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Many sources (such as The Physics of Quantum Mechanics and the answers to this Physics.SE question) warn against conflating

  • a mixed state where $|\psi\rangle$ is $|n\rangle$ with probability $p_n$, and
  • a pure state with a wavefunction $|\psi\rangle = \sum_n \sqrt{p_n}\;|n\rangle$

But in this lecture, it seems that this very mistake is being made. The lecturer assigns (3m20s) a density operator to a single wavefunction $|\psi\rangle$; and later (24m23s) he considers the possibility that this wavefunction may represent a pure or a mixed state.

Is this interpretation a legitimate one, or is the lecture erroneous, or do I misunderstand the lecture?

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    You might be asking too much if you expect people to watch a whole video lecture ... It might help if you could summarize the relevant part of the lecture in the post. Also, this would be more beneficial for the Q+A format of the site. – Norbert Schuch Dec 13 '15 at 00:10
  • There is no problem assigning a density matrix to a single state $|\psi\rangle$, since in this case $\rho=|\psi\rangle\langle \psi|$. – Meng Cheng Dec 13 '15 at 03:54
  • @r-c What Meng said, with emphasis on "the lecturer assigns a density matrix to a pure state, not a wavefunction to a mixed state". A mixed state cannot be represented as a single wavefunction, unless it reduces to a pure state. – udrv Dec 13 '15 at 07:14
  • @NorbertSchuch I don't expect people to watch the whole lecture; I give two pointers to specific moments of the lecture. –  Dec 13 '15 at 08:35
  • @udrv but that's a (pretty boring) special case, right? I don't think that whole lecture is devoted to defining the density operator of a pure state. Besides, see 24m23s in the lecture. –  Dec 13 '15 at 08:37
  • @r-c I did check both segments, but not the entire lecture. My guess is the lecturer is trying to lay the foundation for a discussion that does involve genuine mixed states down the argument and prefers to get the students used to the idea of working with density matrices from the beginning. The 24m23s segment still refers to a pure state as far as I can see. – udrv Dec 13 '15 at 09:03
  • @RC The two moments are potentially out of context. Looking only at those two moments, I fully second udrv: The second segment still refers to the density operator corresponding to a pure state. – Norbert Schuch Dec 13 '15 at 23:14

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