# Tensor product of vector spaces: confusion

1. Feb 18, 2010

### Goldbeetle

Dear all,
I've read the math that defines a tensor product by means of the universal property and I've studied the tensor product construction through a quotient of the free vector space on the cartesian product of two vector spaces. All other constructions of the tensor products are naturally isomorphic to this in a natural way.

The problem I still have is what follows. If we have two concrete vector spaces in quantum mechanics what shall we do? Suppose they are finite dimensional. How do I construct the tensor product in a "real life" situation?

The answer I give to me after all the above math is:

(a) either for some physical reason, or bright idea dreamt up while sleeping, I'm able to come up with a concrete definition of "v tensor w" from to given vectors v and w (then I know, from the above math, that this is unique up to an isomorphism), or
(b) I work somehow more formally, using all the properties of tensor products and their linear mappings, reassured that the above math gives a rigorous foundation to this mathematical tool I'm using.

Do I get anything right?

Kindest regards.
Goldbeetle

Last edited: Feb 18, 2010
2. Feb 18, 2010

### muppet

Hi Goldbeetle,
I can't say I'm familiar with the precise definition of the tensor product of spaces that you've given, but I'll take your word for it that it's isomorphic to that with which I am familiar:

where x denotes the cartesian product and V* is the dual space of V.

The tensor product of the vectors v,w (which are elements of V and W, respectively) with components $$v_i,w_j$$ with respect to some pair of bases is then simply specified by the array $$A_{ij}=v_i w_j$$ with respect to the basis of the tensor product space constructed from the bases chosen for V and W.

The best discussions of tensor products I've ever come across have been in differential geometry texts, and correspondingly in some texts on general relativity.

Caveat: The only time I've ever used the machinery in a QM context is in dealing with products of spin operators in the hyperfine transition, which is slightly less intuitive as your "vectors" are vectors of pauli matrices, and you don't compute the resulting array via matrix products. The computational rule that I was taught in this instance is that the tensor product of two such matrices was a 4x4 matrix comprised of 2x2 blocks consisting of the 2nd matrix multipled by the element of the first matrix corresponding to the block you're talking about. The results of this (linguistically convoluted) procedure can be seen here: http://sps.nus.edu.sg/~tanboons/notes/misc_pauli.pdf

Last edited by a moderator: Apr 24, 2017
3. Feb 18, 2010

### Fredrik

Staff Emeritus
Goldbeetle: I'm not sure what your concern is here. Once you have proved that all constructions that satisfy the universal property are isomorphic, and that a construction exists, then why not just write your tensor product vectors as $\sum_n c_n|\alpha_n\rangle\otimes|\beta_n\rangle$ and be done with it? (You have already proved that it makes sense).

Perhaps the answer you're looking for is that you can write the members of the tensor product space as $\sum_n c_n x_n y_n$ where the c's are complex numbers and the x's and y's are column matrices (containing the components of the vectors in some basis).

Everyone except Goldbeetle: This thread contains a lot of information about the definition/construction of the tensor product space, and this thread (the articles it references) explains why we use the tensor product at all.

4. Feb 18, 2010

### Goldbeetle

For all: the first thread that Frederik indicates has clarified a lot to me on the construction of tensor product spaces. I also suggest to have a look at Jauch - "Foundations of Quantum Mechanics" (from page 175) for physical motivation. The first chapter of Greub's "Multilinear Algebra" contains an extremely complete discussion of tensor product spaces.

Thanks, Fredrik: so you would answer "yes" to the question of item (b) of my post, and would not bother of (a)? In the case of square integrable wave functions the tensor product is very very concrete (we generalize that, actually). So I was wondering whether there are other cases where the tensor product space is so "easily" directly constructable from the factor vector spaces.

Last edited: Feb 18, 2010
5. Feb 18, 2010

### peteratcam

'Tensor' product of vectors is ambiguous, because it sometimes refers to an outer product (which gives an array), whereas you want to turn 2 vectors into one big vector. (I call it the direct product)
$$\begin{pmatrix}a_1\\a_2\end{pmatrix}\otimes\begin{pmatrix}b_1\\b_2\\b_3\end{pmatrix}=\begin{pmatrix}a_1b_1\\a_1b_2\\a_1b_3\\a_2b_1\\a_2b_2\\a_2b_3\end{pmatrix} \qquad \text{({\tt kron(a,b)} in {\sc Matlab}.)}$$
If a and b are normalised, then the thing on the right is also normalised (which is good).
For operators (eg, the operator which acts only on the second spin):

$${\mathbf 1}\otimes\sigma_x=\begin{pmatrix}0&1&0&0\\1&0&0&0\\0&0&0&1\\0&0&1&0\end{pmatrix} \qquad \text{({\tt kron(eye(2),[0 1;1 0])} in {\sc Matlab}.)}$$

Interaction operators are like $$\sigma_x\otimes\sigma_x$$.

Peter