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Second Quantization
Physics — Quantum Mechanics · two-particle-operators-second-quantization
Problem Show that the two-particle operator
F = 1 2 ∑ α ≠ β f ( 2 ) ( x α , x β ) F = \frac{1}{2} \sum_{\alpha \neq \beta} f^{(2)}(\mathbf{x}_\alpha, \mathbf{x}_\beta) F = 2 1 α = β ∑ f ( 2 ) ( x α , x β ) F = 1 2 ∑ i , j , k , m ⟨ i , j ∣ f ( 2 ) ∣ k , m ⟩ a i † a j † a m a k , F = \frac{1}{2} \sum_{i,j,k,m} \langle i,j | f^{(2)} | k,m \rangle\, a_i^\dagger a_j^\dagger a_m a_k, F = 2 1 i , j , k , m ∑ ⟨ i , j ∣ f ( 2 ) ∣ k , m ⟩ a i † a j † a m a k , ⟨ i , j ∣ f ( 2 ) ∣ k , m ⟩ = ∫ d x ∫ d y φ i ∗ ( x ) φ j ∗ ( y ) f ( 2 ) ( x , y ) φ k ( x ) φ m ( y ) . \langle i,j | f^{(2)} | k,m \rangle = \int dx \int dy\, \varphi_i^*(x) \varphi_j^*(y) f^{(2)}(x,y) \varphi_k(x) \varphi_m(y). ⟨ i , j ∣ f ( 2 ) ∣ k , m ⟩ = ∫ d x ∫ d y φ i ∗ ( x ) φ j ∗ ( y ) f ( 2 ) ( x , y ) φ k ( x ) φ m ( y ) . Browse the problem library