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Direct products and direct sums in QM

  1. Mar 22, 2012 #1
    Dear forumers,

    I have a question about taking direct sums and products of state spaces in QM. Picture I have a state space that describes two (indistinguishable) particles which is a direct sum of two one-particles spaces:

    [tex]\epsilon_t = \epsilon_1 \oplus \epsilon_2 [/tex]

    Furthermore, picture that both one-particle state spaces can be represented as products of two space-states for different observables.

    [tex]\epsilon_1 = \epsilon_{o1} \otimes \epsilon_{o2} [/tex]


    [tex]\epsilon_2 = \epsilon_{o3} \otimes \epsilon_{o3} [/tex]

    that is to say

    [tex]\epsilon_t = (\epsilon_{o1} \otimes \epsilon_{o2}) \oplus (\epsilon_{o3} \otimes \epsilon_{o4}) [/tex]

    Now picture that the spaces for o1 (which refers to particles 1) and o3 (which refers to particle 2) are actually the same (they contain all the same states). Is it a stretch to say that since these particles are indistinguishable:

    [tex] (\epsilon_{o1} \otimes \epsilon_{o2}) \oplus (\epsilon_{o1} \otimes \epsilon_{o4})
    = \epsilon_{o1} \otimes (\epsilon_{o2} \oplus \epsilon_{o4}) [/tex]


    I'm new to this fock-space, many particle thing, so pardon me if what I'm doing is totally wrong.
  2. jcsd
  3. Mar 22, 2012 #2


    User Avatar
    Science Advisor

    I don't know if this is the accepted terminology in phisics, but it shouldn't be called 'direct product' it is 'tensor product'. And yes, it is true that it is distrubutive with respect to direct sum.
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