Equivalence Relation to define the tensor product of Hilbert spaces

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 replies · 2K views
victorvmotti
Messages
152
Reaction score
5
I'm following this video on how to establish an equivalence relation to define the tensor product space of Hilbert spaces:

##\mathcal{H1} \otimes\mathcal{H2}={T}\big/{\sim}##

The definition for the equivalence relation is given in the lecture vidoe as

##(\sum_{j=1}^{J}c_j\psi_j, \sum_{k=1}^{K}d_k\varphi_k) \sim \sum_{j=1}^J\sum_{k=1}^Kc_jd_k(\psi_j,\varphi_k)##

But is this correct?

A linear combination of pairs on the right hand side is equivalent to only one pair on the left hand side.

Shouldn't we define the equivalence relation as below so that we have on both sides linear combination of pairs?

##\sum_{i=1}^Ia_i(\sum_{j=1}^{J}c_j\psi_j, \sum_{k=1}^{K}d_k\varphi_k) \sim \sum_{i=1}^I\sum_{j=1}^J\sum_{k=1}^Ka_ic_jd_k(\psi_j,\varphi_k)##
 
on Phys.org
victorvmotti said:
I'm following this video on how to establish an equivalence relation to define the tensor product space of Hilbert spaces:

##\mathcal{H1} \otimes\mathcal{H2}={T}\big/{\sim}##

The definition for the equivalence relation is given in the lecture vidoe as

##(\sum_{j=1}^{J}c_j\psi_j, \sum_{k=1}^{K}d_k\varphi_k) \sim \sum_{j=1}^J\sum_{k=1}^Kc_jd_k(\psi_j,\varphi_k)##

But is this correct?

A linear combination of pairs on the right hand side is equivalent to only one pair on the left hand side.

Shouldn't we define the equivalence relation as below so that we have on both sides linear combination of pairs?

##\sum_{i=1}^Ia_i(\sum_{j=1}^{J}c_j\psi_j, \sum_{k=1}^{K}d_k\varphi_k) \sim \sum_{i=1}^I\sum_{j=1}^J\sum_{k=1}^Ka_ic_jd_k(\psi_j,\varphi_k)##
Your additional ##a_i## are superfluous.