42 3 Neutron Interferometry Comparing the values b n of question 3.3 with this experimental result, and recalling the result of a measurement of µ x for these values, explain why this proves that the state vector of a spin-1/2 particle changes sign under a rotation by an odd multiple of 2π. 3.4 Solutions Section 3.1: Neutron Interferences 3.1.1. The beams ABDC 2 and ACDC 2 interfere. Omitting the propagation factors, one has, at C 2 an amplitude A 2 = α 2 β + β 3 = β(α 2 + β 2 ) . Similarly, for ABDC 3 and ACDC 3 , A 3 =2αβ 2 . The intensities at the two counters are I 2 = R − 4R 2 TI 3 =4R 2 T. 3.1.2. When there is a phase shift δ in C, the above expressions get modified as follows: A 2 = α 2 βe iδ + β 3 = β(α 2 e iδ + β 2 ) ,A 3 = αβ 2 (1 + e iδ ) . The intensities become I 2 = R − 2R 2 T (1 + cos δ) I 3 =2R 2 T (1 + cos δ) . The fact that I 2 + I 3 does not depend on the phase shift δ is a consequence of the conservation of the total number of particles arriving at D. Section 3.2: The Gravitational Effect 3.2.1. This results from elementary trigonometry. (a) Since there is no recoil energy of the silicon atoms to be taken care of, the neutron total energy (kinetic+potential) is a constant of the motion in all the process. This energy is given by E AC = p 2 /2M and E BD =(p − ∆p) 2 /2M + MgH sin φ, hence ∆p  M 2 gH sin φ/p . (b) The velocity √ 2gH is of the order of 0.5 m/s, and the neutron velocity is v = h/Mλ  2700 m/s. The change in velocity ∆v is therefore very small: ∆v = gH/v  2 × 10 −4 m/s for φ = π/2. 3.4 Solutions 43 3.2.2. (a) The gravitational potential varies in exactly the same way along AB and CD. The neutron state in both cases is a plane wave with momen- tum p = h/λ just before A or C. The same Schr¨odinger equation is used to determine the wave function at the end of the segments. This implies that the phases accumulated along the two segments AB and CD are equal. (b) When comparing the segments AC and BD, the previous reasoning does not apply, since the initial state of the neutron is not the same for the two segments. The initial state is exp(ipz/¯h)forAC, and exp[i(p − ∆p)z/¯h]for BD. After travelling over a distance L = AC = BD, the phase difference between the two paths is δ = ∆p L ¯h = M 2 gλd 2 π¯h 2 tan θ sin φ. 3.2.3. >From the previous result, one has δ 2 − δ 1 = Ag(sin φ 2 − sin φ 1 ), where A = M 2 λd 2 tan θ/(π¯h 2 ). Therefore, g = δ 2 − δ 1 A (sin φ 2 − sin φ 1 ) . There are 9 oscillations, i.e. (δ 2 − δ 1 )=18π, between φ 1 = −32 ◦ and φ 2 = +24 ◦ , which gives g  9.8ms −2 . The relative precision of the experiment was actually of the order of 10 −3 . Section 3.3: Rotating a Spin 1/2 by 360 Degrees 3.3.1. Since B is along the z axis, the magnetic Hamiltonian is: ˆ H M = −µ · B 0 = ¯hω 2  10 0 −1  . At time t, the spin state is |Σ(t) = 1 √ 2  e −iωt/2 e +iωt/2  . By a direct calculation of µ or by using Ehrenfest theorem ( d dt µ = 1 i¯h [ ˆ µ, ˆ H]), we obtain: dµ x  dt = ωµ y  dµ y  dt = −ωµ x  dµ z  dt =0. Initially µ x  = µ 0 and µ y  = µ z  = 0; therefore, µ = µ 0 (cos ωt u x +sinωt u y ) . 44 3 Neutron Interferometry 3.3.2. When the neutrons leave the field zone, the probability of finding µ x =+µ 0 is P x (+µ 0 )=|+x|Σ(T )| 2 =cos 2 ωT 2 =cos 2 δ with T = l/v = lMλ/h. 3.3.3. The above probability is equal to 1 if δ = nπ (ωT =2nπ), or B 0 = nb 1 with b 1 = 2π 2 ¯h 2 µ 0 Mlλ =34.5 × 10 −4 T . For δ = nπ the magnetic moment has rotated by 2nπ around the z axis by Larmor precession. 3.3.4. The formulas are similar to those found in question 1.2. The phase of the upper component of the spinor written in the {|+ z , |− z } basis, is shifted by +δ, that of the lower component by −δ: Amplitude at the counter C 2 :e i(p 2 ·r−Et)/¯h β √ 2  β 2 + α 2 e iδ β 2 + α 2 e −iδ  Amplitude at the counter C 3 :e i(p 3 ·r−Et)/¯h αβ 2 √ 2  1+e iδ 1+e −iδ  3.3.5. Since the measuring apparatus is insensitive to spin variables, we must add the probabilities corresponding to S z = ±1, each of which is the modulus squared of a sum of amplitudes. Altogether, we obtain the following intensities of the total neutron flux in the two counters: I 2 = R − 2R 2 T (1 + cos δ) ,I 3 =2R 2 T (1 + cos δ) and I 2 − I 3 = R − 4R 2 T (1 + cos δ) . 3.3.6. There will be a minimum of I 2 −I 3 each time cos δ = +1, i.e. δ =2nπ. This corresponds to a constructive interference in channel 3. On the other hand, there appears a maximum if cos δ = −1, i.e. δ =(2n +1)π,andthis corresponds to a destructive interference in channel 3 (I 3 =0). If δ = nπ, whatever the integer n,oneissure to find the neutrons in the same spin state as in the initial beam. However, the interference pattern depends on the parity of n. The experimental result ∆B =(64±2) ×10 −4 T confirms that if the spin has rotated by 4nπ, one recovers a constructive interference in channel 3 as in the absence of rotation, while if it has rotated by (4n+2)π, the interference in C 3 is destructive. The probability amplitude for the path ACD has changed sign in this latter case, although a spin measurement in this path after the magnet will give exactly the same result as on the incoming beam. 3.4 Solutions 45 References A.W. Overhauser, A.R. Collela, and S.A. Werner, Phys. Rev. Lett., 33, 1237 (1974); 34, 1472 (1975); 35, 1053 (1975). See also D. Greenberger and A.W. Overhauser, Scientific American, May 1980. 4 Spectroscopic Measurement on a Neutron Beam We present here a very precise method for spectroscopic measurements, due to Norman Ramsey. The method, using atomic or molecular beams, can be applied to a very large class of problems. We shall analyse it in the specific case of a neutron beam, where it can be used to determine the neutron magnetic moment with high accuracy, by measuring the Larmor precession frequency in a magnetic field B 0 . A beam of neutrons is prepared with velocity v along the x axis. The beam is placed in a constant uniform magnetic field B 0 directed along the z axis. We write |+ and |− for the eigenstates of the z projection ˆ S z of the neutron spin, and γ for the gyromagnetic ratio of the neutron: ˆ µ = γ ˆ S, ˆ µ being the neutron magnetic moment operator, and ˆ S its spin. The neutrons are initially in the state |−. When they approach the origin, they cross a zone where an oscillating field B 1 (t) is applied in the (x, y) plane. The components of B 1 are B 1x = B 1 e −r/a cos ω(t − z/c) B 1y = B 1 e −r/a sin ω(t − z/c) B 1z =0, (4.1) where r =  x 2 + y 2 . We assume that B 1 is constant (strictly speaking it should vary in order to satisfy ∇·B =0)andthatB 1  B 0 . In all parts of the chapter, the neutron motion in space is treated classically as a linear uniform motion. We are only interested in the quantum evolution of the spin state. 4.1 Ramsey Fringes 4.1.1. Consider a neutron whose motion in space is x = vt, y =0,z =0. What is the Hamiltonian ˆ H(t) describing the coupling of the neutron magnetic moment with the fields B 0 and B 1 ? 48 4 Spectroscopic Measurement on a Neutron Beam Setting ω 0 = −γB 0 and ω 1 = −γB 1 , write the matrix representation of ˆ H(t) in the basis {|+, |−}. 4.1.2. Treating B 1 as a perturbation, calculate, in first order time-dependent perturbation theory, the probability of finding the neutron in the state |+ at time t =+∞ (far from the interaction zone) if it was in the state |− at t = −∞. One measures the flux of neutrons which have flipped their spins, and are in the state |+ when they leave the field zone. This flux is proportional to the probability P −+ that they have undergone the above transition. Show that this probability has a resonant behavior as a function of the applied angular frequency ω. Plot P −+ as a function of the distance from the resonance ω −ω 0 . How does the width of the resonance curve vary with v and a? The existence of this width puts a limit on the accuracy of the measurement of ω 0 , and therefore of γ. Is there an explanation of this on general grounds? 4.1.3. On the path of the beam, one adds a second zone with an oscillating field B  1 . This second zone is identical to the first but is translated along the x axis by a distance b (b  a): B  1x = B 1 e −r  /a cos ω(t − z/c) B  1y = B 1 e −r  /a sin ω(t − z/c) B  1z =0, (4.2) where r  =((x − b) 2 + y 2 ) 1/2 . Show that the transition probability P −+ across the two zones can be expressed in a simple way in terms of the transition probability calculated in the previous question. Why is it preferable to use a setup with two zones separated by a distance b rather than a single zone, as in question 4.1.2, if one desires a good accuracy in the measurement of the angular frequency ω 0 ? What is the order of magnitude of the improvement in the accuracy? 4.1.4. What would be the probability P −+ if one were to use N zones equally spaced by the same distance b from one another? What optical system is this reminiscent of? 4.1.5. Suppose now that the neutrons, still in the initial spin state |−, propagate along the z axis instead of the x axis. Suppose that the length of the interaction zone is b, i.e. that the oscillating field is given by (4.1) for −b/2 ≤ z ≤ +b/2 and is zero for |z| >b/2. Calculate the transition probability P  −+ in this new configuration. For what value of ω is this probability maximum? Explain the difference with the result obtained in question 4.1.2. 4.2 Solutions 49 4.1.6. In practice, the neutron beam has some velocity dispersion around the value v. Which of the two methods described in questions 4.1.3 and 4.1.5 is preferable? 4.1.7. Numerical Application: The neutrons of the beam have a de Broglie wavelength λ n =31 ˚ A. Calculate their velocity. In order to measure the neutron gyromagnetic ratio γ n , one proceeds as in question 4.1.3. One can assume that the accuracy is given by δω 0 = π 2 v b . The most accurate value of the neutron gyromagnetic ratio is currently γ n = −1.912 041 84 (±8.8 × 10 −7 ) q/M p where q is the unit charge and M p the proton mass. In a field B 0 = 1 T, what must be the length b in order to achieve this accuracy? 4.2 Solutions 4.1.1. The magnetic Hamiltonian is ˆ H(t)=− ˆ µ.B = −γ  B 0 ˆ S z + B 1x (t) ˆ S x + B 1y (t) ˆ S y  . Since x = vt, y = z =0, ˆ H(t)=−γ  B 0 ˆ S z + B 1 e −v|t|/a  ˆ S x cos ωt + ˆ S y sin ωt  whose matrix representation is ˆ H(t)= ¯h 2  ω 0 ω 1 exp(−v|t|/a − iωt) ω 1 exp(−v|t|/a +iωt) −ω 0  . 4.1.2. Let |ψ(t) = α(t)|+ + β(t)|− be the neutron state at time t.The Schr¨odinger equation gives the evolution of α and β: i˙α = ω 0 2 α + ω 1 2 e −iωt−v|t|/a β i ˙ β = ω 1 2 e iωt−v|t|/a α − ω 0 2 β. We now introduce the variables ˜α and ˜ β: ˜α(t)=α(t)e iω 0 t/2 ˜ β(t)=β(t)e −iω 0 t/2 , whose evolution is given by 50 4 Spectroscopic Measurement on a Neutron Beam i ˙ ˜α = ω 1 2 e i(ω 0 −ω)t−v|t|/a ˜ β i ˙ ˜ β = ω 1 2 e i(ω−ω 0 )t−v|t|/a ˜α. The equation for ˜α can be formally integrated and it gives ˜α(t)= ω 1 2i  t −∞ e i(ω 0 −ω)t  −v|t  |/a ˜ β(t  )dt  , (4.3) where we have used the initial condition ˜α(−∞)=α(−∞) = 0. Now, since we want the value of α(t)tofirstorderinB 1 , we can replace ˜ β(t  )byits unperturbed value ˜ β(t  ) = 1 in the integral. This gives γ −+ ≡ ˜α(+∞)= ω 1 2i  +∞ −∞ e i(ω 0 −ω)t  −v|t  |/a dt  = ω 1 v ia 1 (ω − ω 0 ) 2 +(v/a) 2 . The transition probability is therefore P −+ = ω 2 1 v 2 a 2 1 [(ω 0 − ω) 2 +(v/a) 2 ] 2 . Fig. 4.1. Transition probability in one zone The width of the resonance curve is of the order of v/a. This quantity is the inverse of the time τ = a/v a neutron spends in the oscillating field. From the uncertainty relation δE.τ ∼ ¯h, when an interaction lasts a finite time τ the accuracy of the energy measurement δE is bounded by δE ≥ ¯h/τ. Therefore, from first principles, one expects that the resonance curve will have a width of the order of ¯h/τ in energy, or 1/τ in angular frequency. 4.2 Solutions 51 4.1.3. In the two-zone case, the transition amplitude (in first order pertur- bation theory) becomes γ −+ = ω 1 2i   +∞ −∞ e i(ω 0 −ω)t−v|t|/a dt +  +∞ −∞ e i(ω 0 −ω)t−|vt−b|/a dt  . If we make the change of variables t  = t − b/v in the second integral, we obtain γ −+ = ω 1 2i  1+ e i(ω 0 −ω)b/v   +∞ −∞ e i(ω 0 −ω)t−v|t|/a dt, which is the same formula as previously but multiplied by 1 + e i(ω 0 −ω)b/v .If we square this expression, in order to find the probability, we obtain P −+ = 4ω 2 1 v 2 a 2 1 [(ω 0 − ω) 2 + v 2 /a 2 ] 2 cos 2  (ω 0 − ω)b 2v  . Fig. 4.2. Ramsey fringes in a two-zone setup The envelope of this curve is, up to a factor of 4, the same as the previous curve. However, owing to the extra oscillating factor, the half-width at half- maximum of the central peak is now of order πv/(2b). The parameter which now governs the accuracy is the total time b/v that the neutron spends in the apparatus, going from one zone to the other. In spectroscopic measurements, it is important to locate the exact position of the maximum of the peak. Multiplying the width of the peak by a factor a/b ( 1sincea  b) results in a major improvement of the measurement accuracy. Of course one could in principle build a single interaction zone of large size ∼ b, but it would be difficult to maintain a well controlled oscillating field over such a large region. From a practical point of view, it is much simpler to use small interaction zones of size a and to separate them by a large distance b. 52 4 Spectroscopic Measurement on a Neutron Beam 4.1.4. It is quite straightforward to generalize the previous results to an arbitrary number of zones: γ −+ = ω 1 2i  1+ e i(ω 0 −ω)b/v + ···+e i(N−1)(ω 0 −ω)b/v  ×  +∞ −∞ e i(ω 0 −ω)t−v|t|/a dt P −+ = ω 2 1 v 2 a 2 1 [(ω 0 − ω) 2 + v 2 /a 2 ] 2 sin 2 [N(ω 0 − ω)b/2v] sin 2 [(ω 0 − ω)b/2v] . As far as amplitudes are concerned, there is a complete analogy with a dif- fraction grating in optics. The neutron (more generally, the particle or the atom) has some transition amplitude t for undergoing a spin flip in a given interaction zone. The total amplitude T is the sum T = t + t e iφ + t e 2iφ + , where e iφ is the phase shift between two zones. 4.1.5. We now set z = vt,andx = y = 0 for the neutron trajectory. This will modify the phase of the field (Doppler effect) ω(t − z/c) → ω(1 − v/c)t =˜ωt with ˜ω = ω (1 − v/c) and we must integrate the evolution of ˜α: i ˙ ˜α = ω 1 2 e i(ω 0 −˜ω)t ˜ β (with ˜ β  1) between t i = −b/(2v)andt f = b/(2v). The transition probability is then P  −+ = ω 2 1 sin 2 [(ω 0 − ˜ω)b/(2v)] (ω 0 − ˜ω) 2 , which has a width of the order of b/v but is centered at ˜ω = ω 0 ⇒ ω = ω 0 1 − v/c  ω 0 (1 + v/c) . Comparing with question 4.1.2, we find that the resonance frequency is dis- placed: The neutron moves in the propagation direction of the field, and there is a first order Doppler shift of the resonance frequency. 4.1.6. If the neutron beam has some velocity dispersion, the experimental result will be the same as calculated above, but smeared over the velocity distribution. In the method of question 4.1.3, the position of lateral fringes, and the width of the central peak, vary with v. A velocity distribution will lead to . find the neutrons in the same spin state as in the initial beam. However, the interference pattern depends on the parity of n. The experimental result ∆B = (64 ±2) ×10 −4 T confirms that if the spin has. is the total time b/v that the neutron spends in the apparatus, going from one zone to the other. In spectroscopic measurements, it is important to locate the exact position of the maximum of the. find that the resonance frequency is dis- placed: The neutron moves in the propagation direction of the field, and there is a first order Doppler shift of the resonance frequency. 4.1 .6. If the neutron