I Tensor Product of Two Hilbert Spaces

[ARoyC]

How to prove that the tensor product of two same-dimensional Hilbert spaces is also a Hilbert space?

I understand that I need to prove the Cauchy Completeness of the new Hilbert space. I am stuck in the middle.

 

[dextercioby]

This is not QM subject, but looks like homework assignment in functional analysis or math. methods in modern physics.
So "stuck in the middle is vague". Show your effort, please.

 

[ARoyC]

This is not QM subject, but looks like homework assignment in functional analysis or math. methods in modern physics.
So "stuck in the middle is vague". Show your effort, please.

I have encountered this problem in Quantum Mechanics.

This is my effort. Please pardon my handwriting.

 

[fresh_42]

That's the wrong start and I think this will become very technical.

We need to show that $H_1\otimes H_2$ is complete. This means we need to show that every converging sequence, or Cauchy sequence in it, converges to a point in $H_1\otimes H_2$.

Now, $v_1\otimes v_2$ is not a typical element in $H_1\otimes H_2$ so we may not assume that elements look like that. A typical element is $\sum_{k\in \mathbb{N}} c_k v_1^{(k)}\otimes v_2^{(k)}$ with only finitely many scalar factors $c_k \neq 0.$ That's why I introduced them. We can write the sum without the $c_k$ but then it becomes more difficult to say that the sum is finite although we sum over potentially infinite bases.

A sequence, therefore, looks like
$$
\left(\sum_{k\in \mathbb{N}} c_{k,n} v_1^{(k,n)}\otimes v_2^{(k,n)}\right)_{n\in \mathbb{N}}
$$

If this is a Cauchy sequence, then the difference between two sequence members becomes as small as we like if we chose the indices high enough, i.e.
\begin{align*}
\left\|\sum_{k\in \mathbb{N}} c_{k,n} v_1^{(k,n)}\otimes v_2^{(k,n)}\, - \,\sum_{k\in \mathbb{N}} c_{k,m} v_1^{(k,m)}\otimes v_2^{(k,m)}\right\| <\varepsilon
\end{align*}
Since only finitely many $c_{k,n}$ and $c_{k,m}$ are different from zero, we can write this difference as
$$
\left\|\sum_{p\in \mathbb{N}} c'_{p} v_1^{(p)}\otimes v_2^{(p)}\right\|<\varepsilon
$$
with new coefficients and over all dyads that occur in either of the previous sums. Now to the crucial part: how is the norm defined? It is induced by the inner product, so the question is: what is the inner product in the tensor space? The answer is
\begin{align*}
\left\|\sum_{p\in \mathbb{N}} c'_{p} v_1^{(p)}\otimes v_2^{(p)}\right\|&=\ldots\\
&=\ldots \\
&=\ldots\\
&= \sqrt{\sum_{r\in \mathbb{N}} |c'_r|\cdot \langle v_1^{(r)}\,|\, v_1^{(r)}\rangle \cdot \langle v_2^{(r)}\,|\,v_2^{(r)} \rangle} < \varepsilon
\end{align*}
where the dots represent a lot of distributive multiplications, re-arrangement of the dyads, and re-indexing. But all factors in the sum under the root are positive, so they are all as small as we want. This means that the $\sqrt{ \sum_p c'_p \langle v_1^{(p)} \,|\, v_1^{(p)} \rangle}$ and $\sqrt{\sum_q c'_q\langle v_2^{(q)} \,|\,v_2^{(q)}\rangle }$ are Cauchy sequences which converge in $H_1$, resp. $H_2,$ say with limits $L_1$ and $L_2.$

Finally, we have to go all the steps backward and show that
\begin{align*}
\left\|\sum_{p\in \mathbb{N}} c'_{p} v_1^{(p)}\otimes v_2^{(p)}\; - \; L_1\otimes L_2\right\| <\varepsilon
\end{align*}

That is the plan. However, I'd rather solve a 1,000-piece puzzle than fill all of the above with the correct epsilontic and all correct indices. Or call for a physicist to do some voodoo with all the indices.

Edit: See posts #8 to #10. $H_1\otimes H_2$ is in general only a pre-Hilbert space. In order that the idea above works we need finite dimension of one of them.

 

[wrobel]

first it would be nice to figure out on the level of definitions what a tensor product of vector spaces is and what a topological tensor product is:)

 

[ARoyC]

That's the wrong start and I think this will become very technical.

We need to show that $H_1\otimes H_2$ is complete. This means we need to show that every converging sequence, or Cauchy sequence in it, converges to a point in $H_1\otimes H_2$.

Now, $v_1\otimes v_2$ is not a typical element in $H_1\otimes H_2$ so we may not assume that elements look like that. A typical element is $\sum_{k\in \mathbb{N}} c_k v_1^{(k)}\otimes v_2^{(k)}$ with only finitely many scalar factors $c_k \neq 0.$ That's why I introduced them. We can write the sum without the $c_k$ but then it becomes more difficult to say that the sum is finite although we sum over potentially infinite bases.

A sequence, therefore, looks like
$$
\left(\sum_{k\in \mathbb{N}} c_{k,n} v_1^{(k,n)}\otimes v_2^{(k,n)}\right)_{n\in \mathbb{N}}
$$

If this is a Cauchy sequence, then the difference between two sequence members becomes as small as we like if we chose the indices high enough, i.e.
\begin{align*}
\left\|\sum_{k\in \mathbb{N}} c_{k,n} v_1^{(k,n)}\otimes v_2^{(k,n)}\, - \,\sum_{k\in \mathbb{N}} c_{k,m} v_1^{(k,m)}\otimes v_2^{(k,m)}\right\| <\varepsilon
\end{align*}
Since only finitely many $c_{k,n}$ and $c_{k,m}$ are different from zero, we can write this difference as
$$
\left\|\sum_{p\in \mathbb{N}} c'_{p} v_1^{(p)}\otimes v_2^{(p)}\right\|<\varepsilon
$$
with new coefficients and over all dyads that occur in either of the previous sums. Now to the crucial part: how is the norm defined? It is induced by the inner product, so the question is: what is the inner product in the tensor space? The answer is
\begin{align*}
\left\|\sum_{p\in \mathbb{N}} c'_{p} v_1^{(p)}\otimes v_2^{(p)}\right\|&=\ldots\\
&=\ldots \\
&=\ldots\\
&= \sqrt{\sum_{r\in \mathbb{N}} |c'_r|\cdot \langle v_1^{(r)}\,|\, v_1^{(r)}\rangle \cdot \langle v_2^{(r)}\,|\,v_2^{(r)} \rangle} < \varepsilon
\end{align*}
where the dots represent a lot of distributive multiplications, re-arrangement of the dyads, and re-indexing. But all factors in the sum under the root are positive, so they are all as small as we want. This means that the $\sqrt{ \sum_p c'_p \langle v_1^{(p)} \,|\, v_1^{(p)} \rangle}$ and $\sqrt{\sum_q c'_q\langle v_2^{(q)} \,|\,v_2^{(q)}\rangle }$ are Cauchy sequences which converge in $H_1$, resp. $H_2,$ say with limits $L_1$ and $L_2.$

Finally, we have to go all the steps backward and show that
\begin{align*}
\left\|\sum_{p\in \mathbb{N}} c'_{p} v_1^{(p)}\otimes v_2^{(p)}\; - \; L_1\otimes L_2\right\| <\varepsilon
\end{align*}

That is the plan. However, I'd rather solve a 1,000-piece puzzle than fill all of the above with the correct epsilontic and all correct indices. Or call for a physicist to do some voodoo with all the indices.

Thank you. I need to delve into the proper functional analysis to do this.

 

[wrobel]

In accordance with the standard definition cite cite , a topological tensor product of normed spaces $X,Y$ consists of finite sums of such elements $x\otimes y,\quad x\in X,\quad y\in Y$ plus the definition of topology in $X\otimes Y$.
$X\hat\otimes Y$ commonly denotes the completion of $X\otimes Y$. In general $X\hat\otimes Y\ne X\otimes Y$ even if $X,Y$ are both Banach spaces. OP claims that Hilbert spaces provide an exception. Strange. At least such a glorious fact must be proved in many texts.

 

[fresh_42]

In accordance with the standard definition cite cite , a topological tensor product of normed spaces $X,Y$ consists of finite sums of such elements $x\otimes y,\quad x\in X,\quad y\in Y$ plus the definition of topology in $X\otimes Y$.
$X\hat\otimes Y$ commonly denotes the completion of $X\otimes Y$. In general $X\hat\otimes Y\ne X\otimes Y$ even if $X,Y$ are both Banach spaces. OP claims that Hilbert spaces provide an exception. Strange. At least such a glorious fact must be proved in many texts.

The critical part of my idea is the conclusion of $|a_n\otimes b_n|<\varepsilon \stackrel{?}{ \Longrightarrow} |a_n|<\varepsilon .$ I thought that it could be handled by taking the maximum of the finitely many $b_n$ into consideration, but that could be wrong. If, then it is there where my idea fails. It is also the point where to construct a counterexample from: let $a_n$ converge so fast to zero that it compensates e.g. an alternating behavior of $b_n.$ If $b_n$ cannot have a limit while $a_n$ and $a_n\otimes b_n$ have, then the statement is false. We can assume an ONB so we only have to deal with the scalar factors $c_n.$

 

[wrobel]

I am looking through the book by Robertson and Robertson. If we apply the general theory to the Hilbert spaces, then the situation seemingly looks in such a way. If $(X,(\cdot,\cdot)_X)$ and $(Y,(\cdot,\cdot)_Y)$ are Hilbert spaces, then there are two standard topologies in $X\otimes Y$. One of them is such that the space $X\hat\otimes Y$ is presented as follows.
$$X\hat\otimes Y\ni w=\sum_{ij}w_{ij}x_i\otimes y_j,\quad \|w\|^2=\sum_{ij}w_{ij}^2<\infty,$$
here $\{x_i\},\{y_i\}$ are orthonormal bases in $X$ and $Y$ respectively. Particularly,
$$u=\sum_iu_ix_i,\quad v=\sum_iv_jy_j\Longrightarrow u\otimes v=\sum_{ij}u_iv_jx_i\otimes y_j$$

 

[fresh_42]

I am looking through the book by Robertson and Robertson. If we apply the general theory to the Hilbert spaces, then the situation seemingly looks in such a way. If $(X,(\cdot,\cdot)_X)$ and $(Y,(\cdot,\cdot)_Y)$ are Hilbert spaces, then there are two standard topologies in $X\otimes Y$. One of them is such that the space $X\hat\otimes Y$ is presented as follows.
$$X\hat\otimes Y\ni w=\sum_{ij}w_{ij}x_i\otimes y_j,\quad \|w\|^2=\sum_{ij}w_{ij}^2<\infty,$$
here $\{x_i\},\{y_i\}$ are orthonormal bases in $X$ and $Y$ respectively. Particularly,
$$u=\sum_iu_ix_i,\quad v=\sum_iv_jy_j\Longrightarrow u\otimes v=\sum_{ij}u_iv_jx_i\otimes y_j$$

I have found the following exercise in my book:

$H_1\otimes H_2$ is complete if and only if either $H_1$ or $H_2$ is finite-dimensional.