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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.

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    $\begingroup$ 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. $\endgroup$ – gj255 Sep 17 '14 at 10:12
  • $\begingroup$ @gj255 Good point, I take it back :) $\endgroup$ – Danu Sep 17 '14 at 11:39
  • $\begingroup$ 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 $\oplus$. $\endgroup$ – user10851 Sep 17 '14 at 17:08
  • $\begingroup$ The first way, $\langle r_1 | \otimes \langle r_2 |$ is way more common, in my experience. $\endgroup$ – DanielSank Sep 17 '14 at 20:54
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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

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    $\begingroup$ This is probably the easiest way to not get lost with factor orderings, but it's important to keep in mind that conventions do vary. $\endgroup$ – Emilio Pisanty Sep 17 '14 at 17:45
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It is a matter of definition of whether you want to revert the order of vector spaces on tensor products or not when going to the complex conjugate vector space, i.e. in physics jargon: from ket-spaces to bra spaces. Different authors use different conventions.

In particular, in the case of super vector spaces with Grassmann-odd elements, in order to minimize sign factors, different conventions are useful for different tasks.

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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 $\mathbb C\otimes\mathbb C=\mathbb C$.

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  • $\begingroup$ 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. $\endgroup$ – Emilio Pisanty Sep 17 '14 at 17:43

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