We are only interested in the magnetic energy states of the chain, due to the magnetic interactions of the N Fe2+ ions, each with spin 1.. We assume that ˆH1 is a small perturbation comp
Trang 1Magnetic Excitons
Quantum field theory deals with systems possessing a large number of de-grees of freedom This chapter presents a simple model, where we study the magnetic excitations of a long chain of coupled spins We show that one can associate the excited states of the system with quasi-particles that propagate along the chain
We recall that, for any integer k:
N
n=1
e2iπkn/N = N if k = pN, with p integer;
= 0 otherwise.
20.1 The Molecule CsFeBr3
Consider a system with angular momentum equal to 1, i.e j = 1 in the basis
|j, m common to ˆ J2 and ˆJ z
20.1.1 What are the eigenvalues of ˆJ2 and ˆJ z?
20.1.2 For simplicity, we shall write |j, m = |σ, where σ = m = 1, 0, −1.
Write the action of the operators ˆJ ±= ˆJ x ± i ˆ J y on the states|σ.
20.1.3 In the molecule Cs Fe Br3, the ion Fe2+ has an intrinsic angular
mo-mentum, or spin, equal to 1 We write the corresponding observable ˆ J , and
we note|σ the eigenstates of ˆ J z The molecule has a plane of symmetry, and
the magnetic interaction Hamiltonian of the ion Fe2+ with the rest of the molecule is
ˆ
H r= D
¯
h2
ˆ
J2
z D > 0
What are the eigenstates of ˆH r and the corresponding energy values? Are there degeneracies?
Trang 2204 20 Magnetic Excitons
20.2 Spin–Spin Interactions in a Chain of Molecules
We consider a one-dimensional closed chain made up with an even number N
of Cs Fe Br3 molecules We are only interested in the magnetic energy states
of the chain, due to the magnetic interactions of the N Fe2+ ions, each with spin 1
We take{|σ1, σ2, · · · , σ N }, σ n = 1, 0, −1, to be the orthonormal basis of
the states of the system; it is an eigenbasis of the operators { ˆ J z n } where ˆJ n
is the spin operator of the n-th ion (n = 1, · · · , N).
The magnetic Hamiltonian of the system is the sum of two terms ˆH =
ˆ
H0+ ˆH1 where
ˆ
H0= D
¯
h2
N
n=1
( ˆJ z n)2
has been introduced in 1.3, and ˆH1is a nearest-neighbor spin–spin interaction term
ˆ
H1= A
¯
h2
N
n=1
ˆ
J n · ˆJ n+1 A > 0
To simplify the notation of ˆH1, we define ˆJ N+1 ≡ ˆJ1.
We assume that ˆH1 is a small perturbation compared to ˆH0 (A D),
and we shall treat it in first order perturbation theory
20.2.1 Show that |σ1, σ2, · · · , σ N is an eigenstate of ˆ H0, and give the cor-responding energy value
20.2.2 What is the ground state of ˆH0? Is it a degenerate level?
20.2.3 What is the energy of the first excited state of ˆH0? What is the
degeneracy d of this level? We shall denote by E1the corresponding eigenspace
of ˆH0, of dimension d.
20.2.4 Show that ˆH1can be written as
ˆ
H1= A
¯
h2
N
n=1
1
2( ˆJ
n
+Jˆn+1
− + ˆJ − n Jˆ+n+1) + ˆJ z n Jˆz n+1
.
20.3 Energy Levels of the Chain
We now work in the subspaceE1 We introduce the following notation
|n, ± = |σ1= 0, σ2= 0, · · · , σ n =±1, σ n+1 = 0, · · · , σ N = 0.
Owing to the periodicity of the chain, we define|N + 1, ± ≡ |1, ±.
Trang 320.3.1 Show that
ˆ
H1|n, ± = A(|n − 1, ± + |n + 1, ±) + |ψ n ,
where|ψ n is orthogonal to the subspace E1.
Without giving the complete form of |ψ n , give an example of one of its
components, and give the energy of the eigenspace of ˆH0to which|ψ n belongs.
20.3.2 Consider the circular permutation operator ˆT , and its adjoint ˆ T † ,
defined by
ˆ
T |σ1, σ2, · · · , σ N = |σ N , σ1, · · · , σ N−1
ˆ
T † |σ1, σ2, · · · , σ N = |σ2, σ3, · · · , σ N , σ1
Write the action of ˆT and ˆ T † on the states|n, ±.
20.3.3 Check that, in the subspace E1, ˆ H1 and A( ˆ T + ˆ T †) have the same
matrix elements
20.3.4 Show that the eigenvalues λ k of ˆT are the N -th roots of unity (we recall that N is assumed to be even):
λ k = e−iq k q k =−π + 2kπ
N k = 0, · · · , N − 1
20.3.5 We seek, inE1, the 2N eigenvectors |q k , ± of ˆ T , each corresponding
to an eigenvalue λ k Each|q k , ± is written
|q k , ± =
n
(a) Write a recursion relation between the coefficients c n
(b) Show that
c n (k) = √1
Ne
is a solution of this recursion relation
(c) Show that the states |q k , ± defined using (20.1) and (20.2) are
ortho-normal
(d) Show that the vectors|q k , ± are also eigenvectors of ˆ T †and ˆT + ˆ T †, and
give the corresponding eigenvalues
(e) Calculate the scalar productn, |q k , (, =±) and write the
expan-sion of the states|n, ± in the basis |q k , ±.
20.3.6 We treat the Hamiltonian ˆH1 of Sect 2 as a perturbation to ˆH0.
We limit ourselves to the first excited level of ˆH0, and we want to calcu-late how the perturbation lifts the degeneracy of this level We recall that,
in the degenerate case, first order perturbation theory consists in diagonaliz-ing the restriction of the perturbdiagonaliz-ing Hamiltonian in the degenerate subspace
of the dominant term ˆH
Trang 4206 20 Magnetic Excitons
(a) Explain why the results of questions 3.3 and 3.5 above allow one to solve this problem
(b) In first order perturbation theory, give the new energy levels which arise from the first excited state of ˆH0, and the corresponding eigenstates (c) Draw qualitatively the energies E(q k ) in terms of the variable q k which
can be treated as a continuous variable, q k ∈ [−π, +π[, if N is very large.
1 What is the degeneracy of each new energy level?
20.4 Vibrations of the Chain: Excitons
We now study the time evolution of the spin chain
20.4.1 Suppose that at time t = 0, the system is in the state
|Ψ(0) =
=±
N−1
k=0
ϕ k |q k , with
=±
N−1
k=0
|ϕ
k |2= 1 Setting ω = 2A/¯ h, write the state |Ψ(t) at a later time t.
20.4.2 We assume that the initial state is |Ψ(0) = |q k , + .
(a) Write the probability amplitude α n (t) and the probability P n (t) of finding
at time t the n-th spin pointing upwards, i.e σ n = +1 and σ m = 0 for
m = n Show that P n (t) is the same for all sites of the chain.
(b) The molecules of the chain are located at x n = na, where a is the lattice spacing Show that the probability amplitude α n (t) is equal to the value
at x = x n of a monochromatic plane wave
Ψ k (x, t) = Ce i(p(q)x −E(q)t)/¯h ,
where C is a constant, q = q k , and x is the abscissa along the chain Express p(q) in terms of q.
(c) Show that Ψ k (x, t) is an eigenstate of the momentum operator ˆ p x = (¯h/i)∂/∂x along the chain.
Show that the value of p(q) ensures the periodicity of Ψ k (x, t), i.e Ψ k (x +
L, t) = Ψ k (x, t), where L = N a is the length of the chain.
(d) Show that, for |q k | 1, Ψ k (x, t) satisfies a Schr¨odinger equation for a
particle of negative mass m, placed in a constant potential; give the value
of m.
20.4.3 In a more complete analysis, one can associate quasi-particles to the
magnetic excitations of the chain These quasi-particles, which we call
“mag-netic excitons”, have an energy E(q k ) and a momentum p(q k)
At very low temperatures, T ≈ 1.4 K, the chain is in the ground state
of ˆH0 If low energy neutrons collide with it, they can create excitons whose
energy and momentum can be determined by measuring the recoil of the
neutrons The experimental result for E(q) as a function of q ∈ [−π, 0] is
given in Fig 20.1
Trang 5Fig 20.1 Experimental measurement of the excitation energy E(q) as a function
of q between −π and 0 The energy scale is in meV (10 −3eV)
(a) Deduce from that data approximate values for D and A.
(b) What do you think of the approximation D A and of the comparison
between theory and experiment? How could one improve the agreement between theory and experiment?
(c) Is it justified to assume that the chain is in its ground state when it
is at thermal equilibrium at 1.4 K? We recall the Boltzmann factor:
N (E2)/N (E1) = exp[−(E2− E1)/kT ], with k = 8.6 × 10 −5 eV K−1.
20.4.4 Consider, at time t = 0, the state
|Ψ(0) =
N−1
k=0
ϕ k |q k , + with
N−1
k=0
|ϕ k |2= 1.
We assume that N 1, that the coefficients ϕ k have significant values only
in a close vicinity of some value k = k0, or, equivalently, q ≈ q0, and that, to
a good approximation, in this vicinity,
E(q) = E(q0) + (q − q0)u0, u0=dE
dq
q=q0
.
Show that the probability P n (t) of finding σ n = +1 at time t is the same as the probability P n (t ) of finding σ n = +1 at another time t whose value will
be expressed in terms of t and of the distance between the sites n and n .
Interpret the result as the propagation of a spin excitation wave along the chain Calculate the propagation velocity of this wave and give its numerical
value for a = 0.7 nm and q0=−π/2.
20.4.5 We now assume that the initial state is|Ψ(0) = |n = 1, +.
Trang 6208 20 Magnetic Excitons
(a) Write the probability P m (t) of finding σ m = +1 at a later time t? (b) Calculate the probabilities P1(t) and P2(t), in the case N = 2, and
inter-pret the result
(c) Calculate P1(t) in the case N = 8 Is the evolution of P1(t) periodic? (d) For N 1, one can convert the above sums into integrals The prob-abilities are then P m (t) ≈ |J m−1 (ωt) |2 where the J n (x) are the Bessel
functions These functions satisfy
|J n (x) |2= 1 and J n = (−) n J −n .
For x 1 we have J n (x) ≈ 2
πx cos(x − nπ/2 − π/4) if x > 2|n|/π, and
J n (x) ≈ 0 if x < 2|n|/π.
Which sites are appreciably reached by the probability wave at a time t such that ωt 1?
(e) Interpret the result as the propagation along the chain of a probability amplitude (or wave) Calculate the propagation velocity and compare it with the result obtained in question 4.4)
20.5 Solutions
Section 20.1: The Molecule CsFeBr 3
20.1.1 The results are: ˆJ2: 2¯h2, ˆ J z : m¯ h ; m = 1, 0, −1.
20.1.2 One has:
J+|1 = 0
J+|0 = ¯h √
2|1
J+| − 1 = ¯h √2|0
J − |1 = ¯h √
2|0
J − |0 = ¯h √
2| − 1
J − | − 1 = 0
20.1.3 The eigenstates are the states |σ The state |0 corresponds to the eigenvalue E = 0, whereas |+ and |−, which are degenerate, correspond to
E = D.
Section 20.2: Spin–Spin Interactions in a Chain of Molecules 20.2.1 It is straightforward to see that
ˆ
H0|σ1, σ2· · · σ N = D
N
n=1
(σ n)2|σ1· · · σ N ,
the corresponding eigenvalue being E = D
σ2
n
20.2.2 The ground state of ˆH0 corresponds to all the σ n equal to zero, so
that E = 0 This ground state is non-degenerate.
20.2.3 The first excited state corresponds to all the σ’s being zero except
one: σ nequal to±1 The energy is D, and the degeneracy 2N, since there are
N possible choices of the non-vanishing σ n, and two values±1 of σ n
20.2.4 J ± = J x ± iJ y A direct calculation leads to the result.
Trang 7Section 20.3: Energy Levels of the Chain
20.3.1 The action of the perturbing Hamiltonian on the basis states is,
set-ting = ±:
ˆ
H1|n, = A (|n − 1, + |n + 1, )
n =n
(|0, · · · 0, σ n = , 0 · · · 0, σ n =−1, σ n +1= +1, 0 · · · 0
+|0 · · · 0, σ n = , 0 · · · 0, σ n = +1, σ n +1=−1, 0 · · · 0)
The vector|ψ = |σ1= 1, σ2=−1, 0 · · · 0, σ n = , 0 · · · 0 belongs to this latter
set; it is an eigenvector of ˆH0 with energy 3D.
20.3.2 The definition of ˆT , ˆ T † and|n, ± implies:
ˆ
T |n, ± = |n + 1, ± ; Tˆ† |n, ± = |n − 1, ±
20.3.3 We therefore obtain
A( ˆ T + ˆ T †)|n, ± = A(|n − 1, ± + |n + 1, ±)
Since
ˆ
H1|n, ± = A(|n − 1, ± + |n + 1, ±) + |ψ n where n , ±|ψ n = 0 ,
ˆ
H1 and A( ˆ T + ˆ T †) obviously have the same matrix elements in the subspace
E1.
20.3.4 Since ˆT N = ˆI, an eigenvalue λ k satisfies λ N k = 1, which proves that
each eigenvalue is an N -th root of unity Conversely, we will see in the following that each N -th root of unity is an eigenvalue.
20.3.5 (a) The corresponding eigenvectors satisfy
|q k , ± =
n
c n |n, ± Tˆ|q k , ± = λ k |q k , ±
therefore one has
n
c n |n + 1, ± = λ k
n
c n |n, ±
Hence the recursion relation and its solution are
λ k c n = c n−1 c n = 1
λ n−1 k
c1= eiq k (n −1) c
1 .
Trang 8210 20 Magnetic Excitons
(b) The normalization condition
n |c n |2= 1 gives N |c1|2= 1 If we choose
c1= eiq k / √
N , the eigenvectors are of unit norm and we recover the solution
given in the text of the problem
(c) The scalar product of|q k , and |q k , is easily calculated:
q k , |q k , = δ , 1
N
n
e2iπn(k −k )/N
= δ , δ k,k
(d) The vectors |q k , ± are eigenvectors of ˆ T † with the complex conjugate
eigenvalues λ ∗
k Therefore they are also eigenvectors of ˆT + ˆ T †with the
eigen-value λ k + λ ∗
k = 2 cos q k=−2 cos(2kπ/N).
(e) From the definition of the vectors, we have
n, |q k , = √1
Ne
iq k n δ
and (directly or by using the closure relation)
|n, ± = √1
N
N−1
k=0
e−iq k n |q k , ±
20.3.6 The restriction of ˆH1 to the subspace E1 is identical to A( ˆ T + ˆ T †)
(question 3.3) InE1, the operator A( ˆ T + ˆ T †) is diagonal in the basis|q k , ±.
Therefore the restriction of ˆH1is also diagonal in that basis The energy levels are
E(q k ) = D + 2A cos(q k ) , (20.3) corresponding to the states|q k , ± As far as degeneracies are concerned, there
is a twofold degeneracy for all levels (the spin value may be +1 or −1) In addition, for all levels except q = −π and q = 0, there is a degeneracy q k ↔
−q k (symmetry of the cosine) Therefore, in general, the degeneracy is 4
Section 20.4: Vibrations of the Chain: Excitons
20.4.1 At time t the state of the chain is (cf (20.3)):
|Ψ(t) = e −iDt/¯h
k
ϕ ke−iωt cos q k |q k ,
20.4.2 We now consider an initial state|q k , ±, evolving as e −iE(q)t/¯h |q k , ±.
(a) We therefore obtain an amplitude
α n (t) = √1
Ne
i(q k n−E(q k )t/¯ h)
and a probability P n (t) = |α n |2= 1
N, which is the same on each site
Trang 9(b) In the expression
α n (t) = √1
Ne
i(q k x n /a−E(q k )t/¯ h) ,
we see that α n (t) is the value at x = x n of the function Ψ k (x, t) =
1
√ N exp[i(px − Et)/¯h] with E(q) = D + ¯hω cos q and p(q) = ¯hq/a.
(c) The function Ψ k (x) is an eigenstate of ˆ p x with the eigenvalue ¯hq k /a Since N is even, we obtain:
eiq k L/a= eiNq k = e2πik = 1 , which proves the periodicity of Ψ k
(d) For|q k | 1, cos q k= 1− q2
k /2 Therefore E = E0+ p2/2m with
E0= D + 2A and m = − ¯h
2
2Aa2 =− ωa¯h2 .
Ψ k then satisfies the wave equation
i¯h ∂ψ
∂t =−¯h
2
2m
∂2ψ
∂x2 + E0ψ ,
which is a Schr¨odinger equation for a particle of negative mass (in solid state physics, this corresponds to the propagation of holes and in field theory, to the propagation of anti-particles)
20.4.3 (a) With the data of the figure which resemble grosso modo the
E(q) drawn in Fig 20.2, one finds D + 2A ∼ 3.2 × 10 −3 eV, and D − 2A ∼ 0.4 × 10 −3 eV Therefore:
D ∼ 1.8 × 10 −3eV A ∼ 0.7 × 10 −3 eV
(b) The approximation D A is poor The theory is only meaningful to order (A/D)2∼ 10% Second order perturbation theory is certainly necessary
to account quantitatively for the experimental curve which has a steeper shape
than a sinusoid in the vicinity of q = −π.
(c) For T = 1.4 K, kT ∼ 1.2 × 10 −4 eV, exp(−(D − 2A)/kT ) ∼ 0.04 To a
few % , the system is in its ground state
20.4.4 Approximating E(q) = E(q0) + (q − q0)u0 in the vicinity of q0, we obtain
α n (t) = √1
Ne
i(q0n−ω0t)
k
ϕ kei(q k −q0 )(n −u0 t/¯ h)
Trang 10212 20 Magnetic Excitons
Fig 20.2 Energy levels
Since the global phase factor does not contribute to the probability, one has
P n (t) = P n (t ) with
t = t + (n − n)¯h
u0 .
This corresponds to the propagation of a wave along the chain, with a group velocity
vg= u0a
¯
h =
a
¯
h
dE dq
q=q0
=− 2a A
¯
h sin q0 . For q0 = −π/2 and a = 0.7 nm, we find vg ∼ 1500 ms −1 One can also
evaluate u0 ∼ 1.2 meV directly on the experimental curve, which leads to
vg∼ 1300 ms −1.
20.4.5 If|Ψ(0) = |n = 1, +, then ϕ+
k = e−iq k / √
N and ϕ −
k = 0
(a) The probability is P m (t) = |m, +|Ψ(t)|2, where
m, +|Ψ(t) = e−iDt/¯h
N
k
eiq k (m −1)e−iωt cos q k
(b) N=2:
There are two possible values for q k : q0 = −π and q1 = 0 This leads to
P1(t) = cos2ωt, P2(t) = sin2ωt These are the usual oscillations of a
two-state system, such as the inversion of the ammonia molecule
(c) N=8:
cos(q k) −1 − √1
2
The probability P1 of finding the excitation on the initial site is
P1(t) = 1
4
cos2(ωt/2) + cos(ωt/ √
2)
2
The system is no longer periodic in time There cannot exist t = 0 for which