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Degenerate spectra, measurement and lack-of-knowledge density matrix interpretation

  1. Sep 12, 2011 #1
    Firstly hello, this is the first time I have posted here (although I have used the site to find info in the past). My query is best illustrated, I think, with an example. Suppose we have some physical system with corresponding state vector
    \left| \psi \right> = a \left| 0 \right> + b \left| 1 \right> + c \left| 2 \right> + d \left| 3 \right> \in \mathbb C^4
    and some physical quantity represented by the operator
    \hat E = E_0 \left| 0 \right> \! \left< 0 \right| + E_1 \left| 1 \right> \! \left< 1 \right|
    + E_2 \left| 2 \right> \! \left< 2 \right| + E_3 \left| 3 \right> \! \left< 3 \right|.

    First suppose that we have that [itex] E_0 = E_1 = E [/itex]. Then the 2-dimensional subspace spanned by [itex] \left| 0 \right>[/itex] and [itex] \left| 1 \right> [/itex] is an eigenspace of [itex]\hat E [/itex] and any measurement with outcome [itex] E [/itex] will leave us with the projection (up to normalization) of [itex] \left| \psi \right> [/itex] onto this subspace. Expressed as a density matrix, the final state is
    \rho = \mathcal N \; \big(
    \left | a \right|^2 \left| 0 \right> \! \left< 0 \right| +
    \left | b \right|^2 \left| 1 \right> \! \left< 1 \right| +
    ab^* \left| 0 \right> \! \left< 1 \right| +
    a^*b \left| 1 \right> \! \left< 0 \right|.

    Now consider the following: we have some experiment that is not accurate enough to distinguish [itex] E_0 [/itex] and [itex] E_1 = E_0 + \epsilon[/itex] but that can distinguish all others (for instance [itex] E_0 [/itex] and [itex] E_1 [/itex] may correspond to very close spectral lines compared to [itex] E_3 [/itex] and [itex] E_4 [/itex]). We perform a measurement, the outcome of which is [itex] E_0 \pm 10\epsilon[/itex]. Then we could argue that the state must have collapsed to either [itex] \left| 0 \right> [/itex] or [itex] \left| 1 \right> [/itex] with probabilities [itex] \left| a \right|^2 [/itex] and [itex] \left| b \right|^2 [/itex]respectively. According to the lack-of-knowledge interpretation of density matrices, the corresponding state (as a density matrix) after measurement is
    \rho' = \mathcal N '\; \big(
    \left | a \right|^2 \left| 0 \right> \! \left< 0 \right| +
    \left | b \right|^2 \left| 1 \right> \! \left< 1 \right|
    where [itex] \mathcal N [/itex] is a normalizing factor.

    The point of this is that (provided my reasoning holds) [itex] \rho [/itex] and [itex] \rho' [/itex] are physically distinct states. But when do we distinguish between the two scenarios? For instance if we have 2 degenerate levels that we know can be split with a magnetic field, do we always have to assume the presence of a magnetic field too weak to measure, or do we assume there is no magnetic field at all? Do we have to distinguish between `true' degeneracy and degeneracy relating to experimental inaccuracy?

    I had a look in various literature and over previous posts in this forum and haven't been able to find an answer to this; I apologize if my search was not sufficiently thorough or if I am missing something obvious.
    Last edited: Sep 12, 2011
  2. jcsd
  3. Sep 12, 2011 #2

    Ken G

    User Avatar
    Gold Member

    Re: Degenerate spectra, measurement and lack-of-knowledge density matrix interpretati

    I would say that the issue here is just what is measurement uncertainty, and why is it present-- is it because you have bad eyesight and cannot read a pointer very accurately, or is it because the experimental pointer you are using is intrinsically imperfect? If the former, then you shouldn't expect to end up with a precise description of the system, your description itself should be inaccurate. If the latter, then you should not expect the system to end up in either one eigenstate or another of the operator you have in mind, because it's not exactly the right operator for that experiment. In effect, you don't even know the true eigenstates of your own measurement if it is not perfect. This latter case is like expecting there to be a tiny magnetic field that you cannot measure precisely enough.

    For either reason above, you end up with an imprecise description of the state of your system, so the two options you give above are not actually distinguishable-- they both fall within the measurement uncertainty.
  4. Sep 12, 2011 #3


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    Science Advisor

    Re: Degenerate spectra, measurement and lack-of-knowledge density matrix interpretati

    woodyhouse, This is exactly the two-slit experiment in disguise. In your first case E0 = E1, the state is described by a wavefunction, i.e. a coherent superposition |ψ> = a|0> + b|1>. You're welcome to square this and write down a density matrix if you want, but it's clear from the one you did write that it's of the form |ψ><ψ|.

    In your second case if E0 and E1 are distinct but there is nothing in your experiment to distinguish them, then a wavefunction description still applies.

    The incoherent superposition only applies if E0 and E1 are distinct and your experiment tells them apart. Not just can tell them apart, but actually does so.
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