1. Consider three nearest-neighbor spins each of magnitude 𝑺 on a line, as shown, coupled by the
Heisenberg interaction:
̂𝒑 . (𝑺
̂𝒑−𝟏 + 𝑺
̂𝒑+𝟏 )
𝑼 = −𝟐𝑱𝑺
Show that the dispersion for spin waves in one dimension is
ℏ𝝎 = 𝟒𝑱𝑺(𝟏 − 𝒄𝒐𝒔 𝒌𝒂).
Dispersion Relation is the relationship between 𝜔 𝑎𝑛𝑑 𝑘. The relation is basically about how the angular
frequency varies with the wave vector. The underline concept is when the spins are inclined at an angle to
each other, they exert torque on each other.
Considering the spin at site p in one dimensional chain, we know from mechanics, the rate of change of
the angular momentum is equal to the torque which acts on the spin.
ℏ
𝑑𝑆𝑝
= 𝜇𝑝 × 𝐵𝑝
𝑑𝑡
(𝟏)
The magnetic moment at site p is given by
𝜇𝑝 = −𝑔𝜇𝐵 𝑆𝑝
(𝟐)
Taking into account the Heisenberg exchange interaction, which says that the interaction is only between
the nearest neighbors. That is, 𝑆𝑝 can interact only with 𝑆𝑝+1 𝑎𝑛𝑑 𝑆𝑝−1 . And the interaction with other
spins will be zero.
Taking the above assumption about the Heisenberg interaction, the exchange energy will be given as
𝑈 = −2𝐽𝑆𝑝 (𝑆𝑝−1 + 𝑆𝑝+1 )
(𝟑)
From equation (2), 𝑆𝑝 is given as
𝑆𝑝 =
−𝜇𝑝
𝑔𝜇𝐵
(𝟒)
Substituting equation (4) in to equation (3)
𝑈=
2𝐽𝜇𝑝
+ 𝑆𝑝+1 )
(𝑆
𝑔𝜇𝐵 𝑝−1
(𝟓)
We also know that energy is given by
𝑈 = −𝜇𝑝 ∙ 𝐵𝑝
(𝟔)
Equating equation (6) and equation (5), we have
−𝜇𝑝 ∙ 𝐵𝑝 =
2𝐽𝜇𝑝
+ 𝑆𝑝+1 )
(𝑆
𝑔𝜇𝐵 𝑝−1
(𝟕)
From equation (7), the value of (𝐵𝑝 ) magnetic field at site p can be given as
𝐵𝑝 =
−2𝐽
+ 𝑆𝑝+1 )
(𝑆
𝑔𝜇𝐵 𝑝−1
Putting equation (8) and equation (2) in to equation (1) gives us
(𝟖)
ℏ
𝑑𝑆𝑝
−2𝐽
= (−𝑔𝜇𝐵 𝑆𝑝 ) × (
+ 𝑆𝑝+1 ))
(𝑆
𝑑𝑡
𝑔𝜇𝐵 𝑝−1
(𝟗)
Simplifying equation (9), we have
𝑑𝑆𝑝 2𝐽
= (𝑆𝑝 × 𝑆𝑝−1 + 𝑆𝑝 × 𝑆𝑝+1 )
𝑑𝑡
ℏ
(𝟏𝟎)
To solve equation (10), let first solve 𝑆𝑝 × 𝑆𝑝−1
In Cartesian coordinate
𝑆𝑝 × 𝑆𝑝−1
𝑖̂
𝑥
= | 𝑆𝑝
𝑥
𝑆𝑝−1
𝑗̂
𝑦
𝑆𝑝
𝑦
𝑆𝑝−1
𝑘̂
𝑆𝑝𝑧 |
𝑧
𝑆𝑝−1
𝑦 𝑧
𝑦
𝑦
𝑦 𝑥
𝑧
𝑥
𝑆𝑝 × 𝑆𝑝−1 = 𝑖̂(𝑆𝑝 𝑆𝑝−1
− 𝑆𝑝𝑧 𝑆𝑝−1 ) − 𝑗̂(𝑆𝑝𝑥 𝑆𝑝−1
− 𝑆𝑝𝑧 𝑆𝑝−1
) + 𝑘̂ (𝑆𝑝𝑥 𝑆𝑝−1 − 𝑆𝑝 𝑆𝑝−1
)
(𝟏𝟏)
Similarly for 𝑆𝑝 × 𝑆𝑝+1
𝑆𝑝 × 𝑆𝑝−1
𝑖̂
= | 𝑆𝑝𝑥
𝑥
𝑆𝑝+1
𝑦
𝑗̂
𝑦
𝑆𝑝
𝑦
𝑆𝑝+1
𝑘̂
𝑆𝑝𝑧 |
𝑧
𝑆𝑝+1
𝑦
𝑦
𝑦
𝑧
𝑧
𝑥
𝑥
𝑆𝑝 × 𝑆𝑝+1 = 𝑖̂(𝑆𝑝 𝑆𝑝+1
− 𝑆𝑝𝑧 𝑆𝑝+1 ) − 𝑗̂(𝑆𝑝𝑥 𝑆𝑝+1
− 𝑆𝑝𝑧 𝑆𝑝+1
) + 𝑘̂ (𝑆𝑝𝑥 𝑆𝑝+1 − 𝑆𝑝 𝑆𝑝+1
)
(𝟏𝟐)
Substituting equation (11) and equation (12) back into equation (10), we have
𝑑𝑆𝑝 2𝐽
𝑦 𝑧
𝑦
𝑦
𝑧
𝑧
𝑧
𝑥
𝑥
= {𝑖̂(𝑆𝑝 (𝑆𝑝−1
+ 𝑆𝑝+1
) − 𝑆𝑝𝑧 (𝑆𝑝−1 + 𝑆𝑝+1 )) − 𝑗̂(𝑆𝑝𝑥 (𝑆𝑝−1
+ 𝑆𝑝+1
) − 𝑆𝑝𝑧 (𝑆𝑝−1
+ 𝑆𝑝+1
))
𝑑𝑡
ℏ
𝑦
𝑦
𝑦
𝑥
𝑥
+ 𝑘̂ (𝑆𝑝𝑥 (𝑆
+ 𝑆 ) − 𝑆𝑝 (𝑆𝑝−1
+ 𝑆𝑝+1
(𝟏𝟑)
))}
𝑝−1
𝑝+1
If we put it in component form
𝑑𝑆𝑝𝑥
𝑑𝑡
=
2𝐽
𝑦 𝑧
(𝑆𝑝 (𝑆𝑝−1
ℏ
𝑦
𝑑𝑆𝑝
𝑦
𝑦
𝑧
+ 𝑆𝑝+1
) − 𝑆𝑝𝑧 (𝑆𝑝−1 + 𝑆𝑝+1 )
(14a)
2𝐽 𝑥 𝑧
𝑧
𝑥
𝑥
+ 𝑆𝑝+1
) − 𝑆𝑝𝑧 (𝑆𝑝−1
+ 𝑆𝑝+1
(𝑆 (𝑆
)
ℏ 𝑝 𝑝−1
(𝟏𝟒𝒃)
𝑑𝑆𝑝𝑧 2𝐽 𝑥 𝑦
𝑦
𝑦
𝑥
𝑥
= (𝑆𝑝 (𝑆𝑝−1 + 𝑆𝑝+1 ) − 𝑆𝑝 (𝑆𝑝−1
+ 𝑆𝑝+1
))
𝑑𝑡
ℏ
(𝟏𝟒𝒄)
𝑑𝑡
=−
Since we assume the spin is in the 𝑧-direction, we need to preserve only the product which contains the 𝑧
product and we need to neglect the product 𝑤ℎ𝑖𝑐ℎ 𝑐𝑜𝑛𝑡𝑎𝑖𝑛𝑠 𝑡ℎ𝑒 𝑥 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑦 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠. Also the
𝑦
amplitude of the excitation is small (𝑖𝑓 𝑆𝑝𝑥 , 𝑆𝑝 ≪ 𝑆), we may obtain an approximate set of linear equation
𝑧
𝑧
by taking all 𝑆𝑝𝑧 = 𝑆𝑝−1
= 𝑆𝑝+1
= 𝑆. Then equation (14) becomes
𝑑𝑆𝑝𝑥
=
𝑑𝑡
2𝐽𝑆
𝑦
(2𝑆𝑝
ℏ
𝑦
𝑑𝑆𝑝
𝑑𝑡
=−
𝑦
𝑦
− 𝑆𝑝−1 − 𝑆𝑝+1 )
2𝐽𝑆
(2𝑆𝑝𝑥
ℏ
𝑥
𝑥
− 𝑆𝑝−1
− 𝑆𝑝+1
)
𝑑𝑆𝑝𝑧
=0
𝑑𝑡
(15a)
(15b)
(𝟏𝟓𝒄)
Assuming the traveling wave solution of equation (15a) and (15b)
𝑆𝑝𝑥 = 𝑢𝑒𝑥𝑝[𝑖(𝑝𝑘𝑎 − 𝜔𝑡)]
𝑦
𝑆𝑝 = 𝑣𝑒𝑥𝑝[𝑖(𝑝𝑘𝑎 − 𝜔𝑡)]
(𝟏𝟔)
(𝟏𝟕)
𝑤ℎ𝑒𝑟𝑒 𝑢, 𝑣 𝑎𝑟𝑒 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 , 𝑝 𝑖𝑠 𝑎𝑛 𝑖𝑛𝑡𝑒𝑔𝑒𝑟, 𝑎𝑛𝑑 𝑎 𝑖𝑠 𝑡ℎ𝑒 𝑙𝑎𝑡𝑡𝑖𝑐𝑒 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
Now, substituting equation (16) and (17) into equation (15a) and (15b),
Equation (15a) becomes
−𝑖𝑢𝜔𝑒 𝑖(𝑝𝑘𝑎−𝜔𝑡) =
2𝐽𝑆
(2𝑣𝑒 𝑖(𝑝𝑘𝑎−𝜔𝑡) − 𝑣𝑒 𝑖((𝑝−1)𝑘𝑎−𝜔𝑡) − 𝑣𝑒 𝑖((𝑝+1)𝑘𝑎−𝜔𝑡) )
ℏ
(𝟏𝟖)
If we open up equation (18), we have
−𝑖𝑢𝜔𝑒 𝑖𝑝𝑘𝑎 𝑒 −𝑖𝜔𝑡 =
2𝐽𝑆
(𝑣(2𝑒 𝑖𝑝𝑘𝑎 𝑒 −𝑖𝜔𝑡 − 𝑒 𝑖𝑝𝑘𝑎 𝑒 −𝑖𝑘𝑎 𝑒 −𝑖𝜔𝑡 − 𝑒 𝑖𝑝𝑘𝑎 𝑒 𝑖𝑘𝑎 𝑒 −𝑖𝜔𝑡 ))
ℏ
(𝟏𝟗)
Collecting the same term on the right side and canceling them with the left side, we have
−𝒊𝝎𝒖 =
𝟐𝑱𝑺
(𝟐
ℏ
− 𝒆−𝒌𝒂 − 𝒆𝒌𝒂 )𝒗 =
𝟒𝑱𝑺
(𝟏
ℏ
− 𝐜𝐨𝐬 𝒌𝒂)𝒗
(20)
Similarly Equation (15b) becomes
2𝐽𝑆
(2𝑢𝑒 𝑖(𝑝𝑘𝑎−𝜔𝑡) − 𝑢𝑒 𝑖((𝑝−1)𝑘𝑎−𝜔𝑡) − 𝑢𝑒 𝑖((𝑝+1)𝑘𝑎−𝜔𝑡) )
ℏ
(𝟐𝟏)
2𝐽𝑆
(𝑢(2𝑒 𝑖𝑝𝑘𝑎 𝑒 −𝑖𝜔𝑡 − 𝑒 𝑖𝑝𝑘𝑎 𝑒 −𝑖𝑘𝑎 𝑒 −𝑖𝜔𝑡 − 𝑒 𝑖𝑝𝑘𝑎 𝑒 𝑖𝑘𝑎 𝑒 −𝑖𝜔𝑡 ))
ℏ
(𝟐𝟐)
−𝑖𝑣𝜔𝑒 𝑖(𝑝𝑘𝑎−𝜔𝑡) =
−𝑖𝑣𝜔𝑒 𝑖𝑝𝑘𝑎 𝑒 −𝑖𝜔𝑡 =
−𝒊𝝎𝒗 = −
𝟐𝑱𝑺
(𝟐 −
ℏ
𝒆−𝒊𝒌𝒂 − 𝒆𝒊𝒌𝒂 )𝒖 = −
𝟒𝑱𝑺
(𝟏 −
ℏ
𝐜𝐨𝐬 𝒌𝒂)𝒖
(23)
Equation (20) and (23) will have non zero solution for 𝑢 𝑎𝑛𝑑 𝑣, if the determinant of the coefficients is
equal to zero
𝑖𝜔
|
−
𝟒𝑱𝑺
(𝟏 − 𝐜𝐨𝐬 𝒌𝒂)
ℏ
𝟒𝑱𝑺
(𝟏 − 𝐜𝐨𝐬 𝒌𝒂)
ℏ
|=0
𝑖𝜔
Solving the determinant, we have
ℏ𝝎 = 𝟒𝑱𝑺(𝟏 − 𝐜𝐨𝐬 𝒌𝒂)
(𝟐𝟒)
Therefore, Equation (24) is the dispersion relation for spin waves in one dimension with nearest-neighbor
interactions.
𝑭𝒊𝒈𝒖𝒓𝒆 𝟏. 𝐷𝑖𝑠𝑝𝑒𝑟𝑠𝑖𝑜𝑛 𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛 𝑓𝑜𝑟 𝑚𝑎𝑔𝑛𝑜𝑛𝑠 𝑖𝑛 𝑎 𝑓𝑒𝑟𝑟𝑜𝑚𝑎𝑔𝑛𝑒𝑡 𝑖𝑛 𝑜𝑛𝑒
𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛 𝑤𝑖𝑡ℎ 𝑛𝑒𝑎𝑟𝑠𝑡 𝑛𝑒𝑖𝑔ℎ𝑏𝑜𝑟 𝑖𝑛𝑡𝑒𝑟𝑎𝑐𝑡𝑖𝑜𝑛𝑠