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Physics
Quantum Physics
Identity Operator for Multiple Particles
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[QUOTE="vanhees71, post: 6038341, member: 260864"] If you have a complete orthonormalized set of vectors ##|u_n \rangle##, then $$\sum_{n=1}^{\infty} |u_n \rangle \langle u_n|=\mathbb{1}.$$ For your case you have single-particle states ##|p_n \rangle## with ##p_n=2 \pi n \hbar/L##, ##n \in \mathbb{Z}## (for the case of particles in a box with periodic boundary conditions; for rigid boundary conditions this makes no sense at all, because you cannot even define a momentum observable to begin with), forming a complete set of one-particle states. If you consider two identical particles the Hilbert space is spanned by the totally symmetrized (antisymmetrized) Kronecker-product states, $$|p_{1j},p_{2k} \rangle=\frac{1}{\sqrt{2}} (|p_{1i} \rangle \otimes |p_{2j} \rangle \pm |p_{2j} \rangle \otimes |p_{1i}).$$ These states are orthonormal, i.e., $$\langle p_{1j},p_{2k}|p_{1j'},p_{2k'} \rangle=\delta_{jj'} \delta_{kk'}.$$ The completeness relation then reads $$\sum_{j,k} |p_{1j},p_{2k} \rangle \langle p_{1j},p_{2k}|=\mathbb{1}.$$ [/QUOTE]
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Quantum Physics
Identity Operator for Multiple Particles
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