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My question is about tensor products. I have learned that the tensor product between two operators is, for example, $A{\otimes}B$. Suppose we have a system $A$ and a system B. Why we can write the bra of single system A as $\langle{a}\rvert_A$=$\langle{a}\rvert_A$$\otimes$$\mathbb{1} $?
Isn't this representation just for operators?

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  • $\begingroup$ No, not really, a vector is a simple tensor, so it can formally be regarded as a tensor like any other. Moreover, significantly, a bra or a ket, as you could study in Dirac's standard Principles of QM book, is an operator acting on a "standard ket", a translationally invariant vacuum, after all! $\endgroup$ – Cosmas Zachos Jan 31 at 16:43
  • $\begingroup$ Where did you get that from? $\langle a | \otimes 1$ makes no sense, you can't tensor a vector with an operator. $\endgroup$ – Javier Jan 31 at 17:28
  • $\begingroup$ Here is where I encountered with it : [link] : (physics.stackexchange.com/questions/276053/…) $\endgroup$ – Farshid Shateri Jan 31 at 17:39
  • $\begingroup$ Consider a vector space $V$ over a field $F$. The field $F$ is a one-dimensional vector space over itself, and there exists a canonical isomorphism $V \cong V \otimes_{F} F$. Bilinearity of the tensor product allows you to pull out $F$-scalars, and you can represent any vector $v \in V$ as $v \otimes 1 \in V \otimes_{F} F$, as a result. This can be extended similarly to obtain a linear homomorphism from the tensor product of two vector spaces as a projection to one of the components via the universal property of tensor products. $\endgroup$ – GodotMisogi Jan 31 at 17:56
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We have a pair of state spaces $A$ and $B$, their combined state space is the tensor product $A\otimes B$. Its elements are of the form $|a\rangle\otimes|b\rangle$ and (limits of) sums of such elements.

Bra's are elements of the dual space $A^\ast$, which are (continuous) linear maps $A\to\mathbb C$. Lets write $\textrm{End}(B)$ for the space of all endomorphisms of $B$ (i.e. operators from $B$ to itself). Then $\langle a|\otimes\mathbf1\in A^\ast\otimes\textrm{End}(B)$.

Now let's see what it does on an element of $|\psi\rangle\otimes |b\rangle\in A\otimes B$:

$$\langle a|\otimes\mathbf{1}(|\psi\rangle\otimes|b\rangle) = \langle a|\psi\rangle\otimes|b\rangle.$$

In other words, $\langle a|\otimes\mathbf{1}$ acts exactly as $\langle a|$ on the factor $A$ and does nothing on $B$, i.e. as if $B$ were not really there. In that sense this element can be identified with the corresponding bra of the single system $A$.

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  • $\begingroup$ You are write. But, I think I had a typo. By 1, I mean identity operator not a 1 bra in second subsystem $\endgroup$ – Farshid Shateri Jan 31 at 17:44
  • $\begingroup$ @FarshidShateri yes, that is how I interpreted it. It maps $|b\rangle$ to itself, not to a number as a bra would have done $\endgroup$ – doetoe Jan 31 at 17:56
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The representation is for vectors too.

Think of it this way: the formalism is for matrices, of any size. Vectors are matrices with a single column.

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  • $\begingroup$ Thank you for your reply. But it seems that it doesnt mean to have atensor product between a vectro and an operator (for example, I have written the tensor product between a vector and Identity operator in my question). I do not have any problem with the tensor product of two vectors as it describes for example a state of a biparite system. $\endgroup$ – Farshid Shateri Jan 31 at 17:15

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