Tensor product of Hilbert spaces

In summary, the tensor product of Hilbert spaces is formed by taking the tensor product of the underlying vector spaces and defining an inner product on it. This inner product is then extended linearly to all elements of the tensor product space. The completion of this space is then taken to obtain a Hilbert space, which is then defined as the tensor product of the original Hilbert spaces. The notion of an orthonormal basis is used to describe the tensor product, which is defined in a similar way as in finite-dimensional vector spaces.
  • #1
Yoran91
37
0
Hi everyone,

I don't quite understand how tensor products of Hilbert spaces are formed.

What I get so far is that from two Hilbert spaces [itex]\mathscr{H}_1[/itex] and [itex]\mathscr{H}_2[/itex] a tensor product [itex]H_1 \otimes H_2 [/itex] is formed by considering the Hilbert spaces as just vector spaces [itex]H_1[/itex] and [itex]H_2[/itex].

Next, there is an inner product on this space, which is defined by
[itex]\langle \phi_1 \otimes \phi_2 \vert \psi_1 \otimes \psi_2 \rangle \equiv \langle \phi_1 \vert \psi_1 \rangle_1 \langle \phi_2 \vert \psi_2 \rangle_2 [/itex] on the simple or pure tensors on this tensor product space. This inner product is extended linearly to an inner product on all elemnets of the tensor product space.

This is where my understanding stops. The 'completion' of this tensor product space is now taken, and the result is a Hilbert space, which is then defined as the tensor product of the Hilbert spaces.
This seems weird to me, because it seems artificial - is the tensor product of Hilbert spaces defined such that its a Hilbert space again?

I don't really understand how taking the completion of a space works. Can anyone provide some insight as to how this works?

Thanks for any help
 
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  • #2
Yes, we want the tensor product of two Hilbert spaces to form a Hilbert space again. If we didn't take the completion, then we would just end up with a pre-Hilbert space. We don't want this, so we take the completion.

The completion of a pre-Hilbert space ##E## is a Hilbert space ##H## with an isometric embedding ##i:E\rightarrow H## such that ##i(E)## is dense in ##H##. If ##i^\prime:E\rightarrow H^\prime## is another such completion, then there is an isometric isomorphism ##f:H\rightarrow H^\prime## such that ##f\circ i = i^\prime##. So the completion is unique up to isometric isomorphism.

A somewhat different description of the tensor product is the following: if ##\{e_i\}_{i\in I}## is an orthonormal basis for ##H_1## and ##\{f_j\}_{j\in J}## is an orthonormal basis for ##H_2##, then ##H_1\otimes H_2## is the Hilbert space with orthonormal basis formed by ##\{e_i\otimes f_j\}_{i\in I, j\in J}##. But this description is dependent of the basis.
 
  • #3
Yes, the tensor product of two vector spaces is a vector space, so when we want to take the tensor product of two Hilbert spaces, we use the fact that Hilbert spaces are vector spaces. The tensor product of those vector spaces is a vector space, and it can be given an inner product in a natural way. But we want the result to be a Hilbert space, so if the result was a vector space that isn't complete, we use its completion instead, and call that the tensor product of the Hilbert spaces.

If you don't understand the tensor product of vector spaces, this thread should be useful.

The simplest example of completion is that the completion of ##\mathbb Q## is ##\mathbb R##. A sequence in ##\mathbb Q## is said to be a Cauchy sequence if for all ε>0, there's an open ball in ##\mathbb Q## with radius ε that contains all but a finite number of terms of the sequence. Let S be the set of Cauchy sequences in ##\mathbb Q##. We define a relation ~ on S by saying that
$$(x_n)_{n=1}^\infty\sim (y_n)_{n=1}^\infty\text{ if and only if }x_n-y_n\to 0.$$ This is an equivalence relation, so each member of S belongs to exactly one equivalence class. I will denote the equivalence class that contains the Cauchy sequence s by .

The the set of all that contain a constant sequence is obviously in bijective correspondence with ##\mathbb Q##. We can define ##\mathbb R## to be the set of all equivalence classes. In other words, for all in S, if contains a constant sequence, we interpret it as a rational number, and if it doesn't, we interpret it as an irrational real number.
 
  • #4
Thanks for the help!

I understand now that the Hilbert tensor product of two Hilbert spaces is 'simply' the completion of the tensor product of the underlying vector spaces with respect to the inner product I gave above.
I have one more question, however. In the discussion of regular tensor products, if [itex]V[/itex] and [itex]W[/itex] are finite dimensional vector spaces, with bases [itex]v_i[/itex] and [itex]w_j[/itex], then the set [itex]v_i \otimes w_j[/itex] is a basis for the tensor product [itex]V\otimes W[/itex].

I read that the same holds if [itex]V[/itex] and [itex]W[/itex] are Hilbert spaces instead of vector spaces. Why is this?

Is there any good source (book) on this?
 
  • #5
Yoran91 said:
In the discussion of regular tensor products, if [itex]V[/itex] and [itex]W[/itex] are finite dimensional vector spaces, with bases [itex]v_i[/itex] and [itex]w_j[/itex], then the set [itex]v_i \otimes w_j[/itex] is a basis for the tensor product [itex]V\otimes W[/itex].

I read that the same holds if [itex]V[/itex] and [itex]W[/itex] are Hilbert spaces instead of vector spaces. Why is this?
Because Hilbert spaces are vector spaces. (Unless I misunderstood the question, it's as simple as that).

Yoran91 said:
Is there any good source (book) on this?
I like this little book: http://books.google.com/books?id=yT56SqF0xpoC&lpg=PP1&pg=PA100#v=onepage&q&f=false
 
  • #6
Yoran91 said:
Thanks for the help!

I understand now that the Hilbert tensor product of two Hilbert spaces is 'simply' the completion of the tensor product of the underlying vector spaces with respect to the inner product I gave above.
I have one more question, however. In the discussion of regular tensor products, if [itex]V[/itex] and [itex]W[/itex] are finite dimensional vector spaces, with bases [itex]v_i[/itex] and [itex]w_j[/itex], then the set [itex]v_i \otimes w_j[/itex] is a basis for the tensor product [itex]V\otimes W[/itex].

I read that the same holds if [itex]V[/itex] and [itex]W[/itex] are Hilbert spaces instead of vector spaces. Why is this?

The notion of basis makes sense in Hilbert space but it is pretty useless in the infinite-dimensional case. The notion that you want to consider is that of an "orthonormal basis". Despite the name, it is somewhat different from the bases in vector spaces.
In vector spaces, you know that ##\{e_i\}_{i\in I}## is a basis if every vector can be written as a finite linear combination of the basis vectors in a unique way. The set ##\{e_i\}_{i\in I}## is then called a (Hamel) basis.
The notion of "orthonormal basis requires all vectors ##\{e_i\}_{i\in I}## to satisfy that they are orthonormal, that is ##<e_i,e_j> = \delta_{ij}##. And furthermore, any vector can be written as a possibly infinite linear combination of the basis vectors.

So a Hamel basis only allows for finite sums. An orthonormal basis allows for infinite sums. With the definition of orthonormal basis, it is indeed true that ##\{e_i\otimes f_j\}_{i\in I, j\in J}## is an orthonormal basis of the tensor product provided that ##\{e_i\}_{i\in I}## and ##\{f_j\}_{j\in J}## are orthonormal bases. If you simply work with Hamel basises, then this fails (because you take the completion, so you add more elements which might not be a linear combination of the basis vectors).

Is there any good source (book) on this?

Fredrik's suggestion is nice. Reed and Simon is another very well-written book.
 
  • #7
Fredrik said:
Because Hilbert spaces are vector spaces. (Unless I misunderstood the question, it's as simple as that).
Ah, now I see what you meant. You thought that the term "Hilbert space" implies that the space is infinite-dimensional. It doesn't. Every finite-dimensional inner product space over ℂ is a Hilbert space.
 

FAQ: Tensor product of Hilbert spaces

What is the definition of tensor product of Hilbert spaces?

The tensor product of two Hilbert spaces is a mathematical operation that combines the elements of each space to create a new, larger Hilbert space. It is denoted by the symbol ⊗ and can be used to describe the composite systems in quantum mechanics.

How is the tensor product of Hilbert spaces calculated?

The tensor product of two Hilbert spaces is calculated by taking the tensor product of the vector spaces and defining a new inner product on the resulting space. This new inner product is constructed in such a way that it satisfies the properties of an inner product, making the resulting space a Hilbert space.

What are the properties of tensor product of Hilbert spaces?

The tensor product of Hilbert spaces has the following properties: it is associative, distributive, and has a unit element. It is also compatible with the inner product on each space, meaning that the inner product of the tensor product space can be calculated using the inner products of the individual spaces.

What is the significance of tensor product of Hilbert spaces in quantum mechanics?

In quantum mechanics, composite systems are described by the tensor product of the individual systems' Hilbert spaces. This allows for the representation of entanglement, a phenomenon where the state of one system is dependent on the state of another system, which is crucial for understanding and describing quantum phenomena.

Can the tensor product of Hilbert spaces be extended to more than two spaces?

Yes, the tensor product of Hilbert spaces can be extended to any finite number of spaces. This is known as the n-fold tensor product and is denoted by ⊗n. It is used to describe composite systems with more than two components, such as multiple particles in quantum mechanics.

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