Tensor Product of Two Hilbert Spaces

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.

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.

dextercioby said:

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.

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.

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:)

fresh_42 said:

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.

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.

wrobel said:

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.##

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,
$$u=\sum_iu_ix_i,\quad v=\sum_iv_jy_j\Longrightarrow u\otimes v=\sum_{ij}u_iv_jx_i\otimes y_j$$

wrobel said:

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.