Let's walk through the step-by-step inversion of the Bogoliubov transformation given in the image. ### **Given transformations (Eq. 22.29):** \[ \begin{align*} \alpha_k^\dagger &= u_k c_k^\dagger - v_k c_{-k} \\ \alpha_{-k}^\dagger &= u_k c_{-k}^\dagger + v_k c_k \\ \alpha_k &= u_k c_k - v_k c_{-k}^\dagger \\ \alpha_{-k} &= u_k c_{-k} + v_k c_k^\dagger \end{align*} \] We are to **invert** these equations to write \( c_k \) and \( c_{-k} \) in terms of \( \alpha \) and \( \alpha^\dagger \): \[ \boxed{ c_k = u_k \alpha_k + v_k \alpha_{-k}^\dagger \\ c_{-k} = u_k \alpha_{-k} - v_k \alpha_k^\dagger } \] --- ### **Step 1: Write the equations for \( \alpha_k \) and \( \alpha_{-k}^\dagger \)** From above: \[ \alpha_k = u_k c_k - v_k c_{-k}^\dagger \tag{1} \] \[ \alpha_{-k}^\dagger = u_k c_{-k}^\dagger + v_k c_k \tag{2} \] --- ### **Step 2: Solve for \( c_k \) and \( c_{-k}^\dagger \)** Let's treat (1) and (2) as two equations for \( c_k \) and \( c_{-k}^\dagger \): Write them as: \[ \begin{align*} \alpha_k &= u_k c_k - v_k c_{-k}^\dagger \\ \alpha_{-k}^\dagger &= v_k c_k + u_k c_{-k}^\dagger \end{align*} \] --- ### **Step 3: Express in matrix form** \[ \begin{pmatrix} \alpha_k \\ \alpha_{-k}^\dagger \end{pmatrix} = \begin{pmatrix} u_k & -v_k \\ v_k & u_k \end{pmatrix} \begin{pmatrix} c_k \\ c_{-k}^\dagger \end{pmatrix} \] --- ### **Step 4: Invert the matrix** The inverse of \( \begin{pmatrix} u_k & -v_k \\ v_k & u_k \end{pmatrix} \) is (since \( u_k^2 + v_k^2 = 1 \) for Bogoliubov coefficients): \[ \frac{1}{u_k^2 + v_k^2} \begin{pmatrix} u_k & v_k \\ - v_k & u_k \end{pmatrix} = \begin{pmatrix} u_k & v_k \\ - v_k & u_k \end{pmatrix} \] if \( u_k^2 + v_k^2 = 1 \). So, \[ \begin{pmatrix} c_k \\ c_{-k}^\dagger \end{pmatrix} = \begin{pmatrix} u_k & v_k \\ - v_k & u_k \end{pmatrix} \begin{pmatrix} \alpha_k \\ \alpha_{-k}^\dagger \end{pmatrix} \] So, \[ \boxed{ c_k = u_k \alpha_k + v_k \alpha_{-k}^\dagger } \] \[ c_{-k}^\dagger = -v_k \alpha_k + u_k \alpha_{-k}^\dagger \] --- ### **Step 5: Now for \( c_{-k} \) and \( c_k^\dagger \)** From the other pair: \[ \alpha_{-k} = u_k c_{-k} + v_k c_k^\dagger \] \[ \alpha_k^\dagger = u_k c_k^\dagger - v_k c_{-k} \] Similarly, write \[ \begin{pmatrix} \alpha_{-k} \\ \alpha_k^\dagger \end{pmatrix} = \begin{pmatrix} u_k & v_k \\ - v_k & u_k \end{pmatrix} \begin{pmatrix} c_{-k} \\ c_k^\dagger \end{pmatrix} \] Invert: \[ \begin{pmatrix} c_{-k} \\ c_k^\dagger \end{pmatrix} = \begin{pmatrix} u_k & -v_k \\ v_k & u_k \end{pmatrix} \begin{pmatrix} \alpha_{-k} \\ \alpha_k^\dagger \end{pmatrix} \] So, \[ \boxed{ c_{-k} = u_k \alpha_{-k} - v_k \alpha_k^\dagger } \] \[ c_k^\dagger = v_k \alpha_{-k} + u_k \alpha_k^\dagger \] --- ## **Final Answers** \[ \boxed{ \begin{aligned} c_k &= u_k \alpha_k + v_k \alpha_{-k}^\dagger \\ c_{-k} &= u_k \alpha_{-k} - v_k \alpha_k^\dagger \end{aligned} } \] These are the inverted relations for the bare fermion operators in terms of the quasiparticle operators.