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This question is regarding a definition of Tensor product of Hilbert spaces that I found in Wald's book on QFT in curved space time. Let's first get some notation straight.

Let $(V,+,*)$ denote a set $V$, together with $+$ and $*$ being the addition and multiplication maps on $V$ that satisfy the vector space axioms. We define the complex conjugate multiplication ${\overline *}:{\mathbb C} \times V \to V$ as $$ c {\overline *} \Psi = {\overline c} * \Psi,~~\forall~~\Psi \in V $$ The vector space formed by $(V,+,{\overline *})$ is called the complex conjugate vector space and is denoted by ${\overline V}$.

Given two Hilbert spaces ${\cal H}_1$ and ${\cal H}_2$ and a bounded linear map $A: {\cal H}_1 \to {\cal H}_2$, we define the adjoint of this map $A^\dagger: {\cal H}_2 \to {\cal H}_1$ as $$ \left< \Psi_2, A \Psi_1 \right>_{{\cal H}_2} = \left< A^\dagger \Psi_2 , \Psi_1 \right>_{{\cal H}_1} $$ where $\left< ~, ~ \right>_{{\cal H}_1}$ is the inner product as defined on ${\cal H}_1$ (similarly for ${\cal H}_2$) and $\Psi_1 \in {\cal H}_1,~\Psi_2 \in {\cal H}_2$. That such map always exists can be proved using the Riesz lemma.

Here the word "bounded" simply means that there exists some $C \in {\mathbb R}$ such that $$\left\| A(\Psi_1) \right\|_{{\cal H}_2} \leq C \left\| \Psi_1 \right\|_{{\cal H}_1}$$ for all $\Psi_1 \in {\cal H}_1$ and where $\left\| ~~ \right\|_{{\cal H}_1}$ is the norm as defined on ${\cal H}_1$ (similarly for ${\cal H}_2$)

Great! Now for the statement. Here it is.

The tensor product, ${\cal H}_1 \otimes {\cal H}_2$, of two Hilbert spaces, ${\cal H}_1$ and ${\cal H}_2$, may be defined as follows. Let $V$ denote the set of linear maps $A: {\overline {\cal H}}_1 \to {\cal H}_2$, which have finite rank, i.e. such that the range of $A$ is a finite dimensional subspace of ${\cal H}_2$. The $V$ has a natural vector space structure. Define the inner product on $V$ by $$ \left< A, B \right>_V = \text{tr}\left( A^\dagger B \right) $$ (The right side of the above equation is well defined, since $A^\dagger B: {\overline {\cal H}}_1 \to {\overline {\cal H}}_1$ has a finite rank). We define ${\cal H}_1 \otimes {\cal H}_2$ to be the Hilbert space completion of $V$. It follows that ${\cal H}_1 \otimes {\cal H}_2$ consists of all linear maps $A: {\overline {\cal H}}_1 \to {\cal H}_2$ that satisfy the Hilbert-Schmidt condition $\text{tr}\left( A^\dagger A \right) < \infty$.

My question is

1. How does this definition of the Tensor product of Hilbert spaces match up with the one we are familiar with when dealing with tensors in General relativity?

PS - I also have a similar problem with Wald's definition of a Direct Sum of Hilbert spaces. I have decided to put that into a separate question. If you could answer this one, please consider checking out that one too. It can be found here. Thanks!

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I figured out the first problem. Removed it from question. –  Prahar Jul 10 '13 at 16:42
    
Concerning tensors, see also physics.stackexchange.com/q/32011/2451 and links therein. –  Qmechanic Jul 10 '13 at 17:38
    
If we consider only vectors, You will have a problem, because an Hilbert space has a positive definite inner product, while the tangent space of the space-time manifold has a pseudo-Riemaniann metric (Minkowski-like metrics) –  Trimok Jul 10 '13 at 18:30

2 Answers 2

up vote 1 down vote accepted

I don't think Wald ever defines a tensor product for infinite dimensional space in his GR text, so I presume your question is about the finite dimensional case where we simply write the tensor product as the vector space over pairs $u_iv_j$ where $u$ and $v$ are a basis. I will show the equivalence in that case.

If we have two finite dimensional Hilbert spaces $H_1$, $H_2$ we can take the orthonormal bases $u_i\in H_1$ , $v_j\in H_2$. Since everything is finite dimensional, everything is finite rank, so the the vector space is just the the space of linear maps from $H_1$ to $H_2$. Take a linear map $A$ and define $a_{ij}= \langle A(u_i),v_j\rangle = \langle u_i,A^{\dagger}(v_j)\rangle$. Using the orthonormality of the bases that means $a_{ij}$ is simply the matrix presentation of $A$, and the vector space is simply the appropriate vector space of matrices. Then we can interpret $Tr(A^\dagger B)$ as the usual matrix trace which gives $\sum_{ij} a_{ij}^*b_{ij}$.

This is equivalent to the usual notation whereby we write tensor products as elements $\sum_{ij}a_{ij}u_i\otimes v_j$. Again the vector space is the appropriately sized matrices. The inner product is defined to be $\langle a\otimes b, c\otimes d\rangle = \langle a,b\rangle\cdot\langle c,d\rangle$. This gives the same result as above after plugging in the basis.

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This is not from his GR text. This is from his text on QFT in curved spacetimes. –  Prahar Jul 10 '13 at 18:25
    
@Prahar: Yeah I know. It's just that you said you wanted to see the definition was equivalent to the "usual" one in GR, but you didn't specify what the usual one is. So I assume you meant the usual finite dimensional notation, as in his GR book –  BebopButUnsteady Jul 10 '13 at 18:28
    
Oh! sorry. My bad. –  Prahar Jul 10 '13 at 18:32

Let there be given a (monoidal) category ${\cal C}$, e.g., the category of finite dimensional vector spaces, the category of Hilbert spaces, etc.

In such a category ${\cal C}$, one typically has the isomorphism

$$\tag{1} {\cal H} \otimes {\cal K} \cong {\cal L}({\cal H}^{*}, {\cal K}), $$

where ${\cal H}^{*}$ is a dual object, and ${\cal L}$ is the pertinent space of morphisms ${\cal H}^{*}\to {\cal K}$.

Often textbooks don't provide the actual definition of a tensor product, which is nevertheless at least partly explained on Wikipedia, but instead cheat by using the isomorphism (1) as a working definition of a tensor product ${\cal H} \otimes {\cal K}$.

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