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Study notes on "Quantum Information and Quantum Optics with Superconducting Circuits"

Derivation of the Lindblad master equation (2.26)

I found the following paper is useful to understand the general derivation of the Lindblad master equation.

It derives the master equation in two different ways. One is starting from the von Neumann equation for the density matrix (the common way of many physicists?):

 \dot\rho_T(t) = -i\left[H_T,\rho_T(t)\right]

The other is starting from the most general expression of time evolution of composite quantum system (CPT-maps derived from Choi-Kraus representation theorem):

 \displaystyle \mathcal V\rho = \sum_iV^\dagger_i\rho V_i

I like the second one as it's based on the pure formalism and requires only assumptions of the quantum information theory! As I found some typos and unclear calculations in the paper, I created my own errata:

enakai00.hatenablog.com

Gauge invariance of \varphi(\mathbf x,t) defined in (3.16)

As the gauge invariance is not explained explicitly in the book, I got confused why \varphi(\mathbf x,t) defined in (3.16) is called "gauge-invariant phase." I refreshed my memory on the gauge field theory, and summarized a rationale behind this definition:

enakai00.hatenablog.com

Derivation of the gauge-independent relation between the phase and the electric field (3.22)

In the book, before deriving the relationship (3.22), it says "We have to complete our derivation to regard more general conditions!" on p.27. However, the assumptions used in the following derivation are not that clear to me. So I tried to write down all the details of the calculations:

enakai00.hatenablog.com

As a result, I found that the following assumptions are required to get (3.22).

  • The charge density n_s = |\psi(\mathbf x,t)|^2 is constant and uniform (independent of  (\mathbf x,\, t)).
  • The currents \mathbf J(\mathbf x,t) are uniform (independent of \mathbf x).

Derivation of the first and second Josephson relation (3.40) - (3.43)

On p.31, it says "As a boundary condition, we impose a current intensity I flowing though the junction." So initially, I misunderstood that (3.37) can be derived from this boundary condition. After struggling with a lengthy calculation:

enakai00.hatenablog.com

I finally realized that:

  • The coefficients \alpha_{\pm} are decided by the usual(?) boundary condition, the continuity of the (gauge-invariant) wavefunction.
  • The boundary condition of the current intensity imposes a relationship between the global phases \varphi_L and \varphi_R of wavefunctions outside the insulator.

Also I got a slightly different result for \alpha_{\pm} from (3.37) in the book (though it doesn't change the following discussions). My result is:

 \displaystyle\alpha_{\pm} = \sqrt{n_s}\frac{e^{i\varphi_R}\pm e^{i\varphi_L}}{e^{\frac{d}{2\xi}}\pm e^{-\frac{d}{2\xi}}} --- (3.37)

Typo in chapter 4

p.57 The critical value is written as 2I_c\cos(\Phi/2\varphi_0) / \varphi_0. The last /\varphi_0 is not necessary.

p.58 (4.56)

 \displaystyle\mathcal L = \frac{1}{2}(C_{J1} + C_{J2})\cdots

Exercise Solutions (Chapter 4)

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Detailed derivation of the Input-Output theory

I found the discussion in Appendix B.3 is rather complicated. So I reconstructed the detailed delibation.

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Typo in chapter 5

p.65 equation after (5.2)

 \displaystyle\frac{1}{2C}\hat q^2 + \frac{1}{2L}\hat\phi^2 (L is required.)

(5.46) should be:

 \displaystyle W_\rho(\mathbf R) = \frac{1}{(2\pi)^{2N}}\int_{\mathbb R^{2N}}e^{i\mathbf Y^T\Omega\mathbf R}{\rm tr}[\rho D(\mathbf Y)]d^{2N}\mathbf y

p.89 (5.62) should be \displaystyle K(t) = \sum_kg_kg_k^*e^{-i\omega_kt} = \cdots

p.92 The Bose-Einstein distribution should be:

 \displaystyle\frac{1}{e^\frac{\hbar\omega_k}{k_BT}-1} --- (5.72)

p.93 (Exercise 5.8) -> (Exercise 5.9) (Right before the equation (5.74))

p.104 The model Hamiltonian of Exercise 5.8 should be:

 \displaystyle H = \omega\hat a^\dagger\hat a +\Omega_0e^{i\omega_d t}\hat a + \Omega_0^*e^{-i\omega_d t}\hat a^\dagger

Exercise Solutions (Chapter 5)

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Convention of the matrix representation for the two level system.

It's not explicitly defined, but I assume the convention is:

 \displaystyle\mid 1\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix},\, \mid 0\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix},\,\rho=\begin{pmatrix}\rho_{11} & \rho_{10} \\ \rho_{01} & \rho_{00}\end{pmatrix}

where \displaystyle \rho_{ij} = \langle i\mid \rho \mid j \rangle

Typo in chapter 6

p.115 (6.19) should be:

 \displaystyle\rho(t) =\begin{pmatrix}\rho_{11}e^{-t/T_1} & e^{-i\Delta t -t/T^*_2}\rho_{10} \\
\rho_{01}e^{+i\Delta t-t/T^*_2} & \rho_{00}+(1-e^{-t/T_1})\rho_{11} \end{pmatrix}

p.127 (6.45)

The characteristic number of Mathieu function should be -2n_g + 2k(n,n_g). Also the variable r in \mathcal M_A(r,q) is not explicitly explained. In this context, it's the same as the characteristic number -2n_g + 2k(n,n_g).

To normalize \psi_n(\varphi) as \displaystyle\int_0^{2\pi}\psi^*_n(\varphi)\psi_n(\varphi)d\varphi = 1, (6.45) should be:

 \displaystyle \psi_n(\varphi) = \frac{e^{in_g\varphi}}{\sqrt{2\pi}}me_{-2n_g+2k(n,n_g)}\left(\frac{\varphi+\pi}{2};\,\frac{E_J}{2E_C}\right)

p.127 (6.47)

The index function (6.47) doesn't work outside n_g \in (-0.5, 0.5). It also generates the wrong order 0\to 2\to 1\to 4\to 3.... The formula in Koch et al. (2007) works for all n_g \in \mathbb R, but still generates the wrong order 0\to 2\to 1\to 4\to 3.... I believe the right one is:

 \displaystyle k(n, n_g) = \sum_{l=\pm 1}\left\{{\rm round}\left(2n_g+\frac{l}{2}\right) {\rm mod}\,2\right\}

        \displaystyle \times\left[{\rm round}(n_g)+l(-1)^m\left\{(m+1)\,{\rm div}\,2\right\}\right]

 where \displaystyle m = n-(-1)^n{\rm sign}(n)

See this for results.

p.132 Figure 6.6, the x label should be \phi/\Psi_0

p.133 (6.54)

 \displaystyle\psi_{L,R}(\varphi) = \langle\hat\varphi\mid L,R\rangle \propto \exp\left[-\frac{1}{2\sigma^2_\varphi}\left(\varphi-\varphi_{L,R}\right)^2\right]

p.136 (6.59) should be:

 \displaystyle H = \frac{1}{2(1/2+\alpha)C_J}q_+^2 + \frac{1}{2C_j}q_-^2 + V(\phi_-,\phi_+)

p.136 (6.60) should be:

 \displaystyle V(\phi_-,\phi_+) = -\alpha E_J\cos\left(\frac{\Phi+\phi_+}{\varphi_0}\right) - 2E_J\cos\left(\frac{\phi_+}{2\varphi_0}\right)\cos\left(\frac{\phi_-}{2\varphi_0}\right)

p.150 Solution of 6.4 (4) should be:

 \displaystyle  \rho(t) = \begin{pmatrix}
P_1 + \left\{\rho_{11}(0)-P_1\right\}e^{-\frac{t}{T_1}} & \rho_{10}(0)e^{-i\Delta t-\frac{t}{2T_1}} \\
\rho_{01}(0)e^{i\Delta t-\frac{t}{2T_1}} & (1-P_1)+\left\{P_1-\rho_{11}(0)\right\}e^{-\frac{t}{T}}
\end{pmatrix}

p.152 Problem 6.10 (1)

 \displaystyle H_0 = 2\omega\mid 2\rangle\langle 2\mid+\omega\mid 1\rangle\langle 1 \mid

 H_{\rm eff} = (2\delta + \alpha)\mid 2\rangle\langle 2\mid + \delta\mid 1\rangle\langle 1 \mid + \epsilon_0(\mid 2 \rangle\langle 1 \mid + \mid 1\rangle\langle 0 \mid +\, {\rm H.c.})

p.153 Problem 6.12 (3) \Phi = 0.5\Phi_0 + \Phi_B.

p.154 Problem 6.15 \epsilon \propto L_J/L

p.155 Problem 6.16 (3) In the limit |J| \ll |\Delta|

Exercise Solution (Chapter 6)

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Derivation of Input-Output relations for waveguide-QED

The equations in "7.2.2 Input-Output Relations" have several typos. I reconstructed the detailed derivation here.

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Typo in chapter 7

p.166 The definition of \xi(x,t,t_0) (after (7.15)) should be:

 \displaystyle \xi(x,t,t_0)=\sum_kg_ke^{ikx-i\omega_k(t-t_0)}\psi_k(t_0)

p.166 (7.16) should be:

 \displaystyle i\partial_tc_1 = \left(\Delta'-i\frac{1}{2}\gamma\right)c_1 + \xi(0,t,t_0)

p.166 The definition of \Psi_\pm(x,t) (before (7.17)) should be:

 \displaystyle\Psi_\pm(x,t) = \frac{1}{d}\sum_{k\in\mathbb R^+}e^{\pm ikx}\psi_k(t)

p.166 (7.17) should be:

 \displaystyle\Psi_\pm(x,t)=\Psi_\pm(x,t,t_0)-i\sum_{k\in\mathbb R^+}\int_{t_0}^tg^*_k\frac{e^\pm ikx-i\omega_k(t-\tau)}{\sqrt{d}}c_1(\tau)\,d\tau

p.169 (7.24) should be:

 \displaystyle \rho_{\rm qb} = \begin{pmatrix}\rho_{11} & \rho_{10} \\ \rho_{01} & \rho_{00}\end{pmatrix} =
\begin{pmatrix} |\beta|^2e^{-\gamma t} & \alpha\beta^*e^{-i\Delta t-\frac{\gamma}{2}t} \\ \alpha^*\beta e^{i\Delta t-\frac{\gamma}{2}t} & 1 + (|\alpha|^2 -1)e^{-\gamma t}\end{pmatrix}

p.170 (7.25) should be:

 \displaystyle i\partial_tc_1(t) = \sqrt{\frac{\gamma v}{2}}\left[\Psi_+^{\rm in}(0,t)+\Psi_-^{\rm in}(0,t)\right]+\left(\Delta'-i\frac{\gamma}{2}\right)c_1(t)

p.170 (7.26) The pre-factor should be \displaystyle\sqrt{\frac{\gamma v}{2}} instead of \displaystyle\sqrt{\frac{\gamma}{2v}}

p.189 (7.57) should be:

 \displaystyle p_- = {\rm tr}(\epsilon_-(\rho)) = (1-M_n){\rm tr}(\rho_{\rm qubit}\mid g\rangle\langle g\mid) + M_n{\rm tr}(\rho_{\rm qubit}\mid e\rangle\langle e\mid)

 \displaystyle p_+ = {\rm tr}(\epsilon_+(\rho)) = M_n{\rm tr}(\rho_{\rm qubit}\mid g\rangle\langle g\mid) + (1-M_n){\rm tr}(\rho_{\rm qubit}\mid e\rangle\langle e\mid)

p.194 Problem 7.1 Show how (7.14) can be... (Not (7.13))

p.194 Problem 7.2 (7.64) should be:

  \displaystyle \partial_tc_1(t) = -i\Delta c_1(t) -\frac{1}{2\pi}\int_0^t\int J(\omega)e^{-i\omega(t-\tau)}c_1(\tau)\,d\omega\,d\tau

Exercise Solution (Chapter 7)

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Typo in chapter 8

p.212 (8.10) should be:

 \displaystyle\mid\pm\rangle = \frac{1}{\sqrt{2}}\mid 1,1\rangle\pm\frac{1}{\sqrt{2}}\mid 0,2\rangle

p.216 The second equation of (8.13) should be:

 \displaystyle s_\alpha = \frac{1}{d}{\rm tr} \left(S^\dagger_\alpha\rho\right) = \langle S_\alpha\rangle_\rho

p.216 (8.14) should be:

 \displaystyle \epsilon(\rho) = (M\mathbf s)^T\cdot\mathbf S = \frac{1}{d}=\sum_{\alpha,\beta=1}^{d^2}M_{\beta,\alpha}{\rm tr}(S_\alpha^\dagger\rho)S_\beta

p.216 The middle part of (8.17) should be:

 \displaystyle\frac{1}{d^2}\sum_{\alpha,\beta}M_{\alpha\beta}S^T_\beta\otimes S_\alpha

p.220 The first equation of (8.24) should be:

 \displaystyle F_{\rm avg} = p + \frac{1-p}{d}

Note that in this section, the depolarization parameter p stands for the probability of not-decaying. So it's the opposite (i.e 1-p) of the depolarization parameter in Table 8.2.

Exercise Solution (Chapter 8)

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Exercise Solution (Chapter 9)

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Gauge Invariance of superconducting circuits wavefunction model

Effective wavefunction describing the flow of superconducting elections (Cooper pairs):

  \displaystyle\psi(\mathbf x, t)\simeq \sqrt{n_s(\mathbf x,t)}e^{i\theta(\mathbf x,t)}

Schrödinger equation for \psi(\mathbf x, t):

  \displaystyle i\hbar\partial_t\psi(\mathbf x, t) = \left[\frac{1}{2m_s}\left\{-i\hbar\nabla-q_s\mathbf A(\mathbf x, t)\right\}^2 + q_sv(\mathbf x,t)\right]\psi(\mathbf x, t) --- (3.5)

I will show that this model is invariant under the gauge transformation:

   \displaystyle\psi(\mathbf x,t) \, \rightarrow\,  \displaystyle\psi'(\mathbf x,t) = e^{-i\frac{q_s}{\hbar}\chi(\mathbf x,t)}\psi(\mathbf x,t)

  \displaystyle \mathbf A(\mathbf x, t)\,\rightarrow\, \mathbf A'(\mathbf x, t) = \mathbf A(\mathbf x, t)-\nabla\chi(\mathbf x, t)

 \displaystyle v(\mathbf x,t)\, \rightarrow\, v'(\mathbf x, t) = v(\mathbf x,t)+\partial_t\chi(\mathbf x, t)

[Proof]

(3.5) is equivalent to

  \displaystyle \left[i\hbar\partial_t - q_sv(\mathbf x,t) - \frac{1}{2m_s}\left\{-i\hbar\nabla-q_s\mathbf A(\mathbf x,t)\right\}^2\right]
\psi(\mathbf x, t) = 0 --- (1)

A simple calculation shows the following commutation relations.

 \displaystyle \left[i\hbar\partial_t,\ e^{i\frac{q_s}{\hbar}\chi(\mathbf x,t)}\right] =-e^{i\frac{q_s}{\hbar}\chi(\mathbf x,t)}q_s\partial_t\chi(\mathbf x,t)

 \displaystyle \left[-i\hbar\nabla,\ e^{i\frac{q_s}{\hbar}\chi(\mathbf x,t)}\right] =e^{i\frac{q_s}{\hbar}\chi(\mathbf x,t)}q_s\nabla\chi(\mathbf x,t)

Hence we have:

 \displaystyle \left[i\hbar\partial_t-q_sv(\mathbf x,t),\ e^{i\frac{q_s}{\hbar}\chi(\mathbf x,t)}\right] =-e^{i\frac{q_s}{\hbar}\chi(\mathbf x,t)}q_s\partial_t\chi(\mathbf x,t) --- (2)

 \displaystyle \left[-i\hbar\nabla-q_s\mathbf A(\mathbf x, t),\ e^{i\frac{q_s}{\hbar}\chi(\mathbf x,t)}\right] =e^{i\frac{q_s}{\hbar}\chi(\mathbf x,t)}q_s\nabla\chi(\mathbf x,t) --- (3)

By using (2), we have:

  \displaystyle \left\{i\hbar\partial_t - q_sv(\mathbf x,t)\right\}\psi(\mathbf x,t)
= \left\{i\hbar\partial_t - q_sv(\mathbf x,t)\right\} e^{i\frac{q_s}{\hbar}\chi(\mathbf x,t)}\psi'(\mathbf x,t)

 \displaystyle = e^{i\frac{q_s}{\hbar}\chi(\mathbf x,t)} \left\{i\hbar\partial_t - q_sv(\mathbf x,t)\right\} \psi'(\mathbf x,t)-e^{i\frac{q_s}{\hbar}\chi(\mathbf x,t)}q_s(\partial_t\chi(\mathbf x,t)) \psi'(\mathbf x,t)

 \displaystyle = e^{i\frac{q_s}{\hbar}\chi(\mathbf x,t)} \left\{i\hbar\partial_t - q_sv'(\mathbf x,t)\right\} \psi'(\mathbf x,t) --- (4)

Similarly, by using (3), we have:

 \displaystyle \left\{-i\hbar\nabla-q_s\mathbf A(\mathbf x,t)\right\}\psi(\mathbf x, t) =  \left\{-i\hbar\nabla-q_s\mathbf A(\mathbf x,t)\right\}e^{i\frac{q_s}{\hbar}\chi(\mathbf x,t)}\psi'(\mathbf x,t)

 \displaystyle = e^{i\frac{q_s}{\hbar}\chi(\mathbf x,t)} \left\{-i\hbar\nabla-q_s\mathbf A(\mathbf x,t)\right\} \psi'(\mathbf x,t)+e^{i\frac{q_s}{\hbar}\chi(\mathbf x,t)}q_s(\nabla\chi(\mathbf x,t)) \psi'(\mathbf x,t)

 \displaystyle = e^{i\frac{q_s}{\hbar}\chi(\mathbf x,t)} \left\{-i\hbar\nabla-q_s\mathbf A'(\mathbf x,t)\right\} \psi'(\mathbf x,t)

Using (3) again:

 \displaystyle \left\{-i\hbar\nabla-q_s\mathbf A(\mathbf x,t)\right\}^2\psi(\mathbf x, t)

 \displaystyle=\left\{-i\hbar\nabla-q_s\mathbf A(\mathbf x,t)\right\}e^{i\frac{q_s}{\hbar}\chi(\mathbf x,t)}\left[
\left\{-i\hbar\nabla-q_s\mathbf A'(\mathbf x,t)\right\} \psi'(\mathbf x,t)
\right]

 \displaystyle=e^{i\frac{q_s}{\hbar}\chi(\mathbf x,t)}\left\{-i\hbar\nabla-q_s\mathbf A(\mathbf x,t)+q_s\nabla\chi(\mathbf x,t)\right\}\left[
\left\{-i\hbar\nabla-q_s\mathbf A'(\mathbf x,t)\right\} \psi'(\mathbf x,t)
\right]

 \displaystyle=e^{i\frac{q_s}{\hbar}\chi(\mathbf x,t)} \left\{-i\hbar\nabla-q_s\mathbf A'(\mathbf x,t)\right\}^2\psi'(\mathbf x, t) --- (5)

From (4) and (5), (1) is equivalent to:

  \displaystyle \left[i\hbar\partial_t - q_sv'(\mathbf x,t) - \frac{1}{2m_s}\left\{-i\hbar\nabla-q_s\mathbf A'(\mathbf x,t)\right\}^2\right]
\psi'(\mathbf x, t) = 0

Now we can show that the phase \varphi(\mathbf x,t) defined by the following equation is gauge-invariant:

 \displaystyle\theta(\mathbf x,t)-\theta(\mathbf x_0,t)=\varphi(\mathbf x,t)-\varphi(\mathbf x_0,t)+\frac{q_s}{\hbar}
\int_{\mathbf x_0}^{\mathbf x}\mathbf A(\mathbf r, t)\cdot d\mathbf l --- (3.16)

First, in terms of the phase \theta(\mathbf x, t), the gauge transformation of \psi(\mathbf x,t) can be written as:

 \displaystyle \theta(\mathbf x,t)\,\rightarrow\,\theta'(\mathbf x,t) = \theta(\mathbf x,t) - \frac{q_s}{\hbar}\chi(\mathbf x,t)

So, by combining with the gauge transformation of \mathbf A(\mathbf x,t), the combination \displaystyle \nabla \theta(\mathbf x,t)-\frac{q_s}{\hbar}\mathbf A(\mathbf x, t) becomes gauge-invariant. By integrating it through a path from \mathbf x_0 to \mathbf x, the following combination gives a gauge-invariant phase \varphi(\mathbf x,t) up to a constant C.

 \displaystyle \varphi(\mathbf x,t) := \theta(\mathbf x,t)-\theta(\mathbf x_0,t)-\frac{q_s}{\hbar}
\int_{\mathbf x_0}^{\mathbf x}\mathbf A(\mathbf r, t)\cdot d\mathbf l + C

By setting C=\varphi(\mathbf x_0,t), we get the expression (3.16)

(Informal) Errata of "A short introduction to the Lindblad Master Equation"

arxiv.org

p.8 Equation (24)

 \displaystyle\rho_E^{(1)}=\langle 0|_2\rho_E| 0\rangle _2 + \langle 1|_2\rho_E| 1\rangle _2
=\frac{1}{2}(|0\rangle\langle 0|_1+|1\rangle\langle 1|_2)

p.10 Equation (30)

 \displaystyle \rho_B = \pmatrix{
0 & 1 & 0 & 0 \\
1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 
}

p.11 Equation (31)

 \displaystyle (\mathbb{1}\otimes T_2)\rho_B = \frac{1}{2}\Bigg\{\pmatrix{1 & 0 \\ 0 & 0}\otimes\pmatrix{0 & 0 \\ 0 & 1} + \pmatrix{0 & 0 \\ 0 & 1}\otimes\pmatrix{1 & 0 \\ 0 & 0}

        \otimes\pmatrix{0 & 0 \\ 1 & 0}\otimes\pmatrix{0 & 0 \\ 1 & 0} + \pmatrix{0 & 1 \\ 0 & 0}\otimes\pmatrix{0 & 1 \\ 0 & 0}\Bigg\}

p.13 Derivation of (45) from (44)

First, in (44), we change the integral variable from s \rightarrow t-s (without any approximation) to get:

 \displaystyle \dot{\hat\rho}(t) = -\alpha^2\int_0^tds {\rm Tr}_E  \Big[ \hat H_I(t) ,\Big[ \hat H_I(t-s),\hat\rho (t)\otimes\hat\rho_E(0) \Big]\Big]

Then, we assume that the kernel in the integration is dominant where s \sim 0 (that is, \hat H_I(t-s) \sim \hat H_I(t)) and extend the upper limit of the integration to infinity. It results in (45).

p.19 Equation (81)

 \displaystyle\frac{d\rho}{dt}=\lim_{\Delta t\to 0}\frac{1}{\Delta t}(\mathcal V(\Delta t)\rho - \rho)
= \lim_{\Delta t\to 0}\frac{1}{\Delta t}\left(\sum_{i,j=1}^{d^2}c_{i,j}(\Delta t)F_i\rho F^\dagger_j-\rho\right)

   \displaystyle=\lim_{\Delta t\to 0}\frac{1}{\Delta t}\Bigg(\sum_{i,j=1}^{d^2-1}\cdots

p.20 Equation (88)

 \displaystyle \frac{d\rho}{dt}=\sum_{i,j}^{d^2-1}g_{i,j}F_i\rho F^\dagger_j + {G,\rho} - i[H,\rho]+\frac{g_{d^2,d^2}}{d}\rho

p.20 Equation (90)

 \displaystyle {\rm Tr}\left[\frac{d\rho}{dt}\right]={\rm Tr}\left[\sum_{i,j=1}^{d^2-1}g_{i,j}F^\dagger_jF_i\rho+2G_2\rho\right]=0