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B Question about states

  1. Apr 2, 2016 #1
    Can a state only be formulated with respect to an observable in consideration? That is to say, does the formulation of the state depend on the particular observable in consideration?

    Thanks.
     
  2. jcsd
  3. Apr 2, 2016 #2
    There's vague terms in your question. Maybe this sounds glib but,

    What is a "formulation" of a state?

    Anyways, a quantum state represents our best (to date) mathematical description of a real life system. It turns out that our best mathematical description of a system in our universe (maybe that's what you meant by "formulation") is one where measurements are probabilities of observables taking on some values, except the catch is that time evolution etc... happens linearly on the complex amplitudes instead of the probabilities (Aww... nature, why do you have to play such a mean trick on us and make novice students confused). I don't know what it would mean to describe a system without reference to observables (without reference to things you can observe).

    More details on what I think you are asking:

    Sometimes, some quantum systems have more states than there are observable values. But think what it means experimentally! It means no matter what you do to observe the state, there are parts of your hilbert space you can't see @ all... which means those extra states that can't be "observed" are redundant, and you can simplify your hilbert space to a smaller one. In other words, those states you put in your hilbert space that don't have distinguishing observable values, you can throw them out and you don't lose any physics... so you should throw them out! So don't complicate your Hilbert space with extra states that are indistinguishable as far as observables are concerned, until your friend in the lab finds a new observable (aha, observable means "real life" things) he can measure.

    CSCO:

    I think what you are asking touches upon the question, how do we know a set of observables completely characterize a system? Look this up, it's prettycool
    https://en.wikipedia.org/wiki/Complete_set_of_commuting_observable [Broken]
     
    Last edited by a moderator: May 7, 2017
  4. Apr 2, 2016 #3

    Nugatory

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    A state is described by a vector in Hilbert space, and like any vector it can be written as the sum of other vectors we call the "basis". There are an infinite number of possible bases, and hence an infinite number of different ways to write any given state.

    However, we usually choose a basis that makes whatever problem we're working on easy to solve, and that's usually the basis made up of the eigenvectors of whatever observable we're concerned with.
     
  5. Apr 2, 2016 #4

    vanhees71

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    There's one caveat. Strictly speaking a pure state is characterized by a normalized vector in Hilbert space modulo a phase, a socalled ray. That becomes important later when it comes, e.g., to the question of which spin quantum numbers are allowed, and since a state is determined by only a ray not by a Hilbert space vector, it's allowed that the rotation around 360 degrees multiplies the state with a phase. As the analysis of the angular-momentum commutation relations show there are indeed half-integer spins allowed by the algebra. Now a angular-momentum component in direction of an axis defines a rotation around this axis, and for a half-integer spin the rotation around 360 degrees multiplies any (allowed) state vector with a factor of -1. But that doesn't matter, because this state vector is in the ray defined by the original state vector and thus describes the same state. So for the very important fact that there are spin-1/2 particles making up the matter around us (and we are made ourselves) it is important to keep in mind that pure states are represented by rays in Hilbert space.

    These rays are a bit unconvenient to work with. But as was explained in the recent threads here in the forum you can describe all states, including pure ones as statistical operators, i.e., self-adjoint positive semidefinite operators of trace 1. The pure states are then precisely those that can be written as
    $$\hat{\rho}=|\psi \rangle \langle \psi|$$
    with ##|\psi \rangle## being a vector normalized to 1. Note that any representant of a ray leads to the same statistical operator, because if
    $$|\psi' \rangle=\exp(\mathrm{i} \varphi) |\psi \rangle, \quad \varphi \in \mathbb{R}$$
    then
    $$\langle \psi'|=\exp(-\mathrm{i} \varphi) \langle \psi|$$
    and thus
    $$\hat{\rho}'=|\psi' \rangle \langle \psi'| = |\psi \rangle \langle \psi|=\hat{\rho},$$
    i.e., the freedom to choose an arbitrary phase to represent the same pure state doesn't matter if you use the statistical operator for the pure state.
     
  6. Apr 2, 2016 #5
    Can I ask a really dumb question here? (It is a B-level thread.) Evidently I am confused about the definition of a "ray" - I always thought a ray started at the origin and extended out to infinity. If you multiply a vector aligned with this ray by -1, wouldn't it then extend in the opposite direction and therefore not be included in the original ray? Thanks for any clarification.
     
  7. Apr 2, 2016 #6

    vanhees71

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    You can as well use the usual definition of projective geometry, i.e., a ray is an equivalence class of vectors
    $$\{\psi \}=\{\alpha |\psi \rangle|\alpha \in \mathbb{C} \}.$$
    Then you use unnormalized vectors as representants of the state, and you have to normalize the vectors later, i.e., when you want to calculate probabilities.
     
  8. Apr 2, 2016 #7
    OK, I think I see, thanks.
     
  9. Apr 2, 2016 #8

    Nugatory

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    That definition works acceptably well (and is intuitive, which is why we use it when we can) for a vector in finite-dimensional space with all of its components real numbers. It doesn't make the same intuitive sense when you're working with complex components (if multiplying by -1 means "opposite direction", why would multiplying by ##i## not mean rotating by 90 degrees given that ##i^2=-1##? But we know that that multiplication doesn't change the ray).

    Thus we use the more formal definition provided by vanhees.
     
  10. Apr 2, 2016 #9
    Yes, I should have realized that complex-valued components would make all the difference. Well, I am less mystified now.
     
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