The Quantum Mechanics Solver 25 potx

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.. Just as for the hyperfine structure of hydrogen, we work in t

246 24 Probing Matter with Positive Muons

m µ c2= 105.66 MeV µ1/h = 67.5 MHz T −1

mec2= 0.511 MeV µ2/h = −1.40 × 104MHz 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, ψ100(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 = E0−2

3

µ0

4π |ψ100(0)|2µˆ1· ˆµ2= E0+A

4 σˆ1· ˆσ2 ,

where the indices 1 and 2 refer respectively to the muon and to the electron,

and where E0 =−mrc2α2/2, with mr being the reduced mass of the (e, µ)

system

24.1.1 Write the matrix representation of the Hamiltonian ˆH in the basis {|σ 1z , σ 2z
, σiz =±}.

24.1.2 Knowing the value of A in the hydrogen atom: A/h = 1420 MHz,

calculate A in muonium We recall that µ1 = q¯ h/(2m µ ) for the muon, µ2 =

−q¯h/(2me) for the electron, µp = 2.79 q¯ h/(2mp) for the proton, where q is the unit charge and mp= 1836.1 me

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

p(t) = q p+(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 = 1/2 in the

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 = E0+A



4 σˆ1· ˆσ2+ D ˆ σ 1zˆσ 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 σx

• 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{|σ1, σ2
} 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) Evaluateψ (t) |ˆσ |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 eiωt 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 eiωt 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)

(arbitrary

units)

g(ω)

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

(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 = E0+A

4 (ˆσ 1x σˆ2x+ ˆσ 1y σˆ2y+ ˆσ 1z σˆ2z ) The matrix representation is therefore

ˆ

H =

⎛

⎜

⎝

⎞

⎟

⎠ ,

where the elements are ordered as:| + +
, | + −
, | − +
, | − −
.

24.1.2 The constant A is related to its value in the hydrogen atom by

A

AH =

|ψ(0)|2

|ψ(0)|2 H

µ1

µp =

|ψ(0)|2

|ψ(0)|2 H

mp

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  AH(mp/2.79 m µ ) and A/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 ˆH consists in the common eigenstates of the total spin

operators ˆS2and ˆS 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 ET= E0+ A/4 singlet state ES= E0− 3A/4

At time t the state is:

|ψ(t)
= e −iE0 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:

ˆ

π+|+
= |+
ˆπ+|−
= 0

ˆ

π −|−
= |−
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

ˆ

π+|1, 1
= |1, 1

ˆ

π+|1, 0
= ˆπ+|0, 0
= √1

2| + −

ˆ

π+|1, −1
= 0 ,

we obtain

ˆ

π+|ψ(t)
= e −i(E0 +A/4)t/¯ h

α | + +
+ β

2



1 + eiAt/¯ h



| + −



.

Squaring the norm of ˆπ+|ψ(t)
, we get

p(t) = |α|2+|β|2 cos2(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.

• 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 | − +
, so that p(t) ≡ p− (t) =

cos2(At/2(¯ h)).

Therefore the result can be cast in the form suggested in the text:

p(t) = q p+(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:

ρ2= 1

2 (|+
+| + |−
−|) ,

so that the initial density operator for the muonium is:

ρ(0) = 1

2| + +
+ + | +12| + −
+ − |

= 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|

+eiAt/¯ 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/¯h1

2+ e

iAt/¯ h1

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{|σ1, σ2
}, the Hamiltonian is written as

ˆ

H = E0+

⎛

⎜

⎝

⎞

⎟

⎠

This Hamiltonian is diagonal in the eigenbasis {|S, m
} of the total spin A

simple calculation shows that the eigenvalues and eigenvectors are

E1= E4= E0+ A  /4 + D |1, 1
and |1, −1

E2= E0+ A  /4 − D |1, 0

E3= E0− 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
+12(|1, 0
+ |0, 0
)

Writing ω i=−Ei /¯ h, we find at time t:

|ψ+(0)
= e√ iω1t

2 |1, 1
+ eiω2t

2 |1, 0
−eiω3t

2 |0, 0

|ψ−(0)
= e√ iω4t

2 |1, −1
+eiω2t

2 |1, 0
+eiω3t

2 |0, 0
,

which can now be converted in the factorized basis:

|ψ+(t)
= e√ iω1t

2 | + +
+eiω2t − e iω3t

2√

2 | + −
+eiω2t+ eiω3t

2√

2 | − +

|ψ− (t)
= e√ iω4t

2 | − −
+eiω2t+ eiω3t

2√

2 | + −
+eiω2t − e iω3t

2√

2 | − +

(b) Since ˆσ 1x |σ1, σ2
= | − σ1, σ2
, the matrix elements ψ± (t) |ˆσ 1x |ψ± (t)

are equal to:

ψ+(t) |ˆσ 1x |ψ+(t)
= 1

2Re

e−iω1 t

eiω2t+ eiω3t

= 1

2 cos

2Dt

¯

h + cos

(A  + 2D)t

¯

h

ψ− (t) |ˆσ 1x |ψ− (t)
= 12Re"

e−iω4 t

eiω2t+ eiω3t

Since ω1= ω4, 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

2ψ ± (t) |ˆσ 1x |ψ ± (t)

Using the result obtained above, we get:

p ± (t) = 12+14 cos2Dt¯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

N0 incoming muons, the number of positrons emitted in the x direction is

dN (t) = N0p(t)e¯ −t/τ dt/τ , where τ 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

f0(ω) = 1

τ

 ∞

0

e(iω −1/τ)t dt = 1

1− iωτ ,

one obtains

f (ω) =1

2f0(ω) +

1 8



f0 ω − 2D

¯

h

+ f0 ω + 2D

¯

h



+1 8



f0 ω − A  + 2D

¯

h

+ f0 ω + A

 + 2D

¯

h



.

254 24 Probing Matter with Positive Muons

The function Re(f0(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 ω1 = −2D/¯h and ω2 = (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 = 92.1 MHz

(b) In general, one expects to see peaks at all frequencies ω i − ωj, and in

particular at ω2−ω3=−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 (ω2− ω3)t in ¯ p(t), which appears in

Fig 24.1

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 Lx 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:

V0(Z) = n



d2r

 +∞

−∞

dz U (

(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) = − C6

d6 ,

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:

the position of the surface of the liquid He bath at rest The wave functions

ψ(r, z) of the H atoms are normalized... 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... 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