Shaoyun Wang
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Physics · 2024-10-15

Scherieffer Wolff Transformation

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1 The different ways to construct effective medium

1.1 The problem and the direct solution

Consider two mass-in-mass elements, j=1,2j=1,2, interacting through a soft spring of constant hh. Each element has an outer mass MjM_{j} and an inner mass mjm_{j}, and a coupling spring of constant kjk_{j}. Calling the displacement of the outer mass UjU_{j} and uju_{j} that of the inner mass, it comes that the system obeys the motion equation

[k1+h−k1−h0−k1k100−h0k2+h−k200−k2k2][U1u1U2u2]=ω2[M1m1M2m2][U1u1U2u2]\left[\begin{array}{cccc} k_{1}+h & -k_{1} & -h & 0 \\ -k_{1} & k_{1} & 0 & 0 \\ -h & 0 & k_{2}+h & -k_{2} \\ 0 & 0 & -k_{2} & k_{2} \end{array}\right]\left[\begin{array}{l} U_{1} \\ u_{1} \\ U_{2} \\ u_{2} \end{array}\right]=\omega^{2}\left[\begin{array}{cccc} M_{1} & & & \\ & m_{1} & & \\ & & M_{2} & \\ & & & m_{2} \end{array}\right]\left[\begin{array}{l} U_{1} \\ u_{1} \\ U_{2} \\ u_{2} \end{array}\right]

For the unperturbated system, the equations are

[k1−k100−k1k10000k2−k200−k2k2][U1u1U2u2]=ω2[M1m1M2m2][U1u1U2u2]\left[\begin{array}{cccc} k_{1} & -k_{1} & 0 & 0 \\ -k_{1} & k_{1} & 0 & 0 \\ 0 & 0 & k_{2} & -k_{2} \\ 0 & 0 & -k_{2} & k_{2} \end{array}\right]\left[\begin{array}{l} U_{1} \\ u_{1} \\ U_{2} \\ u_{2} \end{array}\right]=\omega^{2}\left[\begin{array}{cccc} M_{1} & & & \\ & m_{1} & & \\ & & M_{2} & \\ & & & m_{2} \end{array}\right]\left[\begin{array}{l} U_{1} \\ u_{1} \\ U_{2} \\ u_{2} \end{array}\right]

The eigenvalues are

0, 0, ω1=M1k1+k1m1M1m1, ω2=M2k2+k2m2M2m20, \ 0, \ \omega_1 = \frac{M_{1} k_{1}+k_{1} m_{1}}{M_{1} m_{1}}, \ \omega_2 = \frac{M_{2} k_{2}+k_{2} m_{2}}{M_{2} m_{2}}

and the corresponding eigenvectors are

e1=[1,1,0,0]Te2=[0,0,1,1]Te3=[k1,−k1M1/m1,0,0]Te4=[0,0,k2,−k2M/m2]Te_1 = [1,1,0,0]^T \\ e_2 = [0,0,1,1]^T \\ e_3 = [k_1,-k_1 M_1/m_1,0,0]^T \\ e_4 = [0,0,k_2,-k_2 M/m_2]^T

1.2 First order degenerate pertubation theory

Assume the solution as the superposition of vector e3e_3 and e4e_4

u=xe1+ye2u = x e_1 + y e_2

Subsituting that into Eq. (1), we have

K(xe1+ye2)=ωM(xe1+ye2)K(xe_1+ye_2) = \omega M (x e_1 + y e_2)

Let product of e1e_1 and e2e_2 gives the equations of the coefficients x,yx,y. The equations are

[k12h+M12k1ω14−k1k2h−k1k2hk22h+M22k2ω24][xy]=ω2[M1k12(1+M1m1)00M2k22(1+M2m2)][xy]\left[\begin{array}{cc} k_{1}^{2} h + M_1^2 k_1 \omega_1^4 & -k_{1} k_{2} h \\ -k_{1} k_{2} h & k_{2}^{2} h + M_2^2 k_2 \omega_2^4 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\omega^{2}\left[\begin{array}{cc} M_{1} k_{1}^{2}\left(1+\frac{M_{1}}{m_{1}}\right) & 0 \\ 0 & M_{2} k_{2}^{2}\left(1+\frac{M_{2}}{m_{2}}\right) \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]

If k1=k2k_1=k_2, m1=m2m_1=m_2, M1=M2M_1=M_2, then ω1=ω2\omega_1=\omega_2. And Eq. (7) can be reduced to

[k12h−k1k2h−k1k2hk22h][xy]=(ω2−ω02)[M1k12(1+M1m1)00M2k22(1+M2m2)][xy]\left[\begin{array}{cc} k_{1}^{2} h & -k_{1} k_{2} h \\ -k_{1} k_{2} h & k_{2}^{2} h \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left(\omega^{2}-\omega_0^2\right)\left[\begin{array}{cc} M_{1} k_{1}^{2}\left(1+\frac{M_{1}}{m_{1}}\right) & 0 \\ 0 & M_{2} k_{2}^{2}\left(1+\frac{M_{2}}{m_{2}}\right) \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]

Since this method is the first order degenerate perturbation theory, so the error exhibit O(h2)\mathcal{O}(h^2).

Fig. 1. Frequency spectrum for k1=k2=10,h=1,m1=m2=M1=M2=1k_1 = k_2 = 10, h=1, m_1 = m_2 = M_1 = M_2 = 1. (a) Black lines are from exact solution Eq. (3), and red dash lines are from first degerate perturbation theory. (b) The error is propotional to h2h^2.

Fig. 2. Frequency spectrum for k1=10,k2=15,h=1,m1=m2=M1=M2=1k_1 =10, k_2 = 15, h=1, m_1 = m_2 = M_1 = M_2 = 1. (a) Black lines are from exact solution Eq. (3), and red dash lines are from first degerate perturbation theory. (b) The error is propotional to h2h^2.

1.3 The exact way

We transform the displacement vector to the new vector in the basis of eigenstates, or

[U1u1U2u2]=[10k1010−k1M1/m10010−k2010−k2M2/m2][v1v2v3v4]or u=Uv.\left[\begin{array}{l} U_{1} \\ u_{1} \\ U_{2} \\ u_{2} \end{array}\right] = \left[\begin{array}{cccc} 1 & 0 & k1 & 0 \\ 1 & 0 & -k_1 M_1/m_1 & 0 \\ 0 & 1 & 0 & -k_2 \\ 0 & 1 & 0 & -k_2 M_2/m_2 \end{array}\right] \left[\begin{array}{l} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \end{array}\right] \\ \text{or} \ u = Uv.

Substituting Eq. (9) into Eq. (1), we have

UTKUv=ω2UTMUvU^T K U v = \omega^2 U^T M U v

or

[h−hhk1−hk2−hh−hk1hk2hk1−hk1M12k13+2m1M1k13m12+k12(h+k1)−hk1k2−hk2hk2−hk1k2M22k23+2m2M2k23m22+k22(h+k2)][v1v2v3v4]=ω2(M1+m10000M2+m20000M1k12(M1+m1)m10000M2k22(M2+m2)m2)[v1v2v3v4]\left[\begin{array}{cccc} h & -h & h k_{1} & -h k_{2} \\ -h & h & -h k_{1} & h k_{2} \\ h k_{1} & -h k_{1} & \frac{M_{1}^{2} k_{1}^{3}+2 m_{1} M_{1} k_{1}^{3}}{m_{1}^{2}}+k_{1}^{2}\left(h+k_{1}\right) & -h k_{1} k_{2} \\ -h k_{2} & h k_{2} & -h k_{1} k_{2} & \frac{M_{2}^{2} k_{2}^{3}+2 m_{2} M_{2} k_{2}^{3}}{m_{2}^{2}}+k_{2}^{2}\left(h+k_{2}\right) \end{array}\right] \left[\begin{array}{l} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \end{array}\right] = \omega^2 \left(\begin{array}{cccc} M_{1}+m_{1} & 0 & 0 & 0 \\ 0 & M_{2}+m_{2} & 0 & 0 \\ 0 & 0 & \frac{M_{1} k_{1}^{2}\left(M_{1}+m_{1}\right)}{m_{1}} & 0 \\ 0 & 0 & 0 & \frac{M_{2} k_{2}^{2}\left(M_{2}+m_{2}\right)}{m_{2}} \end{array}\right) \left[\begin{array}{l} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \end{array}\right]

Define H=K−ω2MH = K-\omega^2 M, and rewirte H=[H11H12H21H22]H=\left[\begin{array}{cc} H_{11} & H_{12} \\ H_{21} & H_{22} \end{array}\right] by 2 by 2 block matrices HijH_{ij}. Then the effective matrix of 2 v3,v4v_3,v_4 is

Heff=H22−H21H11−1H12.H_{\text{eff}} = H_{22} - H_{21}H_{11}^{-1}H_{12}.

The spectrum can be obtained from the vanishing of determinant of Eq. (12), or

∣Heff∣=0.|H_{\text{eff}}|=0.

Fig. 3. Frequency spectrum for (a) k1=k2=10,h=1,m1=m2=M1=M2=1k_1 = k_2 = 10, h=1, m_1 = m_2 = M_1 = M_2 = 1. and (b) k1=10,k2=15,h=1,m1=m2=M1=M2=1k_1 =10, k_2 = 15, h=1, m_1 = m_2 = M_1 = M_2 = 1. Black lines are from Eq. (3) and red dash lines are from Eq. (13).

1.4 Schrieffer-Wolff transform

For simplicity, we set m1=m2=M1=M2=1m_1 = m_2 = M_1 = M_2=1, so Eq. (11) can be transformed into

12[h−hh−h−hh−hhh−hh+4k1−h−hh−hh+4k2][v1v2v3v4]=ω2[v1v2v3v4]\frac{1}{2}\left[\begin{array}{cccc} h & -h & h & -h \\ -h & h & -h & h \\ h & -h & h+4 k_{1} & -h \\ -h & h & -h & h+4 k_{2} \end{array}\right] \left[\begin{array}{l} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \end{array}\right] = \omega^2 \left[\begin{array}{l} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \end{array}\right]

We separate Eq. (14) into the leading term and perturbative term, so we have

12[h−h00−hh0000h+4k1−h00−hh+4k2][v1v2v3v4]+[00h−h00−hhh−h00−hh00][v1v2v3v4]=ω2[v1v2v3v4]orKv=ω2v, where K=K0+K1.\frac{1}{2}\left[\begin{array}{cccc} h & -h & 0 & 0 \\ -h & h & 0 & 0 \\ 0 & 0 & h+4 k_{1} & -h \\ 0 & 0 & -h & h+4 k_{2} \end{array}\right] \left[\begin{array}{l} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \end{array}\right] + \left[\begin{array}{cccc} 0 & 0 & h & -h \\ 0 & 0 & -h & h \\ h & -h & 0 & 0 \\ -h & h & 0 & 0 \end{array}\right] \left[\begin{array}{l} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \end{array}\right] = \omega^2 \left[\begin{array}{l} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \end{array}\right] \\ \text{or} \quad K v = \omega^2 v, \ \text{where} \ K = K_0+K_1.

Now we want to do the Schrieffer-Wolff transform to separate the low frequency subspace to the high frequency subspace. We define the unitary transform v=Uwv=\mathcal{U}w where U=eS\mathcal{U}=e^{S} and SS is an anti-Hermitian matrix. After the unitary transform the stiffness matrix is transformed into

K′=eSKe−SK' = e^{S} K e^{-S}

By using the Baker-Campbell-Haussdorf formula , we have

K′=K+[S,K]+12[S,[S,K]]+…K^{\prime}=K+[S, K]+\frac{1}{2}[S,[S, K]]+\ldots

or

K′=K0+K1+[S,K0]+[S,K1]−12[S,[S,K0]]−12[S,[S,K1]]+…K^{\prime}=K_{0}+K_1+\left[S, K_{0}\right]+ [S, K_1]-\frac{1}{2}\left[S,\left[S, K_{0}\right]\right]-\frac{1}{2}[S,[S, K_1]]+\ldots

The Hamiltonian can be made diagonal to first order in K1K_1 by choosing the generator SS such that

[K0,S]=K1\left[K_{0}, S\right]= K_1

This equation always has a definite solution under the assumption that hK1h K_1 is off-diagonal in the eigenbasis of K0K_{0}. Substituting this choice in the previous transformation yields:

K′=K0+12[S,K1]+O(h3)K^{\prime}=K_{0}+\frac{1}{2}[S, K_1]+\mathcal{O}\left(h^{3}\right)

where K′K' is block diagonalized so we finish the separation of low frequency subspace and high frequency subspace. The error after SW transform are O(h3)O\left(h^{3}\right) while it is O(h2)O\left(h^{2}\right) for first degenerate perturbation theory. Actually, the SW transform is the second degenerate pertubation theory which is not commonly used.

To get SS, let’s diagonalize K0K_0 by producting a unitary transform matrix U1\mathcal{U}_1

[12−120012120000E11E1200E21E22]\left[\begin{array}{cccc} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0 & 0 \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 & 0 \\ 0 & 0 & E_{11} & E_{12} \\ 0 & 0 & E_{21} & E_{22} \end{array}\right]

where K1Ei=diEiK_1 E_i = d_i E_i. And K0K_0 is diagonalized as

K0′=[00000h0000d10000d2]K_0' = \left[\begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & h & 0 & 0 \\ 0 & 0 & d_1 & 0 \\ 0 & 0 & 0 & d_2 \end{array}\right]

where

d1=h2+k1+k2−h2+4k12−8k1k2+4k222d2=h2+k1+k2+h2+4k12−8k1k2+4k222.d_1 = \frac{h}{2}+k_{1}+k_{2}-\frac{\sqrt{h^{2}+4 k_{1}^{2}-8 k_{1} k_{2}+4 k_{2}^{2}}}{2}\\ d_2 = \frac{h}{2}+k_{1}+k_{2}+\frac{\sqrt{h^{2}+4 k_{1}^{2}-8 k_{1} k_{2}+4 k_{2}^{2}}}{2}.

And K1K_1 are transformed into

K1′=U1TK1U1K_1' = \mathcal{U}_1^T K_1 \mathcal{U}_1

and it is

[00K11′K12′00K21′K22′K11′K12′00K21′K22′00]\left[\begin{array}{cccc} 0 & 0 & K_{11}' & K_{12}' \\ 0 & 0 & K_{21}' & K_{22}' \\ K_{11}' & K_{12}' & 0 & 0 \\ K_{21}' & K_{22}' & 0 & 0 \end{array}\right]

And the solution of SS is

Sij={Kij′dj−dii≠j0i=jS_{i j}= \left\{\begin{array}{cc} \frac{K_{i j}^{\prime}}{d_{j}-d_{i}} & i \neq j \\ 0 & i=j \end{array}\right.

and then from Eq. (20), we have

Kd′=K0′+12[S,K1′]+O(h3)K'_d =K_{0}'+\frac{1}{2}[S, K_1']+\mathcal{O}\left(h^{3}\right)

Rotating back give the stiffness matrix whose low frequency subspace and high frequency subspace are separated

K′=U1Kd′U1TK' = \mathcal{U}_1 K_d' \mathcal{U}_1^T

Fig. 4. Frequency spectrum for k1=k2=10,h=1,m1=m2=M1=M2=1k_1 = k_2 = 10, h=1, m_1 = m_2 = M_1 = M_2 = 1. (a) Black lines are from exact solution Eq. (3), and red dash lines are from SW transformation. (b) The error is propotional to h3h^3.

Fig. 5. Frequency spectrum for k1=10,k2=15,h=1,m1=m2=M1=M2=1k_1 =10, k_2 = 15, h=1, m_1 = m_2 = M_1 = M_2 = 1. (a) Black lines are from exact solution Eq. (3), and red dash lines are from SW transformation. (b) The error is propotional to h3h^3.