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Tensor product of Hilbert spaces

  1. May 18, 2013 #1
    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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  3. May 18, 2013 #2

    micromass

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    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.
     
  4. May 18, 2013 #3

    Fredrik

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    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.
     
  5. May 19, 2013 #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?
     
  6. May 19, 2013 #5

    Fredrik

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    Because Hilbert spaces are vector spaces. (Unless I misunderstood the question, it's as simple as that).

    I like this little book: http://books.google.com/books?id=yT56SqF0xpoC&lpg=PP1&pg=PA100#v=onepage&q&f=false
     
  7. May 19, 2013 #6

    micromass

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    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).

    Fredrik's suggestion is nice. Reed and Simon is another very well-written book.
     
  8. May 19, 2013 #7

    Fredrik

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    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.
     
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