The Quantum Mechanics Solver 25 potx

246 24 Probing Matter with Positive Muons m µ c 2 = 105.66 MeV µ 1 /h =67.5MHzT −1 m e c 2 =0.511 MeV µ 2 /h = −1.40 × 10 4 MHz T −1 . 24.1 Muonium in Vacuum Muonium is formed by slowing down a beam of µ + , prepared in a given spin state, in a thin metal foil. A sufficiently slow µ + can capture an electron and form a hydrogen-like atom in an excited state. This atom falls into its ground state very quickly (in ∼ 10 −9 s), the muon’s spin state remaining the same during this process. Once it is formed, the muonium, which is electrically neutral, can diffuse outside the metal. We assume that, at t = 0, the state of the muonium atom is the following: • The muon spin is in the eigenstate | + z≡|+ of ˆσ 1z . • The electron spin is in an arbitrary state α|++β|−, with |α| 2 +|β| 2 =1. • The wave fuction Ψ (r) of the system is the 1s wave function of the hydrogen-like system, ψ 100 (r). Just as for the hyperfine structure of hydrogen, we work in the 4 dimen- sional Hilbert space corresponding to the spin variables of the electron and the muon. In this Hilbert space, the spin–spin interaction Hamiltonian is ˆ H = E 0 − 2 3 µ 0 4π |ψ 100 (0)| 2 ˆ µ 1 · ˆ µ 2 = E 0 + A 4 ˆ σ 1 · ˆ σ 2 , where the indices 1 and 2 refer respectively to the muon and to the electron, and where E 0 = −m r c 2 α 2 /2, with m r being the reduced mass of the (e, µ) system. 24.1.1. Write the matrix representation of the Hamiltonian ˆ H in the basis {|σ 1z ,σ 2z ,σ iz = ±}. 24.1.2. Knowing the value of A in the hydrogen atom: A/h = 1420 MHz, calculate A in muonium. We recall that µ 1 = q¯h/(2m µ ) for the muon, µ 2 = −q¯h/(2m e ) for the electron, µ p =2.79 q¯h/(2m p ) for the proton, where q is the unit charge and m p = 1836.1 m e . 24.1.3. Write the general form of an eigenstate of ˆσ 1z with eigenvalue +1: (i) in the basis {|σ 1z ,σ 2z }; (ii) in the eigenbasis of ˆ H. 24.1.4. We assume that, at t = 0, the system is in a state |ψ(0) of the type defined above. Calculate |ψ(t) at a later time. 24.1.5. (a) Show that the operators ˆπ ± =(1±ˆσ 1z )/2 are the projectors on the eigenstates of ˆσ 1z corresponding to the eigenvalues ±1. (b) Calculate for the state |ψ(t) the probability p(t) that the muon spin is in the state |+ at time t. Write the result in the form 24.2 Muonium in Silicon 247 p(t)=qp + (t)+(1− q) p − (t) , where p + (or p − ) is the probability that one obtains if the electron is initially in the eigenstate of σ 2z with eigenvalue +1 (or −1), and where q is a probability, as yet undefined. 24.1.6. In practice, the electronic spins are unpolarized. A rigorous treatment of the problem then requires a statistical description in terms of a density operator. To account for this nonpolarization in a simpler way, we shall set heuristically that the observed probability ¯p(t) corresponds to q =1/2inthe above formula. Using this prescription, give the complete expression for ¯p(t). 24.2 Muonium in Silicon We now form muonium in a silicon crystal sufficiently thick that the muonium does not escape. The muonium stops in an interstitial position inside the crystal lattice, the nearest atoms forming a plane hexagonal mesh around it. The global effect of the interactions between the atoms of the crystal and the muonium atom is to break the spherical symmetry of the spin–spin interaction, but to preserve the rotational symmetry around the z axis perpendicular to the plane of the mesh. We therefore consider the Hamiltonian ˆ H = E 0 + A  4 ˆ σ 1 · ˆ σ 2 + D ˆσ 1z ˆσ 2z , where the constant A  may differ from A since the presence of neighboring atoms modifies the Coulomb potential and, therefore, the wave function at the origin. The constants A  and D will be determined experimentally; their sign is known: A  > 0,D<0. 24.2.1. Calculate the spin energy levels and the corresponding eigenstates of the muonium trapped in the silicon crystal. 24.2.2. We now reconsider the spin rotation experiment with the following modifications: • Initially the µ + spin is now in the eigenstate | + x of σ x . • We want to know the probability of finding the µ + spin in this same eigenstate | + x at time t. One can proceed as in question 1.5: (a) Calculate in the {|σ 1 ,σ 2 } basis the states |ψ + (t) and |ψ − (t) which are initially eigenstates of ˆσ 2z (ˆσ 2z is the projection of the electron spin along the z axis). (b) Evaluate ψ  (t)|ˆσ 1x |ψ  (t),where = ±. 248 24 Probing Matter with Positive Muons (c) Consider the projector ˆπ x =(1+ˆσ 1x )/2 , and deduce from the previous question the probabilities p ± (t). (d) Calculate the measured probability ¯p(t)=(p + (t)+p − (t))/2. 24.2.3. Comparison with experiment: Present day technology in data processing allows one to determine not p(t) itself, but a quantity which is easier to deal with, the characteristic function g(ω)=Re(f (ω)) where f(ω)= 1 τ  ∞ 0 ¯p(t)e −t/τ e iωt dt is the Fourier transform of ¯p(t)e −t/τ /τ. In this expression, the factor e −t/τ /τ is due to the finite lifetime of the µ + (τ ∼ 2.2 µs). We recall that 1 τ  ∞ 0 e −t/τ e iωt dt = 1 1 − iωτ . (a) Figure 24.1a shows the distribution g(ω) as measured in the conditions of question 2.2. Check that this data is compatible with the results found in question 2.2, and deduce from the data the values of A  /h and D/h (we recall that D<0). g(ω) 37,25 54,85 (arbitrary units) g(ω) 0 10 50 100 ω/2π ( ) MHz 010 50 100 ω/2π ( ) MHz 92,1 (a) (b) Fig. 24.1. Experimental variations of the quantity g(ω), defined in the text, with the frequency ν = ω/(2π). (a) In the conditions described in question 2.2, and (b) in another experimental configuration 24.3 Solutions 249 (b) Figure 24.1b is obtained by a slight modification of the previous experi- ment. Can you tell what modification has been made? How can one evalu- ate the position of the third peak, in terms of the constants of the problem? 24.3 Solutions Section 24.1: Muonium in Vacuum 24.1.1. The Hamiltonian is ˆ H = E 0 + A 4 (ˆσ 1x ˆσ 2x +ˆσ 1y ˆσ 2y +ˆσ 1z ˆσ 2z ) . The matrix representation is therefore ˆ H = ⎛ ⎜ ⎜ ⎝ E 0 + A/4000 0 E 0 − A/4 A/20 0 A/2 E 0 − A/40 000E 0 + A/4 ⎞ ⎟ ⎟ ⎠ , where the elements are ordered as: | ++, | + −, |−+, |−−. 24.1.2. The constant A is related to its value in the hydrogen atom by A A H = |ψ(0)| 2 |ψ(0)| 2 H µ 1 µ p = |ψ(0)| 2 |ψ(0)| 2 H m p m µ 1 2.79 . In first approximation, muonium and hydrogen have similar sizes and wave functions, since the muon is much heavier than the electron. Therefore we obtain A  A H (m p /2.79 m µ )andA/h  4519 MHz. The reduced mass correction to the value of the wave function at the origin is straightforward to calculate. It is of the order of 1% and it leads to A h = 4519 (1 − 0.0126) = 4462 MHz . This value is very close to the observed 4463 MHz, the difference being due to relativistic effects. 24.1.3. The state under consideration can be written as |ψ = |+⊗(α|+ + β|−) with |α| 2 + |β| 2 =1. Equivalently, one can write it as |ψ = α| ++ + β|+ −. The eigenbasis of ˆ H consists in the common eigenstates of the total spin operators ˆ S 2 and ˆ S z : triplet states ⎧ ⎨ ⎩ | ++ (| + − + |−+)/ √ 2 |−− singlet state (| + −−|−+)/ √ 2 . 250 24 Probing Matter with Positive Muons Therefore, one also has the representation |ψ = α|1, 1 + β √ 2 (|1, 0 + |0, 0) , where the only constraint on α and β is |α| 2 + |β| 2 =1. 24.1.4. We start from |ψ(0) = |ψ as defined above. The energy levels and the corresponding eigenstates are known: triplet states E T = E 0 + A/4 singlet state E S = E 0 − 3A/4 . At time t the state is: |ψ(t) =e −iE 0 t/¯h  e −iAt/4¯h  α |1, 1 + β √ 2 |1, 0  + β √ 2 e i3At/4¯h |0, 0  . 24.1.5. (a) It is straightforward to check that ˆπ ± are projectors: ˆπ + |+ = |+ ˆπ + |− =0 ˆπ − |− = |− ˆπ − |+ =0. (b) The probability of finding the muon spin in the state |+ at time t is by definition p(t)=ˆπ + |ψ(t) 2 = ψ(t)|ˆπ + |ψ(t) . Using ˆπ + |1, 1 = |1, 1 ˆπ + |1, 0 =ˆπ + |0, 0 = 1 √ 2 | + − ˆπ + |1, −1 =0, we obtain ˆπ + |ψ(t) =e −i(E 0 +A/4)t/¯h  α| ++ + β 2  1+e iAt/¯h  | + −  . Squaring the norm of ˆπ + |ψ(t),weget p(t)=|α| 2 + |β| 2 cos 2 (At/(2¯h)) . There is a periodic modulation of the probability of observing the muon spin aligned with the positive z axis, which can be interpreted as a rotation of the muon spin with frequency ν = A/h. 24.3 Solutions 251 • The probability p + (t) corresponds to the initial state |ψ(0) = |++.This is a stationary state so that p(t) ≡ p + (t) = 1 in this case. • The probability p − (t) corresponds to the initial state |ψ(0) = | + − =(|1, 0 + |0, 0)/ √ 2. There is in this case an oscillation with a 100 % modulation between | + − and |−+,sothatp(t) ≡ p − (t)= cos 2 (At/2(¯h)). Therefore the result can be cast in the form suggested in the text: p(t)=qp + (t)+(1− q) p − (t) , with q = |α| 2 . 24.1.6. When the electronic spins are unpolarized, we obtain following the assumption of the text: ¯p(t)= 3 4 + 1 4 cos(At/¯h) . Note: The rigorous way to treat partially polarized systems is based on the density operator formalism. In the present case the density operator for the unpolarized electron is: ρ 2 = 1 2 (|++| + |−−|) , so that the initial density operator for the muonium is: ρ(0) = 1 2 | ++++| + 1 2 | + −+ −| = 1 2 |1, 11, 1| + 1 4 (|1, 01, 0| + |1, 00, 0| + |0, 01, 0| + |0, 00, 0|) . The density operator at time t is then given by: ρ(t)= 1 2 |1, 11, 1| + 1 4  |1, 01, 0| +e −iAt/¯h |1, 00, 0| +e iAt/¯h |0, 01, 0| + |0, 00, 0|  hence the probability: ¯p(t)=+, +|ρ(t)|+, + + + −|ρ(t)| + − = 1 2 + 1 4  1 2 +e −iAt/¯h 1 2 +e iAt/¯h 1 2 + 1 2  = 3 4 + 1 4 cos(At/¯h) . 252 24 Probing Matter with Positive Muons Section 24.2: Muonium in Silicon 24.2.1. In the factorized basis {|σ 1 ,σ 2 }, the Hamiltonian is written as ˆ H = E 0 + ⎛ ⎜ ⎜ ⎝ A  /4+D 000 0 −A  /4 − DA  /20 0 A  /2 −A  /4 − D 0 00 0A  /4+D ⎞ ⎟ ⎟ ⎠ . This Hamiltonian is diagonal in the eigenbasis {|S, m} of the total spin. A simple calculation shows that the eigenvalues and eigenvectors are E 1 = E 4 = E 0 + A  /4+D |1, 1 and |1, −1 E 2 = E 0 + A  /4 − D |1, 0 E 3 = E 0 − 3A  /4 − D |0, 0 . 24.2.2. (a) The initial states |ψ + (0) and |ψ − (0) are easily obtained in the factorized basis as |ψ + (0) = | + x⊗|+ =(| ++ + |−+)/ √ 2 |ψ − (0) = | + x⊗|−=(| + − + |−−)/ √ 2 . They can be written in the total spin basis {|S, m} as |ψ + (0) = 1 √ 2 |1, 1 + 1 2 (|1, 0−|0, 0) |ψ − (0) = 1 √ 2 |1, −1 + 1 2 (|1, 0 + |0, 0) . Writing ω i = −E i /¯h, we find at time t: |ψ + (0) = e iω 1 t √ 2 |1, 1 + e iω 2 t 2 |1, 0− e iω 3 t 2 |0, 0 |ψ − (0) = e iω 4 t √ 2 |1, −1 + e iω 2 t 2 |1, 0 + e iω 3 t 2 |0, 0 , which can now be converted in the factorized basis: |ψ + (t) = e iω 1 t √ 2 | ++ + e iω 2 t − e iω 3 t 2 √ 2 | + − + e iω 2 t +e iω 3 t 2 √ 2 |−+ |ψ − (t) = e iω 4 t √ 2 |−−+ e iω 2 t +e iω 3 t 2 √ 2 | + − + e iω 2 t − e iω 3 t 2 √ 2 |−+ . 24.3 Solutions 253 (b) Since ˆσ 1x |σ 1 ,σ 2  = |−σ 1 ,σ 2 , the matrix elements ψ ± (t)|ˆσ 1x |ψ ± (t) are equal to: ψ + (t)|ˆσ 1x |ψ + (t) = 1 2 Re  e −iω 1 t  e iω 2 t +e iω 3 t  = 1 2  cos 2Dt ¯h +cos (A  +2D)t ¯h  ψ − (t)|ˆσ 1x |ψ − (t) = 1 2 Re  e −iω 4 t  e iω 2 t +e iω 3 t  . Since ω 1 = ω 4 , the two quantities are equal. (c) The desired probabilities are p ± (t)=ˆπ +x |ψ ± (t) 2 = ψ ± (t)|ˆπ +x |ψ ± (t) or, equivalently, p ± (t)=ψ ± (t)| 1 2 (1 + ˆσ 1x ) |ψ ± (t) = 1 2 + 1 2 ψ ± (t)|ˆσ 1x |ψ ± (t) . Using the result obtained above, we get: p ± (t)= 1 2 + 1 4  cos 2Dt ¯h +cos (A  +2D)t ¯h  . (d) Since p + (t)=p − (t), the result for ¯p(t) is simply: ¯p(t)= 1 2 + 1 4  cos 2Dt ¯h +cos (A  +2D)t ¯h  . 24.2.3. Comparison with Experiment: In practice, the time t corre- sponds to the decay of the µ + , with the emission of a positron e + and two neutrinos. The positron is sufficiently energetic and leaves the crystal. It is emitted preferentially in the muon spin direction. One therefore mea- sures the direction where the positron is emitted as a function of time. For N 0 incoming muons, the number of positrons emitted in the x direction is dN(t)=N 0 ¯p(t)e −t/τ dt/τ,whereτ is the muon lifetime. A convenient way to analyse the signal, and to extract the desired frequen- cies, consists in taking the Fourier transform of the above signal. Defining f 0 (ω)= 1 τ  ∞ 0 e (iω−1/τ)t dt = 1 1 − iωτ , one obtains f(ω)= 1 2 f 0 (ω)+ 1 8  f 0  ω − 2D ¯h  + f 0  ω + 2D ¯h  + 1 8  f 0  ω − A  +2D ¯h  + f 0  ω + A  +2D ¯h  . 254 24 Probing Matter with Positive Muons The function Re(f 0 (ω)) has a peak at ω = 0 whose half width is 1/τ, which corresponds to 100 kHz. (a) The curve of Fig. 24.1 is consistent with this observation. Besides the peak at ω = 0, we find two peaks at ω 1 = −2D/¯h and ω 2 =(A  +2D)/¯h. Assuming that D is negative, which can be confirmed by a more thorough analysis, one obtains 2D/h = −37.25 MHz and A  /h =92.1MHz . (b) In general, one expects to see peaks at all frequencies ω i − ω j ,andin particular at ω 2 −ω 3 = −A  /¯h. In order to observe the corresponding peak, one must measure the µ + spin projection along a direction which is not orthogonal to the z axis. This leads to a term in cos (ω 2 − ω 3 )t in ¯p(t), which appears in Fig. 24.1. 25 Quantum Reflection of Atoms from a Surface This chapter deals with the reflection of very slow hydrogen atoms from a surface of liquid helium. In particular, we estimate the sticking probability of the atoms onto the surface. This sticking proceeds via the excitation of a surface wave, called a ripplon. We show that this probability must vanish at low temperatures, and that, in this limit, the reflection of the atoms on the surface is specular. In all the chapter, the position of a particle is defined by its coordinates r = (x, y) in a horizontal plane, and its altitude z. The altitude z = 0 represents the position of the surface of the liquid He bath at rest. The wave functions ψ(r,z) of the H atoms are normalized in a rectangular box of volume L x L y L z . We write m for the mass of a H atom (m =1.67 10 −27 kg). 25.1 The Hydrogen Atom–Liquid Helium Interaction Consider a H atom above a liquid He bath at rest (cf Fig. 25.1). We model the H-liquid He interaction as the sum of pairwise interactions between the H atom at point (R,Z)(Z>0), and the He atoms at (r,z), with z<0: V 0 (Z)=n  d 2 r  +∞ −∞ dzU(  (R − r) 2 +(Z − z) 2 ) Θ(−z) where n is the number of He atoms per unit volume, and Θ is the Heaviside function. 25.1.1. We recall the form of the Van der Waals potential: U(d)=− C 6 d 6 , which describes the long distance interaction between a H atom and a He atom separated by a distance d. Show that the long distance potential between the H atom and the liquid He bath is of the form: . of the crystal and the muonium atom is to break the spherical symmetry of the spin–spin interaction, but to preserve the rotational symmetry around the z axis perpendicular to the plane of the. ˆσ 1z corresponding to the eigenvalues ±1. (b) Calculate for the state |ψ(t) the probability p(t) that the muon spin is in the state |+ at time t. Write the result in the form 24.2 Muonium in. respectively to the muon and to the electron, and where E 0 = −m r c 2 α 2 /2, with m r being the reduced mass of the (e, µ) system. 24.1.1. Write the matrix representation of the Hamiltonian ˆ H in the