4.2 Solutions 53 a broader central peak and lateral fringes of decreasing amplitude. However the position of the central peak does not depend on the velocity, and it is therefore not shifted if the neutron beam has some velocity dispersion. On the contrary, in the method of question 4.1.5, the position of the cen- tral peak depends directly on the velocity. A dispersion in v will lead to a corresponding dispersion of the position of the peak we want to measure. The first method is highly preferable. 4.1.7. Numerically, for λ n =31 ˚ A, v = h/(M n λ n )  128 m/s. Experimentally, one obtains an accuracy δω 0 /ω 0 = δγ n /γ n =4.6 ×10 −7 .For B = 1 T, the angular frequency is ω 0 = γ n B 0  1.8 × 10 8 s −1 , which gives δω 0 /(2π)  13 Hz and b  2.4m. Actually, one can improve the accuracy considerably by analysing the shape of the peak. In the experiment reported in the reference quoted be- low, the length b is 2 m and the field is B 0 =0.05 T (i.e. an angular frequency 20 times smaller than above). Reference G.L. Greene, N.F. Ramsey, W. Mampe, J.M. Pendlebury, K.F. Smith, W.D. Dress, P.D. Miller, and P. Perrin, Phys. Rev. D20, 2139 (1979). 5 Analysis of a Stern–Gerlach Experiment We analyze a Stern–Gerlach experiment, both experimentally and from the theoretical point of view. In the experimental setup considered here, a mono- chromatic beam of neutrons crosses a region of strongly inhomogeneous mag- netic field, and one observes the outgoing beam. 5.1 Preparation of the Neutron Beam Neutrons produced in a reactor are first “cooled”, i.e. slowed down by cross- ing liquid hydrogen at 20 K. They are incident on a monocrystal, for instance graphite, from which they are diffracted. To each outgoing direction, there cor- responds a well-defined wavelength, and therefore a well-defined momentum. A beryllium crystal acts as a filter to eliminate harmonics, and the vertical ex- tension of the beam is controlled by two gadolinium blocks, which are opaque to neutrons, separated by a thin sheet of (transparent) aluminum of thickness a, which constitutes the collimating slit, as shown in Fig. 5.1. Fig. 5.1. Preparation of the neutron beam 5.1.1. The de Broglie wavelength of these monochromatic neutrons is λ = 4.32 ˚ A. What are their velocity and their kinetic energy? 56 5 Analysis of a Stern–Gerlach Experiment 5.1.2. One observes the impacts of the neutrons on a detector at a distance L = 1 m from the slit. The vertical extension of the beam at the detector is determined by two factors, first the width a of the slit, and second the diffraction of the neutron beam by the slit. We recall that the angular width θ of the diffraction peak from a slit of width a is related to the wavelength λ by sin θ = λ/a. For simplicity, we assume that the neutron beam is well collimated before the slit, and that the vertical extension δ of the beam on the detector is the sum of the width a of the slit and the width of the diffraction peak. Show that one can choose a in an optimal way in order to make δ as small as possible. What is the corresponding width of the beam on the detector? 5.1.3. In the actual experiment, the chosen value is a =5µm. What is the observed width of the beam at the detector? Comment on the respective effects of the slit width a and of diffraction, on the vertical shape of the observed beam on the detector? The extension of the beam corresponds to the distribution of neutron im- pacts along the z axis. Since the purpose of the experiment is not only to observe the beam, but also to measure its “position” as defined by the maxi- mum of the distribution, what justification can you find for choosing a =5µm? Figure 5.2 is an example of the neutron counting rate as a function of z. The horizontal error bars, or bins, come from the resolution of the measur- ing apparatus, the vertical error bars from the statistical fluctuations of the number of neutrons in each bin. The curve is a best fit to the experimental points. Its maximum is determined with an accuracy δz ∼ 5 µm. Fig. 5.2. Measurement of the beam profile on the detector 5.3 The Stern–Gerlach Experiment 57 5.2 Spin State of the Neutrons In order to completely describe the state of a neutron, i.e. both its spin state and its spatial state, we consider the eigenbasis of the spin projection along the z axis, ˆ S z , and we represent the neutron state as |ψ(t) :  ψ + (r,t) ψ − (r,t)  , where the respective probabilities of finding the neutron in the vicinity of point r with its spin component S z = ±¯h/2are d 3 P (r,S z = ±¯h/2,t)=|ψ ± (r,t)| 2 d 3 r. 5.2.1. What are the probabilities P ± (t) of finding, at time t,thevalues±¯h/2 when measuring S z irrespective of the position r? 5.2.2. What is, in terms of ψ + and ψ − , the expectation value of the x com- ponent of the neutron spin S x  in the state |ψ(t)? 5.2.3. What are the expectation values of the neutron’s position r and momentum p in the state |ψ(t)? 5.2.4. We assume that the state of the neutron can be written: |ψ(t) : ψ(r,t)  α + α −  , where the two complex numbers α ± are such that |α + | 2 + |α − | 2 =1.Howdo the results of questions 5.2.2 and 5.2.3 simplify in that case? 5.3 The Stern–Gerlach Experiment Between the slit, whose center is located at the origin (x = y = z = 0), and the detector, located in the plane x = L, we place a magnet of length L whose field B is directed along the z axis. The magnetic field varies strongly with z; see Fig. 5.3. We assume that the components of the magnetic field are B x = B y =0 B z = B 0 + b  z. In what follows we choose 1 B 0 =1Tandb  = 100 T/m. 1 This form violates Maxwell’s equation ∇ · B = 0, but it simplifies the following calculation. With a little modification (e.g. B x =0,B y = −b  y and B y  B z over the region of space crossed by the neutron beam), one can settle this matter, and arrive at the same conclusions. 58 5 Analysis of a Stern–Gerlach Experiment Fig. 5.3. Magnetic field setup in the Stern–Gerlach experiment The magnetic moment of the neutron ˆ µ in the matrix representation that we have chosen for |ψ is ˆ µ = µ 0 ˆ σ , where ˆ σ are the usual Pauli matrices, and µ 0 =1.913 µ N ,whereµ N is the nuclear magneton µ N = q¯h/2M p =5.051 10 −27 J · T −1 . Hereafter, we denote the neutron mass by m. 5.3.1. What is the form of the Hamiltonian for a neutron moving in this magnetic field? Write the time-dependent Schr¨odinger equation for the state |ψ(t). Show that the Schr¨odinger equation decouples into two equations of the Schr¨odinger type, for ψ + and ψ − respectively. 5.3.2. Show that one has d dt  |ψ ± (r,t)| 2 d 3 r =0. What does one conclude as to the probabilities of measuring µ z = ±µ 0 ? 5.3.3. We assume that, at t = 0, at the entrance of the field zone, one has |ψ(0) : ψ(r, 0)  α + α −  and that r =0,p y  = p z  = 0 and p x  = p 0 = h/λ, where the value of the wavelength λ has been given above. The above conditions correspond to the experimental preparation of the neutron beam discussed in Sect. 5.1. Let ˆ A be an observable depending on the position operator ˆ r and the momentum operator ˆ p.WedefinethenumbersA +  and A −  by A ±  = 1 |α ± | 2  ψ ∗ ± (r,t) ˆ Aψ ∗ ± (r,t)d 3 r. 5.4 Solutions 59 What is the physical interpretation of A +  and A − ? Show in particular that |ψ + | 2 /|α + | 2 and |ψ − | 2 /|α − | 2 are probability laws. 5.3.4. Apply Ehrenfest’s theorem to calculate the following quantities: d dt r ±  , d dt p ±  . Solve the resulting equations and give the time evolution of r ±  and p ± . Give the physical interpretation of the result, and explain why one observes a splitting of the initial beam into two beams of relative intensities |α + | 2 and |α − | 2 . 5.3.5. Calculate the splitting between the two beams when they leave the magnet. Express the result in terms of the kinetic energy of the incident neutrons (we recall that L =1mandb  = 100 T/m). Given the experimental error δz in the measurement of the position of the maximum intensity of a beam, i.e. δz =5× 10 −6 m as discussed in question 5.1.3, what is the accuracy on the measurement of the neutron magnetic mo- ment in such an experiment, assuming that the determination of the magnetic field and the neutron energy is not a limitation? Compare with the result of magnetic resonance experiments: µ 0 =(−1.91304184 ±8.8 ×10 −7 ) µ N . 5.3.6. In the same experimental setup, what would be the splitting of a beam of silver atoms (in the original experiment of Stern and Gerlach, the atomic beam came from an oven at 1000 K) of energy E  1.38 × 10 −20 J? The magnetic moment of a silver atom is the same as that of the valence electron |µ e | = q¯h/2m e  9.3 ×10 −24 J.T −1 . 5.3.7. Show that, quite generally, in order to be able to separate the two outgoing beams, the condition to be satisfied is of the form E ⊥ t ≥ ¯h/2 , where E ⊥ is the transverse kinetic energy acquired by the neutrons in the process, and t is the time they spend in the magnetic field. Comment and conclude. 5.4 Solutions Section 5.1: Preparation of the Neutron Beam 5.1.1. We have v = h/(λm)andE = mv 2 /2, which yields v = 916 m s −1 and E =0.438 × 10 −2 eV. 60 5 Analysis of a Stern–Gerlach Experiment 5.1.2. The contribution of diffraction to the beam width is δ diff = L tan θ ∼ Lλ/a. With the simple additive prescription (which can be improved, but this would not yield very different results), we obtain δ = a + Lλ/a which is minimal for a = √ Lλ  21 µm. The spreading of the beam on the detector is then equal to the Heisenberg minimum δ =2 √ Lλ =42µm. The uncertainty relations forbid δ to be less than some lower limit. In other words, the spreading of the wave packet, which increases as a decreases com- petes with the spatial definition of the incoming beam. 5.1.3. For a =5µm, we have δ =91.5 µm. In that case, the effect of diffraction is predominant. The reason for making this choice is that the shape of the diffraction peak is known and can be fitted quite nicely. Therefore, this is an advantage in determining the position of the maximum. However, one cannot choose a to be too small, otherwise the neutron flux becomes too small, and the number of events is insufficient. Section 5.2: Spin State of the Neutrons 5.2.1. P ± (t)=  |ψ ± (r,t)| 2 d 3 r N.B. The normalization condition (total probability equal to 1) is P + + P − =1⇒   |ψ + (r,t)| 2 + |ψ − (r,t)| 2  d 3 r =1. The quantity |ψ + (r,t)| 2 + |ψ − (r,t)| 2 is the probability density of finding the neutron at point r. 5.2.2. By definition, the expectation value of S x is S x  =(¯h/2)ψ|ˆσ x |ψ therefore S x  = ¯h 2   ψ ∗ + (r,t)ψ − (r,t)+ψ ∗ − (r,t)ψ + (r,t)  d 3 r. 5.2.3. Similarly r =  r  |ψ + (r,t)| 2 + |ψ − (r,t)| 2  d 3 r, p = ¯h i   ψ ∗ + (r,t) ∇ψ + (r,t)+ψ ∗ − (r,t) ∇ψ − (r,t)  d 3 r. 5.2.4. If the variables are factorized, we have the simple results: S x  =¯h Re  α ∗ + α −  and r =  r |ψ(r,t)| 2 d 3 r, p = ¯h i  ψ ∗ (r,t) ∇ψ(r,t)d 3 r. 5.4 Solutions 61 Section 5.3: The Stern–Gerlach Experiment 5.3.1. The matrix form of the Hamiltonian is ˆ H = ˆ p 2 2m  10 01  − µ 0 (B 0 + b  ˆz)  10 0 −1  . The Schr¨odinger equation is i¯h d dt |ψ(t) = ˆ H|ψ(t) . If we write it in terms of the coordinates ψ ± we obtain the uncoupled set i¯h ∂ ∂t ψ + (r,t)=− ¯h 2 2m ∆ψ + − µ 0 (B 0 + b  z) ψ + i¯h ∂ ∂t ψ − (r,t)=− ¯h 2 2m ∆ψ − + µ 0 (B 0 + b  z) ψ − or, equivalently i¯h d dt |ψ ±  = ˆ H ± |ψ ±  , with ˆ H ± = − ¯h 2 2m ∆ ∓ µ 0 (B 0 − b  z) . In other words, we are dealing with two uncoupled Schr¨odinger equations, where the potentials have opposite values. This is basically what causes the Stern–Gerlach splitting. 5.3.2. Since both ψ + and ψ − satisfy Schr¨odinger equations, and since ˆ H ± are both hermitian, we have the usual properties of Hamiltonian evolution for ψ + and ψ − separately, in particular the conservation of the norm. The probability of finding µ z = ±µ 0 , and the expectation value of µ z are both time independent. 5.3.3. By definition, we have  |ψ ± (r,t)| 2 d 3 r = |α ± | 2 , where |α ± | 2 is time independent. The quantities |ψ + (r,t)| 2 /|α + | 2 and |ψ − (r,t)| 2 /|α − | 2 are the probability densities for finding a neutron at po- sition r with, respectively, S z =+¯h/2andS z = −¯h/2. The quantities A +  and A −  are the expectation values of the physical quantity A, for neutrons which have, respectively, S z =+¯h/2andS z = −¯h/2. 5.3.4. Applying Ehrenfest’s theorem, one has for any observable d dt A ±  = 1 i¯h|α ± | 2  ψ ∗ ± (r,t)[ ˆ A, ˆ H ± ] ψ ± (r,t)d 3 r. 62 5 Analysis of a Stern–Gerlach Experiment Therefore d dt r ±  = p ± /m and d dt p x±  = d dt p y±  =0 d dt p z±  = ±µ 0 b  . The solution of these equations is straightforward: p x±  = p 0 , p y±  =0, p z±  = ±µ 0 b  t x ±  = p 0 t m = vt , y ±  =0, z ±  = ± µ 0 b  t 2 2m . Consequently, the expectation values of the vertical positions of the neutrons which have µ z =+µ 0 and µ z = −µ 0 diverge as time progresses: there is a separation in space of the support of the two wave functions ψ + and ψ − .The intensities of the two outgoing beams are proportional to |α + | 2 and |α − | 2 . 5.3.5. As the neutrons leave the magnet, one has x = L, therefore t = L/v and ∆z = |µ 0 b  |L 2 /mv 2 = |µ 0 b  |L 2 /2E where E is the energy of the incident neutrons. This provides a splitting of ∆z =0.69 mm. The error on the position of each beam is δz =5µm, that is to say a relative error on the splitting of the beams, or equivalently, on the measurement of µ 0 δµ 0 µ 0  √ 2 δz ∆z ∼ 1.5% , which is far from the accuracy of magnetic resonance measurements. 5.3.6. For silver atoms, one has |µ 0 |/2E =3.4 × 10 −4 T −1 . Hence, in the same configuration, one would obtain, for the same value of the field gradient and the same length L = 1 m, a separation ∆z =3.4 cm, much larger than for neutrons. Actually, Stern and Gerlach, in their first experiment, had a much weaker field gradient and their magnet was 20 cm long. 5.3.7. The condition to be satisfied in order to resolve the two outgoing beams is that the distance ∆z between the peaks should be larger than the full width of each peak (this is a common criterion in optics; by an appro- priate inspection of the line shape, one may lower this limit). We have seen in Sect. 5.1 that the absolute minimum for the total beam extension on the detector is 2 √ Lλ, which amounts to a full width at half-maximum √ Lλ. In other words, we must have: ∆z 2 ≥ Lλ . In the previous section, we have obtained the value of ∆z,and,bysquaring,we obtain ∆z 2 =(µ 0 b  ) 2 t 4 /m 2 ,wheret is the time spent traversing the magnet. On the other hand, the transverse kinetic energy of an outgoing neutron is E ⊥ = p 2 z± /(2m)=(µ 0 b  ) 2 t 2 /(2m). 5.4 Solutions 63 Putting the two previous relations together, we obtain ∆z 2 =2E ⊥ t 2 /m;in- serting this result in the first inequality, we obtain E ⊥ t ≥ h/2 , where we have used L = vt and λ = h/mv. This is nothing but one of the many forms of the time–energy uncertainty relation. The right-hand side is not the standard ¯h/2 because we have considered a rectangular shape of the incident beam (and not a Gaussian). This brings in an extra factor of 2π. Physically, this result is interesting in many respects. (a) First, it shows that the effort that counts in making the experiment feasible is not to improve individually the magnitude of the field gradient, or the length of the apparatus, etc., but the particular combination of the product of the energy transferred to the system and the interaction time of the system with the measuring apparatus. (b) Secondly, this is a particular example of the fundamental fact stressed by many authors 2 that a measurement is never point-like. It has always a finite extension both in space and in time. The Stern–Gerlach experiment is actually a very good example of a measuring apparatus in quantum mechanics since it transfers quantum information – here the spin state of the neutron – into space–time accessible quantities – here the splitting of the outgoing beams. (c) This time–energy uncertainty relation is encountered in most, if not all quantum measurements. Here it emerges as a consequence of the spreading of the wave packet. It is a simple and fruitful exercise to demonstrate rigorously the above property by calculating directly the time evolution of the following expectation values: z ±  ,∆z 2 = z 2 ± −z ±  2 , E T  =  p 2 z± 2m  , z ± p ± + p ± z ±  . 2 See, for instance, L.D. Landau and E.M. Lifshitz, Quantum Mechanics, Pergamon Press, Oxford, 1965. . extension of the beam at the detector is determined by two factors, first the width a of the slit, and second the diffraction of the neutron beam by the slit. We recall that the angular width θ of the diffraction. diffraction, on the vertical shape of the observed beam on the detector? The extension of the beam corresponds to the distribution of neutron im- pacts along the z axis. Since the purpose of the experiment. the beam on the detector? 5.1.3. In the actual experiment, the chosen value is a =5µm. What is the observed width of the beam at the detector? Comment on the respective effects of the slit width