Quantum theory of observation/Entanglement

According to classical physics, the state of a compound system is always determined by the list of the states of its components. Formally, we define the state space of the compound as the cartesian product of the state spaces of the components. If the state of the compound is known exactly, the states of the components are therefore necessarily known with the same accuracy. This is no longer true in quantum physics, because the state space of the compound is the tensor product of the state spaces of the components. For example, if a two-qubits system is in the state $$ \frac {1} {\sqrt 2} (| 00 \rangle + | 11 \rangle) $$ one can not assign state vectors to the qubits taken separately. Each qubit is entangled with the other. This entanglement effect is purely quantum. It has no equivalent in classical physics. It is often considered, since Schrödinger (1935), as the quantum effect par excellence. Entanglement is at the heart of the great mystery of quantum superposition.

This chapter is the most important of the book because quantum entanglement is fundamental to explain the reality of observations.

Definition
The state of a composite system AB ... Z is said to be separable when it is the product of the states of its components:

$$ | \psi_ {AB ... Z} \rangle = | \psi_A \rangle | \psi_B \rangle ... | \psi_Z \rangle $$

for a pure state,

$$ \rho_ {AB ... Z} = \rho_A \rho_B ... \rho_Z $$

for a mixed state.

A state is entangled when it is not separable. It is sometimes called an inseparable state.

Interaction, entanglement and disentanglement
When two parts of a system interact, an initially separable state may become entangled. For example, $$CNOT[\frac{1}{\sqrt 2}(|0\rangle + |1\rangle)|0\rangle] = \frac{1}{\sqrt 2}(|00\rangle + |11\rangle)$$

But interactions do not always lead to entanglement. CNOT does not entangle the states of the computation basis (for two qubits the computation basis is: $$ | 00 \rangle, | 01 \rangle, | 10 \rangle, | 11 \rangle) $$. SWAP is an interaction that never entangles the separable states on which it acts:

$$ SWAP | \alpha \rangle | \beta \rangle = | \beta \rangle | \alpha \rangle $$

When an interaction $$ U $$ transforms a separable state into an entangled state, it is in principle possible to return to the initial separable state, provided that the dynamics of the interaction is reversible, because $$ U ^ {- 1} $$ then represents a possible interaction. Almost all elementary interactions allow such temporal reversibility.

For example :

$$ CNOT \frac {1} {\sqrt 2} (| 00 \rangle + | 11 \rangle) = | 00 \rangle $$

CNOT is its own inverse: $$ CNOT ^ {- 1} = CNOT $$. Any state that has been entangled by CNOT becomes separable again if CNOT is applied a second time. When entanglement is thus destroyed, one can speak of disentanglement, or return to separability.

The Hadamard gate is very useful for modeling entanglement and disentanglement of qubits. From the computation basis, the combination of H on the first qubit followed by CNOT produces the basis of the Bell states:

$$CNOT (H_1 |00\rangle) = \frac{1}{\sqrt 2}(|00\rangle + |11\rangle) = |\beta_{00}\rangle$$

$$CNOT (H_1 |01\rangle) = \frac{1}{\sqrt 2}(|01\rangle + |10\rangle) = |\beta_{01}\rangle$$

$$CNOT (H_1 |10\rangle) = \frac{1}{\sqrt 2}(|00\rangle - |11\rangle) = |\beta_{10}\rangle$$

$$CNOT (H_1 |11\rangle) = \frac{1}{\sqrt 2}(|01\rangle - |10\rangle) = |\beta_{11}\rangle$$

The Bell states $$ | \beta_ {ij} \rangle $$ are the simplest entangled states which can be conceived.

Two systems can get entangled without interacting directly, through a third system. For example, if in a three-qubits system, the second and the third both measure the first, the following can be obtained:

$$U\frac{1}{\sqrt 2}(|0\rangle + |1\rangle)|00\rangle = \frac{1}{\sqrt 2}(|000\rangle + |111\rangle) $$

Similarly, if the second qubit measures the first before being measured by the third, we can obtain:

$$\frac{1}{\sqrt 2}(|0\rangle + |1\rangle)|00\rangle$$ → $$\frac{1}{\sqrt 2}(|00\rangle + |11\rangle)|0\rangle$$ → $$\frac{1}{\sqrt 2}(|000\rangle + |111\rangle)$$

If two qubits are initially entangled and if the first one interacts by a SWAP with a third, then there is transfer of entanglement:

$$SWAP_{13}\frac{1}{\sqrt 2}(|00\rangle + |11\rangle)(\alpha|0\rangle + \beta|1\rangle) = SWAP_{13}\frac{1}{\sqrt 2}[\alpha(|000\rangle + |110\rangle) + \beta(|001\rangle + |111\rangle)]$$

$$= \frac{1}{\sqrt 2}[\alpha(|000\rangle + |011\rangle) + \beta(|100\rangle + |111\rangle)] = \frac{1}{\sqrt 2}(\alpha|0\rangle + \beta|1\rangle)(|00\rangle + |11\rangle) $$

The third qubit thus becomes entangled with the second without having interacted directly with it. The first qubit is disentangled through the SWAP with the third.

Everett relative states
In general, an observation changes the observed system and the measuring apparatus from a separable state to an entangled state. For an ideal measurement:

$$U(\sum_{ij} \alpha_{ij}|i,j\rangle_S)|ready\rangle_A = \sum_{ij} \alpha_{ij}|i,j\rangle_S|i\rangle_A$$

There is no entanglement only if the observed system is in an eigenstate of the measurement.

After the observation, $$\frac{\sum_{j} \alpha_{ij}|i,j\rangle_S}{|\sum_{j} \alpha_{ij}|i,j\rangle_S|}$$ is the state of the observed system relative, in Everett's sense, to the state $$ | i \rangle_A $$ of the measurement apparatus, and vice versa.

More generally, if $$ \sum_ {ij} \alpha_ {ij} | a_i \rangle | b_j \rangle $$ is the state of an AB system where the $$ | a_i \rangle $$ and the $$ | b_j \rangle $$ are any two orthonormal bases of A and B, then $$\frac{\sum_{i} \alpha_{ij} |a_i\rangle}{|\sum_{i} \alpha_{ij} |a_i\rangle|}$$ is the relative state of A with respect to the state $$|b_j\rangle$$ of B, and $$\frac{\sum_{j} \alpha_{ij} |b_j\rangle}{|\sum_{j} \alpha_{ij} |b_j\rangle|}$$ is the relative state of B with respect to the state $$|a_i\rangle$$ of A (Everett 1957).

For mathematical convenience, it is agreed that if the weight of $$ | b_j \rangle $$ is null in $$\sum_{ij} \alpha_{ij} |a_i\rangle|b_j\rangle$$ the state of A relative to $$ | b_j \rangle $$ is the null vector $$ 0 $$. It must be distinguished from $$ | 0 \rangle $$ whose length is one, because the length of $$ 0 $$ is zero. The vector $$ 0 $$ is not a state vector. It is the only element of the vector space of states of a system that can not be identified with state of the system state.

$$|\sum_{i} \alpha_{ij} |a_i\rangle|^2 = \sum_{i} | \alpha_{ij} |^2$$ is the weight of the state $$|b_j\rangle$$ of B in the state $$\sum_{ij} \alpha_{ij} |a_i\rangle|b_j\rangle$$ of AB. Similarly, $$|\sum_{j} \alpha_{ij} |b_j\rangle|^2 = \sum_{j} | \alpha_{ij} |^2$$ is the weight of the state $$|a_i\rangle$$ of A.

A pure state $$ | \psi \rangle $$ of a system AB can always be decomposed as follows:

$$ | \psi \rangle = \sum_i \sqrt {p_i} | a_i \rangle | b_i \rangle $$

where the $$ | a_i \rangle $$ are orthonormal vectors, as well as the $$ | b_i \rangle $$. It is called a Schmidt decomposition. It is unique when the $$ p_i $$ are all different. It contains at least two terms if and only if $$ | \psi \rangle $$ is an entangled state.

For each $$ i $$, $$ | a_i \rangle $$ is the state of A relative to the state $$ | b_i \rangle $$ of B, and vice versa. They have the same weight $$ p_i $$ in $$ | \psi \rangle $$.

$$ \sum_i (\sum_j \alpha_ {ij} | i, j \rangle_S) | i \rangle_A $$ is a Schmidt decomposition of the system SA. According to the Born rule, the weight $$ \sum_j | \alpha_ {ij} | ^ 2 $$ of $$ \sum_j \alpha_ {ij} | i, j \rangle_S $$ and of $$ | i \rangle_A $$ is the probability that the measurement provides the result $$ i $$.

The collapse of the state vector through observation is a disentanglement.
Most textbooks in quantum physics set forth the following two principles:

The evolution of the state vector is described by a unitary operator, or by the Schrödinger equation, as long as the studied system is not observed.

When the system is observed, its initial state $$ | \psi \rangle $$ before the observation becomes one of the states $$\frac{P(i)|\psi\rangle}{|P(i)|\psi\rangle|}$$ where $$P(i) = \sum_{j}|ij\rangle\langle ij|$$ is the projector on the subspace of eigenstates $$ | ij \rangle $$ of the result $$ i $$. It is then said that the state vector collapsed, a kind of quantum jump which moves the system from the state $$ | \psi \rangle $$ to the state $$ \frac {P (i) | \psi \rangle} {| P (i) | \psi \rangle |} $$. We also say that it is a wave function collapse, because a wave function represents the components of a state vector in a position state basis.

After a measurement which has given the result $$ i $$, the relative state of the observed system with respect to the observer system is $$ \frac {P (i) | \psi \rangle} {| P(i) | \psi \rangle |} $$. The collapse of the state vector is therefore the transition from the initial state vector to the state vector relative, in Everett's sense, to the result of the measurement.

The collapse of the state vector through observation is a disentanglement because the observed system passes into an eigenstate of the measurement after the observation. If the measurement is exact, ie if there is only one state of the observed system pointed to by the measurement result, the disentanglement is complete. If the observed system was entangled with its environment, it is no longer so after the measurement. The observation thus destroys any previous entanglement of the observed system with its environment.

Distentanglement through observation explains why quantum physics makes excellent predictions from the calculation on pure states. As interactions between all parts of the Universe occur constantly, and as interactions are often entangling, everything should be entangled with everything, or almost so. Matter never ceases to interact with matter. All material beings generally have a long history of interactions, and therefore of entanglements, with all the other material beings they have encountered. How is it then that one can describe their states by pure states, separated from the rest of the Universe, and make accurate predictions from such a description ?

We know the quantum states of material beings only if we give ourselves conditions of observation which enable us to know them. Since from our points of view our observations are disentangling, we can ignore all entanglements preceding our observations, and thus attribute state vectors to the systems we observe.

Apparent disentanglement results from real entanglement between the observed system and the observer.
The collapse of the state vector through observation can not be described by a unitary operator. We should therefore admit two kinds of evolution, one is unitary and occurs in the absence of observation, the other is not unitary and occurs during a measurement. But a measurement is a natural evolution. The principle of unitary evolution is universal. We assume that it describes all natural processes. We are therefore faced with a contradiction. The quantum jump, the collapse of the state vector, is a natural evolution, but it is not unitary.

Is quantum physics contradictory? Is it not possible to give a unified theory, which describes without contradiction, with the same principles, all the natural processes, whether or not there is observation?

One can believe that quantum physics is only an approximation, that the postulate of unitary evolution is not universal, that a new physics will explain in which cases the evolution is unitary, or almost, and in which other cases collapse of the state vector occurs. But until now, such a new physics, which surpasses and supplants quantum physics does not exist. The various speculations on this subject have never produced real fruits.

The postulate of the reduction of the state vector denies the existence theorem of multiple destinies. If an observation really leads to the collapse of the state vector then we only have a single destiny. The postulate of the collapse of the state vector allows us to preserve our prejudice, that the other destinies do not exist. It is its only justification. There are no others. We introduce a contradiction into the heart of quantum theory because we do not want to believe in other destinies, because we do not accept that reality could be more than what we can directly observe.

To explain our observation results, the postulate of the collapse of the state vector is not necessary. The principle of unitary evolution and the entanglement between the observed system and the observer are sufficient to account for the measured probabilities. To understand it, it suffices to reason on conditional probabilities: what is the probability that an observation provides a result knowing the result of an earlier observation?

Suppose that we do two successive measurements on the same system (Everett 1957), prepared initially in the state $$ | \psi \rangle $$. After the two measurements, the overall system (observed system + measuring devices) is found in the following state:

$$\sum_{ij} (P_2(j) U_S P_1(i) |\psi\rangle)(U_{A_1}|i\rangle_1)|j\rangle_2$$

where the $$ P_1 (i) $$ and $$ P_2 (j) $$ are the projectors associated with the two successive measurements, $$U_S$$ is the evolution operator of the observed system between the two measurements and $$ U_ {A_1} $$ that of the first measuring device.

The probability of obtaining the result $$ j $$ with the second measurement, knowing that $$ i $$ was obtained with the first is:

$$\frac{Pr(ij)}{Pr(i)} = \frac{|P_2(j) U_S P_1(i) |\psi\rangle|^2}{|P_1(i) |\psi\rangle|^2}$$

This is the same probability as that which would be obtained if, immediately after the first measurement, the system had been in the state $$ \frac {P_1 (i) | \psi \rangle} {| P_1 (i) \psi \rangle |} $$.

Once we get the result $$ i $$ everything happens as if the state of the observed system had gone from the state $$ | \psi \rangle $$ to the state $$ \frac {P_1 (i) | \psi \rangle} {| P_1 (i) | \psi \rangle |} $$, as if there had been a collapse of the state vector. The other states of the observed system, the $$ \frac {P_1 (i') | \psi \rangle} {| P_1 (i') | \psi \rangle | }$$ where $$i'$$ is different from $$ i $$, have no influence on subsequent measurements. If they exist, they are obtained in other destinies which have no influence on ours, since they have no influence on our observations.

If one takes quantum physics seriously, if one believes that it correctly describes reality, if one thus accepts the existence theorem of multiple destinies, one does not need the postulate of the collapse of the state vector. The principle of unitary evolution is sufficient to describe reality.

Depending on the point of view chosen, it can be said that the observations are entangling or disentangling. They are entangling because they lead to an entanglement between the observed system and the observer. They are disentangling because they lead to an apparent collapse of the state vector of the observed system.

Disentanglement through observation is only a subjective effect. The observation is disentangling only from the point of view of the observer. From an external point of view, on the contrary, an observation is generally entangling. It is precisely the entanglement between the observed system and the observer that produces the apparent collapse of the state vector, because after an observation an observer can only know the relative state, in Everett's sense, of the observed system. All other states can no longer influence subsequent observations. But it is only from the point of view of the observer that the state vector of the observed system has been reduced to the state relative to the observer. The collapse of the state vector is only a kind of illusion, produced by our viewpoint of an entangled observer, who knows of reality only a tiny part, who can only know one destiny among myriads of others, all equally real.

Dirac's error
Dirac thought he could prove the principle of state vector reduction from the other quantum principles. If his proof had been good, he would have proved that quantum physics, with or without the principle of state vector reduction as an axiom, would be contradictory, because the reduction of the state vector contradicts the Schrödinger equation, and because principles which lead to a consequence which contradicts them are necessarily contradictory.

Here is his reasoning:

''« When we measure a real dynamical variable $$\xi$$, the disturbance involved in the act of measurement causes a jump in the state of the dynamical system. From physical continuity, if we make a second measurement of the same dynamical variable $$\xi$$ immediately after the first, the result of the second measurement must be the same as that of the first. Thus after the first measurement has been made, there is no indeterminacy in the result of the second. Hence, after the first measurement has been made, the system is in an eigenstate of the dynamical variable $$\xi$$, the eigenvalue it belongs to being equal to the result of the first measurement. This conclusion must still hold if the second measurement is not actually made. In this way we see that a measurement always causes the system to jump into an eigenstate of the dynamical variable that is being measured, the eigenvalue this eigenstate belongs to being equal to the result of the measurement. »'' (Dirac 1958, §10, p.36)

Dirac's error is to ignore the entanglement between the measurement system and the measured system. After the first measurement, the measured system is entangled with the system that measured it. To predict the result of the second measurement, this entanglement must be taken into account. The entanglement with the first measurement system is sufficient to explain physical continuity, the identity of the results between two immediately successive measurements of the same dynamic variable. The principle of state vector reduction is therefore not a consequence of physical continuity. Everett was the first to correct this fundamental error.

Can we see non-localized macroscopic states?
If a quantum system can be in the states $$ | 1 \rangle $$ and $$ | 2 \rangle $$, it can also be in the state $$ \alpha | 1 \rangle + \beta | 2 \rangle $$. For example, $$ | 1 \rangle $$ and $$ | 2 \rangle $$ can be two states of the moon at different positions. But the moon is never seen in different positions. A non-localized state of the moon such as $$ \frac {1} {\sqrt 2} (| 1 \rangle + | 2 \rangle) $$ is never observed. The same is true for all macroscopic objects (Schrödinger 1935) even if they are very small. (From the point of view of the biologist, a bacterium is microscopic, but from the point of view of the physicist, it is macroscopic, because it is made up of billions of atoms.)

Vision involves the formation of images. Each point of the image represents a point of an object in the visual field. If the object is not located, it can not have a clear and stable image. But perhaps we could see it sometimes in one position, sometimes in another. The apparent reduction of the state vector during an observation proves that it is not possible. If the object seen is initially unlocalized, it passes, from my point of view, into a localized state as soon as I see it at a defined position. If I repeat the observation several times, I will see it in the same place. The other components of the initial state vector can no longer be observed, once one of the components has been selected by an observation. Due to the entanglement by observation, the simple fact of seeing an object at a defined position suffices to destroy its initial non-localized state.

But this does not prove that it is impossible to observe non-localized macroscopic states (cf. 4.20), it only proves that it is impossible to see them.

The quantum explanation of intersubjectivity
Since the Universe is in an entangled state, one can not in general attribute a truly defined state to each of its parts. The problem is not that we do not know the states of these parts, but that they are not defined, that they simply do not exist, at least according to the theory. However we know the Universe always by observing its parts. We can not know anything about it apart from the parts we observe. And when we have observed them we believe we know how they really are. Why say that we know their real state, whereas according to the theory such a state does not even exist ?

The response of quantum physics is essentially relativistic, in the sense that the observed reality is always relative to the observers. The beings actually present in the Universe do not have a single state defined absolutely, invariantly, ie the same for all observers. The theory does not attribute to them a single state, but many states, because the real state of a being is always relative to the state of another being.

If A, B and C are three beings, the state of A relative to a state of B is in general different from the state of A relative to a state of C. We conclude that B and C each have their reality. The representations of the world of all observers should therefore always be different. How is it then that we can agree on the same reality that we all observe?

Everett (1957) showed that communication between observers is sufficient to observe the same reality:

Suppose that A is observed by B which is then observed by C such that the information possessed by B on A is transmitted to C.

We argue on a simplified model: the states $$ | a_i \rangle $$ of A are the eigenstates associated with the pointer states $$ | b_i \rangle $$ of B, which are themselves the eigenstates associated with the pointer states $$ | c_i \rangle $$ of C.

Starting from the initial state $$ \sum_i \alpha_i | a_i \rangle $$ of A, one obtains, after the observation of A by B, the state $$ \sum_i \alpha_i | a_i \rangle | b_i \rangle $$ of AB. After the observation of B by C, one obtains the state $$ \sum_i \alpha_i | a_i \rangle | b_i \rangle | c_i \rangle $$ of ABC. Before the communication between B and C, the state vector of A relative to a state of C is not defined, because A is entangled with B. We show below (cf. 4.15) that it can be defined as a mixed state, but it is not a state vector. After the communication between B and C, the state $$ | a_i \rangle $$ of A relative to a state $$ | c_i \rangle $$ of C is the same as the state of A relative to the state $$ | b_i \rangle $$ of B relative to the state $$ | c_i \rangle $$ of C. The reality of A is therefore the same for B and C. Quantum theory thus explains the possibility of intersubjectivity in an essentially relativistic universe.

Einstein, Bell, Aspect and the reality of quantum entanglement
Suppose that two entangled qubits in the state $$\frac{1}{\sqrt 2}(|00\rangle + |11\rangle)$$ are very far from each other. One can therefore make a measurement on one without touching the other. If we measure the first qubit, if it has $$ | 0 \rangle $$ and $$ | 1 \rangle $$ for eigenstates, we can deduce from the result of the measurement the state $$ | 0 \rangle $$ or $$ | 1 \rangle $$ of the second qubit, which is very far from the measuring instrument. Einstein concludes that this state must represent an element of reality which existed before the measurement (Einstein, Podolsky & Rosen 1935). An immediate instantaneous action of the first qubit or of the measuring instrument on the second qubit is excluded. It is contrary to the principles of physics which he has greatly contributed to establish. But the quantum equations do not assign a defined state to the second qubit before the measurement. According to Einstein, therefore, they do not completely describe reality, there must be hidden variables, that is, real quantities ignored by quantum theory, which must complete the quantum description of reality, necessarily incomplete.

Einstein could not believe that quantum physics gives a complete description of reality because he did not want to give up the principle of the separability of reality. All classical physics respects the principle that the state of the system is always to be identified with the list of states of its parts. To speak of an entangled state, of a defined state of a system in which the parts have no definite state, has no meaning in classical physics.

The existence of these hidden variables, postulated by Einstein, remained very hypothetical until Bell understood in 1964 how to put this hypothesis to the test of experience (Bell 1988). The experiment was made (Aspect, Grangier & Roger 1982, Gisin, Tittel, Brendel & Zbinden 1998, Gisin 2014) and the result is very clear: Einstein was wrong to believe that reality is necessarily separable. The entangled quantum states really exist and they completely describe reality, as far as we know.

One can understand the reasoning of Bell and the results obtained by Aspect by studying a system with two entangled qubits and considering two measuring instruments for each of them.

Let $$|x^+\rangle = \frac{1}{\sqrt 2}(|0\rangle + |1\rangle)$$ and $$|x^-\rangle = \frac{1}{\sqrt 2}(|0\rangle - |1\rangle)$$ be two new basis vectors (the names $$x^+$$ and $$x^-$$ come from the spin 1/2 theory).

{$$|0\rangle,|1\rangle$$} and {$$|x^+\rangle,|x^-\rangle$$} are the two bases of eigenstates of the measuring instruments of the first qubit.

{$$|v^+\rangle,|v^-\rangle$$} are {$$|w^+\rangle,|w^-\rangle$$} are the two bases of eigenstates of the measuring instruments of the second qubit.

At each iteration of the experiment, Alice chooses one of the two instruments and applies it to her qubit, Bob does the same on the other qubit. There are therefore four possible experiences, which may have as results :


 * $$0v^+, 0v^-, 1v^+, 1v^-$$


 * $$0w^+, 0w^-, 1w^+, 1w^-$$


 * $$x^+v^+, x^+v^-, x^-v^+, x^-v^-$$


 * $$x^+w^+, x^+w^-, x^-w^+, x^-w^-$$

It seems that the Bell state $$\frac{1}{\sqrt 2}(|00\rangle + |11\rangle)$$ favors the basis {$$|0\rangle,|1\rangle$$} but it is an illusion :

$$\frac{1}{\sqrt 2}(|x^+x^+\rangle + |x^-x^-\rangle)=\frac{1}{2 \sqrt 2}[ (|0\rangle + |1\rangle)(|0\rangle + |1\rangle) + (|0\rangle - |1\rangle)(|0\rangle - |1\rangle)] = \frac{1}{\sqrt 2}(|00\rangle + |11\rangle)$$

If one choses :

$$|v^+\rangle=cos(\pi/8)|0\rangle + sin(\pi/8)|1\rangle$$

$$|v^-\rangle=sin(\pi/8)|0\rangle - cos(\pi/8)|1\rangle$$

$$|w^+\rangle=cos(3\pi/8)|0\rangle + sin(3\pi/8)|1\rangle$$

$$|w^-\rangle=sin(3\pi/8)|0\rangle - cos(3\pi/8)|1\rangle$$

one can positively correlate the eight following couples: $$0$$ with $$v^+$$ and $$w^-$$, $$x^+$$ with $$v^+$$ and $$w^+$$, $$1$$ with $$v^-$$ and $$w^+$$, $$x^-$$ with $$v^-$$ and $$w^-$$:

$$Pr(0v^+) = \frac{1}{2} | \langle 0v^+| ( |00\rangle + |11\rangle ) |^2 = \frac{1}{2} cos^2(\pi/8) > Pr(0) Pr(v^+)= \frac{1}{4}$$

The same result is found for the other seven couples. Their sum is greater than 3:

$$4 cos^2(\pi/8) \approx 3.414$$

Bell understood that such a result, predicted by quantum physics and empirically verifiable, is incompatible with the principle of separability of reality.

Since we have:

$$\frac{1}{\sqrt 2}(|00\rangle + |11\rangle) = \frac{1}{\sqrt 2}(|v^+v^+\rangle + |v^-v^-\rangle) = \frac{1}{\sqrt 2}(|w^+w^+\rangle + |w^-w^-\rangle)$$,

if Alice measured on the first qubit the states $$ | v^+ \rangle $$ and $$ | v^- \rangle $$, she could deduce Bob's results from hers, if he measured too the states $$ | v^+ \rangle $$ and $$ | v^- \rangle $$ on the second qubit. The same holds for $$ | w^+ \rangle $$ and $$ | w^- \rangle $$. According to Einstein's criterion, there should therefore be elements of reality which determine Bob's results. As the roles of Alice and Bob are symmetrical, there should also be elements of reality that determine Alice's results. We should therefore be able to distribute the experiments into sixteen groups which correspond to the sixteen possible combinations of elements of reality. If the experiment is well done, these sixteen groups should always occur with the same probabilities. We can not measure these probabilities because quantum physics forbids measuring the four elements $$0$$ or $$1$$, $$x^+$$ or $$x^-$$, $$v^+$$ or $$v^-$$, and $$w^+$$ or $$w^-$$, of a combination. But if we accept Einstein's reasoning based on the postulate of separability, we must accept the existence of these sixteen probabilities.

The eight measurable probabilities of correlated couples can be considered as the expectation values of eight random variables. For example $$ Pr (0v^+) $$ is the expectation value of the variable which equals 1 if $$ 0v^+ $$ is observed and zero otherwise. The sum of these eight variables is also a random variable whose expectation value is the sum of the eight expectation values of the summed variables. It is possible to verify on each of the sixteen combinations that this sum is always less than or equal to three. For example, it is three for the combination $$ 0x^+ v^+ w^+ $$. Hence its expectation value is necessarily less than or equal to three. The principle of separability thus contradicts the quantum prediction which has been confirmed empirically.

Copresence without possible encounter
Is it possible to be in the same place without being able to meet?

Let two beings A and B be able to interact when they are nearby. $$|0\rangle_A$$ and $$|0\rangle_B$$ are states of A and B when they are in a place 0, $$|1\rangle_A$$ and $$|1\rangle_B$$ for another place 1. It is assumed that if they are in different places, they can not interact. So we have :

$$ U (| 0 \rangle_A | 1 \rangle_B) = (U| 0 \rangle_A)(U | 1 \rangle_B) $$

$$ U (| 1 \rangle_A | 0 \rangle_B) = (U| 1 \rangle_A)(U | 0 \rangle_B) $$

If they are initially in the entangled state $$\frac{1}{\sqrt 2}(|0\rangle_A|1\rangle_B+|1\rangle_A|0\rangle_B)$$, then they will not interact with each other.

Proof : $$U[\frac{1}{\sqrt 2}(|0\rangle_A|1\rangle_B+|1\rangle_A|0\rangle_B)] =\frac{1}{\sqrt 2}[(U|0\rangle_A)(U|1\rangle_B)+(U|1\rangle_A)(U|0\rangle_B)]$$. End of proof.

However, there is a probability $$ \frac {1} {2} $$ to find A in the place 0, similarly for B. They are therefore both present in the place 0, at least partially, but they can not meet. Similarly for the place 1.

For example Alice and Bob have agreed to an appointment, but for security reasons they have not defined the place in advance. They use an entangled pair, assumed in the state $$\frac{1}{\sqrt 2}(|0\rangle_a|0\rangle_b+|1\rangle_a|1\rangle_b)$$. Just before the rendezvous, each must observe his or her part of the pair and decide accordingly the place where they will meet. But unbeknownst to Alice and Bob, Charles is jealous and replaced the pair with another, $$ \frac {1} {\sqrt 2} (| 0 \rangle_a | 1 \rangle_b + | 1 \rangle_a | 0 \rangle_b) $$. Then Alice and Bob will go to their rendezvous without being able to meet there while being both present.

Entangled space-time
Parts of a quantum system are generally conceived as material beings, particles, atoms, molecules, or larger objects. But at a more fundamental level, it seems that elementary particles and their compounds should be considered as forms of excitation of space, or of the quantum vacuum. The parts of the system are then conceived as regions of space. The state space of the whole space is the tensor product of the state spaces of the regions in which it has been decomposed. The state vector of the universe thus describes the entanglement between all regions of space (Wallace 2012). But it describes the multiple destinies of all beings in the universe. Most of these destinies are separated and can never meet, while they occur in the same space and sometimes in the same places. This poses a difficulty of principle: why say of two destinies that they pass by the same place if they can not meet there?

This makes sense because even if A and B can not meet in a certain place when they both pass through it, they can still meet each a third being C present in this place. C will never meet A and B at the same time, only one or the other, but not both. The initial destiny of C will bifurcate into two destinies, one where he meets A, the other B.

Everett's thesis is sometimes stated in a way that seems a little absurd: at each observation the universe of the observer would separate into several independent universes that would each correspond to a possible result. We then conceive the evolution of the universe as a tree in which each branch can be divided into several branches which can then be divided, and so on. This image of the tree is sometimes useful (cf. chapter 6) but it incorrectly suggests that there would be several spaces and several times, several branches of the flow of time and several spaces where they flow. Everett's thesis does not say anything like that. It takes quantum physics as it is, without adding any hypothesis. It therefore admits that there is a single space-time. It only notes that quantum physics attributes to material beings not one but many destinies which are all entangled in the same space-time.

Action, reaction and no cloning
During the CNOT interaction it seems that the control qubit acts on the target qubit but not the reverse. This appearance is false. If the initial state of the control qubit is $$\alpha|0\rangle + \beta|1\rangle$$, the final state after the measurement is:

$$ CNOT (\alpha | 0 \rangle + \beta | 1 \rangle) = \alpha | 00 \rangle + \beta | 11 \rangle) $$

When $$ \alpha $$ and $$\beta$$ are both nonzero, the control qubit does not remain in its initial state. It gets entangled with the target qubit.

In fact, there is a hidden symmetry in the CNOT quantum gate. With an appropriate change of basis, the first qubit becomes the target and the second the control (Nielsen & Chuang 2010): Let $$|x^+\rangle = \frac{1}{\sqrt 2}(|0\rangle + |1\rangle)$$ and $$|x^-\rangle = \frac{1}{\sqrt 2}(|0\rangle - |1\rangle)$$ be two new basis vectors (the names $$x^+$$ and $$x^-$$ come from the spin 1/2 theory). In this new basis, the $$ CNOT $$ operator is determined by:

$$CNOT|x^+x^+\rangle = |x^+x^+\rangle$$

$$CNOT|x^-x^+\rangle = |x^-x^+\rangle$$

$$CNOT|x^+x^-\rangle = |x^-x^-\rangle$$

$$CNOT|x^-x^-\rangle = |x^+x^-\rangle$$

The calculation is very simple. For example :

$$ CNOT|x^+x^-\rangle = CNOT[\frac{1}{2}(|0\rangle + |1\rangle)(|0\rangle - |1\rangle)] $$ $$ = CNOT[\frac{1}{2}(|00\rangle - |01\rangle + |10\rangle - |11\rangle)] $$

$$ = \frac{1}{2}(|00\rangle - |01\rangle + |11\rangle - |10\rangle) $$

$$ = \frac{1}{2}(|0\rangle - |1\rangle)(|0\rangle - |1\rangle) = |x^-x^-\rangle $$

A classical CNOT gate copies the information carried by the control bit to the target bit. It is a cloning of information. At first glance it seems that a quantum gate CNOT does the same thing with qubits. But this appearance is false. The quantum information carried by the control qubit is not copied to the target qubit. The cloning of quantum information is actually forbidden by quantum principles (Dieks 1982, Wooters & Zurek 1982). For quantum cloning, there should be an interaction U such that:

$$ U | \phi \rangle | ready \rangle = | \phi \rangle | \phi \rangle $$

for all states $$ | \phi \rangle $$ of the cloned system. In particular, for two orthogonal states $$ | \phi_1 \rangle $$ and $$ | \phi_2 \rangle $$:

$$ U | \phi_1 \rangle | ready \rangle = | \phi_1 \rangle | \phi_1 \rangle $$

$$ U | \phi_2 \rangle | ready \rangle = | \phi_2 \rangle | \phi_2 \rangle $$

But we do not have $$U(|\phi_1\rangle+|\phi_2\rangle)|ready\rangle = (|\phi_1\rangle+|\phi_2\rangle)(|\phi_1\rangle+|\phi_2\rangle)$$ because the interaction is necessarily entangling:

$$U(|\phi_1\rangle+|\phi_2\rangle)|ready\rangle = |\phi_1\rangle|\phi_1\rangle + |\phi_2\rangle|\phi_2\rangle$$

Entanglement is responsible for the impossibility of quantum cloning.

The ideal measurement of entangled states
If we observe separately the parts of an entangled system, we necessarily destroy their entanglement, since we are led to attribute to each a determinate state. How then can we observe entangled states without destroying them?

We want to make an ideal measurement whose eigenstates are the Bell states of a two-qubit system AB.

Two observer qubits C and D are used to measure the Bell states of the first two qubits A and B.

A first measurement method involves an interaction of A and B as follows (Kaye & Laflamme 2007):

$$U_1|00\rangle = |x^+ 0\rangle$$

$$U_1|01\rangle = |x^+ 1\rangle$$

$$U_1|10\rangle = |x^- 1\rangle$$

$$U_1|11\rangle = |x^- 0\rangle$$,

where $$|x^+\rangle = \frac{1}{\sqrt 2}(|0\rangle + |1\rangle)$$ and $$|x^-\rangle = \frac{1}{\sqrt 2}(|0\rangle - |1\rangle)$$

$$U_1$$ consists of operating the CNOT gate on AB and then the Hadamard H gate on A:

$$ U_1 = H_A CNOT_ {AB} $$

$$ U_1 $$ destroys the entanglement of the Bell states:

$$U_1|\beta_{00}\rangle = |00\rangle$$

$$U_1|\beta_{01}\rangle = |01\rangle$$

$$U_1|\beta_{10}\rangle = |10\rangle$$

$$U_1|\beta_{11}\rangle = |11\rangle$$

Then C and D measure the states $$ | 0 \rangle $$ and $$ | 1 \rangle $$ of A and B:

$$ U_2 = CNOT_ {AC} CNOT_ {BD} $$

More explicitly:

$$U_2|0000\rangle = |0000\rangle$$

$$U_2|0100\rangle = |0101\rangle$$

$$U_2|1000\rangle = |1010\rangle$$

$$U_2|1100\rangle = |1111\rangle$$

The evolution $$ U_1 $$ followed by $$ U_2 $$ therefore gives for the Bell states of AB:

$$ U_2 U_1 | \beta_ {00} \rangle | 00 \rangle = | 0000 \rangle $$

$$ U_2 U_1 | \beta_ {01} \rangle | 00 \rangle = | 0101 \rangle $$

$$ U_2 U_1 | \beta_ {10} \rangle | 00 \rangle = | 1010 \rangle $$

$$ U_2 U_1 | \beta_ {11} \rangle | 00 \rangle = | 1111 \rangle $$

This is not yet an ideal measurement, because A and B have been disturbed. For them to return to their initial state, it is necessary to operate $$U_1^{-1} = CNOT_{AB}^{-1} H_A^{-1} = CNOT_{AB} H_A $$ because CNOT and H are their own inverses.

The complete ideal measurement is therefore defined by:

$$U = U_1^{-1} U_2 U_1 = CNOT_{AB} H_A CNOT_{AC} CNOT_{BD} H_A CNOT_{AB}$$

$$U|\beta_{00}\rangle |00\rangle = |\beta_{00}\rangle|00\rangle$$

$$U|\beta_{01}\rangle |00\rangle = |\beta_{01}\rangle|01\rangle$$

$$U|\beta_{10}\rangle |00\rangle = |\beta_{10}\rangle|10\rangle$$

$$U|\beta_{11}\rangle |00\rangle = |\beta_{11}\rangle|11\rangle$$

It is actually an ideal measurement of the Bell states of AB.

This method of measurement requires that A and B interact to be disentangled (if they are initially entangled), then they are measured separately, then they interact once more to get entangled again. If A and B are distant and can not interact directly, is it still possible to make an ideal measurement of their entangled states?

The following measurement process produces such a result. First, C is used to measure successively A and B with a CNOT interaction:

$$ V_1 = CNOT_ {BC} CNOT_ {AC} $$

D is then used to measure A and B with an interaction W which measures the states $$ | x ^ + \rangle $$ and $$ | x ^ - \rangle $$:

$$ W | x ^ + 0 \rangle = | x ^ + 0 \rangle $$

$$ W | x ^ + 1 \rangle = | x ^ + 1 \rangle $$

$$ W | x ^ -0 \rangle = | x ^ -1 \rangle $$

$$ W | x ^ -1 \rangle = | x ^ -0 \rangle $$

The measurement of A and B by D is defined by:

$$ V_2 = W_ {BD} W_ {AD} $$

One can verify (the calculation is simple but a little tedious) that one gets an ideal measurement of the Bell states. Explicitly:

$$ V | \beta_ {00} \rangle | 00 \rangle = | \beta_ {00} \rangle | 00 \rangle $$

$$ V | \beta_ {01} \rangle | 00 \rangle = | \beta_ {01} \rangle | 10 \rangle $$

$$ V | \beta_ {10} \rangle | 00 \rangle = | \beta_ {10} \rangle | 01 \rangle $$

$$ V | \beta_ {11} \rangle | 00 \rangle = | \beta_ {11} \rangle | 11 \rangle $$

where $$ V = V_2 V_1 = W_ {BD} W_ {AD} CNOT_ {BC} CNOT_ {AC} $$

So we have:

$$ V = SWAP_ {CD} U $$

Why does not the measurement of entangled states enable us to observe our other destinies?
If I observe, by a CNOT measurement, a qubit prepared in the state $$ \frac {1} {\sqrt 2} (| 0 \rangle + | 1 \rangle) $$, the observation leads to the state vector $$ \frac {1} {\sqrt 2} (| 00 \rangle + | 11 \rangle) $$ where the second qubit registers the result of the measurement. $$ \frac {1} {\sqrt 2} (| 00 \rangle + | 11 \rangle) $$ can be interpreted as a superposition of two destinies, one where I get the result $$ 0 $$, the other where I get the result $$ 1 $$. If we take seriously quantum physics these destinies both exist. The observation of the qubit makes my destiny bifurcate into two separate destinies, but I know of only one. Is it possible, however, to verify the existence of this other destiny? If it exists, we would like to see it. Can we make a television that shows us other destinies?

It seems that the ideal measurement of entangled states enables us to observe these other destinies. After observing the first qubit, I place myself and it in front of a measuring device which detects our state of entanglement. In order to be sure that such an experiment (observation of a qubit prepared in the state $$ \frac {1} {\sqrt 2} (| 0 \rangle + | 1 \rangle) $$ and ideal measurement of our entangled state) always puts us into an entangled state, I repeat it many times. If the entanglement measuring device informs me each time that we are in the expected entangled state, I shall have confirmation that other destinies exist.

There is, however, one problem. When I make an observation, I am able to memorize it. So there is a part of me that keeps this information. At least it takes a qubit which remains in the same state. But the ideal measurement of the entanglement of the two qubits can disturb them. An ideal measurement does not disrupt the observed system only if it is in an eigenstate of the measurement.

Suppose that the observed qubit is initially in the state $$ | 0 \rangle $$, this observation leads to the state vector $$ | 00 \rangle $$. There is only one measurement result and therefore no bifurcation in several destinies. Since this state is not entangled, it is not an eigenstate of the ideal measurement of entangled states. If I place myself in front of a device that performs such a measurement, I will be disturbed. My qubit which recorded the first observation will not be able to retain its information. The proposed protocol, intended to verify the existence of another destiny, can not therefore function if I have properly memorized the result of my first observation.

This shows that we can not observe a destiny in which we have memorized observations different from those we have memorized in this destiny.

The apparent collapse of the state vector is enough to prove that we can not observe the other destinies, since after an observation everything happens as if the undetected components of the state vector no longer exist: they can no longer have effect on subsequent observations. The above reasoning clarifies this conclusion. It is the memorization of the results of observation which prevents us from observing destinies in which we have obtained other results.

Reduced density operators
A reduced density operator represents the maximum information about a part of a system as long as the state of the remainder is ignored.

When the state $$ | \psi \rangle $$ of a system AB is given as a Schmidt decomposition:

$$ | \psi \rangle = \sum_i \sqrt {p_i} | a_i \rangle | b_i \rangle $$

where the $$ | a_i \rangle $$ are orthonormal vectors, as well as the $$ | b_i \rangle $$, the reduced density operators of $$ | \psi \rangle \langle \psi | $$ are :

$$ \rho_A = \sum_i p_i | a_i \rangle \langle a_i | $$

and

$$ \rho_B = \sum_i p_i | b_i \rangle \langle b_i | $$

In general, the reduced density operators are obtained by assigning to the basis states of a part of a system probabilities equal to their weight in the state of the system:

If AB is in the state $$\sum_{ij} \alpha_{ij} |a_i\rangle|b_j\rangle$$ then:

$$\rho_A = \sum_i ( \sum_{j} |\alpha_{ij}|^2 ) |a_i\rangle \langle a_i|$$

and

$$\rho_B = \sum_j ( \sum_{i} |\alpha_{ij}|^2 ) |b_j\rangle \langle b_j|$$

$$ \rho_A $$ is sufficient to determine all the probabilities of the observation results on A alone, when it is separated from B. We can therefore consider that it represents the mixed state of A.

The reduced density operators can be defined with the partial trace operation:

$$ \rho_A = Tr_B (\rho_ {AB}) $$

and

$$ \rho_B = Tr_A (\rho_ {AB}) $$

where $$ Tr_X (O) $$ is the partial trace on $$ X $$ of an operator $$ O $$.

It can be defined as follows:

Let be $$O = \sum_{ijkl} O_{ijkl}|a_i\rangle|b_j\rangle \langle a_k|\langle b_l|$$

where the $$ | a_i \rangle $$ are a basis of the state space of A, and the $$ | b_i \rangle $$ a basis of the state space of B, then:

$$ Tr_A (O) = \sum_ {jlm} O_ {mjml} | b_j \rangle \langle b_l | $$

and

$$ Tr_B (O) = \sum_ {ikm} O_ {imkm} | a_i \rangle \langle a_k | $$

In particular :

$$ Tr_A (O_A \otimes O_B) = Tr (O_A) O_B $$

and

$$ Tr_B (O_A \otimes O_B) = Tr (O_B) O_A $$

If a $$ \rho_ {AB} $$ density operator is separable ($$ \rho_ {AB} = \rho_A \otimes \rho_B $$) then $$ \rho_A $$ and $$ \rho_B $$ are its reduced density operators.

Whether separable or not, $$ \rho_ {AB} $$ is always sufficient to determine the reduced operators $$ \rho_A $$ and $$ \rho_B $$. It can be concluded that the state of AB is sufficient to determine the states of A and B. With reduced density operators the states of the parts can be determined from the state of the whole. But this is not enough to make the mystery of entanglement disappear, because if it represents an entangled state, $$\rho_{AB}$$ is not determined by $$\rho_A$$ and $$\rho_B$$.

For example, if $$|\psi\rangle = \frac{1}{\sqrt 2}(|0\rangle_A|0\rangle_B + |1\rangle_A|1\rangle_B)$$ then:

$$\rho_A = \frac{1}{2}(|0\rangle_A \langle 0|_A + |1\rangle_A \langle 1|_A)$$

and

$$\rho_B = \frac{1}{2}(|0\rangle_B \langle 0|_B + |1\rangle_B \langle 1|_B)$$

But $$ | \psi \rangle \langle \psi | $$ is entangled while $$ \rho_A \otimes \rho_B $$ is separable.

Relative density operators
Let $$ \rho_ {AB} $$ be the density operator of the system AB.

Let $$ \rho'_B $$ be any given density operator of B:

$$ \rho_B'= \sum_i p_i | b_i \rangle \langle b_i | $$

The state $$ \rho'_A $$ of A relative to the state $$ \rho'_B $$ of B when AB is in the state $$ \rho_ {AB} $$ is defined by:

$$ \rho'_A = \sum_i p_i | a_i \rangle \langle a_i | $$

where the $$ | a_i \rangle $$ are the states of A relative to the states $$ | b_i \rangle $$ of B.

We can then define the relative states, generally mixed, of any part of the Universe with respect to any pure or mixed state of any other.

Let A and B be two parts of the Universe and E their environment. The state of the Universe is a density operator $$ \rho_ {ABE} $$. We can then define the state $$ \rho'_ {AE} $$ of AE relative to any state $$ \rho'_B $$ of B. By partial trace one obtains then the state $$ \rho'_A $$ of A relative to the state $$ \rho'_B $$ of B:

$$ \rho'_A = Tr_E (\rho'_ {AE}) $$

Why don't entangled pairs enable us to communicate?
Let A and B be two parts of an AB system. It is assumed that they have interacted in the past and that AB is now in an entangled state where they are very far from each other. If we observe only one of the parts, and if we reason on the reduction of the state vector as if it were a real effect, we conclude that there should be an instantaneous action at a distance from the observed part on the unobserved one, because the state of B after the measurement is different from its previous state. It is then surprising that this effect, supposed to be real, does not enable us to communicate instantaneously at a distance. If there was really action, there should be a possibility of communication.

In general, whether the action is instantaneous or delayed, the entangled pairs never enable us to communicate in the way suggested by the reduction of the state vector. From this point of view quantum physics is not distinguished from classical physics. For there to be communication, or transport of information, there must be an action or an interaction which travels at the speed of light or at a lower speed. When a distant part of an entangled pair is observed, there is no interaction with the unobserved part. No measurable effect of the former on the latter can be detected.

Formally, this lack of communication between parts results in the invariance of the reduced density operator of the unobserved part at the end of the observation. The unobserved part does not change its state, it is not disturbed by the observation of the other part. It is understood that the non-perturbed state is a mixed state. A reduced density operator determines all the probabilities of the measurements performed on a part (see 2.8) as long as there is no information on the results of the measurements performed on the other parts. As an observation does not change the reduced density operators of unobserved parts, it can have no measurable effect on them.

Decoherence through entanglement
When a system is in a pure state, a superposition such as $$ \sum_ {i} \alpha_ {i} | i \rangle $$ is said to be coherent. It's almost a pleonasm. A true quantum superposition is always coherent. But a mixture of states $$ | i \rangle $$ with probabilities $$ p_i $$ is sometimes incorrectly called an incoherent superposition.

The formalism of density operators specifies this difference. The density operator of a coherent superposition is:

$$(\sum_i \alpha_i|i\rangle)(\sum_i \alpha_{i}^*\langle i|) = \sum_{ij} \alpha_i \alpha_{j}^*|i\rangle \langle j|$$ while for an incoherent superposition it is:

$$ \sum_ {i} p_i |i\rangle \langle i| $$

The non-diagonal elements $$ \alpha_i \alpha_ {j} ^ * $$ of the density matrix of a coherent superposition thus make all the difference with an incoherent superposition. They are sometimes called the coherences of this density matrix. $$\alpha_i \alpha_{j}^*$$ is the coherence between the states $$|i\rangle$$ and $$|j\rangle$$.

When a measurement is ideal, the eigenstates of the measurement are not disturbed by the observation. On the other hand, any coherent superposition of eigenstates associated with distinct eigenvalues ​​is disrupted: Let a system initially in the state $$ \sum_ {i} \alpha_ {i} | i \rangle_S $$, where it has been assumed that the measurement results each have a single eigenstate, in order to simplify the writings. After the observation it is entangled with the measuring apparatus, because the final state of the global system is $$ \sum_ {i} \alpha_ {i} | i \rangle_S | i \rangle_A $$.

After the measurement, the state of the observed system can be considered a mixture of the states associated with all the possible results. For an ideal measurement, we already know, with the Born rule, the density operator which represents this mixture, but it is worth doing the calculation in detail to compare it with interactions more general than ideal measurements:

$$\rho^f_{S} = Tr_A[(\sum_i \alpha_i|i\rangle_S|i\rangle_A)(\sum_i \alpha_{i}^*\langle i|_S\langle i|_A)] = Tr_A(\sum_{ij} \alpha_i \alpha_{j}^* |i\rangle_S|i\rangle_A \langle j|_S\langle j|_A)$$

$$= \sum_{ij} \alpha_i \alpha_{j}^* Tr_A(|i\rangle_S|i\rangle_A \langle j|_S\langle j|_A) = \sum_{ij} \alpha_i \alpha_{j}^* |i\rangle_S\langle j|_S Tr(|i\rangle _A\langle j|_A)$$

$$= \sum_{i} \alpha_i \alpha_{i}^* |i\rangle_S \langle i|_S$$

where the last equality comes from the fact that the $$ | i \rangle_A $$ are orthonormal.

Since the non-diagonal elements of the density matrix have disappeared, the coherence of the initial superposition has been destroyed. This effect of disturbance by measurement is called decoherence (Zurek 2003). In the particular case of an ideal measurement, the coherence of the superpositions of eigenstates associated with distinct results is completely destroyed by the measurement. Their decoherence is complete. On the other hand, the coherence of the superpositions of eigenstates associated with the same result is not destroyed by an ideal measurement. The subspaces of such eigenstates are free of decoherence.

If a system remains isolated, without interactions with its environment, there is no decoherence, because the evolution of an isolated system is unitary:

$$U(\sum_{i} \alpha_{i}|i\rangle) = \sum_{i} \alpha_{i}U|i\rangle$$

It follows that the coherences $$\alpha_i \alpha^*_j$$ are conserved quantities, in the following sense: if a system is isolated the coherence between the final states $$U|i\rangle$$ and $$U|j\rangle$$ is equal to the coherence between the initial states $$ | i \rangle $$ and $$ | j \rangle $$.

If a system S is disturbed by an environment E, the coherences are no longer necessarily conserved. They depend on the entanglement of S with its environment. It is sometimes believed that interactions or disturbances are sufficient to cause decoherence, but this is not entirely true. The SWAP gate for example is an interaction without entanglement. If the two qubits which interact with SWAP are in pure states, they remain in pure states. There is no loss of coherence, because there is no entanglement. Decoherence is measured by the degree of entanglement with the environment.

Let S be a system subject to the influence of an environment E. Let us consider two orthogonal states $$ | j \rangle_S $$ and $$ | k \rangle_S $$ of S. What is the degree of coherence $$ | C_ {jk} | ^ 2 $$ between these two states?

The $$ | i \rangle_S $$ is an orthonormal basis of S which contains $$ | j \rangle_S $$ and $$ | k \rangle_S $$. The $$ | l \rangle_E $$ are any orthonormal basis of E. Any pure state of the system SE can then be determined by:

$$ | \psi \rangle = \sum_ {il} \alpha_ {il} | i \rangle_S | l \rangle_E $$ $$ = \sum_ {i} | i \rangle_S \sum_ {l} \alpha_ {il} | l \rangle_E $$ $$ = \sum_ {i} | i \rangle_S | \phi (i) \rangle_E $$

where $$ | \phi (i) \rangle_E = \sum_ {l} \alpha_ {il} | l \rangle_E $$

The reduced density operator of S is:

$$\rho_S = Tr_E(|\psi\rangle \langle \psi|) = Tr_E(\sum_{ili'l'} \alpha_{il} {\alpha^*}_{i'l'} |i\rangle_S |l\rangle_E \langle i'|_S \langle l'|_E)$$

$$= \sum_{ili'} \alpha_{il} {\alpha^*}_{i'l} |i\rangle_S \langle i'|_S$$

We can deduce : $$C_{jk} = \sum_l \alpha_{jl} {\alpha^*}_{kl}$$ $$=\langle \phi(j)|_E\phi(k)\rangle_E$$

$$|C_{jk}|^2$$ is therefore equal to $$|\langle \phi(j)|_E\phi(k)\rangle_E|^2$$. It is a measure of the quantum resemblance between $$|\phi(j)\rangle_E$$ et $$|\phi(k)\rangle_E$$. The more $$|\phi(j)\rangle_E$$ and $$|\phi(k)\rangle_E$$ are different, the more the coherence between $$|j\rangle_S$$ and $$|k\rangle_S$$ is decreased by the influence of the environment. If $$|\phi(j)\rangle_E$$ is orthogonal to $$|\phi(k)\rangle_E$$ the decoherence between $$|j\rangle_S$$ et $$|k\rangle_S$$ is complete and the entanglement between S et E is maximal. If $$|\phi(j)\rangle_E$$ is equal to $$|\phi(k)\rangle_E$$ the coherence between $$|j\rangle_S$$ and $$|k\rangle_S$$ is maximal and there is no entanglement between S and E for these two states of S. Entanglement is actually responsible for decoherence.

The Feynman Rules
When a system is not in an eigenstate of an ideal measurement, observation produces an entangled state between the observed system and the measuring apparatus:

$$ U (\sum_ {i} \alpha_ {i} | i \rangle_S) | ready \rangle_A = \sum_ {i} \alpha_ {i} | i \rangle_S | i \rangle_A $$

where it has been assumed that the $$ i $$ results have only one eigenstate $$ | i \rangle_S $$ to simplify the writings.

The reduced density operator which represents the mixed state of the system observed after the measurement is therefore:

$$ \sum_ {i} | \alpha_i | ^ 2 | i \rangle_S \langle i | _S $$

We conclude that an observation makes the observed system pass from a pure state to a mixed state when it is not an eigenstate of the measurement.

This explains the Feynman Rules for calculating probabilities in quantum physics (Feynman 1966):

Let a system which passes from an initial state $$ a $$ to a final state $$ b $$ via intermediate states $$ i $$.

If the intermediate states $$ i $$ are not observed, we must sum the probability amplitudes $$[A(a$$→$$i)A(i$$→$$b)]$$ of all paths $$a$$→$$i$$→$$b$$ and then take the squared modulus of this sum to find the probability that the system will go from $$ a $$ to b:

$$Pr(a$$→$$b)= |\sum_i A(a$$→$$i)A(i$$→$$b)|^2$$

If the intermediate states $$ i $$ are observed, we must sum the probabilities $$[Pr(a$$→$$i)Pr(i$$→$$b)]$$ of all paths $$a$$→$$i$$→$$b$$ to find the probability that the system will go from $$ a $$ to $$ b $$:

$$Pr(a$$→$$b)= \sum_i Pr(a$$→$$i)Pr(i$$→$$b)$$

where $$Pr(x$$→$$y)=|A(x$$→$$y)|^2$$

The first rule is typical of quantum physics. It results from the postulate of unitary evolution. The second rule is a classical rule of addition of probabilities. It results from decoherence through observation. When intermediate states $$ i $$ are observed, the observed system is represented by a mixture of states. The classical rule of addition of probabilities must then be applied. Entanglement by observation thus explains Feynman's second rule.

The a posteriori reconstitution of interference patterns
The a posteriori reconstitution of interference patterns illustrates in a striking way the relativity, in the Everett sense, of the observed states with respect to the observer:

In a Mach-Zehnder interferometer, an observer qubit is placed on one of the photon trajectories. The evolution of the system can then be described by the operator:

$$H_1 CNOT_{12} H_1 |00\rangle= \frac{1}{2}(|00\rangle + |10\rangle + |01\rangle - |11\rangle)$$

$$=\frac{1}{2}[(|0\rangle + |1\rangle)|0\rangle + (|0\rangle - |1\rangle)|1\rangle = |\psi\rangle$$

where the first qubit represents the photon and the second, the observer.

The reduced density operator of the photon at the output of the second beam splitter is:

$$\frac{1}{2}(|0\rangle \langle 0| + |1\rangle \langle 1|)$$

The photon therefore has a probability 1/2 of being absorbed by each of the detectors. The observation of the path followed by the photon thus destroyed the effect of interference between the two paths.

Then we apply the Hadamard gate to the observer qubit:

$$H_{2} |\psi\rangle =\frac{1}{2\sqrt2}[(|0\rangle+|1\rangle)(|0\rangle+|1\rangle)+(|0\rangle)-|1\rangle)(|0\rangle-|1\rangle)]$$

$$=\frac{1}{\sqrt2}(|00\rangle+|11\rangle)$$

The relative states of the first qubit with respect to the second are therefore $$ | 0 \rangle $$ with respect to $$ | 0 \rangle $$ and $$ | 1 \rangle $$ with respect to $$ | 1 \rangle $$. If we observe the second qubit, we can deduce the state of the first qubit. In this way it is possible to reconstruct an interference pattern by linking the observations of the two qubits. Such a reconstruction can be made even if the second qubit is observed long after the first.

It is sometimes said that the interference patterns are destroyed because of the disturbance by the probes placed on the followed paths. This is not entirely accurate since these patterns can be reconstructed after these disturbances have occurred. The probes do not necessarily destroy the interference patterns, but only the conditions of their observation. The entanglement of the object observed with the probes changes these conditions because the observed states are always states relative, in the sense of Everett, to the observer.

The fragility of non-localized macroscopic states
A single photon is enough to destroy a non-localized macroscopic state, because of the decoherence through entanglement.

Suppose for example, a non-localized macroscopic state $$\frac{1}{\sqrt 2}(|1\rangle_M + |2\rangle_M)$$ and an incident photon in the state $$|0\rangle_p$$. The photon may or may not be localized.

Suppose it is initially located near $$ | 1 \rangle_M $$. After the interaction, the following state is obtained:

$$\frac{1}{\sqrt 2}(|1\rangle_p|1'\rangle_M + |0\rangle_p|2\rangle_M)$$

M being macroscopic, its localized state $$ | 1 \rangle_M $$ is very little affected by the impact of the photon. $$ | 1 '\rangle_M $$ is almost equal to $$ | 1 \rangle_M $$, in the sense that their scalar product is not very different from 1. But this does not prevent the destruction of the non-localized state, because $$ | 1 \rangle_p $$ is in general very different from $$ | 0 \rangle_p $$. If $$ \langle 0_p | 1 \rangle_p $$ is zero, the decoherence between the states $$|1\rangle_M$$ et $$|2\rangle_M$$ is complete. The non-localized state is thus completely destroyed, because of the entanglement with a single photon.

If the photon is not initially localized, the interaction leads to the following state:

$$\frac{1}{\sqrt 2}(|1\rangle_p|1'\rangle_M + |2\rangle_p|2'\rangle_M)$$

$$|1\rangle_p$$ is usually very different from $$ | 2 \rangle_p $$. Again, if $$ \langle 2_p | 1 \rangle_p $$ is equal to zero, the decoherence is complete.

As long as the macroscopic objects are not in an ultra-cold vacuum, they are permanently impacted by very many particles, photons and molecules of the ambient air. It might be that only one of these particles is not sufficient to cause a complete decoherence of a non-localized state, if $$ \langle 2_p | 1 \rangle_p $$ is not close enough to zero. But a small number of incident particles is always sufficient for the decoherence to be complete. The decoherence of non-localized macroscopic states is therefore a very powerful and very rapid effect. It is almost impossible to escape. The larger the objects, the more sensitive they are to this destruction of their non-localized states. Microscopic beings, on the other hand, particles and small molecules, are much less sensitive because they can propagate over great distances without interacting, or with very little interacting, with other particles or molecules.

Even in an ultra-cold vacuum, non-localized macroscopic states undergo a very rapid decoherence if they are not ultra-cold themselves, because they emit photons with which they are entangled.

Non-localized macroscopic states are very fragile, because generally, macroscopic objects can not be isolated, or not very well, and because their non-localized states are destroyed by the entanglement with the environment. Localized macroscopic states are not as fragile, because the interactions are always local. Everything happens as if the macroscopic objects were permanently observed by their environment. Since the interactions are local, the localized states can be eigenstates of this observation. If it was an ideal measurement, they would not be disturbed. The locality of interactions selects the localized states among all the states of a macroscopic object because they are the most robust, the less disturbed by their environment (Joos, Zeh & ... 2003, Zurek 2003, Schlosshauer 2007).

Experiments of the "Schrödinger's cat" type
Quantum principles do not prohibit the observation of quantum superpositions of macroscopic states. They only predict that such observations are very difficult, because it is necessary to isolate the observed system to protect it from decoherence by its environment.

Schrödinger invented an unfortunately rather sinister thought experiment to illustrate the very counterintuitive character of the principle of quantum superposition, which he helped to discover, since this principle necessarily accompanies his famous equation. A cat is enclosed in a box equipped with a diabolical device: a radioactive atom is coupled to a vial of poison which will only spread if the atom decays. It is assumed that the half-life of the atom is one hour and that at the beginning of the experiment it has been verified that the atom has not disintegrated. The experiment consists in leaving the cat enclosed for an hour and then opening the box (Schrödinger 1935).

If the box is completely isolated from its environment, this experiment can be described with a unitary operator on a system of a qubit and a qutrit. The qubit represents the atom, which can be in the state $$ | 0 \rangle $$ if it has disintegrated, and $$ | 1 \rangle $$ otherwise. The qutrit represents all the rest of the box: the disintegrating product, if present, the poison vial, the cat and the box itself. $$|alive_0\rangle$$ is the state of the box and its contents, except the atom, at the beginning of the experiment when the cat is alive, | alive \rangle, the state of the box and its contents except the atom at the end of the experiment if the cat is alive, $$ | dead \rangle $$, if it is dead. The experiment is then described by:

$$U|1, alive_0\rangle = \frac{1}{\sqrt 2}(|1, alive\rangle + |0, dead\rangle)$$

The state of the system at the end of the experiment is therefore an entangled state between the atom, the cat and the rest of the box, where the cat is simultaneously dead and alive.

Any experiment which prepare and observe a state such as $$\frac{1}{\sqrt 2}(|alive\rangle + |dead\rangle)$$ for a macroscopic system is called an experiment of the "Schrödinger's cat" type. Obviously it is not necessary to kill a cat.

Schrödinger's original experiment does not prepare the state $$\frac{1}{\sqrt 2}(|alive\rangle + |dead\rangle)$$ of the cat, but only an entangled state between the cat and its environment. In order to prepare the state $$\frac{1}{\sqrt 2}(|alive\rangle + |dead\rangle)$$, one should completely isolate the cat from its environment and put the diabolic device in its belly. But a living being can not survive in ultracold vacuum, so it can never be in a state like $$\frac{1}{\sqrt 2}(|alive\rangle + |dead\rangle)$$.

Schrödinger's thought experiment shows that in principle nothing prevents the application of the principle of quantum superposition to a macroscopic system, provided that it can be perfectly isolated.

An experiment of the "Schrödinger cat" type has never yet been realized with a truly macroscopic system, because we do not know how to isolate them sufficiently from their environment. On the other hand, it can be realized in various ways with mesoscopic objects, small systems of particles, atoms or molecules.

The excitation modes of an electromagnetic cavity are the quantum states of the photons which it contains. Haroche and his collaborators are able to manufacture a cavity where they can prepare, manipulate and observe quite freely the quantum states they imagine. Cavity states are prepared and observed using giant atoms which function as microscopic probes. They have thus realized a "Schrödinger's cat" type experiment with an ultra-cold cavity containing several photons.

Some quantum states of the electromagnetic field resemble very much classical states, especially if they contain on average many photons. They are called Glauber states.

A mode of the electromagnetic field is mathematically similar to a harmonic oscillator, because the field oscillates periodically like any other oscillator. In quantum physics the accessible energies of a harmonic oscillator are quantified. They correspond to the number of photons excited in a mode of an electromagnetic cavity. When the energy of an oscillator is very large with respect to the energy difference between two successive states, it is possible to construct quantum states which resemble very much the classical states of an oscillator. They can have precisely defined position and momentum. These are the Glauber states. They are not states with a definite number of photons, the Fock states, which are purely quantum, which have no equivalent in classical physics, even if they contain a very large number of photons. Glauber states are also called "coherent states".

A Glauber state can be prepared simply by briefly coupling the initially empty cavity to a conventional source of electromagnetic radiation. One can arrange for the giant atom to disrupt the field in the cavity by causing a phase shift of the Glauber state, as if an oscillator had been moved without changing its energy, much as if the atom had given a shot on the field. According to its initial state $$ | g \rangle $$ or $$ | e \rangle $$, the atom causes a phase shift in opposite directions (Haroche & Raimond 2006, pp.377- 378). The interaction between the atom and the cavity is described by:

$$U|g\rangle|G_0\rangle= |g\rangle|G_\phi\rangle$$

$$U|e\rangle|G_0\rangle= e^{-i\phi}|e\rangle|G_{-\phi}\rangle$$

where $$|G_0\rangle$$ is the initial Glauber state and $$|G_\phi\rangle$$ the Glauber state obtained by a phase shift of $$ \phi $$.

If the atom is prepared in the state $$\frac{1}{\sqrt 2}(|g\rangle+|e\rangle)$$ one obtains thus :

$$U \frac{1}{\sqrt 2}(|g\rangle + |e\rangle)|G_0\rangle = \frac{1}{\sqrt 2}(|g\rangle|G_\phi\rangle + e^{-i\phi}|e\rangle|G_{-\phi}\rangle)$$

Since the conditions of the experiment enable to reach values ​​of $$ \phi $$ greater than a radian, the state obtained can be considered as a state of the "Schrödinger's cat" type for the system atom + field. In order to prepare the state $$\frac{1}{\sqrt 2}(|G_\phi\rangle+|G_{-\phi}\rangle)$$ of the cavity, it is sufficient to put the atom initially in the state $$\frac{1}{\sqrt 2}(|g\rangle+|e\rangle)$$ and then observe it at the output of the cavity in one of the states $$\frac{1}{\sqrt 2}(|g\rangle+e^{-i\phi}|e\rangle)$$ or $$\frac{1}{\sqrt 2}(|g\rangle-e^{-i\phi}|e\rangle)$$.

If, for example, it has been observed at output in the state $$\frac{1}{\sqrt 2}(|g\rangle+e^{-i\phi}|e\rangle)$$

the state of the cavity relative to the observation is:

$$\frac{1}{\sqrt 2}(|G_\phi\rangle+|G_{-\phi}\rangle)$$

since

$$U \frac{1}{\sqrt 2}(|g\rangle + |e\rangle|G_0\rangle) = \frac{1}{\sqrt 2}(|g\rangle|G_\phi\rangle + e^{-i\phi}|e\rangle|G_{-\phi}\rangle)$$

$$= \frac{1}{2\sqrt2}[(|g\rangle+e^{-i\phi}|e\rangle)(|G_\phi\rangle+|G_{-\phi}\rangle) + (|g\rangle-e^{-i\phi}|e\rangle)(|G_\phi\rangle-|G_{-\phi}\rangle)]$$

In order to verify that the desired "Schrödinger's cat" state has been properly prepared, the observation system uses homodyne detection (Haroche & Raimond 2006, pp.374-375).

By proceeding this way Haroche and his collaborators have prepared many "Schrödinger's cat" states and they have observed their fragility. They were able to verify empirically the theory of decoherence through entanglement with the environment. The tiny perturbations by the ultracold and very well isolated cavity are sufficient to destroy very quickly the "cat of Schrödinger" states even if they contain a small number of photons.

The following thought experiment, modeled on the previous one, shows how to prepare and observe a non-localized macroscopic state of a mirror. Experimenting poses great difficulties, because the mirror must be perfectly isolated from its environment. Any interaction between the mirror and its environment can destroy the non-localized state that one wishes to observe.

We assume that a state $$ | 1 \rangle_m $$ of a mirror is in unstable equilibrium. If it is struck by a photon in the $$ | 1 \rangle_p $$ state, it falls into the $$ | 0 \rangle_m $$ state, while the reflected photon passes into the state $$ | 2 \rangle_p $$. If the mirror is initially in the state $$ | 0 \rangle_m $$, it does not affect the photon. The following measurement process is thus obtained:

$$ U | 1 \rangle_m | 1 \rangle_p = | 0 \rangle_m | 2 \rangle_p $$

$$ U | 0 \rangle_m | 1 \rangle_p = | 0 \rangle_m | 1 \rangle_p $$

To prepare the non-localized macroscopic state $$ \frac {1} {\sqrt 2} (| 0 \rangle_m + | 1 \rangle_m) $$ we make use of a photon prepared (by means of a beam splitter for example) in a state $$ \frac {1} {\sqrt 2} (| 0 \rangle_p + | 1 \rangle_p) $$ where $$ | 0 \rangle_p is a state in which the photon does not interact with the mirror. Having prepared the mirror in the state we get:

If the photon is then detected (with a beam splitter followed by a photodetector) in the state, we have prepared the state of the mirror. With this non-localized macroscopic state and a new photon in the state we get:

If the experiment is repeated many times with a detector capable of detecting the photon in the state, this will always be detected. It can be concluded that the non-localized state of the mirror was observed.

This example is mainly theoretical. Experimenting poses great difficulties, because the mirror must be perfectly isolated from its environment. Any interaction between the mirror and its environment can destroy the non-localized state that one wishes to observe.

Experiments of the "Schrödinger's cat" type are therefore in principle feasible. It is possible in principle to observe superpositions of different macroscopic states.

Is the existence theorem of multiple destinies empirically verifiable?
In the experiment imagined above, the mirror in unstable equilibrium can be considered as an observation system, designed to detect and record the presence of a photon. This experiment makes it possible to verify the existence of the multiple destinies of an observation system. We deduce that the existence theorem of multiple destinies is empirically verifiable. But this surprising conclusion is accompanied by severe restrictions.

The observer system must be completely isolated from its environment, so in ultracold ultrahigh vacuum. If such isolation is not perfect, decoherence by the environment is enough to destroy the state of superposition of destinies whose existence one wishes to verify. We can imagine chambers capable of perfectly isolating a living being, but it is practically infeasible. Even the isolation of very small systems, much less sensitive to their environment, is usually very difficult.

Another fundamental reason prevents us from observing the multiple destinies of a living being. For a superposition of two observational results to be observable, these results must be recorded in a reversible way, so that they can be "forgotten" by the observer system. In the above experiment, the final state of the mirror did not keep track of previous states  and  by which it has simultaneously passed. This is general. For a state to be observable, we need a detector and an initial state  such that  lead to, where  is the state of the detector when it detected | \psi_0 \rangle $$, and $$ | \psi_1 \rangle $$ is the final state of the observed system. If $$ | \psi_0 \rangle $$ is a superposition of observation results, $$ \frac {1} {\sqrt 2} (| a \rangle + | b \rangle) $$ for example, the results obtained $$ a $$ and $$ b $$ can not both be stored in the state $$ | \psi_1 \rangle $$. At least one of the two results, and perhaps both, has been cleared. We can conclude that when observations are irreversibly recorded, the superposition of different destinies is not observable. As the destinies of living beings are successions of processes and irreversible observations, their superpositions are not observable. The existence of the multiple destinies of living beings is therefore not empirically verifiable, because of the irreversibility of living processes.

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