Tensor Product of Two Hilbert Spaces

In summary: 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\
  • #1
ARoyC
56
11
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.
 
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  • #2
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.
 
  • #3
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.

Effort.jpeg
 
  • #4
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.
 
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  • #5
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:)
 
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  • #6
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.
 
  • #7
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.
 
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  • #8
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.##
 
  • #9
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,
$$u=\sum_iu_ix_i,\quad v=\sum_iv_jy_j\Longrightarrow u\otimes v=\sum_{ij}u_iv_jx_i\otimes y_j$$
 
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  • #10
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.
 
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