Direct products and direct sums in QM

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SUMMARY

The discussion focuses on the mathematical representation of state spaces for indistinguishable particles in quantum mechanics (QM). It establishes that the state space can be expressed as a direct sum of one-particle spaces, specifically using the notation ε_t = ε_1 ⊕ ε_2. The one-particle state spaces are represented as tensor products, ε_1 = ε_{o1} ⊗ ε_{o2} and ε_2 = ε_{o3} ⊗ ε_{o4}. The key conclusion is that for indistinguishable particles, the expression (ε_{o1} ⊗ ε_{o2}) ⊕ (ε_{o1} ⊗ ε_{o4}) = ε_{o1} ⊗ (ε_{o2} ⊕ ε_{o4}) holds true, confirming the distributive property of tensor products over direct sums.

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Amok
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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]

and

[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})<br /> = \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.
 
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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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