# I Tensor product in QM?

1. Apr 15, 2016

### pellman

A particle in a 1-D Hilbert space would have position basis states $|x \rangle$ where $\langle x' | x \rangle = \delta(x'-x)$ A 3-D Hilbert space for one particle might have a basis $| x,y,z \rangle$ where $\langle x', y', z' | x,y,z \rangle = \delta(x'-x) \delta (y-y') \delta(z-z')$ . Would it be correct to write $| x,y,z \rangle = | x \rangle \otimes | y \rangle \otimes | z \rangle$ ? Why or why not?

Call the 1-D Hilbert space $H_1$ and the 3-D Hilbert space $H_3$. Is this question equivalent to asking is $H_3 = H_1 \otimes H_1 \otimes H_1$?

2. Apr 15, 2016

### andrewkirk

Yes. To see why, it's easier to look at it the other was around. The full Hilbert space is defined as the tensor product of the the three 'component' Hilbert spaces, so that a basis for the full space is the set of all tensor products of basis elements of those three component spaces: $| x,y,z \rangle = | x \rangle \otimes | y \rangle \otimes | z \rangle$.

We then define the inner product on the product space in the most natural way, as:
$$\langle x,y,z|x',y',z'\rangle \equiv \langle x|x'\rangle\times\langle y|y'\rangle\times\langle z|z'\rangle$$
and extending linearly. We need to confirm that this obeys the inner product rules, but that's pretty easy to do.

It then follows that
$$\langle x,y,z|x',y',z'\rangle \equiv \langle x|x'\rangle\times\langle y|y'\rangle\times\langle z|z'\rangle\equiv\delta(x'-x) \delta (y-y') \delta(z-z')$$

The three Hilbert spaces are isomorphic to one another, but they are not the same space, as they relate to different physical phenomena. Hence it is more accurately represented by saying that thre three spaces are $H_a,H_b,H_c$, with $H_a\cong H_b\cong H_c$ and the full Hilbert space is $H_3\equiv H_a\otimes H_b\otimes H_c$.

3. Apr 16, 2016

### pellman

What does $\cong$ mean here?