Density Matrix

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  • #1
Nusc
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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 [tex]| \psi_i^{~} \rangle[/tex] and [tex]|\phi_{i}^{~}\rangle[/tex] generate the same density matrix iff [tex]|\psi_{i}^{~}\rangle = \sum_{j} u_{ij} |\phi_{j}^{~}\rangle[/tex] assuming that [tex]| \psi_i^{~}\rangle [/tex] is not necessarily normalized.

What if [tex]| \psi_i^{~}\rangle [/tex] is normalized and [tex]| \phi_i^{~}\rangle[/tex] not independent?

Would the necessary condition for which [tex] p = | \psi_i \rangle \langle \psi_i |= q = | \phi_j \rangle \langle \phi_j | [/tex] require that you have [tex]|\psi_{i}^{~}\rangle = \sum_{j} u_{ij} |\phi_{j}^{~}\rangle[/tex] ?
 
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  • #2
Nusc
760
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We know for normalized states psi and phi that [tex] p = | \psi_i \rangle \langle \psi_i |= q = | \phi_j \rangle \langle \phi_j | [/tex] iff [tex] \sqrt{p_{i}} | \psi_i \rangle = \sum_j u_{ij} \sqrt{q_j} | \phi_j \rangle [/tex]
 

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