# Projection on Tensor Product of Hilbert Space

Let $H$ be a Hilbert space of finite dimension $n \ge 2$. Let $\{v_1,...,v_n\}$ be an orthonormal basis of $H$. Consider the tensor product $V \otimes V$ and the projections $e_1$ onto the subspace $\mathbb{C}(v_1+\cdots+v_n) \otimes V$ of $V \otimes V$, and $e_2$ onto the subspace $\oplus_{j=1}^n \mathbb{C}(v_j \otimes v_j)$ of $V \otimes V$.

I'm beginning to play with tensor products, so that I'd like to see why the following is the case. For all $j,k \in \{1,\ldots,n\}$, we have

$$e_1(v_j \otimes v_k) = \frac{1}{n} \sum_{l=1}^n v_l \otimes v_k,$$

$$e_2(v_j \otimes v_k) = \delta_{jk} (v_j \otimes v_k).$$

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Let $w = \frac{1}{\sqrt{n}} \sum_{k=1}^n v_k$. Then $\{w\}$ is an orthonormal basis for $E := \mathbb{C}(v_1+\cdots+v_n)$, and hence $\{w \otimes v_1, \dotsc, w \otimes v_k\}$ is an orthonormal basis for $E \otimes V$, so that the orthogonal projection onto $E \otimes V$ is $$P_{E \otimes V} = \sum_{k=1}^n \left|w \otimes v_k \right\rangle \left\langle w \otimes v_k \right|.$$ Since $\left\langle w \mid v_k \right\rangle = \tfrac{1}{\sqrt{n}}$ for each $k$, it follows that $$P_{E \otimes V} (v_i \otimes v_j) = \sum_{k=1}^n \left|w \otimes v_k \right\rangle \left\langle w \otimes v_k \right|(v_i \otimes v_j)\\ = \sum_{k=1}^n \langle w \mid v_i \rangle \langle v_k \mid v_j \rangle(w \otimes v_k) = \frac{1}{\sqrt{n}} w \otimes v_j \\ = \left(\frac{1}{n} \sum_{l=1}^n v_l \right) \otimes v_j$$ for any $i$ and $j$, as required.

Now, let $F = \oplus_{k=1}^n \mathbb{C}(v_k \otimes v_k)$. Then $F$ admits the orthonormal basis $\{v_k \otimes v_k\}_{k=}^n$, so that the orthogonal projection onto $F$ is $$P_F = \sum_{k=1}^n \left| v_k \otimes v_k \right\rangle \left\langle v_k \otimes v_k \right|,$$ so that $$P_F(v_i \otimes v_j) = \sum_{k=1}^n \left| v_k \otimes v_k \right\rangle \left\langle v_k \otimes v_k \right|(v_i \otimes v_j)\\ = \sum_{k=1}^n \left\langle v_k \mid v_i \right\rangle\left\langle v_k \mid v_j \right\rangle v_k \otimes v_k\\ = \sum_{k=1}^n \delta_{ki}\delta_{kj} v_k \otimes v_k\\ = \delta_{ij} v_j \otimes v_j\\ = \delta_{ij} v_i \otimes v_j.$$

Of course, since $\{v_k \otimes v_k\}_{j=1}^n$ is a subset of the orthonormal basis $\{v_i \otimes v_j\}_{i,j=1}^n$, it follows directly that $v_i \otimes v_j \in F := \operatorname{span}\{v_k \otimes v_k\}_{j=1}^n$ if and only if $v_i \otimes v_j \in \{v_k \otimes v_k\}_{j=1}^n$, if and only if $i=j$, whilst $v_i \otimes v_j \in F^\perp = \{v_k \otimes v_k \mid 1 \leq k \leq n\}^\perp$ if and only if $v_i \otimes v_j \notin \{v_k \otimes v_k\}_{j=1}^n$, if and only if $i \neq j$, which directly yields $P_F (v_i \otimes v_j) = \delta_{ij} v_i \otimes v_j$.

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What do you mean by $\left|w \otimes v_k \right\rangle$ and $\left\langle w \otimes v_k \right|$? –  ragrigg Mar 18 '13 at 7:14
Sorry about that, it's notation borrowed from physics (i.e., bra-ket notation) for lack of anything better. If $V$ is an inner product space, then for $v,w \in V$, $\left|v\right\rangle \left\langle w\right| \in B(V)$ is the rank $1$ operator defined by $\left|v\right\rangle\left\langle w\right|(x) := \langle w \mid x \rangle v$. –  Branimir Ćaćić Mar 18 '13 at 7:38
Thanks for the clarification. I've added a second projection to my original question. For $e_2$, do you take $\{ \oplus_{l=1}^n (v_l \otimes v_l) \}$ as your basis? The symbol $\oplus$ is confusing me. –  ragrigg Mar 18 '13 at 9:18
No. The $\oplus$ in this case is an internal direct sum, since $\{v_j \otimes v_j\}_{j=1}^n$ is an orthonormal set, so that $$\mathbb{C} v_1 \otimes v_1 + \cdots + \mathbb{C} v_n \otimes v_n = \oplus_{k=1}^n \mathbb{C} v_k \otimes v_k.$$ In particular, $\{v_j \otimes v_j\}_{j=1}^n$ is therefore an orthonormal basis for $\oplus_{k=1}^n \mathbb{C} v_k \otimes v_k$. –  Branimir Ćaćić Mar 18 '13 at 9:21
I see. Thanks, again. –  ragrigg Mar 18 '13 at 9:29