International Journal of Theoretical and Mathematical Physics 2014, 4(5): 202-219 DOI: 10.5923/j.ijtmp.20140405.04

On the Measurement Problem

Raed M. Shaiia

Department of Physics, Damascus University, Damascus, Syria

Abstract In this paper, we will see that we can reformulate the purely classical theory, using similar language to the one used in . This leads us to reformulate quantum mechanics itself using this different way of understanding , which in turn will yield a new interpretation of quantum mechanics. In this reformulation, we still prove the existence of none classical phenomena of quantum mechanics, such as quantum superposition, quantum entanglement, the uncertainty principle and the collapse of the wave packet. But, here, we provide a different interpretation of these phenomena of how it is used to be understood in quantum physics. The advantages of this formulation and interpretation are that it induces the same experimental results, solves the measurement problem, and reduces the number of axioms in quantum mechanics. Besides, it suggests that we can use new types of Q-bits which are more easily to manipulate. Keywords Measurement problem, Interpretation of quantum mechanics, Quantum computing, Quantum mechanics, Probability theory

Measurement reduced the wave packet to a single wave and 1. Introduction a single momentum [3, 4]. The ensemble interpretation: This interpretation states Throughout this paper, Dirac's notation will be used. that the wave function does not apply to an individual In quantum mechanics, for example [1] in the experiment system – or for example, a single particle – but is an of measuring the component of the spin of an electron on abstract mathematical, statistical quantity that only applies the Z-axis, and using as a basis the eigenvectors of σˆ3 , the to an ensemble of similarly prepared systems or particles [5] state vector of the electron before the measurement is: [6]. The many-worlds interpretation: It asserts the objective ψα=ud + β (1) reality of the universal wavefunction and denies the actuality of wavefunction collapse. Many-worlds implies However, after the measurement, the state vector of the that all possible alternative histories and futures are real, electron will be either or , and we say that the u d each representing an actual "world" (or "") [7, 8]. state vector of the electron has collapsed [1]. The main Consistent histories: This interpretation of quantum problem of a measurement theory is to establish at what mechanics is based on a consistency criterion that then point of time this collapse takes place [2]. allows to be assigned to various alternative Some physicists interpret this to mean that the state histories of a system such that the probabilities for each vector is collapsed when the experimental result is history obey the rules of classical probability while being registered by an apparatus. But the composite system that is consistent with the Schrödinger equation. In contrast to constituted from such an apparatus and the electron has to some interpretations of quantum mechanics, particularly the be able to be described by a state vector. The question then Copenhagen interpretation, the framework does not include arises when will that state vector be collapsed [1, 2]? "wavefunction collapse" as a relevant description of any Many interpretations of quantum mechanics have been physical process, and emphasizes that measurement theory offered to deal with this problem, and here, we will list a is not a fundamental ingredient of quantum mechanics brief summary of some of them: [9, 10]. The Copenhagen Intrepretation: When a measurement of De Broglie–Bohm theory: In addition to a wavefunction the wave/particle is made, its wave function collapses. In on the space of all possible configurations, it also postulates the case of momentum for example, a wave packet is made an actual configuration that exists even when unobserved. of many waves each with its own momentum value. The evolution over time of the configuration (that is, of the positions of all particles or the configuration of all fields) is * Corresponding author: [email protected] (Raed M. Shaiia) defined by the wave function via a guiding equation. The Published online at http://journal.sapub.org/ijtmp evolution of the wave function over time is given by Copyright © 2014 Scientific & Academic Publishing. All Rights Reserved Schrödinger's equation [11, 12].

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Relational quantum mechanics: it treats the state of a results and not both. That means that one of the elementary quantum system as being observer-dependent, that is, the events only will happen: either {}H or {}T , but not both state is the relation between the observer and the system at once [23]. [13, 14]. What if the state vectors were nothing but another Transactional interpretation: describes quantum representation of events in the sense of the usual classical interactions in terms of a standing wave formed by both probability theory? What if the state vector before retarded ("forward-in-time") waves, in addition to advanced measurement is just the representation of a , ("backward-in-time") waves [15, 16]. and the result we get after measurement is just an ordinary Stochastic interpretation: it involves the assumption of , in a similar manner to the coin example, spacetime stochasticity, the idea that the small-scale and in this sense the measurement is an experiment in the structure of spacetime is undergoing both metric and sense of the word used in probability theory? If we could topological fluctuations ('s reformulate classical probability theory in such a way that "quantum foam"), and that the averaged result of these allows the representation of ordinary events by vectors, then fluctuations recreates a more conventional-looking metric at this will lead to an entirely different understanding of the larger scales that can be described using classical physics, underlying mathematics of quantum mechanics, and hence along with an element of nonlocality that can be described to quantum mechanics itself. And this is the aim of this using quantum mechanics [17, 18]. paper. Von Neumann–Wigner interpretation: It is an interpretation of quantum mechanics in which consciousness is postulated to be necessary for the 2. An Alternative Way to Formulate completion of the process of quantum measurement Classical Probability Theory [19, 20]. The many minds interpretations: They are a class of “no In this section we focus on the reformulation of collapse” interpretations of quantum mechanics, which is probability theory, then we use this formulation in next considered to be a universal theory. This means that they section to reformulate quantum mechanics. We reformulate assert that all physical entities are governed by some classical probability theory in a similar language to that is version of quantum theory, and that the physical dynamics used in quantum mechanics. Later on, we show that this of any closed system (in particular, the entire universe) is formulation reduces the number of postulates used in governed entirely by some version, or generalization, of the quantum mechanics. Schrödinger equation [21]. First we will start by considering finite sample spaces. In this paper, we will trod a different route in trying to Here it will be presented an outline of the method to be solve the measurement problem. used in this formulation: Ω This subject of measurement in quantum mechanics was Having an experiment with a finite sample space , studied by many prominent scientists, including Heisenberg, There is always a finite dimensional Hilbert space H von Neumann, Wigner and van Kampen [22]. And more with a dimension equal to the number of the elementary events of the experiment. recent studies such as the one done by Theo Nieuwenhuizen (Institute of Physics, UvA) and his colleagues Armen Then, we can represent each event by a vector in H Allahverdyan (Yerevan Physics Institute) and Roger Balian using the following method: (IPhT, Saclay), found that altogether, nothing else than I the square of the norm of a vector representing an event standard quantum theory appears required for understanding is equal to the probability of the event. ideal measurements. The statistical formulation of quantum II Given an orthonormal basis of H , we represent each mechanics, though abstract and minimalist, is sufficient to elementary event by a vector parallel to one of these explain all relevant features. Since alternative basis vectors, such that no different elementary events interpretations involve unnecessary assumptions of one kind are represented by parallel vectors, and the square of or another [22]. the norm of the representing vector is equal to the In this paper, we will tread a different route in trying to probability of the elementary event. solve the measurement problem. And we will start by III Then every event is represented by the vector sum of contemplating, thoroughly, classical probability theory at the elementary events that constitute it. first. From (I) we see that the vector ϕ representing the When we toss a coin for one time, the sample space of impossible event must be the zero vector because: this simple experiment is [23]: 2 Ω={HT ,} (2) ϕϕ= ϕ =p() ϕ = 0 (3) Of course, as it is known, this does not mean that the coin So: has all of these possibilities at once, it is merely a statement ϕ = 0 (4) about the possible outcomes of the experiment. And after doing the experiment, we will get just one of these two And the vector representing the sample space must be

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normalized, because: N ϕ =00 = ∑ ui (17) ΩΩ =p()1 Ω = ⇒ Ω=1 (5) i=1 Furthermore, we know that the probability of an event is As a result of (II) and (III) we see that Ω is represented equal to the sum of the probabilities of the elementary by: events that constitute it [23], for example if: N A={ ab , ,..., d } (6) Ω=∑ cuii (18) = So: i 1 And we see that: pA( )= pa ({ }) + pb ({ }) ++ ... pd ({ }) (7) NN So, are (I), (II) and (III) consistent with this rule? 2 ΩΩ =∑∑cii = pp =1 = () Ω (19) Actually they are. To see that, let us suppose that sample ii=11 = space is: As an example that helps clarifying the former ideas, let Ω= {uu12 , ,..., uN } (8) us take the experiment to be throwing a fair die and the And let us take some event A ≠ ϕ to be: result to be the number appearing on top of it after it stabilizes on a horizontal surface. Au={kl ,..., u } (9) The sample space of this experiment is: N Ω={1,2,...,6} (20) Let us take the orthonormal basis {}ui i=1 to represent the elementary events such that: 6 Let us take {}i i=1 to be a orthonormal basis in Hilbert {}u is represented by i space, so: 2 * 6 cui i: p ({ u i }) = c i = cc ii (10) ∑ ii= I (21) N i=1 Since {}ui i=1 is a orthonormal basis, then it satisfies: = δ N ij ij (22) ∑ uuii= I (11) Let us choose to represent the elementary event {}i by i=1 the vector: 1 uui j= δ ij (12) i (23) 6 Where I is the identity operator. That means Ω is represented by: According to (III), A must be represented by: 6 1 A= ckk u ++... cu ll (13) Ω= i (24) ∑ 6 Where in the last equation we did not write i=1 l and the event A ={2, 5} for example is represented by the A= ∑ cuii because that would mean that vector: ik= 11 A =25 + (25) A={ uukk ,++12 , u k ,..., u l } which may not be the case in 66 general. but the event ϕ is represented by the zero vector. And by adopting the notation:

αkku++... αα ll uu ≡ k k ++... α l u l (14) 2.1. The Algebra of Events ∗∗ 2.1.1. The Intersection of Two Events αkku++... αα ll uu ≡ k k ++... α l u l (15) The intersection of two events is an event constituted of We have: the common elements of the two events [23] [24]. So, it A A= ckk u ++... cu ll c kk u ++... cu ll must be represented by the vector sum of the vectors representing the common elementary events of the two 22 vectors. =ck ++... cp lk = ++... p l (16) For example, in the die experiment mentioned above, if Obviously, we see that ϕ is represented using this we took the two events: AB={1, 2,3}, = {2,3, 4} then basis as: AB ={2, 3}.

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So: = ABAB (36) 11 1 A =++123 (26) So: 66 6 1 1 1 ABA= ABAB  (37) | = |2 + |3 + |4 (27) 6 6 6 And: And: 𝐵𝐵 ⟩ ⟩ ⟩ ⟩ √ 11√ √ ABB= ABAB  (38) AB =23 + (28) 66 As a result: Now, we will prove a little result. Supposing the sample AΩΩ = AA Ω Ω (39) space of some experiment is:   N But since AA Ω= , so: Ω= cu (29) ∑ ii AΩ= AA (40) i=1 We see that for any event A , we have: Or: A  ϕϕ= and AAϕϕ= = 0 because ϕ = 0 . AΩ= pA() (41) Now let us take two arbitrary events A and B So: (,)AB≠≠ϕϕ And let us suppose they are represented by: A Ω≥0 (42) And: A= ckk u ++... cu ll (30) A BA≥ 0 (43) B= cmm u ++... cu nn (31) Then we have: Because ABA= ABAB ≥ 0 . And:

AB= ckk u ++... cu ll c mm u ++... c nn u AΩ=Ω A = AA = pA() (44) ** We can write the former results, since the probability of =cckm uu k m++... cc kn uu k n some event is equal to the square of its norm, by the **following manner: ++... cclm uu l m++... cc ln uu l n = ⇒ ** ABAB ABA  =cck mδδ km ++... cc k n kn (32) ** pAB()= ABA (45) ++... cclmlmδδ ++... cc lnln We see that: We notice that if AB = ϕ then all of deltas are zeros ABAB= ABA  ⇒ so AB = 0 . ∗ So, we get this result: ABAB= AAB  ⇒ A B=⇒=ϕ AB 0 so AB⊥ (33) ABAB= AAB  ⇒ We see that for any two events A and B , we can ABA= AAB (46) write the event A as: Finally, we can see also that, if we have two events: = ++ A= AB + A11:( A  AB ) A ckk u... cu ll (47) (34) = ϕ = ,(A1  AB ) A B= cmm u ++... cu nn (48)

But since A1 () AB= ϕ then as we see according Then we can write the intersection of them as: to (33) that: A B= B A = ukk Bu ++... uBu ll AA1  B= 0 (35) =u Au ++... u Au (49) So: mm nn

ABA= ABA( 1 + AB ) 2.1.2. The Difference of Two Events

=ABA1 + ABAB  Let us take two events A and B . We saw from (34)

206 Raed M. Shaiia: On the Measurement Problem

that we can write the event A as: 2.1.4. The We know that the complementary event A′ of an event A= A + AB 1 A is given by [23] [24]: Where A1 is an event constituted of elements not AA′ = Ω \ present in B but belong to A so: So: AA′ =Ω\ =Ω−Ω A =Ω− A AB\ = A1 That means: So we have: A= AB\ + A B AA′ =Ω− (53) So: We can directly verify that: AB\ = A − A B (50) A′′ A= Ω− A Ω− A =1 − pA () = pA (′ ) We can see immediately that: ABAB\\=−−( A) ( A B) ( A) ( A B) 3. Observables =−( A) ( ABA) −−( A) ( ABAB) Let us suppose we have a system, and we want to do an experiment with it, which has the sample space: =−−+AAABA AAB  ABAB  Ω={uu1 ,...,N } (54) =−−+pA()()()() pA B pA B pA B Or equivalently: =−=pA()()(\) pA B pA B N 2.1.3. The Union of Two Events Ω=∑ cuii (55) i=1 We know that the union of two events is an event constituted of the elements belonging exclusively to the first N Where as we saw, since {}ui i=1 is a orthonormal one, the elements belonging exclusively to the second one basis: and the common elements between the two [23] [24]. So, it must be represented by: N ∑ uuii= I and uui j= δ ij AB=++ AB\\ AB BA i=1 Which we can write as: Now, if we take any N real numbers:

AB=+++−⇒ AB\\ AB  BA AB  AB λλ12, ,..., λN Then we can consider them to be the eignvalues of a AB=+− A B AB (51) Hermitian operator Aˆ which is represented in this basis Noting that: by the matrix:

BA\0 B= diag(λλ1 ,...,N ) Because B (\) AB= ϕ we can directly verify that: N Clearly, the vectors of the ordered basis {}ui i=1 are ABAB the eignvectors of Aˆ which satisfy: =( A) +−( B) () AB( A) +−( B) () AB ˆ Aui= λ ii u (56) =+−pA() pB () pA ( B ) = pA ( B ) From the fact that this is true for any lambdas, in other In a side note, we can prove that: words the values of λi are arbitrary, then the vectors of AB= A B\ A + A B N (  ) the basis {}ui i=1 are the eignvectors of an infinite =AB\ A + AA B number of Hermitian operators in Hilbert space. Not even just that, but since this is true for any lambdas, =+=0 AA B AA B then whenever we assign real numbers to elementary events, we can consider them to be the eignvalues of some So: Hermitian operator in Hilbert space corresponding to the ABBA= = AABBABpAB =  = ()(52) N eignvectors {}ui i=1.

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And since the observable is by definition a function from Or equivalently: the elementary events to real numbers [1], then we can N represent any observable we define on the system, by a Ω=∑ cuii (60) Hermitian operator in Hilbert space. i=1 But we have to be careful here: all the observables we have talked about have the same set of eignvectors, and we Now we will divide all the experiments we can do on the will call them compatible observables, and if we take any system into classes of experiments. Each class is composed two of them, we find that their commutator is zero, because of experiments that have the same number of outcomes (the same number of elementary events). So the experiment that they have the same eignvectors. we talked about is one member of the class C where If we take one of these observables, let it be Aˆ , which is N represented by: CN is the class of experiments that have N elementary events by definition. diag(λλ1 ,...,N ) We will name our experiment E1 . We saw that for the Then we can think of the experiment as giving us one experiment E , we can define an infinite number of eignvalue of the observable. And since this is true for every 1 compatible observables (Hermitian operators) which all one of the compatible observables with Aˆ as we saw, then N it is clear that compatible observables can be measured have the same eignvectors {}ui i=1 . Let us now take simultaneously together with a single experiment, which is another experiment from the same class C which we the experiment we talked about. N Now, let us suppose that: will call E2 . What we mean by another experiment on the system is that we cannot do E and E simultaneously. λλij= =... = λ k = λ 1 2  (57) Let us suppose the sample space of E is: g values 2 Γ={tt , ,..., t } Which means that λ is degenerate with a degeneracy 12 N (61) g .. That means we will get λ in the experiment if we get Now in general, the of E2 may ui , u j , ... , or uk . In other words, if the event: be radically different from that of E1 . So, how are we B={ uuij , ,..., u k } happened. going to represent the events of E2 by vectors? Well, So: since we can represent Γ by a vector in any N-dimensional Hilbert space, then we can represent it in the p()λ = pB () = BB ⇒ same Hilbert space that we used to represent Ω . We can use another basis in this space and use another vector p()λ = cu + c u ++... c u cu + c u ++... c u ii j j kk ii j j kk (different from Ω ) to represent Γ , or we can use the 222 =ccij + ++... c k ⇒ same basis and a different vector from Ω to represent Γ , or we can use a different basis (different from 222 pu()λ = Ω + u Ω ++... u Ω (58) N Ω Γ ij k {}ui i=1 ) and the same vector to represent , So we see, that if λ was degenerate, then it is where the components of Ω on the new basis are which probability is equal to the probability of the projection of determine the probabilities of the events of E2 . All these Ω on the subspace spanned by the eigenvectors approaches are valid, but we will choose the last one (we corresponding to λ . So, each eignvalue of the observable could also have worked in a different Hilbert space all is represented by a subspace so to speak. together). Of course we can represent E2 with the same But what about other observables we can define on the basis and the same vector for the sample space, if it has the system corresponding to other experiments? Those same probability distribution of E1 . But to distinguish experiments may have in general totally different E as an experiment that cannot be done simultaneously probability distributions, which means the probabilities of 2 their elementary events are different of those in the with E1 , we will represent its elementary events by a experiment we have talked about in the beginning of this different basis. section. Even more, the number of the elementary events Now for E2 , we have: may be different. N Let us suppose we have a system. And let us assume that Ω= bt we intend to do some experiment on the system which has ∑ jj (62) the sample space: j=1

Ω={uu1 ,...,k } (59) As in the case of E1 , here, we can define an infinite

208 Raed M. Shaiia: On the Measurement Problem

number of observables (Hermitian operators), all can define the observable Aˆ to take the value 1 for Heads, N and the value -1 for Tails. The second experiment, E , is compatible, and which have the eignvectors {}t j j=1 , 2 throwing the coin in a special way, that makes it stabilizes and those observables are associated with . Since all of E2 on its edge on some horizontal surface. We suppose that the them have the same set of eignvectors, then the commutator edge of the coin is half painted. We can define an of any two of them is zero. observable Bˆ to take the value 1 if we looked at the coin But if we take one of them, let it be Bˆ , and take an from above and saw the edge either all painted or all not ˆ ˆ observable A associated with E1 , then since A and painted, and -1 if we saw it partially painted. We see that ˆ N we cannot do both E1 and E2 simultaneously, so: B do not have the same eignvectors (because {}ui i=1 [,]ABˆ ˆ ≠ 0. is not {}t N ), then [,]ABˆ ˆ ≠ 0. j j=1 *(Notice that for some class of experiments CN , once we used a vector to represent the sample space of some And since E1 and E2 cannot be done simultaneously, experiment, then we have to ask ourselves, can we use it to ˆ ˆ then we cannot A and B simultaneously, represent all the sample spaces of all the experiments of this because each observable is defined in terms of the class that can be done on the system? experiment it is associated with. So we call them Clearly, when we want to represent the sample space of incompatible. some experiment by a vector in Hilbert space, we can Now, before we continue, let us take some examples of choose any Hilbert space that has the right dimensionality, some compatible and incompatible observables. and any orthonormal basis in it to represent the elementary 1- Compatible observables: events, and a vector to represent the sample space that In the experiment of throwing the die, we can define the satisfies that the squared norms of its components are the first observable to be the number appearing on the top side probabilities of elementary events. But after this if we want of the die, and the second observable to be the square of the this vector to represent all the experiments of this class, we number appearing on the top side of the die. have to choose the bases representing those experiments carefully. Or we can choose the bases that represent all Let us call the first Aˆ and the second Bˆ . We have: experiments in Hilbert space, then look for the vector that 6 can be used as a sample space vector for all of them. Ω= 1 i (63) ∑ 6 And as we will see in the future, not every vector we use i=1 to represent the sample space of some experiment of some Where: class, satisfies this for all experiments of the given class. So, we will call any vector that actually satisfies this condition, 6 meaning it represents the sample space of all possible ij= δ and ii= I (64) ij ∑ experiments of a given class that can be done on the system, i we will call it the state vector of the system because it gives We have: us the information about any experiment we can do on the system for a given class of experiments. And from now on, Aiˆ = ii (65) throughout this paper, when we use the term "state vector", And: we mean it in this particular sense. We will talk more about this later). 2 Biˆ = i i (66) Now, what if we take two experiments from different classes, say CN and CM where MN≠ ? So Aˆ is represented by: Let us suppose that the experiment E1 is from the class diag(1,2,3.4,5,6) (67) CN and E2 is from the class CM . Here we will ˆ While B is represented by: represent each experiment in its own Hilbert space: E1 in diag(1,4,9,16, 25,36) (68) H1 and E2 in H2 . We see that: Let us suppose we represent the sample space of E1 in [,]Aˆ Bˆ =−= ABˆ ˆ BAˆ ˆ 0 (69) H1 by: 2- Incompatible observables: N Ω=cu ∈ H (70) Let us take a coin. We will imagine two ideal 11∑ ii = experiments that we can do with it. In the first, let us call it i 1 E1 , we toss the coin and it stabilizes on a horizontal And that we represent the sample space of E2 in H2 surface and the top side of it is either Heads or Tails. We by:

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M 222 p= cb = c b = pu( )( pt ) (79) Ω=22∑ btjj ∈ H (71) ij i j i j j=1 Well, it is the same probability of the event {(utij , )} Now, let us suppose that we do both experiments on the in the experiment E which has a (.NM ) elementary system. Even more, we will suppose that doing either experiment does not affect the probability distribution of the events. From the above we see that we can represent the other, whatever the order of doing the two experiments, sample space of the composite experiment E by a vector was. Ω in the Hilbert space HH12⊗ where: When we do E1 , we have the sample space: Ω=Ω12 ⊗Ω (80) Ω=11{uu ,...,N } (72) Where the elementary event {(utij , )} is represented Let us suppose that after doing E1 we do E2 and that by cb ut . for E2 the sample space is: i j ij Now, what if doing one experiment affects the Ω={tt ,..., } (73) 21 M probability distribution of the other? Here, we can still write Let us take the composite experiment E that is doing the sample space of the composite experiment as: E then E on the system. The sample space of this 1 2 Ω=Ω12 ×Ω experiment is: ={(ut11 , ),...,( ut 1 ,M ),...,( u N , t1 ),...,( u NM , t )} (81) Ω={(ut11 , ),( ut 12 , ),...,( uNM , t )}  (74) Because the outcomes of the two experiments remain the NM⋅ same, but what is changing, is probability distributions. So, Where we know that, if the probability of {}ui in E1 we cannot write: p({ uij , t })= pu { i } pt { j } because is pu() and the probability of {}t in E is pt() i j 2 j finding ui is not independent from finding t j in then the probability of {(utij , )} is [24] : general. Still, the outcomes of the composite experiment are p= pu(ij ). pt ( ) (75) NM⋅ elementary events, so we have to represent Ω in a Hilbert space with ⋅ dimensionality. And any That is interesting, because if we take the vector space: NM Hilbert space with this dimensionality will do. So, we can HH=12 ⊗ H (76) represent Ω in:

and take the vector Ω in it which is: HH=12 ⊗ H And since we can choose any orthonormal basis in it, and NM  NM⋅ Ω=Ω ⊗Ω =cu ⊗ b t since {}utij are orthonormal basis, so we can 12∑∑ii jj i=1 ij=11= represent Ω by: NM. Ω=f ut ∈ H ⊗ H = H = ∑ cbij u i⊗ t j ∑ ij, i j 12 (82) ij, ij,

NM. Where ΩΩ =1 and fij, ut i j representing the ≡ cb ut (77) ∑ i j ij elementary event {(ut , )}, thus: ij, ij Where by definition: 2 p({( uti , j )}) = f ij, (83) ut≡⊗ u t ij i j (78) We can generalize this to any number of experiments. First of all, we see that the dimensionality of H is P.S. when we define the sample space of some experiment, it is not necessary that we really do the (.NM ). experiment, but it just describes a potential experiment. If we considered Ω to be the vector representing the Now, let us ask ourselves a question: is the state vector sample space of some experiment that has (.NM ) unique? Can we represent it for a given class, with another elementary events, then the probability of the elementary vector/vectors? If it is not unique, then we must find the same event utij is: probabilities for all experiments of this class that we can do

210 Raed M. Shaiia: On the Measurement Problem

on the system, whether we used Ω or –if exist- the other here. Let us take another experiment of the same class. state vector/vectors that can be used as state vectors. E2 We know that it must be represented by another basis, let us Let us suppose that for an experiment E1 , the state N vector of the system is written as: say {}t j j=1 . We must have: N N Ω=∑ cull (84) Ω= bt (92) l=1 ∑ jj j=1 Let us replace the vector Ω by the vector Ω′ We have: which is: NN N Ω=b t = cu (93) Ω=′ ′′ = ∑∑jj ll ∑ cl u l: c l cz ll (85) jl=11= l=1 Where: Where z are complex numbers which we will write in l NN the form: bj= t j Ω=∑∑ tu jl u lΩ= tuc jll (94) iθl + ll=11= zll= Ae:, A l ∈∈ Rθ l R (86) For Ω′ we must have: Where: i = −1 . for Ω′ to be a state vector for the system, then all the probabilities of the elementary events N (so all the probabilities of all events since the probability of Ω=′′∑ btjj (95) an event is the sum of the probabilities of its elementary j=1 events) of any experiment from this class must be the same So: as given by Ω . So, the probabilities of the elementary N events of E do not change. 1 b′′j= t j Ω=∑ tu jl u lΩ ′ So the following equation must hold: l=1 ′ 22= (87) NN ccll iθl = ∑∑tj uc ll′ = t j uce ll (96) So: ll=11= 2 iθl 2 We saw that the probabilities of the events of E1 do not Aell c= c l ⇒ change. But to reach our goal, which is that we want Ω′ 22 2 to be a state vector too, then the probabilities of the events Acll= c l (88) of E2 must not change. So, we must have: And because the former condition is true even if we 22 choose the experiment to satisfy: cl ≠ 0 for all values of bb′jj= cl , because our choice of E1 is arbitrary, we must have: bb′′**= bb 2 jj jj Al =1 NN  θθ −  So we have the condition:  t u ceiilk u t c* e  ∑∑jll kjk  lk=11 =  Al =1 (89)  NN  Which means that: =  tuc ut c*  ∑∑jll kjk iθl lk=11 =  zel = (90) And that: So we must have: **i()θθlk− iθl t u u t cc e = t u u t cc cllll′ = cz = ce (91) ∑∑j l k j lk j l k j lk(97) lk,,lk But that is not enough, because the condition that probabilities must not change must be true for any other The former equation must be true for any ck and cl experiment from the same class we can do on the system because we are speaking of arbitrary experiments with and not just E1 , because we are talking about state vectors arbitrary probability distributions. It is true when we fix the

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basis whatever cl were. So their coefficients must be the But, according to the note (*) , the state vector after the same (we fixed the two bases and can define an infinite measurement i.e. after doing the second experiment (after number of experiments on them): reading the result), must be able to represent this experiment. So it must give the same probabilities for the i()θθlk− tujl ute kj = tujl ut kj⇒ elementary events of this experiment. So we must have: pu({l }) = δ kl (105) i()θθlk− tujl ut kj[ e −= 1] 0 (98) Which means the state vector after the measurement must It must be true for all experiments so for all bases even be of the form: N when: tujl≠ 0 for any l and j . So: iiθθ Ω=′ euk = e∑δik u i (106) θθ− i=1 eik()l =1 we can use this vector to represent any other sample So: space of any experiment of the same class CN , which we iiθθlk ee= ⇒=θθlk (99) do after the measurement, because it is the state vector after And that is for any l and k . So we have: the measurement. So, if the state vector before we do some experiment was zzlk= (100) given by (103), and after that, we did the experiment and So we see that: got the result uk , then the state vector of class CN after the measurement becomes: = = = ≡ iθ (101) zz12... zN e iθ Ω=′ euk (107) So: Of course we can choose any value for θ including NN 0 iiθθ so we can write the state vector after the measurement as: Ω=′ ∑∑cel u l = e c ll u ⇒ ll=11= Ω=′ uk (108) Ω=′ eiθ Ω (102) We can call this a collapse in the state vector. But we also see that there is nothing mysterious here, for we just So, for Ω′ to be a state vector too, it must be of the have a change in probability distribution after the former form. From the above we see that we can multiply measurement. We can see that another way to express the above is, that Ω by any pure phase without changing anything. if Aˆ is an observable that the experiment measures (as we have mentioned that means a Hermitian operator that has N 4. The Collapse of the State Vector {}ui i=1 as its eigenvectors), then the system after the Let us suppose that we have a system. We want to do on measurement will be in an eigen state of Aˆ corresponding it an experiment E1 of the class CN . That means the to the eigenvalue of it that we will measure. state vector of it is: N 5. Entangled States Ω= cuii (103) ∑ Now, how do we represent composite systems? i=1 Let us at first take two non-interacting systems. Let us do Let us suppose that the result of the experiment was , uk an experiment E1 on the first one of the class C1N which means that the elementary event has {}uk ( C1N is the class of experiments we can do on the first happened. Let us suppose that we want to do another system with N outcomes). Its sample space will be of the experiment now on the system from the same class, after we form: did the first one. Well, one such experiment could be just reading the Ω=11{uu,..., N } (109) result of the former experiment. Since the result was uk and we will suppose the state vector of it is: then definitely we will find the result uk . So we can N represent the sample space of this experiment by: Ω=11∑ cuii ∈ H (110) N i=1 ∑δikuu i= k (104) And let us suppose that the sample space of the second i=1

212 Raed M. Shaiia: On the Measurement Problem

experiment E2 on the second system of the class C2M is: From all that we see that we can use: ψ =Ω=Ω ⊗Ω ∈HH = ⊗ H (119) Ω=21{tt,..., M } (111) 1 2 12

and we will suppose its state vector is to represent Ω with cbi j ut ij representing the M elementary event {(,)utij} . Ω=22∑ btjj ∈ H (112) j=1 And since we are working in a new space altogether, we can use this vector to be the state vector of the composite We know that the sample space of the composite system, taking into account that we have to be careful after experiment E of doing E1 and E2 together is: it, to choose the bases representing other experiments of the class C on the composite system in the right way. Ω=Ω12 ×Ω NM. We can call ψ a product state. = {(ut11 , ),...,( ut 1 ,M ),...,( u N , t1 ),...,( u NM , t )} (113) We can take as an example of the above two We see that E has NM. outcomes. non-interacting fair coins. The sample space of the Even more, since the two systems do not interact with experiment of tossing a first coin is: each other, the then probability of the result (,)utij is: Ω=1 {HT, } (120)

p{(,) uij t} = pu{ i} . pt{ j} (114) and let us suppose that its state vector before measurement was: Where pu{ i } is the probability of getting ui in the 11 Ω=11H + TH ∈ (121) experiment E1 , while pt{ j } is the probability of getting 22 t j in the experiment E2 . And for the second coin: We know that we can represent Ω in any Hilbert space Ω=2 {HT, } (122) of dimensionality NM. by a vector that gives the elementary events of E the former probabilities. And we suppose too that its state vector: If we take the vector: 11 Ω=22H + TH ∈ (123) ψ =Ω1 ⊗Ω 2 ∈HH = 12 ⊗ H (115) 22 We see that H has the dimensionality NM. . And the sample space of the composite system (tossing Furthermore: the two coins together) is: Ω={(HH , ), ( HT , ), ( TH , ), ( TT , )} (124) NM ψ =Ω ⊗Ω =cu ⊗ b t 12∑∑ii jj So the state vector for the composite system of the two ij=11= coins is:

NM..NM Ω=Ω12 ⊗Ω =∑∑cbij u i ⊗≡ t j cb ij ut ij (116) 1 111 ij,,ij =HH +++ HT TH TT (125) 2 222 We see that: P.S: we see that writing sample space as a vector is more 222 ψψ =∑∑cbij = c i b j ⇒ expressive, because it does not show only the results, but it ij,, ij shows their probabilities too. Now: what if the two systems were interacting with each ψψ = pu. pt= p (,) u t = 1 other? ∑∑{ i} { j} { ij} (117) ij,,ij Here, we can still write the sample space as given by the equation (113), and that is if the outcomes of the two So, we can use ψ as a representation of the sample experiments remain the same, but what interaction is space of some experiment with NM. outcomes. changing, is probability distributions. So, we cannot write: p{(,) uij t} = pu{ i} . pt{ j} because finding ui is not Furthermore, the basis vector utij can represent an elementary event with a probability: independent from finding t j in general. 222 Still, the outcomes of the composite experiment are p= cbijij = c b = p{ u ij}. p{ t} = p{ (,) u ij t } (118) NM. elementary events, so we have to represent Ω in a

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Hilbert space with NM. dimensionality. And any Hilbert 221 cc= = (134) space with this dimensionality will do. So, we can represent HT2 Ω in: The state vector of the composite system (which can be HH=12 ⊗ H (126) thought of as the sample space of the experiment of And since we can choose any orthonormal basis in it, and observing the composite system of coin – coin tosser) as we know, can be written as: since { utij} are orthonormal basis, and we are working Ω=c1 HOH + c 234 HO THT + c TO + c TO (135) in a whole new space HH=12 ⊗ H different from both But we know that we can only get either HO or H1 and H2 , thus we can represent the state vector of the H composite system by: TOT , so: Ω=f ut ∈ H ⊗ H = H ∑ ij, i j 12 (127) cc23= = 0 (136) ij, Even more, the probability of the observer sees it Heads is the same as the probability of getting Heads. The same Where ΩΩ =1 and f ut representing the ij, i j goes for tails. So: elementary event {(,)utij} . 221 cc14= = (137) We see that the product state is a special case of the 2 former formula when fij, = cb i j. So:

We call any state Ω=∑ fij, ut i j an entangled Ω=c1 HOHT + c 4 TO ≠Ω 12 ⊗Ω (138) ij, So, Ω is an entangled state. In the same way, if we state if it is not a product state. Now we can talk about the composite (system-observer) have a system with a state vector: system. N Ω=11∑ cuii ∈ H (139) i=1 6. The Observer-system Composite System Then each cuii corresponds to an elementary event of the observer which has the same probability, so it can be Let us start by an example, then generalize. Let us take represented by where: the composite system of (coin-coin tosser). The sample bOii space of the coin is: 22 bcii= (140) Ω=1 {HT, } (128) and the state vector of the observer is: And we will suppose its state vector is: N 11 Ω=22∑bOii ∈ H (141) Ω=1 HT + (129) 22 i=1 The observer (coin tosser) may get two results: observer And the state vector of the composite system (which is seeing the coin Heads, or observer seeing the coin Tails. So, the sample space of the experiment that is observing the the sample space of the experiment which is the observer composite system) is of the form: observing the coin is: Ω= d uO∈ H ⊗ H ∑ ij, i j 12 (142) Ω=2 {OOHT, } (130) ij, And let us suppose its state vector is: But since the probability of the event dij, uO i j is Ω=2 cHH O + cO TT (131) zero when ij≠ so: Bearing in mind that: Ω= d uO ∑ i ii (143) pO{ H } = pH{ } (132) i And that: And we have: 2 (144) pO{ T } = pT{ } (133) p{(, uii O )} = pu{ i} = pO{ i} = c i So, we have:

214 Raed M. Shaiia: On the Measurement Problem

We can write Ω as: So: ˆ† ˆ Ω=d uO ≠Ω ⊗Ω Ω(0)U ( tU ) ( t ) Ω= (0) 1 (150) ∑ i ii 12 (145) i We see that we can accomplish that by choosing Uˆ to Where: be unitary: 222 † dbciii= = (146) Uˆ () tUˆ () t= I (151) So, the measurement is an entanglement between the Where I is the identity operator. So we will choose Uˆ system and the observer. to be a unitary operator. Furthermore, from (148) we see that: 7. Time Evolution of Systems UIˆ (0) = (152) The physical state of the system might change with time, It is a well known fact [1] that from the equations (151) so that means the state vector describing it might change and (152) in addition to (148) we can deduce that: with time, because the probability distributions of d Ω experiments might change with time. We will talk about iH =ˆ Ω (153) time evolution of closed systems at first. First of all, what is dt the definition of a closed system? We will adopt the Where: following definition: ˆ † ˆ (154) A closed system is a system which satisfies that its HH= characteristics are independent of time, meaning, when we which means that Hˆ is a Hermitian operator. study the system, it does not matter where we choose the Now, what about unclosed systems? origin of time, as long as we do not do a measurement on it. We have to find an equation that describes the system in Since the number of outcomes is the same in any moment general, whether closed or not, and which becomes identical of time we want to do the experiment, so at time t we can to (153) when the system is closed. For this, we can define represent the state vector of the experiment in the same an operator Hˆ for each system that satisfies: vector space that we represented the state vector of it at ˆ † ˆ time t0 . 1- HH= We know that each observable is represented by a d Ω 2- iH =ˆ Ω Hermitian operator, and the experiment we do to measure it dt has its events represented by orthonormal basis that is the ˆ eigenvectors of this operator. We will keep the bases 3- When the system is closed, H is deduced from a representing all the experiments the same, and see how unitary evolution with time of the state vector. must the state vector change with time to keep satisfying that it is the state vector for the closed system. So, for the N 8. Treating a Simple System as a ˆ observable A that u j are its eigenvectors we Composite System { } j=1 can keep them the same, but then the state vector might If we have a system, and E1 is an experiment of the change in general. So the state vector at time t is of the class CN that we can do on it, and the state vector for the form: experiments of this class for this system is: N Ω=()t ctu () (147) N ∑ jj Ω= ∈ j=1 11∑ cuii H (155) i=1 Where t is the time elapsed after the moment t0 = 0 . While E2 is another experiment we can do on the We will search for a linear operator Uˆ such that, if we system, and it is of the class CM where NM≠ . Where start the system in any initial state Ω(0) , then its state we assume that the state vector of this system for this class after a time t is given by: of experiments is: Ω=()t Utˆ () Ω (0) (148) M Ω=22∑ btjj ∈ H (156) But since the state vector is a sample space vector, hence j=1 it is normalized, so: Each experiment has its probability distribution, which ΩΩ()tt () = 1 (149) we will assume it is independent from the other experiment.

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As we saw, the sample space vector of the composite +∞ experiment E which we can do on the system which is ψ = ∫ dxc() x x (162) doing E1 then doing E2 on it is: −∞ NM Because then: Ω=Ω ⊗Ω =cu ⊗ b t (157) +∞ 12∑∑ii jj ij=11= x′′ψ = ∫ dxc() x x x And since it is a vector in a NM. dimensional Hilbert −∞ space, then we can consider it as the state vector of the +∞ system for the class CNM. ensuring of course that we =∫ dxcx()(δ x −= x′′ ) cx ( ) (163) represent the experiments of this class by the bases that −∞ ensures that Ω is a state vector. So: But from the above we see that we can think of the +∞ system as equivalent to two separate non-interacting ψψ= dx x x (164) systems where the state vector of the first is given by (155), ∫ and the state vector of the second is given by (156). So the −∞ state vector of the composite system will be given by (157), b since they are non-interacting: We can define ∫ dx x x where: a 9. Continuous Probability Distributions ab,,∈ R { −∞ +∞} , as: For any ψ : What if we want the experiment to tell us about the position of some particle? Clearly, the way we used when bb talking about representing events by vectors is of no use dx x xψψ= dx x x (165) here, because we are dealing with continuous probability ∫∫ distributions. So, we have to update our tools a little. aa We will work first with a special example, then bb generalize. Let our example be a particle moving on a line. φφdx x x= dx x x (166) Here, we will represent events by vectors in an infinite ∫∫ aa dimensional Hilbert space, because we have infinite values of x . And in this space, we will represent the observable So: ∀ ψ : x by the operator Xˆ , which satisfies: +∞ +∞ Xˆ x= xx (158) ψ= dx x xψ= dx x x ψψ= = I ψ ⇒ ∫∫ −∞ −∞ And we will demand the inner product in this space to satisfy: +∞ = xx′′=δ () x − x (159) ∫ dx x x I (167) −∞ Furthermore, we demand this inner product to satisfy So we have: also: +∞ For any ψ and any well behaved complex function φψφψφ=I = ∫ dx x x ψ fx( ) : −∞ bb +∞ ψψdxf( x) x= dxf( x) x (160) ∫∫ = ∫ dxφψ x x (168) aa −∞ bb With all of that being said, now we can represent events dxf( x) xψψ= dxf( x) x (161) as the following: ∫∫ aa We represent the event: Where: ab,,∈ R { −∞ +∞} . A={ x: x ∈ ( ab , ) ... ( cd , )} (169) We can write any ψ as: (where ab, ,..., cd , ∈ R { −∞, +∞} and the intervals

216 Raed M. Shaiia: On the Measurement Problem

(,ij ) might be open, closed or half open/half closed and We know that when xA∈ we have: xA′ = 0 so we have a finite number of them, let us call it N ) by a substituting in (176) gives: vector A in some infinite dimensional Hilbert space x∈ A ⇒ x Ω= xA (177) H as follows: Using the former equation in (175) we get: 1. pA()= AA bd 2. x∉⇒ A xA =0 p() A=ΩΩ++ΩΩ dx x x... dx x x We know that: ∫∫ ac +∞ bd A= I A = dx x x A 22 ∫ =dx Ω( x ) ++ ... dx Ω ( x ) (178) −∞ ∫∫ ac Thus: Where: ab Ω=Ω A=∫∫ dx x x A+ dx x x A ++... ()xx (179) −∞ a So the probability of x being in the interval d +∞ (ab , ) ... ( cd , ) is: ++dx x x A dx x x A (170) bd ∫∫ 222 cd ∫∫dxΩ( x ) ++ ... dx Ω ( x ) so Ω()x is the But all integrals except the ones in which are over ac intervals spanned by the event A are equal to zero probability density. because in the intervals that do not satisfy that we have We can generalize this process to any continuous xA= 0 so: observable. And if we want to measure another incompatible observable, we do as we did in the case of bd discrete variables, meaning we represent it by another basis. A=∫∫ dx x x A ++... dx x x A (171) It is fairly easy to generalize for three dimensions. ac Also here, we will represent events by vectors in an infinite dimensional Hilbert space. And we will suppose Where we are integrating only over intervals spanned by  A and this is the meaning we will give to the former that there is in this space the vectors { r } which satisfy: equation throughout this paper. We can see that:    rr′′=δ () r − r (180) +∞ Ω=dx x x Ω Where: ∫ (172)    −∞ ∫∫∫ fr()(δτ r−= rd′′ ) fr ( ) (181)  Since: Where the integration over the whole of space and r is ′ = Ω (173) the position vector of the particle, and dτ is the AA\  infinitesimal volume around r . Furthermore we demand That yields: the inner product in this space to satisfy: Ω=AA + ′ (174) For any ψ and any well behaved complex function  Now: fr(): +∞     p() A= A A = ∫ dx A x x A ψ∫∫∫d τ fr() r= ∫∫∫ d τψ fr() r (182) −∞  (175) ττ bd  = dx A x x A+... ++ dx A x x A   ∫∫ dτ fr() r ψτψ= d fr() r (183) ac ∫∫∫ ∫∫∫ ττ But from (174) we know that: Where τ may be all of space or only some part of it. Ω=AA +′ ⇒  And we define ∫∫∫ drrτ where τ may be the xΩ= xA + xA′ (176) τ whole of space or only some part of it, as follows: for any

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ψ : 22 Ω=αHHT + ααα T:1 HT+ = (189)    Now, after tossing the coin, we have some definite result, drrτ ψ= drr τψ (184) ∫∫∫ ∫∫∫ either Heads or Tails. And as we saw before, we can ττ represent the sample space vector of any experiment of the  same class that we can do on the coin after tossing, i.e. the   state vector after the measurement, by (if the result was say ψdrr τ = dτψ rr (185) ∫∫∫ ∫∫∫ Heads): ττ Ω=′ H (190) From all of the above we can prove, in a similar manner to what we did in the one dimensional case, that: Which means that if we read the result of tossing, we will For any ψ and φ we have: find it definitely Heads. In the same way, before turning the magnetic field on,  ψ= ∫∫∫ drr τψ (186) there is no meaning of the question "is the component of the spin of the electron up or down?" but the state vector before  φψ= ∫∫∫ d τ φ rr ψ (187) the measurement (which represents the sample space) of measuring the component is of the form:  ∫∫∫ drrτ = I (188) 22 ψα=uud + α d:1 α ud += α (191) Where the integration extends over all of space, and I But after turning on the magnetic field, we will have a is the identity operator. definite result, let us say up, so the state vector after the With these tools, we can continue exactly as we did in the measurement can be written as: one dimensional case. ψ ′ = u (192) 10. Quantum Mechanics And that represents the sample space for any experiment of the same class after the measurement. Now with these concepts at hand, we need no assumptions in quantum mechanics, just we need to apply them. We will take as an example the component of the spin 11. What about Bell's Theorem? of an electron along the z-axis [1]. If we created an electron somehow, in general, when we In fact this new interpretation is in total agreement with turn on a magnetic field in the z-direction, we might find Bell's theorem. the component of the spin of the electron either up or down As we saw when we spoke of incompatible observables, with a certain probability depending on the initial state. If they are in this interpretation, observables that cannot be the electron was originally up or originally down, then we measured in the same experiment. And that any two find it after turning on the magnetic field certainly up/down observables that their commutator is not zero are as is known [1]. But there are states that we sometimes get incompatible. up and other times get down so they are different than the Furthermore, we saw that if the sample space of the electron being up or being down before we turn on the experiment is not already an eigenvector of the observable magnetic field [1]. So how do we explain that according to we want to measure, then there is no meaning of the the new formulation of classical probability theory? In this question "what was this property before doing the new interpretation of quantum mechanics, we see that experiment?". So, when we have two electrons in the singlet quantum mechanics is just a statistical theory in the same state, and since according to this interpretation, the state way classical experiments are. To understand the vector in the singlet state is nothing more than a experiment of measuring the component of the spin of an representation of the sample space of the experiment of electron, we will compare it to a classical experiment which measuring the spins of the electrons, and this sample space is measuring the Headness or the Tailsness of a coin. What is represented by: we mean by the statement: the result of the experiment is 1 Heads/Tails, is that after we toss the coin, the upper surface ψ =( ud − du ) (193) of the coin after it stabilizes on a horizontal surface is 2 Heads/Tails. So, before tossing the coin, i.e. when it is still It means that we cannot say before measuring the spins in my hand, there is no meaning of the question "is the coin that the spins were opposite because it is like saying the Heads or Tails?". All we can talk about is the sample space coin is Heads while it is still in my hand. of the experiment. And in this case we saw that the state Also, the three components of the spin of an electron are vector of the coin (which represents the sample space) is of represented by incompatible observables, so they cannot be the form: measured together. And since there is no meaning of what

218 Raed M. Shaiia: On the Measurement Problem

is the component before measuring it, so we cannot say that find that they simultaneously assume opposite spin the three components of the spin of the electron can have components along the Z-axis (for short we will say they values together. assume opposite spins). But we know from special relativity From all of the above we see that Bell's theorem is that simultaneity is relative. So, according to which actually in favour of this interpretation and supports it. observer this word “simultaneously” is referring [25]? In fact, we have no contradiction between special relativity and quantum mechanics here at all. The reason is, that here, 12. Implications we should not talk of two events –the first spin assumed the value up and the other assumed the value down (not Of course, we see that if this interpretation is true, then it necessarily in this order)- rather, here we have one event can replace some of the currently existing interpretations of which is doing the experiment in the sense of the word used quantum mechanics. Furthermore, this interpretation needs in probability theory (because as we saw the outcomes of an no special requirements in quantum physics and sees it as experiment are not recognised until we do it). And doing an inevitable result if we consider all properties of particles the experiment in this interpretation is doing the to be statistical, i.e. if nature is indeterministic, and hence measurement. And when we do an experiment in the overwhelming number of quantum mechanical axioms probability theory and get some particular result, then we become results of this new formulation of classical can say that the result is one event that comes from doing probability theory and not axioms derived from the experiment. And we can in a looser language say that experiments. this event (getting the result) is the same as doing the Since space is represented by operators (position experiment. In the same sense, we say that after doing the operators) in quantum mechanics, so before measuring them, measurement on the two electrons in the singlet state, then they have no meaning, according to this interpretation. Still, what we get is one event and not two events. But we know the system that we are studying exists. So, according to this from special relativity that different observers agree on interpretation, and in order for it to be true, space must be events, but in general disagree on how to mark them with an emergent property. So, there must be deeper ways to spatial and temporal coordinates. So this event –the collapse describe matter. And the laws of quantum mechanics are the of the state vector of the composite system (which means statistical laws governing this un spatial state. So, maybe that after measurement the electrons assume opposite this state has its own laws that may be non statistical and spins)- is an event that all observers agree that it happens in that are even deeper than quantum mechanical laws. this way. What they could disagree on is where or when it Moreover, if we take two entangled particles, let us say two happened. electrons in the singlet state, so the state vector is given by Another result of this interpretation is that it proves that if the equation (193). we have any indeterministic universe, which can be This means that before the measurement they are neither described using probability theory, and in which the (the first up and the second down), nor (the first down and properties of its elementary particles are indeterministic in the second up), as the coin when it is still in my hand is this manner, then the laws that govern this kind of universe neither Heads nor Tails. must be those of quantum mechanics. And hence the answer If we separated these two electrons by a great distance, to the question: “why are the laws of our universe the way then do the measurement, so measuring one will surely they are?” may be: “because our universe is affect the other because their spins will become opposite. indeterministic”. How this happened? Well, this link between them that is deeper than space which we have talked about, suggests that it may provide the answer by allowing particles to exist and ACKNOWLEDGEMENTS interact in a level deeper than space itself. Also, with this new understanding of quantum concepts, I wish to acknowledge Diaa Fadel and Ammar Atrash for we see that quantum systems are not as different as we their help in typing this document. thought before from the classical systems. The only difference is that we assume that all properties of quantum systems are statistical, but we do not tend to think about classical systems that way. Anyway, we can apply the REFERENCES concepts of superposition, entanglement and the collapse of state vectors on some aspects of classical systems as we saw, [1] http://theoreticalminimum.com/courses/quantum-entangleme which imply that we can use some classical systems as nt/2006/fall. Q-bits that are much more easily to manipulate, and hence [2] Quantum Mechanics - an Introduction, 4th ed. - W. Greiner. may make quantum computation more practical. [3] http://abyss.uoregon.edu/~js/21st_century_science/lectures/l There is another problem that we can bypass using this ec15.html. interpretation. If we take two electrons in the singlet state which is given by (193). Then when we measure them, we [4] https://www.princeton.edu/~achaney/tmve/wiki100k/docs/C

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