Let's proceed step by step as requested. --- ## **Step 1: Derive the field equations for \( X^\mu \) and \( \^\mu \).** ### **Action:** \[ S = \int d\xi \left( \frac{1}{2} \dot{X}^\mu \dot{X}_\mu - i \psi^\mu \dot{\psi}_\mu \right) \] #### **For \( X^\mu \):** Vary the action with respect to \( X^\mu \): \[ \delta S = \int d\xi \left( \dot{X}_\mu \delta \dot{X}^\mu \right) \] Integrate by parts: \[ = -\int d\xi \, \ddot{X}_\mu \delta X^\mu \] So, the equation of motion is: \[ \boxed{\ddot{X}^\mu = } \] **Explanation:** This is the equation for a free particle: the second derivative of position with respect to the parameter \( \xi \) vanishes. --- #### **For \( \psi^\mu \):** Vary the action with respect to \( \psi^\mu \): \[ \delta S = \int d\xi \left( -i \delta \psi^\mu \dot{\psi}_\mu \right) \] Integrate by parts (and use the anticommuting property of fermions): \[ = i \int d\xi \, \dot{\psi}^\mu \delta \psi_\mu \] So, the equation of motion is: \[ \boxed{\dot{\psi}^\mu = } \] **Explanation:** This means the Majorana fermion field \( \psi^\mu \) is constant along the worldline parameter \( \xi \). --- ## **Step 2: Show that the action is invariant under the global supersymmetry transformations.** Given transformations: \[ \delta X^\mu = i \epsilon \psi^\mu, \qquad \delta \psi^\mu = \frac{1}{2} \epsilon \dot{X}^\mu \] Calculate the variation of the action under these transformations: ### **Variation of the Lagrangian:** \[ \delta S = \int d\xi \, \delta \mathcal{L} \] where \[ \mathcal{L} = \frac{1}{2} \dot{X}^\mu \dot{X}_\mu - i \psi^\mu \dot{\psi}_\mu \] Calculate each term: #### **First Term:** \[ \delta \left( \frac{1}{2} \dot{X}^\mu \dot{X}_\mu \right) = \dot{X}_\mu \delta \dot{X}^\mu = \dot{X}_\mu \frac{d}{d\xi}(i \epsilon \psi^\mu) = i \epsilon \dot{X}_\mu \dot{\psi}^\mu \] #### **Second Term:** \[ \delta\left( -i \psi^\mu \dot{\psi}_\mu \right) = -i (\delta \psi^\mu \dot{\psi}_\mu + \psi^\mu \delta \dot{\psi}_\mu ) \] \[ = -i \left( \frac{1}{2} \epsilon \dot{X}^\mu \dot{\psi}_\mu + \psi^\mu \frac{d}{d\xi} \left( \frac{1}{2} \epsilon \dot{X}_\mu \right) \right) \] \[ = -i \left( \frac{1}{2} \epsilon \dot{X}^\mu \dot{\psi}_\mu + \frac{1}{2} \epsilon \psi^\mu \ddot{X}_\mu \right) \] ### **Combine terms:** \[ \delta \mathcal{L} = i \epsilon \dot{X}_\mu \dot{\psi}^\mu - i \frac{1}{2} \epsilon \dot{X}^\mu \dot{\psi}_\mu - i \frac{1}{2} \epsilon \psi^\mu \ddot{X}_\mu \] \[ = i \frac{1}{2} \epsilon \dot{X}_\mu \dot{\psi}^\mu - i \frac{1}{2} \epsilon \psi^\mu \ddot{X}_\mu \] The second term can be rewritten (by integrating by parts and ignoring boundary terms): \[ - i \frac{1}{2} \epsilon \psi^\mu \ddot{X}_\mu = i \frac{1}{2} \epsilon \dot{\psi}^\mu \dot{X}_\mu \] So, \[ \delta \mathcal{L} = i \frac{1}{2} \epsilon \dot{X}_\mu \dot{\psi}^\mu + i \frac{1}{2} \epsilon \dot{\psi}^\mu \dot{X}_\mu = i \epsilon \dot{X}_\mu \dot{\psi}^\mu \] But this is a total derivative: \[ i \epsilon \dot{X}_\mu \dot{\psi}^\mu = \frac{d}{d\xi}(i \epsilon \dot{X}_\mu \psi^\mu) \] **Explanation:** The action changes by a total derivative under the supersymmetry transformation, which does not affect the equations of motion. Thus, the action is invariant. --- ## **Final Answer** ### **Step 1:** - The field equations are: \[ \boxed{\ddot{X}^\mu = \qquad \dot{\psi}^\mu = } \] ### **Step 2:** - The action is invariant under the global supersymmetry transformations given, since its variation is a total derivative. --- Let me know if you want the solution to part 3!

Let's proceed with the entire solution, including part 3. --- ## **Step 1: Derive the field equations for \( X^\mu \) and \( \psi^\mu \).** ### **Action:** \[ S = \int d\xi \left( \frac{1}{2} \dot{X}^\mu \dot{X}_\mu - i \psi^\mu \dot{\psi}_\mu \right) \] #### **For \( X^\mu \):** Vary the action with respect to \( X^\mu \): \[ \delta S = \int d\xi \left( \dot{X}_\mu \delta \dot{X}^\mu \right) \] Integrating by parts gives: \[ = -\int d\xi \, \ddot{X}_\mu \delta X^\mu \] Thus, the equation of motion is: \[ \boxed{\ddot{X}^\mu = 0} \] **Explanation:** This indicates that the particle described by \( X^\mu \) moves uniformly, reflecting the free particle motion in Minkowski space. --- #### **For \( \psi^\mu \):** Vary the action with respect to \( \psi^\mu \): \[ \delta S = \int d\xi \left( -i \delta \psi^\mu \dot{\psi}_\mu \right) \] Integrating by parts (and using the anticommuting property of fermions): \[ = i \int d\xi \, \dot{\psi}^\mu \delta \psi_\mu \] Thus, the equation of motion is: \[ \boxed{\dot{\psi}^\mu = 0} \] **Explanation:** This indicates that the Majorana fermion field \( \psi^\mu \) is constant along the worldline parameter \( \xi \). --- ## **Step 2: Show that the action is invariant under the global supersymmetry transformations.** Given transformations: \[ \delta X^\mu = i \epsilon \psi^\mu, \qquad \delta \psi^\mu = \frac{1}{2} \epsilon \dot{X}^\mu \] Calculate the variation of the action under these transformations: ### **Variation of the Lagrangian:** \[ \delta S = \int d\xi \, \delta \mathcal{L} \] where \[ \mathcal{L} = \frac{1}{2} \dot{X}^\mu \dot{X}_\mu - i \psi^\mu \dot{\psi}_\mu \] ### **Calculate each term:** #### **First Term:** \[ \delta \left( \frac{1}{2} \dot{X}^\mu \dot{X}_\mu \right) = \dot{X}_\mu \delta \dot{X}^\mu = \dot{X}_\mu \frac{d}{d\xi}(i \epsilon \psi^\mu) = i \epsilon \dot{X}_\mu \dot{\psi}^\mu \] #### **Second Term:** \[ \delta\left( -i \psi^\mu \dot{\psi}_\mu \right) = -i (\delta \psi^\mu \dot{\psi}_\mu + \psi^\mu \delta \dot{\psi}_\mu ) \] \[ = -i \left( \frac{1}{2} \epsilon \dot{X}^\mu \dot{\psi}_\mu + \psi^\mu \frac{d}{d\xi} \left( \frac{1}{2} \epsilon \dot{X}_\mu \right) \right) \] \[ = -i \left( \frac{1}{2} \epsilon \dot{X}^\mu \dot{\psi}_\mu + \frac{1}{2} \epsilon \psi^\mu \ddot{X}_\mu \right) \] ### **Combine terms:** \[ \delta \mathcal{L} = i \epsilon \dot{X}_\mu \dot{\psi}^\mu - i \frac{1}{2} \epsilon \dot{X}^\mu \dot{\psi}_\mu - i \frac{1}{2} \epsilon \psi^\mu \ddot{X}_\mu \] \[ = i \frac{1}{2} \epsilon \dot{X}_\mu \dot{\psi}^\mu - i \frac{1}{2} \epsilon \psi^\mu \ddot{X}_\mu \] The second term can be rewritten (by integrating by parts and ignoring boundary terms): \[ - i \frac{1}{2} \epsilon \psi^\mu \ddot{X}_\mu = i \frac{1}{2} \epsilon \dot{\psi}^\mu \dot{X}_\mu \] Thus, \[ \delta \mathcal{L} = i \epsilon \dot{X}_\mu \dot{\psi}^\mu \] This is a total derivative: \[ i \epsilon \dot{X}_\mu \dot{\psi}^\mu = \frac{d}{d\xi}(i \epsilon \dot{X}_\mu \psi^\mu) \] **Explanation:** The action changes by a total derivative under the supersymmetry transformation, which does not affect the equations of motion. Thus, the action is invariant. --- ## **Step 3: Show that the commutator \([ \delta_1, \delta_2 ]\) gives a translation by an amount \( d \xi \).** Let \( \delta_1 \) and \( \delta_2 \) be two infinitesimal supersymmetry transformations with parameters \( \epsilon_1 \) and \( \epsilon_2 \): \[ \delta_1 = \epsilon_1 \delta, \quad \delta_2 = \epsilon_2 \delta \] ### **Calculate the commutator:** \[ [\delta_1, \delta_2] = \delta_1 \delta_2 - \delta_2 \delta_1 \] #### **For \( X^\mu \):** Using the transformations: \[ \delta_1 X^\mu = i \epsilon_1 \psi^\mu, \quad \delta_2 X^\mu = i \epsilon_2 \psi^\mu \] Then: \[ \delta_1 \delta_2 X^\mu = i \epsilon_1 \delta_2 \psi^\mu = i \epsilon_1 \left( \frac{1}{2} \epsilon_2 \dot{X}^\mu \right) \] \[ \delta_2 \delta_1 X^\mu = i \epsilon_2 \delta_1 \psi^\mu = i \epsilon_2 \left( \frac{1}{2} \epsilon_1 \dot{X}^\mu \right) \] Thus, \[ [\delta_1, \delta_2] X^\mu = \frac{i}{2}(\epsilon_1 \epsilon_2 - \epsilon_2 \epsilon_1) \dot{X}^\mu \] Since \( \epsilon_1 \) and \( \epsilon_2 \) are Grassmann parameters, \( \epsilon_1 \epsilon_2 = - \epsilon_2 \epsilon_1 \): \[ [\delta_1, \delta_2] X^\mu = \epsilon_1 \epsilon_2 \dot{X}^\mu \] ### **Result:** The commutator results in a transformation that corresponds to a translation in the parameter \( \xi \): \[ [\delta_1, \delta_2] = \xi^\mu \partial_\mu \] This indicates that the action of two consecutive supersymmetry transformations translates the fields in the parameter space. **Explanation:** The commutator of the two supersymmetry transformations leads to a transformation that corresponds to translations in the Minkowski space, reflecting the underlying symmetry of the theory. --- ## **Final Answer** ### **Step 1:** - The field equations are: \[ \boxed{\ddot{X}^\mu = 0 \quad \text{and} \quad \dot{\psi}^\mu = 0} \] ### **Step 2:** - The action is invariant under the global supersymmetry transformations. ### **Step 3:** - The commutator \([ \delta_1, \delta_2 ]\) gives a translation in space-time. --- Let me know if you need further clarifications!