Graduate Equivalence Relation to define the tensor product of Hilbert spaces

[victorvmotti]

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?

$\sum_{i=1}^Ia_i(\sum_{j=1}^{J}c_j\psi_j, \sum_{k=1}^{K}d_k\varphi_k) \sim \sum_{i=1}^I\sum_{j=1}^J\sum_{k=1}^Ka_ic_jd_k(\psi_j,\varphi_k)$

 

[PeroK]

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.