1. Oct 28, 2008

### KFC

For a system consists of two particle, says two spins, the wavefunction for that system is a direct product of each individual states. And the corresponding operator is also a direct product of each individual operators. I know this is a procedure to construct such a space, but can anybody told me why in this way? Why direct product? and how do we know the new space is still a Hilbert space?

2. Oct 29, 2008

### malawi_glenn

Well this might be a quite wacko answer, but it follows from axioms and theorems of Hilbert Spaces.

Sum of two hilbert spaces is a hilbert space.

3. Oct 29, 2008

### strangerep

Strictly speaking, you want a "tensor product" of the two single-particle spaces, not
the "direct product". See:

http://en.wikipedia.org/wiki/Direct_product
http://en.wikipedia.org/wiki/Tensor_product

The tensor product ensures the correct linearity properties (correct superposition
of 2-particle states).

A tensor product of 2 vector spaces yields another vector space. An
Hermitian inner product can be defined on the tensor product space
in this case easily enough. That almost gets you a Hilbert space.
For the final step (relating to the Hilbert space being "complete"),
one appeals to product topology (relating to the topologies of
the component spaces) to define "convergence" sensibly.

4. Oct 29, 2008

### Haelfix

Incidentally the confusion between tensor product, direct product, semidirect product, etc etc all applied to different mathematical objects (fields, rings, groups, spaces, modules etc etc), is one of the worst abuses of language I know off, and caused me a lot of grief back in the day.

Not only is it confusing to keep track off, in some cases there really is only surface level similarities between the concepts when applied.

In physics in particular, this can be particularly irratating, b/c its not always obvious upon inspection what various quantities actually are.

5. Oct 29, 2008

### KFC

Oh, this is really confusing. Because many textbook just said direct product. I just look it up in some mathematics book, and I guess the 'direct product' in some textbook is actually an abbreviation of 'tensor direct product'

6. Oct 29, 2008

### Fredrik

Staff Emeritus
If it's not too much work for you, would you mind writing a short summary of when the different products are used and why? I actually still find these things a bit confusing. E.g. in the construction of a Fock space, the n-particle spaces of each species are supposed to be tensor products, right? And the Fock space itself is a direct sum of those? Is a "direct product" ever used at all?

7. Oct 30, 2008

### strangerep

I'm guessing those were physics textbooks? Those tend to be insufficiently rigorous
in such matters.

I doubt that, but perhaps different authors have different conventions.
If it's a mathematics book, there should be a rigorous definition of the
concepts somewhere (or at least a reference to another math book containing
one).

Roughly, one could think of "direct product" of two vector spaces
V and W as the cartesian product $V\times W$. If v,w are vectors
in V, W respectively, then the pair (v,w) is in $V\times W$. However,
(2v, w) and (v, 2w) are distinct elements of $V\times W$, whereas
in the tensor product $V\otimes W$ these two are considered equivalent
(ie a single vector in the tp space). That's one of the properties
you want for a 2-particle Hilbert space: $2(\psi_1\psi_2)$ , $(2\psi_1)\psi_2$
and $\psi_1 (2 \psi_2)$ should all be the same state physically.

Last edited: Oct 30, 2008
8. Oct 30, 2008

### atyy

Isn't $2\psi_1$ the same state as $\psi_1$?

9. Oct 30, 2008

### atyy

10. Oct 30, 2008

### KFC

R. Shankar, Principles of Quantum mechanics