition is similar to the observation of a probabilistic system, except that we need to take the absolute square of the amplitude to get the probability and that the state after measuring is called post-measurement-state (p.m.s.).
Definition 4.2 (Measuring a quantum system) Given a quantum State \(\psi \in \mathbb{C}^n\), we will get the outcome \(i\) with a probability \(|\psi_i|^2\). The post-measurement-state (p.m.s.) is then \[
e_i = \begin{pmatrix} 0 \\ \vdots \\ 0 \\ 1 \\ 0 \\ \vdots \\ 0 \end{pmatrix} \leftarrow \text{1 at the $i$-th position}.
\] This is called a complete measurement in the computational basis.
With this similarity to the probabilistic observation in the definition, one might assume that measuring a quantum state has also no impact on the system. This is not the case, measuring a quantum state changes the system! We can see this effect with an example:
Let \(\psi = \begin{pmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{pmatrix}\) be a quantum state and \(H=\frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}\) be a unitary transformation. We look at two different cases: First we apply \(H\) immediately and then measure the system. As a second case, we do a measurement before the application of the \(H\) unitary and then a measurement after applying it.
Measure the final state
We first calculate the state after applying \(H\): \[
H\psi = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}.
\]
Measuring this state will get the outcome \(0\) with probability \(\Pr[0] = |\psi_0|^2 = 1\) and have the post-measurement-state \(\begin{pmatrix} 1 \\ 0 \end{pmatrix}\). The outcome \(1\) can never occur, i.e. \(\Pr[1] = |\psi_1|^2 = 0\)
Measure the initial and the final state
Measuring \(\psi\) with no further unitary matrices applied can have the outcome \(0\) or \(1\). We will look at the final measurement for each case:
The first measurement will have outcome \(0\) with probability \(\Pr[0] = |\psi_0|^2 = \frac{1}{2}\) and the post-measurement-state will be \(\begin{pmatrix} 1 \\ 0 \end{pmatrix}\). \(H\) applied to this post-measurement-state will be \(H\begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{pmatrix}\). When measuring this state, we will get the outcome \(0\) with probability \(\Pr[0] = |\frac{1}{\sqrt{2}}|^2 = \frac{1}{2}\) and outcome \(1\) with with probability \(\Pr[1] = |\frac{1}{\sqrt{2}}|^2 = \frac{1}{2}\).
The outcome \(1\) will appear at the initial state with probability \(\Pr[1] = |\psi_1|^2 = \frac{1}{2}\) and the post-measurement-state will be \(\begin{pmatrix} 0 \\ 1 \end{pmatrix}\). \(H\) applied to this post-measurement-state will be \(H\begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} \end{pmatrix}\). When measuring this state, we will get the outcome \(0\) with probability \(\Pr[0] = |\frac{1}{\sqrt{2}}|^2 = \frac{1}{2}\) and outcome \(1\) with with probability \(\Pr[1] = |-\frac{1}{\sqrt{2}}|^2 = \frac{1}{2}\).
So independent of the outcome of the first measurement, at the second measurement the outcome \(0\) and \(1\) have a probability of \(\frac{1}{2}\). This shows that when measuring before applying \(H\), we will receive different probabilities for the second measurement, then when measuring only at the end. This proves that measurements can change the system.
Elitzur–Vaidman bomb tester
Next we examine an application that would not be possible with classical methods: the Elitzur-Vaidman bomb tester.
The situation is that a box is given and we want to determine whether it contains a bomb. To test this, a photon can be sent through the box. If a bomb is present, it is wired to a photon detector. If this detects a photon, the bomb explodes.
This means that if no bomb is present, the photon can pass through the box, but if a bomb is present, the photon is detected and the bomb explodes. The detector is perfect (it never misses a photon entering the box).
Surprisingly, using quantum superposition, it is possible to detect the presence of a bomb without it exploding.
To understand how the bomb detector work, we first need to introduce a concept called a beam splitter:
A beam splitter is basically a semi-transparent mirror. In particular, when a photon comes in on the up path, it could come out on the up or down path. And if it comes in on the down path, it could come out on the up or down path as well.
Quantum mechanically, the photon could even come in in a superposition between up and down, denoted by a quantum state \(\begin{pmatrix} \alpha \\ \beta \end{pmatrix}\), where \(\alpha\) is the amplitude of the up classical possibility, and \(\beta\) of the down classical possibility. Then it exits the beam splitter in a superposition \(\begin{pmatrix} \gamma \\ \delta \end{pmatrix}\) of up and down, where: \[
\begin{pmatrix} \gamma \\ \delta \end{pmatrix} = B \begin{pmatrix} \alpha \\ \beta \end{pmatrix},\quad B \coloneqq \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}\end{pmatrix}.
\]
Note that the specific matrix \(B\) depends on the precise physical setup. Here we chose one that makes the calculation more pleasant. We will later introduce \(B\) as Hadamard-gate.
Now we can describe the bomb tester:
We begin with a photon in the up state, represented by \(\begin{pmatrix} 1 \\ 0 \end{pmatrix}\). The second state is called down, represented by \(\begin{pmatrix} 0 \\ 1 \end{pmatrix}\).
This photon in that up state is sent through a first beam splitter, which creates a superposition and sends the photon in a different direction depending on its state. Thus, after the beam splitter the photon is in the state \[
B \begin{pmatrix} 1 \\ 0 \end{pmatrix}
= \begin{pmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{pmatrix}.
\]
Thus, the photon is in an equal superposition of the up state and the down state. The down path passes through the box and potentially the bomb, while the up state bypasses the box.
At this point, it makes a difference whether the bomb is present or not. Firstly, we consider the case where there is no bomb:
In this case the photon can pass through both paths without any measurement. It therefore reaches at a second beam splitter in any case. This is identical to the first, which means that the photon is then in the following state afterwards:
\[
B \begin{pmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{pmatrix}
= \begin{pmatrix} 1 \\ 0 \end{pmatrix}.
\]
Then we measure on which path the photon is. This results in the following probability distribution:
| State up |
1 |
| State down |
0 |
| Bomb explodes |
0 |
Now we consider the case where a bomb is present. In this case, by observing whether the bomb explodes, we effectively measure whether the photon took the up/down path. It is \(\Pr[\mathit{down}] = \Pr[\mathit{up}] = \left( \frac{1}{\sqrt{2}} \right) ^2 = \frac{1}{2}\). If the photon is in the down state, the photon passes the box and the bomb explodes. If the photon is in the up state \(\begin{pmatrix} 1 \\ 0 \end{pmatrix}\) (the post-measurement state after the outcome “no explosion”), it bypasses the box and reaches the second beam splitter. After this it is in the state: \[
B \begin{pmatrix} 1 \\ 0 \end{pmatrix}
= \begin{pmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{pmatrix}.
\]
Then we measure on which the photon exits. This is again an equal superposition between both states, which leads to the following probability distribution:
| State up |
1 |
1/4 |
| State down |
0 |
1/4 |
| Bomb explodes |
0 |
1/2 |
Thus, if the photon is in the down state at the end, we know that a bomb is present without the bomb exploding. While a probability of \(1/2\) of exploding is not very satisfactory, we nonetheless have achieved something striking: With probability \(1/4\), we “observe” that there is a bomb, without the bomb ever having been touched by the photon. This is impossible classically.
Notice that the setup can be improved (details omitted) so that the probability distribution is as follows:
| State up |
1 |
≈ 0 |
| State down |
0 |
≈ 1 |
| Bomb explodes |
0 |
≈ 0 |
So it can almost certainly be found out whether a bomb is present without it exploding.
