0
$\begingroup$

I am learing about quantum mechanics, and I am quite unclear about the the role of measurement and the density operator.

Let $\mathcal H$ be a Hilbert space, $\psi(t) \in \mathcal H$, $O$ be a bounded hermitian operator (“observable”) with discrete eigenvalues $o_n$.

Immediately after a measurement at time $t=t_0$ yielding $o_i$, the state will be projected into the Eigenspace of the operator for that corresponding eigenvalue: $$\left\{ \phi: \|\phi\| = 1 \:\wedge\: O\phi = o_i \phi \right\} = \{\phi_{i\lambda}: \lambda \in \{0..d_i\}\}$$ …if $o_i$ is $d_i$-degenerate.

  1. Is that a correct formulation of “being projected onto the eigenspace”?

Thus, our $\psi' = \sum\limits_{\lambda=0}^{d_i}\langle\psi(t_0),\, \phi_{i\lambda} \rangle$

  1. Is it correct, that for non-degenerate eigenvalues, the resulting state must be pure, since we project only against $\phi_{i0}$ being an eigenstate itself and not a superposition of multiple eigenstates?

  2. What the heck is the density matrix? I often read about it and heard definitions, but what deeply confuses me is that in certain literature, it is not mentioned at all (see an example). What role does it have? Why do we need it? And why don't we, appearently?

This kind of question probably has been asked before, but I hope I could explain specifically what my knowledge base and confusion is. Other questions would mostly assume I understood the density matrix, and others still don't explain why/how quantum mechanics can be done without it, apparently.

$\endgroup$
5
  • $\begingroup$ Density matrix and pure and mixed states en.m.wikipedia.org/wiki/Density_matrix .The off diagonal elements of the density matrix represent the ability of the system to exhibit interference. $\endgroup$
    – user108787
    Jul 24 '16 at 18:08
  • $\begingroup$ I would suggest asking the questions about the density matrix separately. $\endgroup$ Jul 24 '16 at 21:29
  • 1
    $\begingroup$ 1. Formulation correct, but $\psi' = \sum_{\lambda=0}^{d_i} \langle \phi_{i\lambda}, \psi(t_0) \rangle |\phi_{i\lambda}\rangle$. 2. Yes. 3. First courses in QM are usually done without mentioning the density matrix simply to avoid complications and confusions. As soon as you progress to, say, statistical physics, condensed matter, etc, the density matrix becomes a central concept. $\endgroup$
    – udrv
    Jul 25 '16 at 3:41
  • $\begingroup$ For a quick intro to its nature you may be interested in this answer of mine, physics.stackexchange.com/questions/204100/…. It was prompted by a different question, but I think it touches on the basics you are looking for. $\endgroup$
    – udrv
    Jul 25 '16 at 3:42
  • $\begingroup$ Density matrix is essentially a momentum map, it's not a description of the 'state' in the Hilbert state space. Using density matrix can help to explore the geometrical structure of QM. $\endgroup$
    – XXDD
    Jul 26 '16 at 3:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.