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Schmidt decomposition and bipartite system

  1. Jan 14, 2010 #1


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    I asked a question about Schmidt decomposition in one of the math forums, but haven't gotten any replies yet. Link. The question is about why the state of a bipartite system (a system that consist of two component systems) can be expressed in the form

    [tex]\sum_{n=1}^\infty \sqrt{p_n}|a_n\rangle\otimes|b_n\rangle[/tex]

    where the pn can be interpreted as probabilities. If you know something about this (the infinite-dimensional case), please take a look at my question.

    I never know what forum to use for questions about functional analysis with applications to QM. "Calculus & Analysis" should be the right one (since functional analysis is analysis), but that's not the sort of question other people post there. "Linear & Abstract Algebra" would also make sense since functional analysis is just linear algebra on vector spaces that aren't required to be finite-dimensional, and "Quantum Physics" would make sense too, since that subject is the reason I'm asking.
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  3. Jan 14, 2010 #2


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    I'm not exactly sure what the real question is, but... perhaps a start is to recognize
    that the tensor product vector can be expressed as a matrix (non-square in
    general if the component spaces are of different dimensions). The problem
    then reduces to an application of singular value decomposition (SVD).
    (In the square matrix case, we're basically dealing with the spectral theorem,
    but SVD applies more generally.)

    SVD is mentioned on the Wiki entry you quoted in the other forum, and
    Wiki's SVD page:


    also covers some more general Hilbert space cases, and has
    quite a lot of references.

  4. Jan 14, 2010 #3


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    Yes, I noticed that the Wikipedia entries about Schmidt decomposition and the singular value decomposition contain everything I might want to know about the finite-dimensional case. I haven't studied them carefully yet, but I probably will. I started the thread because I wanted to see a statement and a proof of the theorem for the infinite-dimensional case. (The finite-dimensional case should be a trivial corollary of that).

    The stuff that atyy linked to looks interesting. I'll take a closer look at it a few hours from now.
  5. Jan 16, 2010 #4


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    The explanation in the document that atyy recommended was pretty easy to understand. For physics nerds like us, probably much easier than the proof based on singular value decomposition. This is how it goes: Any member of [itex]\mathcal H_A\otimes\mathcal H_B[/itex] can be expressed in a form that looks like the Schmidt form.

    [tex]|\psi\rangle =\sum_{i\mu}a_{i\mu}|i\rangle\otimes|\mu\rangle =\sum_i|i\rangle\otimes\bigg(\sum_\mu a_{i\mu}|\mu\rangle\bigg) =\sum_i|i\rangle\otimes|\tilde i\rangle[/tex]

    Now the trick is to choose [itex]\{|i\rangle\}[/itex] to be the basis in which the reduced density matrix for system A is diagonal.

    [tex]\rho_A=\sum_i p_i|i\rangle\langle i|[/tex]

    [tex]\rho_A=\mbox{Tr}_B(|\psi\rangle\langle\psi|)=\mbox{Tr}_B\Big(\sum_{ij}|i\rangle\langle j|\otimes |\tilde i\rangle\langle\tilde j|\Big)=\sum_{ij}\langle\tilde j|\tilde i\rangle |i\rangle\langle j|[/tex]


    [tex]\langle\tilde j|\tilde i\rangle=p_i\delta_{ij}[/tex]

    which means that the [itex]|\tilde i\rangle[/itex] states are orthogonal and can be normalized by defining

    [tex]|i'\rangle=\sqrt{p_i}|\tilde i\rangle[/tex]

    and this gives us the final result

    [tex]|\psi\rangle=\sum_i \sqrt{p_i}|i\rangle\otimes|i'\rangle[/tex]

    For now, this is good enough for me. I don't know if this is a valid proof for the infinite-dimensional case (in particular, does the second equality in the first line hold?), but it's at least a very nice proof of the finite-dimensional case.
    Last edited: Jun 13, 2013
  6. Jan 16, 2010 #5


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    I would think "yes". The tensor product construction starts from an ordinary
    Cartesian product, and then mods out by certain equivalence relations which
    basically guarantee an intuitive linearity in the product space. See, e.g.,


    (Look at the paragraph that says
    Those linearity equivalence relations cover the step you were unsure about, (afaict).

    (For more gory detail, one could talk about Tychonoff product topology which is
    used to define a useful topology on Cartesian products of inf-dim spaces, and then
    invoke the fact that addition and scalar multiplication are continuous
    operations in the component spaces to extend a similar property to the tensor
    product space. But that's probably too much information, right? :-)
  7. Jan 16, 2010 #6


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    I don't think they do. They don't seem to be saying anything about infinite "linear combinations".

    This definition of the tensor product is probably more useful:
    I think of this as the definition of the tensor product of vector spaces, and the stuff you referenced as a "construction", or a "proof of existence". The definition implies that the map [itex](x,y)\mapsto x\otimes y[/itex] is bilinear, so the map [itex]y\mapsto x\otimes y[/itex] must be linear, but I think it also needs to be continuous, which it is if and only if it's bounded. Maybe that's implied by the definition. I haven't really thought about it yet.

    An additional complication is that the above is just the definition of the tensor product of two vector spaces. The definition of the tensor product of two Hilbert spaces has an additional step: completion. So even the first equality in #4 isn't really valid. The set of expressions of the form


    doesn't fill up all of [itex]\mathcal H_1\otimes\mathcal H_2[/itex]. It's just a dense subset.

    As George Jones likes to say, "yup". :smile:

    Edit: Yup, the map [itex]y\mapsto x\otimes y[/itex] is continuous for every x. It's quite easy to see this. Define [tex]\tilde x:\mathcal H_B\rightarrow\mathcal H_B[/tex] by [tex]\tilde x(y)=x\otimes y[/tex] for all y. We have

    [tex]\|\tilde x(y)\|^2=\|x\otimes y\|^2=\langle x\otimes y,x\otimes y\rangle=\langle x,x\rangle\langle y,y\rangle=\|x\|^2\|y\|^2[/tex]

    So [tex]\tilde x[/tex] is bounded, and therefore continuous. And that ensures that the second step in this calculation is valid:

    [tex]|i\rangle\otimes\bigg(\sum_{\mu=0}^\infty a_{i\mu} |\mu\rangle\bigg) =|i\rangle\otimes\bigg(\lim_{N\rightarrow\infty} \sum_{\mu=0}^N a_{i\mu}|\mu\rangle\bigg) =\lim_{N\rightarrow\infty} \bigg(|i\rangle\otimes\bigg(\sum_{\mu=0}^N a_{i\mu}|\mu\rangle\bigg)\bigg)[/tex]

    [tex]=\lim_{N\rightarrow\infty}\sum_{\mu=0}^N a_{i\mu}\big(|i\rangle\otimes|\mu\rangle\big)=\sum_{\mu=0}^\infty a_{i\mu}\big(|i\rangle\otimes|\mu\rangle\big)[/tex]

    That stuff I said about a tensor product of Hilbert spaces needing "completion" doesn't seem to complicate anything significant. It just makes a proper statement of the theorem a bit awkward, since we shouldn't be saying that [itex]|\psi\rangle[/itex] is equal to an expression of the Schmidt type. We should be saying that we can choose the [itex]a_{i\mu}[/itex] to make the expression of that type arbitrarily close to [itex]|\psi\rangle[/itex].

    Hm, this actually looks like a complete proof of the (countable) infinite-dimensional case. I suppose there could be something wrong with other details, e.g. the fact that I assumed that there exists a basis in which [itex]\rho_A[/itex] is diagonal. I don't know the relevant theorems well enough yet, but I hope I'll be able to get through a book on functional analysis before the summer.
    Last edited: Jun 13, 2013
  8. Jan 16, 2010 #7


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  9. Jan 17, 2010 #8


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    The Wiki definition is applicable to a tensor product of 2 inf-dim vector spaces,
    since it only uses elementary properties of vector spaces.

    I think it's actually equivalent to the Wiki version. The "mod'ing out by
    equivalence relations" that takes you from the cartesian product to
    the tensor product is equivalent to the abstract linear mapping in the
    definition you gave.

    Right. That's exactly why I ventured to mention product topology. :-)

    A subtlety occurs if you have an infinite tensor product, i.e., a tensor product of an
    infinite number of component spaces. That's where Tychonoff product topology comes in.
    It's subtly different from ordinary product topology, but is generally regarded as
    the "right" topology for such cases, since it preserves various useful topological
    properties, e.g., compactness.

    Can't you get that bit out of the spectral theorem?

    Think about getting a copy of the Schaum Outline of General Topology
    by Lipschutz. It's quite cheap and more easily readable than any other
    introductory topology book I've seen. It explains the Tychonoff
    subtleties quite succinctly. I think it would be a useful book on the shelf
    of someone like yourself who's obviously interested in much more
    mathematical detail than the average physicist. :-)
  10. Jan 17, 2010 #9


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    I think I didn't express myself very clearly. I meant that the explicit construction of the tensor product space doesn't contain anything that enables me to see (without doing the stuff I added to #6 when I edited it) that the second equality in #4 holds when there's an infinite number of terms in my sum.

    Yes, I know. The thread I quoted contains most of what I know about tensor products. I'll probably keep adding stuff to it as I learn more. (I think it's already a pretty good place to learn about tensor products, for people who don't know this stuff already).

    I'll keep that in mind. It's not relevant here though, since the Schmidt decomposition isn't even valid for a tensor product of three Hilbert spaces.

    Yes, I'm assuming that I could, if I had made the effort to learn exactly what it says. I've been planning to study functional analysis in detail for a long time, so I haven't even bothered to learn exactly what the most interesting theorems say yet.

    Thanks for the tip. I'll consider it. For now I'm quite happy with one of my functional analysis books ("Functional Analysis: Spectral Theory", by V.S. Sunder), which covers a lot of this sort of stuff in the appendix.
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