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Direct product and direct sum

  1. Dec 23, 2008 #1
    I wasn't sure where to put this since this is under group theory. I am having a little bit trouble understand when to use direct product vs direct sum. One question I have about this is is that if you have two vector spaces that are orthogonal to each other (and example of this might be two different Hilbert spaces in quantum mechanics, where one space is all the spacial wavefunctions of bosons and the other is all the spatial wave functions of fermions due to the fact that one is symmetric and the other is antisymmetric) will the direct product of these two spaces be 0 and what will the direct sum yield (continuing on with the quantum mechanics example, will is yield all possible linear combinations of states since any state can be broken up into a symmetric and antisymmetric part). Also if anyone could state their answer to my dilema in terms of quantum mechanics states that would be most appreciated. I thank all the people who reply to this post in advance. (Also let me know if I have any terminology in the wrong context).
     
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  3. Dec 23, 2008 #2
    Direct product and direct sum are really the same thing. Direct product is used for multiplicative groups, and direct sum is used for additive (abelian) groups. Both of them assign a group structure on the Cartesian product of two groups; the difference is only whether the groups are written multiplicatively or additively.
     
  4. Dec 23, 2008 #3

    Fredrik

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    The direct sum of two vector spaces is defined here. The direct sum of two Hilbert spaces is defined on the same page. The only additional step is to define the inner product. There's no difference between the direct sum and the direct product of a finite number of Hilbert spaces. The concepts are different when dealing with infinitely many Hilbert spaces. The difference is explained here.

    If H1 is the set of possible states of particle 1, and H2 is the set of possible states of particle 2, then the set of two-particle states is the tensor product of H1 and H2. The direct sum is only used to combine the Hilbert spaces of "n-particle states" for all n into a single Hilbert space.

    It should also be mentioned that all complex infinite-dimensional separable Hilbert spaces are isomorphic to each other. These constructions aren't really done to get the Hilbert space we want, but to get operators with the properties we want.
     
  5. Dec 23, 2008 #4

    Hurkyl

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    In general, the difference between a product structure and a coproduct (i.e. sum) structure is the direction it relates to its components.

    When you have a product structure, you have 'projections' [itex]A \times B \to A[/itex] and [itex]A \times B \to B[/itex], as well as the fact that any pair of maps [itex]C \to A[/itex] and [itex]C \to B[/itex] can be lifted to a map [itex]C \to A \times B[/itex] in exactly one way.

    When you have a sum structure, you have 'inclusions' [itex]A \to A + B[/itex] and [itex]B \to A + B[/itex], as well as the fact that any pair of maps [itex]A \to C[/itex] and [itex]B \to C[/itex] can be pushed to a map [itex]A + B \to C[/itex] is exactly one way.

    The thing that makes linear algebra really cool is that [itex]A \times B \cong A + B[/itex] for vector spaces and other similar structures, as adriank said. (Though this doesn't remain true when you compare an infinite product of structures to an infinite sum of structures)
     
  6. Nov 25, 2010 #5
    A direct answer to your direct question:

    "...one space is all the spacial wavefunctions of bosons and the other is all the spatial wave functions of fermions ... will the direct product of these two spaces be 0" ?

    No, the direct product will be a hilbert space of all the spatial wave functions of fermions.

    "... and what will the direct sum yield (continuing on with the quantum mechanics example, will it yield all possible linear combinations of states since any state can be broken up into a symmetric and antisymmetric part ?"

    Yes indeed.


    More generally:
    Direct product is suitable to consruct eigenvectors of two (or more) different particles (or degrees of freedom).

    Direct sum is suitable to describe hilbert space of two (or more) smaller orthogonal hilbert spaces related to the same particle (or degree of freedom). e.g. each smaller hilbert space may be the vector space of different eigenvalue.

    I don't know how to prove or even mathematically define the above statements, so they might be false, but they are probably correct.
     
  7. Nov 25, 2010 #6
    I forgot to mention, the difference between sum and product stems from the definition of the norm of the constructed hilbert space. In direct sum (product) the norm in the new space is the sum (product) of the norms in the smaller spaces.
     
  8. Nov 25, 2010 #7

    disregardthat

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    Note that they differ for over infinite domains, direct sums [tex]\bigoplus_{n=1}^{\infty} G_n[/tex] can only have a finite number of components different from the identity, but elements of [tex]\prod_{n=1}^{\infty} G_n[/tex] can have an infinite number of components of any kind. That means for an element a different from the identity, (a,a,a,...) is an element of the infinite direct product, but not the infinite direct sum.
     
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