Tensor (Direct) Products vs. Direct Sums

Written: 14/06/2012 Last edited: 02/07/2012 Filed under: Specialist Topics, Quantum Physics, Mathematical Physics

A common theme in quantum mechanics and multi-particle systems is manipulation of vector spaces - usually spaces of physical states of a system.

I've come across a little confusion between outer products or tensor products (specifically tensor direct products) and direct sums in this area in particular, so just a quick comparison to make the difference clear:

Notation: U, V two (for simplicity finite-dimensional) vector spaces with bases u⁽¹⁾, ..., u^(l) and v⁽¹⁾, ..., v^(m).

	Tensor product	Direct sum
What it looks like:	$W = U \otimes V$	$W = U \oplus V$
Basis elements:	$w^{(i j)} = u^{(i)} \otimes v^{(j)} \equiv u^{(i)}v^{(j)}$	$w^{(i)} = \begin{cases}u^{(i)} & \text{if }1\le i\le l \\ v^{(i-l)} & \text{if }l<i\le(l+m) \end{cases}$
Typical element:	$w = 4 u^{(1)}\otimes v^{(3)} - u^{(2)}\otimes v^{(1)}$	$w = 5 u^{(1)} - 2 v^{(5)}$
Dimensions:	$n = lm$ basis elements	$n = l + m$ basis elements
Physics:	You need to know information (of different types) from both U and V to describe the system	U and V represent two alternative (groups of) possibilities for the system, so that you can have a system definitely in a U-type state, for example
QM example:	U specifies spin of a particle, V specifies spin of a second (or the position of the first)	U contains all states which are symmetric under interchange of two particles, V contains all states which are antisymmetric U contains states in which the particle is in {x ≤ 0}, V those in which it is in {x > 0}

More Confusion: Direct Products

Something I fell for when I was writing this down the first time is that there is both a direct product $U \times V$ and a tensor (direct) product/outer product $U \otimes V$ . A direct product is - incredibly confusingly - basically the same as a direct sum in this case.

A direct product is $U \times V = \{(u,v) : u\in U, v\in V\}$ - so each element involves specifying one element of U and one element of V. In the specific case described above - where u and v can be added together - this looks exactly the same as (is isomorphic to) the direct sum. We just write (u,v) ≡ u + v. This is quite general, as you can see in the Wolfram article on direct products.

If you're wondering what the relationship between a tensor product and a direct product is, it's slightly subtle. The tensor product arises from the free vector space on the direct product $U \times V$ . The idea is that we define one basis vector ('direction') per element of $U \times V$ , so there is a basis $e_{(u,v)}$ . This is a huge space. (If non-trivial, it is of uncountably infinite dimension here, since each real multiple of one vector corresponds to a new dimension.)

Then we identify basis vectors in a particular way (discussed properly in the Wikipedia article on tensor products) so that addition and scalar multiplication of the elements of $U \times V$ corresponds to addition and multiplication in $U \otimes V$ - that is

$\begin{align*} e_{(v_1 + v_2, w)} &\sim e_{(v_1, w)} + e_{(v_2, w)}\\ e_{(v, w_1 + w_2)} &\sim e_{(v, w_1)} + e_{(v, w_2)}\\ ce_{(v, w)} &\sim e_{(cv, w)} \sim e_{(v, cw)} \end{align*}$

This gets rid of a huge load of basis elements, and leaves us (it turns out!) with the tensor product.

See also: On Wikipedia, the tensor product of Hilbert spaces article provides some more formalism.

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Tensor (Direct) Products vs. Direct Sums

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Tensor (Direct) Products vs. Direct Sums

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Tensor (Direct) Products vs. Direct Sums

Clearing up a little confusion