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I State Representation in QM and Vector Spaces

  1. Dec 23, 2016 #1
    Hello Forum,

    The state of a quantum system is indicated by##\Psi## in Dirac notation.
    Every observable (position, momentum, energy, angular momentum, spin, etc.) corresponds to a linear operator that acts on ##\Psi##.Every operator has its own set of eigenstates which form an orthonormal basis that can be used to expand the state ##\Psi##. The same state ##\Psi## can be represented also using the eigenstates of any other operator. Some operators have a continuous spectrum and some a discrete spectrum.

    I understand that ##\Psi## lives in the vectors space of square integrable functions and that every linear vector space has an infinite number of possible bases all having the same number of basis vectors. Each bases contains as many linearly independent vectors as the dimension of the linear vector space. Among these bases, there is also the basis formed by the eigenstates of a specific operator.

    Here my question: does the state ##\Psi##, which describes the system, live in many different vector spaces simultaneously and is each operator associated to each different vector space? For instance, the state ##\Psi## could be expanded as a weighted sum of several energy eigenstates which implies that the energy basis is multidimensional and the vector is in that case living in a multidimensional vector space. But if we consider, for instance, the spin operator, there are only two eigenstates so the basis is just a two dimensional basis and the state ##\Psi## becomes the weighted sum of those two eigenstates. The spin basis cannot live in the same vector space as the energy basis or the basis of any other operator...

    How do things work?

    Thanks!!
     
  2. jcsd
  3. Dec 23, 2016 #2

    PeroK

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    The simple answer to your question is that the Hilbert Space of state vectors is dependent on the quantum system under consideration. If you consider the energy of a particle, then the Hilbert Space is infinite dimensional. If you consider the spin of a particle, the Hilbert Space is finite dimensional. And, if you consider both the energy and spin together, you have a more complex Hilbert Space that is a product of the previous two: each vector in the product space represents the combination of energy and spin of the particle.

    Note that although you labelled the thread as Intermediate, I guessed from your question that you might need a fairly basic answer.
     
  4. Dec 23, 2016 #3
    Thanks Perok.

    Let me make sure I understand: the system is in a certain and particular abstract vector state ##\Psi##. We want to measure a specific observable at a time and depending on the particular observable we intend to measure, the linear vector space in which ##\Psi## lives can be finite dimensional or infinite dimensional.
    ##\Psi## becomes a linear superposition of the eigenstates of the operator corresponding to the observable. So, in a sense, does the operator (which corresponds to the observable) determine the type and dimensionality of the linear vector space? If later we want to measure a different observable, the linear vector space becomes different.

    In which cases would we want to consider two observables (like spin and energy) together at the same time? Would the representation of ##\Psi## be a linear superposition of the eigenstates of both operators? Let's say there are two observables, A and B, having basis sets ##a_{ii}## and ##b_{ii}## for the two vector spaces. We can introduce the product space (tensor direct product) and the operator product...Is that what is going on?

    In general, are operators in QM tensor quantities? For finite dimensional spaces, operators become matrices with a finite number of elements but if the space is infinite dimensional, the operator cannot be a matrix, I guess....

    thanks,
     
  5. Dec 24, 2016 #4

    PeroK

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    You probably need a course in linear algebra to straighten out your thoughts on all this. Vectors are vectors and operators are functions that map vectors to vectors. Any vector can be represented in any basis. It seems to be a common misconception that the basis and the vector depend on each other.

    Operators don't become matrices. They can be represented by matrices - again a basic course in linear algebra is needed.

    You can have infinite dimensional matrices, although they don't in general have such nice properties as finite matrices.

    You might want to leave looking at the tensor product of vector spaces until you've got the basics.

    A final point is that it's not so much the observable that determines the Hilbert space but the system and the set of observables you are interested in. If you are only interested in spin, hence spin-related observables, then you can use a simple finite dimensional Hilbert space. The specific Hilbert space depends on the number and type of the particles in your system.
     
  6. Dec 24, 2016 #5

    vanhees71

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    Indeed the observable algebra determines the representation of the Hilbert space. The abstract Hilbert space is in fact almost always the same separable Hilbert space. Mathematically there is only one such Hilbert space (up to equivalence, i.e., up to different representations, e.g., as square integrable complex-valued functions ("wave mechanics") or the square-summable complex sequences ("matrix mechanics").

    The most simple example, usually treated first in QM 1, is the Heisenberg algebra for a particle moving in one-dimensional space. The observable algebra is generated by position and momentum, and everything is defined through the one non-trivial commutator relation
    $$[\hat{x},\hat{p}]=\mathrm{i} \hbar \hat{1}.$$
    You find the derivation of "wave mechanics" from this axiomatics in a recent posting of mine

    https://www.physicsforums.com/threads/delta-function-ordering.897876/#post-5650324
     
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