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Why can't QM just be built on a model of statespace over a background timespacespace

  1. Jul 8, 2004 #1
    I just read a few articles introducing the basic concept of QM.

    I found QM is built as a model of a statespace over a background of 3-dim space.

    Observables are just Hermitian operators on this statespace.

    Position is an Hermitian operator on this statespace as well. The classical position is basically the expectational value of this operator.

    Momentum is a Hermitian operator as the generator of space translation.

    If we just make the background as a timespace manifold, Hamiltonian can be the generator of time translation instead of time evolvement. Why do we need this time evolvement in place to make things more complicated?

    Also, I found that the concept of two particles shall not occupy the same state in this statespace is natural. I don't see why only Fermions need to comply to this Pauli Exclusion Principle.

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  3. Jul 8, 2004 #2

    Tom Mattson

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    The state space is needed because the eigenstates of the operators span the state space, not 3D Euclidean space. The space whose basis is the (maximal) set of linearly independent eigenstates is the state in which the operators live. In turn, those basis eigenstates are functions of the 3D position vector. That is why both are needed.

    It's a result not from QM, but from QFT. It's a consequence of merging quantum theory with special relativity. The spin-statistics theorem is still a little beyond me right now, but a result is that the creation and annihilation operators for fermion fields anticommute:


    So, when we construct the number operator (N=b+b) from them, we get an operator whose only eigenvalues are 0 and 1, hence the Pauli Exclusion Principle.
  4. Jul 13, 2004 #3

    I think your answer is good.

    Actually, I was just start getting used to this quantum math. of non-equal communtation rule and operators instead of simple numbers or vectors.

    In my mind, my model was actualy a little different from the standard QM defnition of state.

    I thought we can assume all particles live in a physical universe and each one occupies its own point ( or place or state or something ) in this universe and they can not occupy the same point. I think QM probably call this "the Hilbert space". That was why I thought no two particles shall occupy the same state. I think the issue here is I did not understand "state" correctly.

    I originally thought this space is the statespace. Now I realized from more readings that the statespace in QM is probably a projection of this original universe to a projection superplane. I see a book seems to define it as operator on the Hilbert space. I still need to read more to reconcile how this model is actually constructed.

    Any way, by this projection view, I can see why some particles can project into the same point on this projection superplane and some others do not.

    Or this might not even be correct to QM. QM seems to construct the statespace from the experimentation results and then come to make a direct product of the statespaces to construct the background universe ( Please do not confuse this "background universe" to the "background timespace" ) .

    My view is rather the experiment extracted the projected attributes on different projection superplane and so we can guess the original universe's structure in at least an isomophical way.

  5. Jul 13, 2004 #4
    Based on Leon Takhtajan's "Lectures on QM" pages 35 thru 37, the universe we are investigating shall be the "set of states" not the "space of states".

    Here, this universe is modeled as a subset, which is formed by all positive operators with trace=1, of the ideal on L(H).

    The Hilbert space is not the physical universe itself but a construct to support the operation we investigate.

    That's how I interprete these pages.
  6. Jul 14, 2004 #5
    Can the "Pure State" be viewed as the state only allowed for a single particle?

    The "Mixed state" can be viewed as superpositions of many "pure states" and normalized to fit the total probability to be one. So, the " mixed state" is more likely for measuring an ensemble of particles.
  7. Jul 14, 2004 #6
    I got these from this site posted by Eye_In_The_Sky, if you are interested.

    http://www.math.sunysb.edu/~leontak/book.pdf [Broken]

    From pages 35 thru 44, without following thru its vigrous proofs, I hope my interpretation is correct.

    In page 44, the "stationary" pure states seem to be the stable pure states, whose energies are like the energy coresponding to the energy describes by the quantum number n,l,m,s etc... . A particle probably can fit into a A "non-stationary pure state" but will be unstable.
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  8. Jul 14, 2004 #7

    Tom Mattson

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    In QM, we don't talk about the one point at which each particle exists, unless an infinitely precise measurement of the particles' position is made (in which case the momentum is infinitely uncertain). Apart from that kind of measurement, we talk about the value of the wavefunction at a point, not the location of a particle at a point.

    Right, this is where the distinction between fermion and boson comes in.

    The state space is a vector space in the abstract sense of the word. The eigenfunctions of an operator are like the basis vectors i, j, and k in "normal" 3-space.

    The set of all the linearly independent states is the basis for the space of states. So, from our study of the set of states, we can say things about the space of states.

    Are those operators supposed to be observables? If so, then the "trace=1" condition is not right. The Pauli spin matrices correspond to observables, and they are traceless.

    I agree with that.

    You bet! I'm always looking for free e-books.

    The set of stationary states is just a subset of the set of pure states. Stationary states are characterized by a well-defined E (quantum number n). Pure states are characterized by well-defined L, Lz, and S. You can have a stationary state that is not well-defined in Lz, for example, and is thus not pure.

    The word "stable" doesn't occur once in that link. How do you define it?
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  9. Jul 15, 2004 #8
    Hi, Tom,

    Sorry. I used more pragmatic words than math. correct terms.

    I shall have said "element" in a "set", instead a "point" in a space. I did not even force this universe to be Hilbert space or vector space. In QM, I think it was formulated as a product of groups or vector spaces and that is what I tried to find out how QM came to the conclusion of this is the right way to formulate that.

    Take an example, L^2(R, dq) space from page 54 thru 57 as the Hilbert Space, the Hamiltonian eigenfunctions span the H-space; so any \psi belonging to the L^2(R,dq) is a linear combination of them.

    Now, taking a \psi from it, this is not a state. First, its norm might not be one. From a ray-type defintion of state, this \psi belongs to the equivalence class of {c\Psi}. So we will normalize it to make an \alpha.\psi whose norm os one.

    Then the projection operator P_\alpha\psi is now the pure state defined in this book.

    Any way, this arbitrary \psi is not a stationary one, because it's not one of eigensolutions.

    By the way, it never made clear how to define P_\psi, or maybe I skipped it. My guess is that P_\psi is defined as P_\psi(\psi) = \psi and P_psi(a) =0 elsewhere.

    Also, this L^2(R,dq) is not the "set" representing the universe either; Physicist found other factors such as "spin" so they will need to expand it to make a more complete picture of this "set".

  10. Jul 15, 2004 #9


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  11. Jul 16, 2004 #10

    Thanks for the reply. I totaly agree with your statments.

    Two other points might make something clearer:

    1. Time is not an operator in QM, even though you really can see the time evolvement opertor is analogous to momentum operator. I think that's what 's responsible for saying QM is not completely built on a model of timespace.

    2. The timespace interexchangeability proposed by SR and GR were not applied to QM and that's where I saw people like to call QM as nonrelativistic quantum theory.

    One thing to correct in my posts earlier.

    By definition, The 'stationary' state is just a state with a time evolvement as just a phase factor.

    There is a theorem saying that the Hamiltonian eigenfunctions is stationary.
    I meixed them up a little bit here.

    Backed to what I said about earlier, there shall be only a partiticle allowed in a "pure state". Reily's statement of one-to-one coorespondence basically said that one particle is only allowed in the combination of spin and position operator. This is really part of what I am saying. My view is when you observe two electrons in one background timespace referenced by the same frame. The two electrons will have different \psi if your isimorphical construct ( if its not a Hilbert space ) includes both "spin" and "coordinate". Actually two protons shall not have the same "coordinate" \psi. But two electrons are allowed to have the same "coordinate" \psi if they have different spins. The third electron then will have to have a different "coordinate" psi then.

    Any way, I noticed that in a ray-definition of state that two phases of a stationary state seem to be regarded the same, if our ray is defined as the equivalence class of {c\psi | c is a constant complex number }.

    In Leon's definition, the two phases can be regarded as two different states.

    Now, I can construct a mixed state from pure states, it is basically the sum of [tex] w_i *P_\psi_i [/tex] , i= 1, 2, ... N, and the sum of
    [tex] w_i [/tex] needs to be one.

    If by this defintion, we will see there are many particles allowed in the same "pure state". Since QM's statespace is not the completely expanded space but rather a projected subplane, I guess that's OK for it. But that's not in my mind what I am looking for.

    Last edited: Jul 16, 2004
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