Exercise 3.1 Discussion

$\def\abs#1{|#1}\def\i{\mathbf {i}}\def\ket#1{|{#1}\rangle}\def\bra#1{\langle{#1}|}\def\braket#1#2{\langle{#1}|{#2}\rangle}\mathbf{Exercise\ 3.1}$

Let $V$ be a vector space with basis $\{(1,0,0), (0,1,0), (0,0,1) \}$. Give two different bases for $V\otimes V$.

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Let $U$ and $W$ both be vector spaces over field $\mathbb{F}$, with $B_{U} = \big\{u_{1}, \ldots, u_{m}\big\}$ a basis for $U$ and $B_{W} = \big\{w_{1}, \ldots, w_{n}\big\}$ a basis for $W$. Then, the set $B_{U\otimes W} = \big\{u_{i} \otimes w_{j} | 1 \leq i \leq m \,, 1 \leq j \leq n \big\}$ is a basis for the tensor product space $U \otimes W$.

Thus, two different bases for $V \otimes V$ can be obtained using two different bases for $V$. Let $B$ be the basis given in the problem description, with $b_{1} = (1,0,0)$, $b_{2} = (0,1,0)$, and $b_{3} = (0,0,1)$. Let a second basis for $V$ be the set $B' = \big\{b_{1}', b_{2}', b_{3}' \big\}$ where $b_{1}' = \tfrac{1}{\sqrt{2}}(1,0,1)$, $b_{2}' = \tfrac{1}{\sqrt{2}}(1,0,-1)$, and $b_{3}' = (0,1,0)$.

With these two bases for $V$, it is possible to form the two following bases for $V \otimes V$:

(1)Noting that

(2)it is possible to write the $\mathcal{B}_{B,B'}$ basis exclusively in terms of the $b_{i}$ basis vectors:

(3)ReplyOptions