Physics Capstone Karsten Gimre May 19, 009
Abstract Quantum mechanics is, without a doubt, one of the major advancements of the 0th century. It can be incredibly difficult to understand the fundamental differences between the two opposing views of nature: classical mechanics and quantum mechanics. In this paper, we construct a betting game to illustrate the fundamental differences between the classical and quantum mechanical views of physics. A computer simulation indicates that over the average of twenty trials of the game, a quantum physicist can earn more than ten million times as much money as a classical physicist.
1 Introduction Quantum physics introduces a new view of reality that shakes the old classical view at its foundations in a variety of ways. It s often difficult to wrap your mind around the differences; quantum physics is weird, but it s not easy to pinpoint the exact nature of the weirdness. This paper describes a thought experiment that highlights one aspect of this weirdness known as entanglement in a concrete way that has practical consequences. By harnessing the power of quantum entanglement, one can consistently win money from an opponent who uses only classical correlations. Similar attempts have been made by authors such as [ to replicate quantum effects to macroscopic and observable events. One reason to be interested in entanglement is the profound philosophical implications it has for our understanding of reality. This thought experiment/game helps readers confront this in a tangible way. Entanglement also has practical applications which have begun to be developed. Specifically, quantum entanglement can be viewed as a resource that allows us to do things that would be impossible in a classical reality [1. The concrete experimental system we ll work with in this paper is the Stern Gerlach apparatus, described in the following following section. Background Consider a stream of particles traveling through an inhomogeneous magnetic field; namely, the magnetic field is described by 1 B(x, y, z) = αxî + (B 0 + αz)ˆk in which α describes how much B spreads out in the î direction and B 0 describes the overall magnitude of the field. This setup describes the Stern-Gerlach apparatus, and the mathematical prediction of what happens to the magnetic dipole is derived in the following sections..1 Classical Approach Classically, the force on a particle in the stream is described by F = (µ B) Then since the magnetic dipole moment µ is defined by µ = γs where γ is the gyromagnetic ratio and S is the spin angular momentum, F = γα( S x î + S zˆk) 1 For conceptual simplicity, there would ideally only be a ˆk component in the magnetic field, but that would violate the Maxwell equation B = 0. The gyromagnetic ratio describes the ratio of the magnetic dipole moment of a system of particles to the angular momentum of the system, and is a constant throughout the Stern- Gerlach process. 3
But according to the Larmor precession of S around B 0, which describes how the spin of an atom will rotate around a magnetic field axis, S x = sin α cos (γb 0t) which means that S x oscillates at a frequency of γb 0. So then on average the x component of the spin cancels, so the net force is in the z-direction: specifically, F z = γαs z. Classically, spin can be oriented in any direction, so that S z could be anything within a certain range of values; hence, F z would vary continuously from atom to atom and if a stream of elementary particles were directed through the magnetic field, they would be deflected into a continuous stream.. Quantum Approach Now operating in the quantum perspective, the Hamiltonian of the particle, as defined as H = µ B, is here 0 for t < 0 H(t) = γ(b 0 + αz)s z for 0 t T 0 for t > T where t is defined so that t = 0 corresponds to the time when the particle enters the magnetic field and t = T is when it leaves the field. Now suppose that the particle is elementary, so that it has spin 1/. Then suppose that when t < 0, it is in the state χ(t) = aχ + + bχ Once it enters the magnetic field, its state will evolve according to Schrödinger s equation, so that when 0 t T, in which χ(t) = aχ + e ie + t / + bχ e ie t / E ± = γ(b 0 + αz) So then once t = T, corresponding to when the particle leaves the magnetic field, χ(t) = (ae iγt B0/ χ + )e i(αγt/)z + (be iγt B0/ χ )e i(αγt/)z Since the Hamiltonian is zero once t T, this is the state that the particle will remain in at all times following its exit from the magnetic field. Since the orthonormal eigenfunctions of the momentum operator are 1 π e ipx/ where the eigenvalues p are the observable values for momentum, it is now clear to see that the observable momenta in the case of the Stern-Gerlach apparatus 4
αγt are ±, which is positive if the particle is spin up in the z-direction and negative if the particle is spin down. So in the quantum derivation, the magnetic field splits the beam of elementary particles into two separate beams, one for spin-up particles and one for spin-down. This prediction has been verified by numerous experiments. The Stern-Gerlach apparatus is summarized in the following diagram, including the classical and quantum predictions: Figure 1: The Stern-Gerlach apparatus. [5 5
.3 Multiple Stern-Gerlach Apparatuses Now consider multiple Stern-Gerlach apparatuses in series: Figure : Multiple Stern-Gerlach apparatuses. [5 In figure (a), there is a second Stern-Gerlach apparatus oriented parallel to the first. Since the first apparatus split the particles into spin-up particles and spin-down particles, only spin-up particles will enter the second apparatus, so all particles entering that apparatus will be deflected upwards. In Figure (b), though, the second apparatus is rotated by 90. The particles entering it are only spin-up along the z-axis, but they have a random spin along the x-axis. The same is true for any perpendicular axes. The natural question that then arises is this: what happens if the second apparatus is not rotated by a right angle? What if, for example, the second apparatus were aligned with the line z = x: that is, a 45 rotation? To answer this, first we have to derive some more mathematical predictions. 3 Mathematical Formulation 3.1 State Vectors, Spin Matrices, and Observables Before we can derive the predictions for spin measurements, we need to formulate some mathematical ideas about how to express quantum mechanical measurements and observations mathematically. Quantum mechanics is unique from classical mechanics in that observation of a certain quantum system plays a key role in determining how that system will evolve. Specifically, we mathematically represent the act of observation as operators and observables. Observables are, quite simply, things that we can observe by measurement. Examples of observables include position, momentum, kinetic energy, and so on. Let a generic 6
observable be denoted by Q. For each observable we associate a quantum operator, represented by ˆQ. Suppose we perform this measurement on a particle in the state Ψ. Then the result of the measurement must be one of the eigenvalues of ˆQ, and furthermore the probability of measuring some eigenvalue λ is the absolute square of the λ component of Ψ, if Ψ is expressed in an orthonormal basis of eigenvectors of ˆQ. 3 In our example of the Stern-Gerlach apparatus, Q is the measured spin and ˆQ is called a spin matrix. Specifically, measurement along x, y, and z axes are called the Pauli spin matrices, and are defined respectively by S x = [ 0 1 1 0 S y = [ 0 i i 0 S z = [ 1 0 0 1 Note that the eigenvalues of each of these matrices is ±, which means that the possible measurements of spin are ±, which is consistent with the properties of spin. 3. A Single Particle Consider an axis in the xz-plane that is an angle θ from the z-axis. Then the spin matrix, denoted here by S θ, associated with that axis is a linear combination of the S x and S z, where the weights on each part of the combination are given by the component of the new axis along the previous axes: S θ = S z cos θ + S x sin θ = [ 1 0 0 1 [ cos θ sin θ cos θ + [ 0 1 1 0 sin θ = sin θ cos θ The eigenvalues of S θ, the values λ for which there exists a solution v to Av = λv, are ±, which are what they should be: since eigenvalues of the observable S θ correspond to the possible measurements. The eigenvectors corresponding to those eigenvalues are, respectively, [ [ cos θ sin θ v + = sin θ v = cos θ Now, if we want to measure the spin along the θ-axis after measuring the spin along the z-axis, as described above, the initial vector from before measuring along the θ-axis is [ 1 v = 0 3 Eigenvalues of a matrix A are values λ such that A λi = 0, where I is the identity matrix. The eigenvector v corresponding to an eigenvalue is a vector such that Av = λv. 7
To write v as a linear combination of v ±, it s straightforward to determine that the coefficients are: v = v + cos θ v sin θ (1) What this means in the context of the problem at hand is that if we measure the z-spin of a particle and get, and we then measure the spin in an axis rotated by an angle of θ, the probability of getting again is cos θ and the probability of getting is sin θ. Similarly, if we first measured along the z-axis, then the probability of measuring in the rotated axis is cos θ and the probability of measuring is sin θ. 3.3 Entangled Particles Now we consider the the decay of a singlet state with total spin 0 into two spin- 1 particles. Clearly, to conserve angular momentum, the two new particles must have opposite spin. Label the two particles with A and B: if we first measure the spin of A along some axis and we then measure the spin of B along the same axis, the spins must be opposite. If instead we measure the spin of B along a rotated axis, then the spin of B would not necessarily be opposite to what was measured: as derived in the section above, the probability of measuring opposite spins for A and B would be cos θ, where θ is the angle between the axis for measurement of A and the axis for measurement of B. 4 The Experiment Suppose that a team of two (X and Y) wants to convince a third person (A) that they have in their possession a certain necklace. The necklace they desire happens to be impossible: it must have an even number of beads, adjacent beads must have different colors (there are two possible colors for each bead), and the beads at either end of the necklace are the same. The way in which they shall attempt to convince A is this: 1. A will ask X the color of a certain bead at position i on the necklace.. A will then ask Y the color of either bead i + 1 or i 1. If X and Y can answer in a non-contradictory manner, it would certainly seem to A that they each possess such an impossible necklace. Moreover, if, over a repeated trial (each time with a different necklace), they can answer correctly each time, it would seem to prove that they have the impossible necklace [3, [4. 4.1 Classical Physics Suppose that the team in question is a team of classical physicists. If they want to win (to convince A that they know how to build the necklace), then their best 8
strategy is to decide ahead of time what they will answer for each bead. Since it is impossible to build such a necklace, the best they can do is to decide on an answer set with one flaw in it. For example, they agree to have one adjacent pair of beads be the same color. If there are N beads on the necklace, then in a certain trial, the probability that they will fail is clearly 1 N. Then the probability that they will succeed is 1 1 N. It may seem that the classical physicists have good odds of winning. After all, if there are 100 beads on the necklace, then they have a 99% chance of winning. But if there are repeated trials say, 500 of them then the odds of them winning each one are (99%) 50 = 0.657%. Indeed, as N with 5N trials, the odds of the classical physicist winning them all is e 5 = 0.674%. 4. Quantum Physics Quantum physicists, on the other hand, have no predetermined answers. What they could do to win is to take two entangled particles such as the two particles that result from a spin-0 singlet state, as described in Section 3.3 above. Define a set of axes: again, for N beads on the necklace, define an axis that makes an angle θ i = πi N with the z-axis. An example of this when N = 6 is shown below in Figure 3. Figure 3: Axes for measurement when N = 6. When A asks X for the color of bead i, X will measure the spin of his particle along the θ i axis. Spin up will correspond to a certain, predetermined color, and spin down will correspond to the other possible color. When A asks Y about the color of a certain bead (i ± 1), he will also measure the spin along that certain axis and answer accordingly. As derived in the theory section above, on a single trial, the odds that they will succeed are cos π N. If there are 100 beads on the necklace, the odds of success on a single trial are cos π 00 = 99.975%. Over the course of 500 trials, the odds of success on all of them are then (99.975%) 500 = 88.393%. Clearly, the quantum physicists have a tremendous advantage over the classical physicists. 9
4.3 A Computer Simulation Input: Starting amount of money (D), number of beads on a necklace (n), and the number of trials (m) Output: Amount of money for each physicist after each trial begin C[1 := D Q[1 := D for i from 1 to m do Cl := rand(1:c[i) Qu := rand(1:q[i) if (n-1)/n < rand(0:1) then C[i+1 := C[i - Cl else C[i+1 := C[i + Cl end if cos(pi/(n))^ < rand(0:1) then Q[i+1 := Q[i - Qu else Q[i+1 := Q[i + Qu end end Result: C[m, Q[m end The above algorithm is not an exact replica of the experiment described above. Rather than testing how many consecutive times each team wins, it is a betting game: each team starts with a certain amount of money, D. On each game one iteration of the program there are m trials. On each one of the m trials, each team bets a certain amount of money: Cl for the classical team and Qu for the quantum team. If the classical team can correctly answer the question involving the necklace, then they earn Cl, and if they answer incorrectly, they lose that amount. The same is true for the quantum team, except they stand to lose or gain Qu. In the program, Qu and Cl are random nonnegative integers that are less than the current amount of money the team in question has. The values are random to ignore any effects of optimal betting; the results would be comparatively the same anyways. Over the course of a single game, the necklace presumably changes, but each time it has a specific number of beads, n. Specifically, if they each start with $10 with 100 beads on the supposed necklace on each of the 500 trials, on average, the quantum team will have approximately $8 10 86 and the classical team will have approximately 5 10 79 : only 0.00001% of the quantum amount. Although both of these values seem rather large, it s because success is measured on a trial-to-trial basis, and the probability of success per trial for both the classical and the quantum teams are close to 1. The multiplicative difference between values, here 10 7, is still valid. 10
References [1 Isaac Chuang and Michael Nielsen. Quantum Computation and Quantum Information. [ David Mermin. Is the moon there when nobody looks? Reality and the quantum theory. [3 Lev Viadman. Tests of Bell Inequalities. [4 Hans Christian von Baeyer. Information: The New Language of Science. [5 Wikipedia.org. Wikipedia: Stern-Gerlach Experiment. 11