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I have a question regarding the slide:
http://theory.physics.helsinki.fi/~kvanttilaskenta/Lecture3.pdf
On page 18-21 it gives the proof of the theorem that | \psi_i^{~} \rangle and |\phi_{i}^{~}\rangle generate the same density matrix iff |\psi_{i}^{~}\rangle = \sum_{j} u_{ij} |\phi_{j}^{~}\rangle assuming that | \psi_i^{~}\rangle is not necessarily normalized.
What if | \psi_i^{~}\rangle is normalized and | \phi_i^{~}\rangle not independent?
Would the necessary condition for which p = | \psi_i \rangle \langle \psi_i |= q = | \phi_j \rangle \langle \phi_j | require that you have |\psi_{i}^{~}\rangle = \sum_{j} u_{ij} |\phi_{j}^{~}\rangle ?
http://theory.physics.helsinki.fi/~kvanttilaskenta/Lecture3.pdf
On page 18-21 it gives the proof of the theorem that | \psi_i^{~} \rangle and |\phi_{i}^{~}\rangle generate the same density matrix iff |\psi_{i}^{~}\rangle = \sum_{j} u_{ij} |\phi_{j}^{~}\rangle assuming that | \psi_i^{~}\rangle is not necessarily normalized.
What if | \psi_i^{~}\rangle is normalized and | \phi_i^{~}\rangle not independent?
Would the necessary condition for which p = | \psi_i \rangle \langle \psi_i |= q = | \phi_j \rangle \langle \phi_j | require that you have |\psi_{i}^{~}\rangle = \sum_{j} u_{ij} |\phi_{j}^{~}\rangle ?
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