# Tensor product notation convention?

For two particle state, the Dirac ket is writren as $$\lvert\textbf{r}_1\rangle \otimes \lvert\textbf{r}_2 \rangle.$$ Then how do we write its bra vector, $$\langle\textbf{r}_1\rvert \otimes \langle\textbf{r}_2\rvert ~~\text{or}~~\langle\textbf{r}_2\rvert \otimes \langle\textbf{r}_1\rvert ~~\text{?}$$ Is there any rule or convention? I'm just asking the order of bra vector.

• Actually I think one should not preserve 'symmetry'. I would have written the rule for taking the inner product of two tensor product states as this: $(\langle a| \otimes \langle b |) (|c \rangle \otimes |d \rangle) = \langle a | c \rangle \langle b | d \rangle$. And taking a look at (en.wikipedia.org/wiki/Tensor_product_of_Hilbert_spaces) this implies we shouldn't reverse the direction. – gj255 Sep 17 '14 at 10:12
• @gj255 Good point, I take it back :) – Danu Sep 17 '14 at 11:39
• Do note that the tensor product $\otimes$ is not the same as the direct product $\times$, the latter of which is really more akin to the direct sum $\oplus$. – user10851 Sep 17 '14 at 17:08
• The first way, $\langle r_1 | \otimes \langle r_2 |$ is way more common, in my experience. – DanielSank Sep 17 '14 at 20:54

The way I imagine it is that the left side of the direct product is exclusively reserved for hilbert space 1 and the right side is for Hilbert space 2. So that the total hilbert space you are working in is written as: $$H=H_1⊗H_2$$ And so when you have a wavefunction in H you write: $$|\psi\rangle = |r_1\rangle \otimes |r_2 \rangle$$ And then: $$|\psi\rangle^\dagger = (|r_1\rangle \otimes |r_2 \rangle)^\dagger = |r_1\rangle^\dagger \otimes |r_2 \rangle^\dagger$$ $$\langle\psi| = \langle r_1| \otimes \langle r_2|$$ Hence there is no way $r_1$ and $r_2$ can swap places
Remember that by definition of the tensor $$(a_1\otimes b_1)(a_2\otimes b_2)=(a_1a_2)\otimes(b_1b_2),$$ and that $\mathbb C\otimes\mathbb C=\mathbb C$.
• This is really a matter of convention, and the equality you give need not really hold. If $a_j\in\mathcal H_j$, and $⟨·,·⟩_{ij}:\mathcal H_i\times\mathcal H_j\to C$ are bilinear (sesquilinear) forms, then the tensor bilinear (sesquilinear) form $⟨·,·⟩_\otimes:\mathcal H_1\otimes\mathcal H_2\times\mathcal H_3\otimes\mathcal H_4\to C$ can be defined equally well as $$⟨a_1\otimes a_2,a_3\otimes a_4⟩_\otimes=⟨a_1,a_3⟩⟨a_2,a_4⟩$$ and as $$⟨a_1\otimes a_2,a_3\otimes a_4⟩_\otimes=⟨a_1,a_4⟩⟨a_2,a_3⟩.$$ It all depends on what the product is inside of that $)($, and how it is defined. – Emilio Pisanty Sep 17 '14 at 17:43