Today s Outline - April 18, C. Segre (IIT) PHYS Spring 2017 April 18, / 23

Transcription

1 Today s Outline - April 18, 2017 C. Segre (IIT) PHYS Spring 2017 April 18, / 23

2 Today s Outline - April 18, 2017 The Einstein, Podolsky, Rosen paradox C. Segre (IIT) PHYS Spring 2017 April 18, / 23

3 Today s Outline - April 18, 2017 The Einstein, Podolsky, Rosen paradox Bell s inequality C. Segre (IIT) PHYS Spring 2017 April 18, / 23

4 Today s Outline - April 18, 2017 The Einstein, Podolsky, Rosen paradox Bell s inequality The EPR experiment C. Segre (IIT) PHYS Spring 2017 April 18, / 23

5 Today s Outline - April 18, 2017 The Einstein, Podolsky, Rosen paradox Bell s inequality The EPR experiment Homework Assignment #11: Chapter 11:7,9,12,14,18,20 due Tuesday, April 25, 2017 C. Segre (IIT) PHYS Spring 2017 April 18, / 23

6 Today s Outline - April 18, 2017 The Einstein, Podolsky, Rosen paradox Bell s inequality The EPR experiment Homework Assignment #11: Chapter 11:7,9,12,14,18,20 due Tuesday, April 25, 2017 Final Exam: Wednesday, May 3, :00-16:00, 036 Rettaliata Engineering C. Segre (IIT) PHYS Spring 2017 April 18, / 23

7 Quantum paradoxes and other fun stuff About your cat, Mr. Schrödinger I have good news and bad news... C. Segre (IIT) PHYS Spring 2017 April 18, / 23

8 Einstein Podolsky Rosen paradox.desc RI PT ION OF P H YSI CAL REALITY of lanthanum is 7/2, hence the nuclear magnetic moment as determined by this analysis is 2.5 nuclear magnetons. This is in fair agreement with the value 2.8 nuclear magnetons determined, from La III hyperfine structures by the writer and N. S. Grace. 9 ' M. F. Crawford and N. S. Grace, Phys. Rev. 4'7, 536 (1935). This investigation was carried out under the supervision of Professor G. Breit, and, I wish to thank him for the invaluable advice and assistance so freely given. I also take this opportunity to acknowledge the award of a Fellowship by the Royal Society of Canada, and to thank the University of Wisconsin and the Department of Physics for the privilege of working here. MAY 15, 1935 PH YSI CAL REVI EW VOLUM E 4 7 Can Quantum-Mechanical Description of Physical Reality Be Considered Complete'? A. EINsTEIN, B. PQDoLsKY AND N. RosEN, Institute for Advanced Study, Princeton, New Jersey (Received March 25, 1935) In a complete theory there is an element corresponding to each element of reality. A sufficient condition for the reality of a physical quantity is the possibility of predicting it with certainty, without disturbing the system. In quantum mechanics in the case of two physical quantities described by non-commuting operators, the knowledge of one precludes the knowledge of the other. Then either (1) the description of reality given by the wave function in A NY serious consideration of a physical theory must take into account the distinction between the objective reality, which is independent of any theory, and the physical concepts with which the theory operates. These concepts are intended to correspond with the objective reality, and by means of these concepts we picture this reality to ourselves. In attempting to judge the success of a physical theory, we may ask ourselves two questions: (1) "Is the theory correct?" and (2) "Is the description given by the theory complete?" It is only in the case in which positive answers may be given to both of these questions, that the concepts of the theory may be said to be satisfactory. The correctness of the theory is judged by the degree of agreement between the conclusions of the theory and human experience. This experience, which alone enables us to make inferences about reality, in physics takes the form of experiment and measurement. It is the second question that we wish to consider here, as applied to quantum mechanics. quantum mechanics is not complete or (2) these two quantities cannot have simultaneous reality. Consideration of the problem of making predictions concerning a system on the basis of measurements made on another system that had previously interacted with it leads to the result that if (1) is false then (2) is also false. One is thus led to conclude that the description of reality as given by a wave function is not complete. Whatever the meaning assigned to the term conzp/eee, the following requirement for a complete theory seems to be a necessary one: every element of the physical reality must have a counter part in the physical theory We shall ca. 11 this the condition of completeness. The second question is thus easily answered, as soon as we are able to decide what are the elements of the physical reality. The elements of the physical reality cannot be determined by a priori philosophical considerations, but must be found by an appeal to results of experiments and measurements. A comprehensive definition of reality is, however, unnecessary for our purpose. We shall be satisfied with the following criterion, which we regard as reasonable. If, without in any way disturbing a system, we can predict with certainty (i.e., with probability equal to unity) the value of a physical quantity, then there exists an element of physical reality corresponding lo this physical quantity. It seems to us that this criterion, while far from exhausting all possible ways of recognizing a physical reality, at least provides us with one 778 E I NSTE I N, PODOLSKY AN D ROSE N such way, whenever the conditions set down in In accordance with quantum mechanics we can it occur. Regarded not as a necessary, but only say that the relative probability that a merely as a sufficient, condition of reality, this measurement of the coordinate will give a result criterion is in agreement with classical as well as lying between a and b is quantum-mechanical ideas of reality. To illustrate the ideas involved let us consider P(a, b) = PPdx= I dx=b a. the quantum-mechanical description of the (6) behavior of a particle having a single degree of freedom. The fundamental concept of the theory Since this probability is independent of a, but is the concept of state, which is supposed to be depends only upon the difference b a, we see completely characterized by the wave function that all values of the coordinate are equally P, which is a function of the variables chosen to probable. describe the particle's behavior. Corresponding A definite value of the coordinate, for a particle in the state given by Eq. (2), is thus not to each physically observable quantity A there is an operator, which may be designated by the predictable, but may be obtained only by a same letter. direct measurement. Such a measurement however disturbs the particle and thus alters its If P is an eigenfunction of the operator A, that is, if state. After the coordinate is determined, the A/=a g, particle will no longer be in the state given by Eq. (2). The usual conclusion from this in where a is a number, then the physical quantity quantum mechanics is that when the momentnm A has with certainty the value a whenever the of a particle is known, its coordhnate has no physical particle is in the state given by P. In accordance reali ty. with our criterion of reality, for a particle in the More generally, it is shown in quantum mechanics that, if the operators corresponding to state given by P for which Eq. (1) holds, there is an element of physical reality corresponding two physical quantities, say A and B, do not to the physical quantity A. Let, for example, commute, that is, if AB/BA, then the precise 'p e (pre/ p) ppg knowledge of one of them precludes such (2) a knowledge of the other. Furthermore, any where h is Planck's constant, po is some constant attempt to determine the latter experimentally number, and x the independent variable. Since will alter the state of the system in such a way the operator corresponding to the momentum of as to destroy the knowledge of the first. the particle is From this follows that either (1) t'he guanturnmechanical description of rea1ity given by the wave p = (h/2rri) 8/Bx, function is not cornplele or (2) when the operators we obtain corresponding.to two physical qlantities do not p' = pp = (h/2iri) 8$/Bx = p pp (4) commute the two quantifies cannot have simultaneous reality. For if both of them had simultaneous reality and thus definite values these Thus, in the state given by Eq. (2), the momentum has certainly the value pp. It thus has values would enter into the complete description, meaning to say that the momentum of.the particle in the state given by Eq. (2) is real. then the wave function provided such a complete according to the condition of completeness. If On the other hand if Eq. (1) does not hold, description of reality, it would contain these we can no longer speak of the physical quantity values; these would then be predictable. This A having a particular value. This is the case, for not being the case, we are left with the alternatives stated. example, with the coordinate of the particle. The operator corresponding to it, say g, is the operator In quantum mechanics it is usually assumed of multiylication by the independent variable. that the wave function does contain a complete Thus, description of the physical reality of the system in the state to which it corresponds. At first Can quantum-mechanical description of physical reality be considered complete?, A. Einstein, B. Podolsky, and N. Rosen, Physical Review 47, (1935). C. Segre (IIT) PHYS Spring 2017 April 18, / 23

9 Einstein Podolsky Rosen paradox If, without in any way disturbing a system, we can predict with certainty (i.e., with probability equal to unity) the value of a physical quantity, then there exists an element of physical reality corresponding to this physical quantity. C. Segre (IIT) PHYS Spring 2017 April 18, / 23

10 Einstein Podolsky Rosen paradox If, without in any way disturbing a system, we can predict with certainty (i.e., with probability equal to unity) the value of a physical quantity, then there exists an element of physical reality corresponding to this physical quantity. If ψ is an eigenfuction of an operator A, then we know its expectation value, a C. Segre (IIT) PHYS Spring 2017 April 18, / 23

11 Einstein Podolsky Rosen paradox If, without in any way disturbing a system, we can predict with certainty (i.e., with probability equal to unity) the value of a physical quantity, then there exists an element of physical reality corresponding to this physical quantity. If ψ is an eigenfuction of an operator A, then we know its expectation value, a Aψ = aψ C. Segre (IIT) PHYS Spring 2017 April 18, / 23

12 Einstein Podolsky Rosen paradox If, without in any way disturbing a system, we can predict with certainty (i.e., with probability equal to unity) the value of a physical quantity, then there exists an element of physical reality corresponding to this physical quantity. If ψ is an eigenfuction of an operator A, then we know its expectation value, a for example take the momentum operator Aψ = aψ C. Segre (IIT) PHYS Spring 2017 April 18, / 23

13 Einstein Podolsky Rosen paradox If, without in any way disturbing a system, we can predict with certainty (i.e., with probability equal to unity) the value of a physical quantity, then there exists an element of physical reality corresponding to this physical quantity. If ψ is an eigenfuction of an operator A, then we know its expectation value, a for example take the momentum operator Aψ = aψ p = i x C. Segre (IIT) PHYS Spring 2017 April 18, / 23

14 Einstein Podolsky Rosen paradox If, without in any way disturbing a system, we can predict with certainty (i.e., with probability equal to unity) the value of a physical quantity, then there exists an element of physical reality corresponding to this physical quantity. If ψ is an eigenfuction of an operator A, then we know its expectation value, a for example take the momentum operator and the eigenfuction Aψ = aψ p = i x C. Segre (IIT) PHYS Spring 2017 April 18, / 23

15 Einstein Podolsky Rosen paradox If, without in any way disturbing a system, we can predict with certainty (i.e., with probability equal to unity) the value of a physical quantity, then there exists an element of physical reality corresponding to this physical quantity. If ψ is an eigenfuction of an operator A, then we know its expectation value, a for example take the momentum operator and the eigenfuction Aψ = aψ p = i x ψ = e ip 0x/ C. Segre (IIT) PHYS Spring 2017 April 18, / 23

16 Einstein Podolsky Rosen paradox If, without in any way disturbing a system, we can predict with certainty (i.e., with probability equal to unity) the value of a physical quantity, then there exists an element of physical reality corresponding to this physical quantity. If ψ is an eigenfuction of an operator A, then we know its expectation value, a for example take the momentum operator and the eigenfuction pψ = i x eip 0x/ Aψ = aψ p = i x ψ = e ip 0x/ C. Segre (IIT) PHYS Spring 2017 April 18, / 23

17 Einstein Podolsky Rosen paradox If, without in any way disturbing a system, we can predict with certainty (i.e., with probability equal to unity) the value of a physical quantity, then there exists an element of physical reality corresponding to this physical quantity. If ψ is an eigenfuction of an operator A, then we know its expectation value, a for example take the momentum operator and the eigenfuction pψ = i x eip 0x/ = i ( ) i p 0 e ip 0x/ Aψ = aψ p = i x ψ = e ip 0x/ C. Segre (IIT) PHYS Spring 2017 April 18, / 23

18 Einstein Podolsky Rosen paradox If, without in any way disturbing a system, we can predict with certainty (i.e., with probability equal to unity) the value of a physical quantity, then there exists an element of physical reality corresponding to this physical quantity. If ψ is an eigenfuction of an operator A, then we know its expectation value, a for example take the momentum operator and the eigenfuction pψ = i x eip 0x/ = i Aψ = aψ p = i x ψ = e ip 0x/ ( ) i p 0 e ip0x/ = p 0 ψ C. Segre (IIT) PHYS Spring 2017 April 18, / 23

19 Einstein Podolsky Rosen paradox If, without in any way disturbing a system, we can predict with certainty (i.e., with probability equal to unity) the value of a physical quantity, then there exists an element of physical reality corresponding to this physical quantity. If ψ is an eigenfuction of an operator A, then we know its expectation value, a for example take the momentum operator and the eigenfuction pψ = i x eip 0x/ = i thus the momentum in state ψ is said to be real Aψ = aψ p = i x ψ = e ip 0x/ ( ) i p 0 e ip0x/ = p 0 ψ C. Segre (IIT) PHYS Spring 2017 April 18, / 23

20 Einstein Podolsky Rosen paradox If Aψ = aψ does not hold, however, A cannot be said to have a particular value as we know from the position operator q C. Segre (IIT) PHYS Spring 2017 April 18, / 23

21 Einstein Podolsky Rosen paradox If Aψ = aψ does not hold, however, A cannot be said to have a particular value as we know from the position operator q qψ = xe ip 0x/ C. Segre (IIT) PHYS Spring 2017 April 18, / 23

22 Einstein Podolsky Rosen paradox If Aψ = aψ does not hold, however, A cannot be said to have a particular value as we know from the position operator q qψ = xe ip 0x/ aψ C. Segre (IIT) PHYS Spring 2017 April 18, / 23

23 Einstein Podolsky Rosen paradox If Aψ = aψ does not hold, however, A cannot be said to have a particular value as we know from the position operator q qψ = xe ip 0x/ aψ P(a, b) = b a ψ ψ dx C. Segre (IIT) PHYS Spring 2017 April 18, / 23

24 Einstein Podolsky Rosen paradox If Aψ = aψ does not hold, however, A cannot be said to have a particular value as we know from the position operator q qψ = xe ip 0x/ aψ P(a, b) = b a ψ ψ dx = b a dx C. Segre (IIT) PHYS Spring 2017 April 18, / 23

25 Einstein Podolsky Rosen paradox If Aψ = aψ does not hold, however, A cannot be said to have a particular value as we know from the position operator q qψ = xe ip 0x/ aψ P(a, b) = b a ψ ψ dx = b a dx = b a C. Segre (IIT) PHYS Spring 2017 April 18, / 23

26 Einstein Podolsky Rosen paradox If Aψ = aψ does not hold, however, A cannot be said to have a particular value as we know from the position operator q qψ = xe ip 0x/ aψ P(a, b) = b a ψ ψ dx = b there is an equal probability of measuring any value of the position a dx = b a C. Segre (IIT) PHYS Spring 2017 April 18, / 23

27 Einstein Podolsky Rosen paradox If Aψ = aψ does not hold, however, A cannot be said to have a particular value as we know from the position operator q qψ = xe ip 0x/ aψ P(a, b) = b a ψ ψ dx = b there is an equal probability of measuring any value of the position a dx = b a this is a specific instance of the fact that if two operators do not commute ([A, B] 0) then precise knowledge of one precludes such knowledge of the other C. Segre (IIT) PHYS Spring 2017 April 18, / 23

28 Einstein Podolsky Rosen paradox If Aψ = aψ does not hold, however, A cannot be said to have a particular value as we know from the position operator q qψ = xe ip 0x/ aψ P(a, b) = b a ψ ψ dx = b there is an equal probability of measuring any value of the position a dx = b a this is a specific instance of the fact that if two operators do not commute ([A, B] 0) then precise knowledge of one precludes such knowledge of the other the authors thus conclude that C. Segre (IIT) PHYS Spring 2017 April 18, / 23

29 Einstein Podolsky Rosen paradox If Aψ = aψ does not hold, however, A cannot be said to have a particular value as we know from the position operator q qψ = xe ip 0x/ aψ P(a, b) = b a ψ ψ dx = b there is an equal probability of measuring any value of the position a dx = b a this is a specific instance of the fact that if two operators do not commute ([A, B] 0) then precise knowledge of one precludes such knowledge of the other the authors thus conclude that 1 the quantum-mechanical description of reality given by the wave function is not complete, or C. Segre (IIT) PHYS Spring 2017 April 18, / 23

30 Einstein Podolsky Rosen paradox If Aψ = aψ does not hold, however, A cannot be said to have a particular value as we know from the position operator q qψ = xe ip 0x/ aψ P(a, b) = b a ψ ψ dx = b there is an equal probability of measuring any value of the position a dx = b a this is a specific instance of the fact that if two operators do not commute ([A, B] 0) then precise knowledge of one precludes such knowledge of the other the authors thus conclude that 1 the quantum-mechanical description of reality given by the wave function is not complete, or 2 when the operators corresponding to two physical quantities do not commute the two quantities cannot have simultaneous reality C. Segre (IIT) PHYS Spring 2017 April 18, / 23

31 Einstein Podolsky Rosen paradox Suppose that we have two systems, I and II which interact over a time period 0 t T C. Segre (IIT) PHYS Spring 2017 April 18, / 23

32 Einstein Podolsky Rosen paradox Suppose that we have two systems, I and II which interact over a time period 0 t T if we know the state of each system before t = 0, then by using the Schrödinger equation, we can calculate the state of the combined system I+II at any subsequent time, specifically t > T and we call it Ψ C. Segre (IIT) PHYS Spring 2017 April 18, / 23

33 Einstein Podolsky Rosen paradox Suppose that we have two systems, I and II which interact over a time period 0 t T if we know the state of each system before t = 0, then by using the Schrödinger equation, we can calculate the state of the combined system I+II at any subsequent time, specifically t > T and we call it Ψ suppose that a physical quantity A has eigenvalues C. Segre (IIT) PHYS Spring 2017 April 18, / 23

34 Einstein Podolsky Rosen paradox Suppose that we have two systems, I and II which interact over a time period 0 t T if we know the state of each system before t = 0, then by using the Schrödinger equation, we can calculate the state of the combined system I+II at any subsequent time, specifically t > T and we call it Ψ suppose that a physical quantity A has eigenvalues a 1, a 2, a 3,... C. Segre (IIT) PHYS Spring 2017 April 18, / 23

35 Einstein Podolsky Rosen paradox Suppose that we have two systems, I and II which interact over a time period 0 t T if we know the state of each system before t = 0, then by using the Schrödinger equation, we can calculate the state of the combined system I+II at any subsequent time, specifically t > T and we call it Ψ suppose that a physical quantity A has eigenvalues and eigenfunctions a 1, a 2, a 3,... C. Segre (IIT) PHYS Spring 2017 April 18, / 23

36 Einstein Podolsky Rosen paradox Suppose that we have two systems, I and II which interact over a time period 0 t T if we know the state of each system before t = 0, then by using the Schrödinger equation, we can calculate the state of the combined system I+II at any subsequent time, specifically t > T and we call it Ψ suppose that a physical quantity A has eigenvalues and eigenfunctions a 1, a 2, a 3,... u 1 (x 1 ), u 2 (x 1 ), u 3 (x 1 ),... C. Segre (IIT) PHYS Spring 2017 April 18, / 23

37 Einstein Podolsky Rosen paradox Suppose that we have two systems, I and II which interact over a time period 0 t T if we know the state of each system before t = 0, then by using the Schrödinger equation, we can calculate the state of the combined system I+II at any subsequent time, specifically t > T and we call it Ψ suppose that a physical quantity A has eigenvalues and eigenfunctions then the system-wide wavefunction as a function of x 1 is a 1, a 2, a 3,... u 1 (x 1 ), u 2 (x 1 ), u 3 (x 1 ),... C. Segre (IIT) PHYS Spring 2017 April 18, / 23

38 Einstein Podolsky Rosen paradox Suppose that we have two systems, I and II which interact over a time period 0 t T if we know the state of each system before t = 0, then by using the Schrödinger equation, we can calculate the state of the combined system I+II at any subsequent time, specifically t > T and we call it Ψ suppose that a physical quantity A has eigenvalues and eigenfunctions then the system-wide wavefunction as a function of x 1 is a 1, a 2, a 3,... u 1 (x 1 ), u 2 (x 1 ), u 3 (x 1 ),... Ψ(x 1, x 2 ) = ψ n (x 2 )u n (x 1 ) n=1 C. Segre (IIT) PHYS Spring 2017 April 18, / 23

39 Einstein Podolsky Rosen paradox Suppose that we have two systems, I and II which interact over a time period 0 t T if we know the state of each system before t = 0, then by using the Schrödinger equation, we can calculate the state of the combined system I+II at any subsequent time, specifically t > T and we call it Ψ suppose that a physical quantity A has eigenvalues and eigenfunctions then the system-wide wavefunction as a function of x 1 is where ψ n (x 2 ) are the coefficients of the expansion in u n (x 1 ) a 1, a 2, a 3,... u 1 (x 1 ), u 2 (x 1 ), u 3 (x 1 ),... Ψ(x 1, x 2 ) = ψ n (x 2 )u n (x 1 ) n=1 C. Segre (IIT) PHYS Spring 2017 April 18, / 23

40 Einstein Podolsky Rosen paradox Suppose that we have two systems, I and II which interact over a time period 0 t T if we know the state of each system before t = 0, then by using the Schrödinger equation, we can calculate the state of the combined system I+II at any subsequent time, specifically t > T and we call it Ψ suppose that a physical quantity A has eigenvalues and eigenfunctions then the system-wide wavefunction as a function of x 1 is where ψ n (x 2 ) are the coefficients of the expansion in u n (x 1 ) a 1, a 2, a 3,... u 1 (x 1 ), u 2 (x 1 ), u 3 (x 1 ),... Ψ(x 1, x 2 ) = ψ n (x 2 )u n (x 1 ) if A is now measured and is found to have value a k then the first system must be left in state u k (x 1 ) and the second system must be, therefore found in state ψ k (x 2 ) n=1 C. Segre (IIT) PHYS Spring 2017 April 18, / 23

41 Einstein Podolsky Rosen paradox Suppose that we have two systems, I and II which interact over a time period 0 t T if we know the state of each system before t = 0, then by using the Schrödinger equation, we can calculate the state of the combined system I+II at any subsequent time, specifically t > T and we call it Ψ suppose that a physical quantity A has eigenvalues and eigenfunctions then the system-wide wavefunction as a function of x 1 is where ψ n (x 2 ) are the coefficients of the expansion in u n (x 1 ) a 1, a 2, a 3,... u 1 (x 1 ), u 2 (x 1 ), u 3 (x 1 ),... Ψ(x 1, x 2 ) = ψ n (x 2 )u n (x 1 ) if A is now measured and is found to have value a k then the first system must be left in state u k (x 1 ) and the second system must be, therefore found in state ψ k (x 2 ) so that the combined state is Ψ = ψ k (x 2 )u k (x 1 ) n=1 C. Segre (IIT) PHYS Spring 2017 April 18, / 23

42 Einstein Podolsky Rosen paradox If instead of operator A, we choose to expand the overall wavefunction in terms of the eigenfunctions of B, then we have C. Segre (IIT) PHYS Spring 2017 April 18, / 23

43 Einstein Podolsky Rosen paradox If instead of operator A, we choose to expand the overall wavefunction in terms of the eigenfunctions of B, then we have b 1, b 2, b 3,... C. Segre (IIT) PHYS Spring 2017 April 18, / 23

44 Einstein Podolsky Rosen paradox If instead of operator A, we choose to expand the overall wavefunction in terms of the eigenfunctions of B, then we have b 1, b 2, b 3,... v 1 (x 1 ), v 2 (x 1 ), v 3 (x 1 ),... C. Segre (IIT) PHYS Spring 2017 April 18, / 23

45 Einstein Podolsky Rosen paradox If instead of operator A, we choose to expand the overall wavefunction in terms of the eigenfunctions of B, then we have b 1, b 2, b 3,... v 1 (x 1 ), v 2 (x 1 ), v 3 (x 1 ),... Ψ(x 1, x 2 ) = ϕ s (x 2 )v s (x 1 ) s=1 C. Segre (IIT) PHYS Spring 2017 April 18, / 23

46 Einstein Podolsky Rosen paradox If instead of operator A, we choose to expand the overall wavefunction in terms of the eigenfunctions of B, then we have b 1, b 2, b 3,... v 1 (x 1 ), v 2 (x 1 ), v 3 (x 1 ),... Ψ(x 1, x 2 ) = ϕ s (x 2 )v s (x 1 ) if B is measured and found to be b r, then the combined system can be said to be in the state Ψ = ϕ r (x 2 )v r (x 1 ) and the second system must be in state ϕ r (x 2 ) s=1 C. Segre (IIT) PHYS Spring 2017 April 18, / 23

47 Einstein Podolsky Rosen paradox If instead of operator A, we choose to expand the overall wavefunction in terms of the eigenfunctions of B, then we have b 1, b 2, b 3,... v 1 (x 1 ), v 2 (x 1 ), v 3 (x 1 ),... Ψ(x 1, x 2 ) = ϕ s (x 2 )v s (x 1 ) if B is measured and found to be b r, then the combined system can be said to be in the state Ψ = ϕ r (x 2 )v r (x 1 ) and the second system must be in state ϕ r (x 2 ) Thus, as a consequence of two different measurements made on System I, System II can be left in states with two different wave functions, even when it is far away from, and not interacting with System I s=1 C. Segre (IIT) PHYS Spring 2017 April 18, / 23

48 Einstein Podolsky Rosen paradox If instead of operator A, we choose to expand the overall wavefunction in terms of the eigenfunctions of B, then we have b 1, b 2, b 3,... v 1 (x 1 ), v 2 (x 1 ), v 3 (x 1 ),... Ψ(x 1, x 2 ) = ϕ s (x 2 )v s (x 1 ) if B is measured and found to be b r, then the combined system can be said to be in the state Ψ = ϕ r (x 2 )v r (x 1 ) and the second system must be in state ϕ r (x 2 ) Thus, as a consequence of two different measurements made on System I, System II can be left in states with two different wave functions, even when it is far away from, and not interacting with System I Thus one can assign two different wave functions, ψ k (x 2 ) and ϕ r (x 2 ), to the same reality (System II after interaction with System I) s=1 C. Segre (IIT) PHYS Spring 2017 April 18, / 23

49 Einstein Podolsky Rosen paradox Suppose that ψ k (x 2 ) and ϕ r (x 2 ) are eigenfunctions of two non-commuting operators, P and Q with eigenvalues p k and q r C. Segre (IIT) PHYS Spring 2017 April 18, / 23

50 Einstein Podolsky Rosen paradox Suppose that ψ k (x 2 ) and ϕ r (x 2 ) are eigenfunctions of two non-commuting operators, P and Q with eigenvalues p k and q r By measuring either A or B on System I, we are able to predict with certainty, and without disturbing System II, either the value of P (p k ) or Q (q r ) C. Segre (IIT) PHYS Spring 2017 April 18, / 23

51 Einstein Podolsky Rosen paradox Suppose that ψ k (x 2 ) and ϕ r (x 2 ) are eigenfunctions of two non-commuting operators, P and Q with eigenvalues p k and q r By measuring either A or B on System I, we are able to predict with certainty, and without disturbing System II, either the value of P (p k ) or Q (q r ) According to our criterion for reality, in the first case, P must be an element of reality and in the second case, Q must be an element of reality but ψ k (x 2 ) and ϕ r (x 2 ) were shown to be part of the same reality and this leads to a contradiction of the postulate that when the operators corresponding to two physical quantities do not commute the two quantities cannot have simultaneous reality C. Segre (IIT) PHYS Spring 2017 April 18, / 23

52 Einstein Podolsky Rosen paradox Suppose that ψ k (x 2 ) and ϕ r (x 2 ) are eigenfunctions of two non-commuting operators, P and Q with eigenvalues p k and q r By measuring either A or B on System I, we are able to predict with certainty, and without disturbing System II, either the value of P (p k ) or Q (q r ) According to our criterion for reality, in the first case, P must be an element of reality and in the second case, Q must be an element of reality but ψ k (x 2 ) and ϕ r (x 2 ) were shown to be part of the same reality and this leads to a contradiction of the postulate that when the operators corresponding to two physical quantities do not commute the two quantities cannot have simultaneous reality Thus, the authors conclude that the quantum-mechanical description of reality given by the wave function is not complete C. Segre (IIT) PHYS Spring 2017 April 18, / 23

53 Einstein Podolsky Rosen paradox Suppose that ψ k (x 2 ) and ϕ r (x 2 ) are eigenfunctions of two non-commuting operators, P and Q with eigenvalues p k and q r By measuring either A or B on System I, we are able to predict with certainty, and without disturbing System II, either the value of P (p k ) or Q (q r ) According to our criterion for reality, in the first case, P must be an element of reality and in the second case, Q must be an element of reality but ψ k (x 2 ) and ϕ r (x 2 ) were shown to be part of the same reality and this leads to a contradiction of the postulate that when the operators corresponding to two physical quantities do not commute the two quantities cannot have simultaneous reality Thus, the authors conclude that the quantum-mechanical description of reality given by the wave function is not complete If this realist version of quantum mechanics is correct, the complete description of reality must include some local hidden variable(s) which specify the state of the system completely C. Segre (IIT) PHYS Spring 2017 April 18, / 23

54 Problem 12.1 Consider a two-level system, φ a and φ b, with φ i φ j = δ ij Prove that the (entangled) two-particle state α φ a (1) φ b (2) + β φ b (1) φ a (2) cannot be expressed as a product ψ r (1) ψ s (2) for any one-particle states ψ r and ψ s C. Segre (IIT) PHYS Spring 2017 April 18, / 23

55 Problem 12.1 Consider a two-level system, φ a and φ b, with φ i φ j = δ ij Prove that the (entangled) two-particle state α φ a (1) φ b (2) + β φ b (1) φ a (2) cannot be expressed as a product ψ r (1) ψ s (2) for any one-particle states ψ r and ψ s Start by assuming that the entangled state can be so expressed C. Segre (IIT) PHYS Spring 2017 April 18, / 23

56 Problem 12.1 Consider a two-level system, φ a and φ b, with φ i φ j = δ ij Prove that the (entangled) two-particle state α φ a (1) φ b (2) + β φ b (1) φ a (2) cannot be expressed as a product ψ r (1) ψ s (2) for any one-particle states ψ r and ψ s Start by assuming that the entangled state can be so expressed α φ a (1) φ b (2) + β φ b (1) φ a (2) = ψ r (1) ψ s (2) C. Segre (IIT) PHYS Spring 2017 April 18, / 23

57 Problem 12.1 Consider a two-level system, φ a and φ b, with φ i φ j = δ ij Prove that the (entangled) two-particle state α φ a (1) φ b (2) + β φ b (1) φ a (2) cannot be expressed as a product ψ r (1) ψ s (2) for any one-particle states ψ r and ψ s Start by assuming that the entangled state can be so expressed α φ a (1) φ b (2) + β φ b (1) φ a (2) = ψ r (1) ψ s (2) expand ψ r and ψ s in terms of the two basis states C. Segre (IIT) PHYS Spring 2017 April 18, / 23

58 Problem 12.1 Consider a two-level system, φ a and φ b, with φ i φ j = δ ij Prove that the (entangled) two-particle state α φ a (1) φ b (2) + β φ b (1) φ a (2) cannot be expressed as a product ψ r (1) ψ s (2) for any one-particle states ψ r and ψ s Start by assuming that the entangled state can be so expressed α φ a (1) φ b (2) + β φ b (1) φ a (2) = ψ r (1) ψ s (2) expand ψ r and ψ s in terms of the two basis states ψ r = A φ a + B φ b C. Segre (IIT) PHYS Spring 2017 April 18, / 23

59 Problem 12.1 Consider a two-level system, φ a and φ b, with φ i φ j = δ ij Prove that the (entangled) two-particle state α φ a (1) φ b (2) + β φ b (1) φ a (2) cannot be expressed as a product ψ r (1) ψ s (2) for any one-particle states ψ r and ψ s Start by assuming that the entangled state can be so expressed α φ a (1) φ b (2) + β φ b (1) φ a (2) = ψ r (1) ψ s (2) expand ψ r and ψ s in terms of the two basis states ψ r = A φ a + B φ b ψ s = C φ a + D φ b C. Segre (IIT) PHYS Spring 2017 April 18, / 23

60 Problem 12.1 α φ a (1) φ b (2) + β φ b (1) φ a (2) = [A φ a (1) + B φ b (1) ] [C φ a (2) + D φ b (2) ] C. Segre (IIT) PHYS Spring 2017 April 18, / 23

61 Problem 12.1 α φ a (1) φ b (2) + β φ b (1) φ a (2) = [A φ a (1) + B φ b (1) ] [C φ a (2) + D φ b (2) ] = AC φ a (1) φ a (2) + AD φ a (1) φ b (2) + BC φ b (1) φ a (2) + BD φ b (1) φ b (2) C. Segre (IIT) PHYS Spring 2017 April 18, / 23

62 Problem 12.1 α φ a (1) φ b (2) + β φ b (1) φ a (2) = [A φ a (1) + B φ b (1) ] [C φ a (2) + D φ b (2) ] = AC φ a (1) φ a (2) + AD φ a (1) φ b (2) + BC φ b (1) φ a (2) + BD φ b (1) φ b (2) comparing the two sides of the equation: C. Segre (IIT) PHYS Spring 2017 April 18, / 23

63 Problem 12.1 α φ a (1) φ b (2) + β φ b (1) φ a (2) = [A φ a (1) + B φ b (1) ] [C φ a (2) + D φ b (2) ] = AC φ a (1) φ a (2) + AD φ a (1) φ b (2) + BC φ b (1) φ a (2) + BD φ b (1) φ b (2) comparing the two sides of the equation: α = AD C. Segre (IIT) PHYS Spring 2017 April 18, / 23

64 Problem 12.1 α φ a (1) φ b (2) + β φ b (1) φ a (2) = [A φ a (1) + B φ b (1) ] [C φ a (2) + D φ b (2) ] = AC φ a (1) φ a (2) + AD φ a (1) φ b (2) + BC φ b (1) φ a (2) + BD φ b (1) φ b (2) comparing the two sides of the equation: α = AD 0 = AC C. Segre (IIT) PHYS Spring 2017 April 18, / 23

65 Problem 12.1 α φ a (1) φ b (2) + β φ b (1) φ a (2) = [A φ a (1) + B φ b (1) ] [C φ a (2) + D φ b (2) ] = AC φ a (1) φ a (2) + AD φ a (1) φ b (2) + BC φ b (1) φ a (2) + BD φ b (1) φ b (2) comparing the two sides of the equation: α = AD β = BC 0 = AC C. Segre (IIT) PHYS Spring 2017 April 18, / 23

66 Problem 12.1 α φ a (1) φ b (2) + β φ b (1) φ a (2) = [A φ a (1) + B φ b (1) ] [C φ a (2) + D φ b (2) ] = AC φ a (1) φ a (2) + AD φ a (1) φ b (2) + BC φ b (1) φ a (2) + BD φ b (1) φ b (2) comparing the two sides of the equation: α = AD β = BC 0 = AC 0 = BD C. Segre (IIT) PHYS Spring 2017 April 18, / 23

67 Problem 12.1 α φ a (1) φ b (2) + β φ b (1) φ a (2) = [A φ a (1) + B φ b (1) ] [C φ a (2) + D φ b (2) ] = AC φ a (1) φ a (2) + AD φ a (1) φ b (2) + BC φ b (1) φ a (2) + BD φ b (1) φ b (2) comparing the two sides of the equation: α = AD β = BC 0 = AC 0 = BD at least two of the coefficients (A, B, C, or D) must be zero to satisfy the two equations on the right, but then either α or β must also be zero which is a condition excluded by the initial definition of the entangled state C. Segre (IIT) PHYS Spring 2017 April 18, / 23

68 Bell s inequality On the Einstein Podolsky Rosen paradox, J.S. Bell, Physics 1, (1964). C. Segre (IIT) PHYS Spring 2017 April 18, / 23

69 Bell s inequality Consider an experiment wherein a pair of spin one-half particles are prepared in a singlet spin state and are moving freely in opposite directions. C. Segre (IIT) PHYS Spring 2017 April 18, / 23

70 Bell s inequality Consider an experiment wherein a pair of spin one-half particles are prepared in a singlet spin state and are moving freely in opposite directions. Suppose that two apparati are prepared to measure the spin state of each particle, σ 1 and σ 2 by means of Stern-Gerlach magnets which do not interfere with the other particle. C. Segre (IIT) PHYS Spring 2017 April 18, / 23

71 Bell s inequality Consider an experiment wherein a pair of spin one-half particles are prepared in a singlet spin state and are moving freely in opposite directions. Suppose that two apparati are prepared to measure the spin state of each particle, σ 1 and σ 2 by means of Stern-Gerlach magnets which do not interfere with the other particle. If we measure the component σ 1 â and get +1 C. Segre (IIT) PHYS Spring 2017 April 18, / 23

72 Bell s inequality Consider an experiment wherein a pair of spin one-half particles are prepared in a singlet spin state and are moving freely in opposite directions. Suppose that two apparati are prepared to measure the spin state of each particle, σ 1 and σ 2 by means of Stern-Gerlach magnets which do not interfere with the other particle. If we measure the component σ 1 â and get +1 then we can presume that σ 2 â = 1 without doing the measurement, the result is predetermined. C. Segre (IIT) PHYS Spring 2017 April 18, / 23

73 Bell s inequality Consider an experiment wherein a pair of spin one-half particles are prepared in a singlet spin state and are moving freely in opposite directions. Suppose that two apparati are prepared to measure the spin state of each particle, σ 1 and σ 2 by means of Stern-Gerlach magnets which do not interfere with the other particle. If we measure the component σ 1 â and get +1 then we can presume that σ 2 â = 1 without doing the measurement, the result is predetermined. Assume that this predetermination is characterized by parameters λ (a local hidden variable). C. Segre (IIT) PHYS Spring 2017 April 18, / 23

74 Bell s inequality Consider an experiment wherein a pair of spin one-half particles are prepared in a singlet spin state and are moving freely in opposite directions. Suppose that two apparati are prepared to measure the spin state of each particle, σ 1 and σ 2 by means of Stern-Gerlach magnets which do not interfere with the other particle. If we measure the component σ 1 â and get +1 then we can presume that σ 2 â = 1 without doing the measurement, the result is predetermined. Assume that this predetermination is characterized by parameters λ (a local hidden variable). The result, A, of measuring σ 1 â is then determined by â and λ C. Segre (IIT) PHYS Spring 2017 April 18, / 23

75 Bell s inequality Consider an experiment wherein a pair of spin one-half particles are prepared in a singlet spin state and are moving freely in opposite directions. Suppose that two apparati are prepared to measure the spin state of each particle, σ 1 and σ 2 by means of Stern-Gerlach magnets which do not interfere with the other particle. If we measure the component σ 1 â and get +1 then we can presume that σ 2 â = 1 without doing the measurement, the result is predetermined. Assume that this predetermination is characterized by parameters λ (a local hidden variable). The result, A, of measuring σ 1 â is then determined by â and λ and the results, B, of measuring σ 2 ˆb is then determined by ˆb and λ. A(â, λ) = ±1, B(ˆb, λ) = ±1 C. Segre (IIT) PHYS Spring 2017 April 18, / 23

76 Bell s inequality Result B for particle 2 is assumed not to depend on â and neither does A depend on ˆb. C. Segre (IIT) PHYS Spring 2017 April 18, / 23

77 Bell s inequality Result B for particle 2 is assumed not to depend on â and neither does A depend on ˆb. If ρ(λ) is the probability distribution of λ, then the expectation value of the product of σ 1 â and σ 2 ˆb is C. Segre (IIT) PHYS Spring 2017 April 18, / 23

78 Bell s inequality Result B for particle 2 is assumed not to depend on â and neither does A depend on ˆb. If ρ(λ) is the probability distribution of λ, then the expectation value of the product of σ 1 â and σ 2 ˆb is P(â, ˆb) = ρ(λ)a(â, λ)b(ˆb, λ) dλ C. Segre (IIT) PHYS Spring 2017 April 18, / 23

79 Bell s inequality Result B for particle 2 is assumed not to depend on â and neither does A depend on ˆb. If ρ(λ) is the probability distribution of λ, then the expectation value of the product of σ 1 â and σ 2 ˆb is and it should be equivalent to the quantum expectation value P(â, ˆb) = ρ(λ)a(â, λ)b(ˆb, λ) dλ C. Segre (IIT) PHYS Spring 2017 April 18, / 23

80 Bell s inequality Result B for particle 2 is assumed not to depend on â and neither does A depend on ˆb. If ρ(λ) is the probability distribution of λ, then the expectation value of the product of σ 1 â and σ 2 ˆb is and it should be equivalent to the quantum expectation value P(â, ˆb) = ρ(λ)a(â, λ)b(ˆb, λ) dλ ( σ 1 â)( σ 2 ˆb) = â ˆb C. Segre (IIT) PHYS Spring 2017 April 18, / 23

81 Bell s inequality Result B for particle 2 is assumed not to depend on â and neither does A depend on ˆb. If ρ(λ) is the probability distribution of λ, then the expectation value of the product of σ 1 â and σ 2 ˆb is and it should be equivalent to the quantum expectation value P(â, ˆb) = ρ(λ)a(â, λ)b(ˆb, λ) dλ ( σ 1 â)( σ 2 ˆb) = â ˆb The essence of Bell s paper is that this last statement is not possible because of the presence of the hidden local variable(s) λ C. Segre (IIT) PHYS Spring 2017 April 18, / 23

82 Aside: Problem 4.50 Suppose we have 2 spin 1 2 fermions in a singlet configuration. If S a1 is the component of the first particle s spin angular momentum in the direction â and S b2 is the component of the second particle s spin angular momentum in the ˆb direction. Compute S a1 S b2 C. Segre (IIT) PHYS Spring 2017 April 18, / 23

83 Aside: Problem 4.50 Suppose we have 2 spin 1 2 fermions in a singlet configuration. If S a1 is the component of the first particle s spin angular momentum in the direction â and S b2 is the component of the second particle s spin angular momentum in the ˆb direction. Compute S a1 S b2 The singlet state can be written as C. Segre (IIT) PHYS Spring 2017 April 18, / 23

84 Aside: Problem 4.50 Suppose we have 2 spin 1 2 fermions in a singlet configuration. If S a1 is the component of the first particle s spin angular momentum in the direction â and S b2 is the component of the second particle s spin angular momentum in the ˆb direction. Compute S a1 S b2 The singlet state can be written as 0 0 = 1 2 ( ) C. Segre (IIT) PHYS Spring 2017 April 18, / 23

85 Aside: Problem 4.50 Suppose we have 2 spin 1 2 fermions in a singlet configuration. If S a1 is the component of the first particle s spin angular momentum in the direction â and S b2 is the component of the second particle s spin angular momentum in the ˆb direction. Compute S a1 S b2 The singlet state can be written as if we choose â ẑ and put ˆb in the x-z plane 0 0 = 1 2 ( ) C. Segre (IIT) PHYS Spring 2017 April 18, / 23

86 Aside: Problem 4.50 Suppose we have 2 spin 1 2 fermions in a singlet configuration. If S a1 is the component of the first particle s spin angular momentum in the direction â and S b2 is the component of the second particle s spin angular momentum in the ˆb direction. Compute S a1 S b2 The singlet state can be written as if we choose â ẑ and put ˆb in the x-z plane 0 0 = 1 2 ( ) S a1 = S z1 C. Segre (IIT) PHYS Spring 2017 April 18, / 23

87 Aside: Problem 4.50 Suppose we have 2 spin 1 2 fermions in a singlet configuration. If S a1 is the component of the first particle s spin angular momentum in the direction â and S b2 is the component of the second particle s spin angular momentum in the ˆb direction. Compute S a1 S b2 The singlet state can be written as if we choose â ẑ and put ˆb in the x-z plane 0 0 = 1 2 ( ) S a1 = S z1 S b2 = cos θs z2 + sin θs x2 C. Segre (IIT) PHYS Spring 2017 April 18, / 23

88 Aside: Problem 4.50 Suppose we have 2 spin 1 2 fermions in a singlet configuration. If S a1 is the component of the first particle s spin angular momentum in the direction â and S b2 is the component of the second particle s spin angular momentum in the ˆb direction. Compute S a1 S b2 The singlet state can be written as if we choose â ẑ and put ˆb in the x-z plane 0 0 = 1 2 ( ) S a1 = S z1 S b2 = cos θs z2 + sin θs x2 S a1 S b2 0 0 = C. Segre (IIT) PHYS Spring 2017 April 18, / 23

89 Aside: Problem 4.50 Suppose we have 2 spin 1 2 fermions in a singlet configuration. If S a1 is the component of the first particle s spin angular momentum in the direction â and S b2 is the component of the second particle s spin angular momentum in the ˆb direction. Compute S a1 S b2 The singlet state can be written as 0 0 = 1 2 ( ) if we choose â ẑ and put ˆb in the S a1 = S z1 x-z plane S b2 = cos θs z2 + sin θs x2 S a1 S b2 0 0 = 1 [S z1 (cos θs z2 + sin θs x2 )] ( ) 2 C. Segre (IIT) PHYS Spring 2017 April 18, / 23

90 Aside: Problem 4.50 Suppose we have 2 spin 1 2 fermions in a singlet configuration. If S a1 is the component of the first particle s spin angular momentum in the direction â and S b2 is the component of the second particle s spin angular momentum in the ˆb direction. Compute S a1 S b2 The singlet state can be written as 0 0 = 1 2 ( ) if we choose â ẑ and put ˆb in the S a1 = S z1 x-z plane S b2 = cos θs z2 + sin θs x2 S a1 S b2 0 0 = 1 [S z1 (cos θs z2 + sin θs x2 )] ( ) 2 = 1 2 [S z1 1 (cos θs z2 2 ) ] C. Segre (IIT) PHYS Spring 2017 April 18, / 23

91 Aside: Problem 4.50 Suppose we have 2 spin 1 2 fermions in a singlet configuration. If S a1 is the component of the first particle s spin angular momentum in the direction â and S b2 is the component of the second particle s spin angular momentum in the ˆb direction. Compute S a1 S b2 The singlet state can be written as 0 0 = 1 2 ( ) if we choose â ẑ and put ˆb in the S a1 = S z1 x-z plane S b2 = cos θs z2 + sin θs x2 S a1 S b2 0 0 = 1 [S z1 (cos θs z2 + sin θs x2 )] ( ) 2 = 1 2 [S z1 1 (cos θs z2 2 + sin θs x2 2 ) ] C. Segre (IIT) PHYS Spring 2017 April 18, / 23

92 Aside: Problem 4.50 Suppose we have 2 spin 1 2 fermions in a singlet configuration. If S a1 is the component of the first particle s spin angular momentum in the direction â and S b2 is the component of the second particle s spin angular momentum in the ˆb direction. Compute S a1 S b2 The singlet state can be written as 0 0 = 1 2 ( ) if we choose â ẑ and put ˆb in the S a1 = S z1 x-z plane S b2 = cos θs z2 + sin θs x2 S a1 S b2 0 0 = 1 [S z1 (cos θs z2 + sin θs x2 )] ( ) 2 = 1 2 [S z1 1 (cos θs z2 2 + sin θs x2 2 ) S z1 1 (cos θs z2 2 )] C. Segre (IIT) PHYS Spring 2017 April 18, / 23

93 Aside: Problem 4.50 Suppose we have 2 spin 1 2 fermions in a singlet configuration. If S a1 is the component of the first particle s spin angular momentum in the direction â and S b2 is the component of the second particle s spin angular momentum in the ˆb direction. Compute S a1 S b2 The singlet state can be written as 0 0 = 1 2 ( ) if we choose â ẑ and put ˆb in the S a1 = S z1 x-z plane S b2 = cos θs z2 + sin θs x2 S a1 S b2 0 0 = 1 [S z1 (cos θs z2 + sin θs x2 )] ( ) 2 = 1 2 [S z1 1 (cos θs z2 2 + sin θs x2 2 ) S z1 1 (cos θs z2 2 + sin θs x2 2 )] C. Segre (IIT) PHYS Spring 2017 April 18, / 23

94 Aside: Problem 4.50 (cont.) S a1 S b2 0 0 = 1 2 [S z1 1 (cos θs z2 2 + sin θs x2 2 ) S z1 1 (cos θs z2 2 + sin θs x2 2 )] C. Segre (IIT) PHYS Spring 2017 April 18, / 23

95 Aside: Problem 4.50 (cont.) S a1 S b2 0 0 = 1 2 [S z1 1 (cos θs z2 2 + sin θs x2 2 ) S z1 1 (cos θs z2 2 + sin θs x2 2 )] = 1 2 { 1 [ cos θ 2 + sin θ 2 ] 2 4 } + 1 [cos θ 2 + sin θ 2 ] C. Segre (IIT) PHYS Spring 2017 April 18, / 23

96 Aside: Problem 4.50 (cont.) S a1 S b2 0 0 = 1 2 [S z1 1 (cos θs z2 2 + sin θs x2 2 ) S z1 1 (cos θs z2 2 + sin θs x2 2 )] = 1 2 { 1 [ cos θ 2 + sin θ 2 ] 2 4 } + 1 [cos θ 2 + sin θ 2 ] = 2 [ cos θ 1 ( ) sin θ 1 ] ( ) 2 C. Segre (IIT) PHYS Spring 2017 April 18, / 23

97 Aside: Problem 4.50 (cont.) S a1 S b2 0 0 = 1 2 [S z1 1 (cos θs z2 2 + sin θs x2 2 ) 0 0 S a1 S b2 0 0 = 2 4 S z1 1 (cos θs z2 2 + sin θs x2 2 )] = 1 2 { 1 [ cos θ 2 + sin θ 2 ] 2 4 } + 1 [cos θ 2 + sin θ 2 ] = 2 [ cos θ 1 ( ) sin θ 1 ] ( ) 2 cos θ C. Segre (IIT) PHYS Spring 2017 April 18, / 23

98 Aside: Problem 4.50 (cont.) S a1 S b2 0 0 = 1 2 [S z1 1 (cos θs z2 2 + sin θs x2 2 ) 0 0 S a1 S b2 0 0 = 2 4 S z1 1 (cos θs z2 2 + sin θs x2 2 )] = 1 2 { 1 [ cos θ 2 + sin θ 2 ] 2 4 } + 1 [cos θ 2 + sin θ 2 ] = 2 [ cos θ 1 ( ) sin θ 1 ] ( ) 2 cos θ = 2 4 cos θ C. Segre (IIT) PHYS Spring 2017 April 18, / 23

99 Aside: Problem 4.50 (cont.) S a1 S b2 0 0 = 1 2 [S z1 1 (cos θs z2 2 + sin θs x2 2 ) 0 0 S a1 S b2 0 0 = 2 4 S z1 1 (cos θs z2 2 + sin θs x2 2 )] = 1 2 { 1 [ cos θ 2 + sin θ 2 ] 2 4 } + 1 [cos θ 2 + sin θ 2 ] = 2 [ cos θ 1 ( ) sin θ 1 ] ( ) 2 cos θ = 2 4 cos θ = 2 4 â ˆb C. Segre (IIT) PHYS Spring 2017 April 18, / 23