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B Superposition at absolute zero

  1. Jun 27, 2016 #1
    I googled a bit about this and managed to find this


    Since I cant make much out of them, maybe except this interesting phrase in the first one

    of all it is intriguing that even at
    absolute zero a coherent superposition of states can get destroyed by zero point fluctuations

    But this, I dont understand what they mean by this:

    "At absolute zero the quantum system
    can only lose energy to the
    cold environment"

    And another question : From what I know, the more atoms you try to put in superposition, you need to get them closer to absolute zero, so can one deduce that theoretically at absolute zero you could have the whole universe in a superposition state ?
  2. jcsd
  3. Jun 28, 2016 #2


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    For a closed system, if you start in a pure state, you'll always stay in a pure state. In the Schrödinger picture, if ##|\psi_0 \rangle## is the vector, representing a pure state, at later times the vector is (I use natural units with ##\hbar=k_{\text{B}}=c=1##)
    $$|\psi(t) \rangle=\exp(-\mathrm{i} \hat{H} t) |\psi_0 \rangle.$$
    This has nothing to do with temperature, which makes only sense if you have a system in (maybe local) thermal equilibrium. In the canonical ensemble this state is described by the statistical operator
    $$\hat{\rho}_{\text{can}}=\frac{1}{Z} \exp \left (-\frac{\hat{H}}{T} \right ), \quad Z=\mathrm{Tr}\exp \left (-\frac{\hat{H}}{T} \right ) .$$
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