Let's solve each exercise step by step. --- ## **Exercise 1** ### a) Write the operator \( c_p^\dagger c_q \) in normal form. **Normal ordering** means all **creation operators** (\( c^\dagger \)) to the **left** of all **annihilation operators** (\( c \)). - \( c_p^\dagger c_q \) is already in normal order. **Final Answer:** \[ c_p^\dagger c_q \text{ (already in normal form)} \] --- ### b) Write the operator \( c_p^\dagger c_q^\dagger c_r c_s \) in normal form. - Arrange creation operators (\( c^\dagger \)) to the left, annihilation operators (\( c \)) to the right. - \( c_p^\dagger c_q^\dagger c_r c_s \) is already in normal form (all creations to the left). **Final Answer:** \[ c_p^\dagger c_q^\dagger c_r c_s \text{ (already in normal form)} \] --- ## **Exercise 2** ### a) Write the Hamiltonian \( H \) in normal form and extract the Hartree-Fock potential. The Hamiltonian is: \[ H = \sum_{\alpha \beta} \langle \alpha |T| \beta \rangle c^\dagger_\alpha c_\beta + \frac{1}{4} \sum_{\alpha\beta\gamma\delta} \langle \alpha\beta|V|\gamma\delta \rangle c^\dagger_\alpha c^\dagger_\beta c_\delta c_\gamma \] Both terms are already in normal order. **To extract the Hartree-Fock potential:** - The Hartree-Fock mean-field potential arises from the two-body term by contracting the operators (using Wick's theorem). - Contracting one pair of creation with one pair of annihilation operators gives single-particle terms (the Fock operator). The normal-ordered form: \[ H = \sum_{\alpha\beta} \langle \alpha |h^{\text{HF}}| \beta \rangle c_\alpha^\dagger c_\beta + \frac{1}{4} \sum_{\alpha\beta\gamma\delta} \langle \alpha\beta|V|\gamma\delta \rangle :c_\alpha^\dagger c_\beta^\dagger c_\delta c_\gamma: + E_0 \] where \( h^{\text{HF}} \) is the Hartree-Fock single-particle Hamiltonian, and \( E_0 \) is the reference energy. **Final Answer:** \[ H = \sum_{\alpha\beta} \langle \alpha |T| \beta \rangle c^\dagger_\alpha c_\beta + \frac{1}{4} \sum_{\alpha\beta\gamma\delta} \langle \alpha\beta|V|\gamma\delta \rangle c^\dagger_\alpha c^\dagger_\beta c_\delta c_\gamma \] The Hartree-Fock potential is included in the single-particle term when you contract one creation and one annihilation operator in the two-body term. --- ### b) Evaluate \( \langle 0|H|0 \rangle \). - In the vacuum state, all operators annihilate the vacuum except for contractions that reduce to a number. - Only the fully contracted terms contribute. \[ \langle 0 | c_\alpha^\dagger c_\beta | 0 \rangle = 0 \] \[ \langle 0 | c_\alpha^\dagger c_\beta^\dagger c_\delta c_\gamma | 0 \rangle = 0 \] So, **Final Answer:** \[ \langle 0|H|0\rangle = 0 \] --- ## **Exercise 3** Write the operator \( c_p^\dagger c_q^\dagger c_r c_s :: c_i^\dagger c_j^\dagger :: \) in normal form. - The double colons \( :: \) indicate the normal ordering of \( c_i^\dagger c_j^\dagger \), which is \( c_i^\dagger c_j^\dagger \). - The full operator is a product of normal-ordered operators, which means all creation operators to the left. So: \[ c_p^\dagger c_q^\dagger c_r c_s c_i^\dagger c_j^\dagger \] Now, commute \( c_i^\dagger, c_j^\dagger \) to the left past \( c_r, c_s \): \[ = c_p^\dagger c_q^\dagger c_i^\dagger c_j^\dagger c_r c_s \] (Each time you commute past a fermion operator, you pick up a minus sign. Commuting \( c_i^\dagger \) past \( c_r \) gives \( - c_r c_i^\dagger \), etc., but since all are creation operators, their ordering doesn't matter except for sign.) The fully normal-ordered form is: \[ c_p^\dagger c_q^\dagger c_i^\dagger c_j^\dagger c_r c_s \] But, to maintain proper antisymmetrization, note that swapping creation operators introduces minus signs if any indices are equal. **Final Answer:** \[ c_p^\dagger c_q^\dagger c_i^\dagger c_j^\dagger c_r c_s \] This is the normal-ordered form. --- ## **Summary of Answers** ### **Exercise 1** a) \( c_p^\dagger c_q \) (already normal) b) \( c_p^\dagger c_q^\dagger c_r c_s \) (already normal) ### **Exercise 2** a) Hamiltonian is already in normal form; Hartree-Fock potential comes from contracting the two-body term. b) \( \langle 0|H|0\rangle = 0 \) ### **Exercise 3** Normal form: \( c_p^\dagger c_q^\dagger c_i^\dagger c_j^\dagger c_r c_s \)