# Tensor product of kets and bras

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?

• 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! – Cosmas Zachos Jan 31 '19 at 16:43
• Where did you get that from? $\langle a | \otimes 1$ makes no sense, you can't tensor a vector with an operator. – Javier Jan 31 '19 at 17:28
• Here is where I encountered with it : [link] : (physics.stackexchange.com/questions/276053/…) – Farshid Shateri Jan 31 '19 at 17:39
• 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. – GodotMisogi Jan 31 '19 at 17:56

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 $$A$$.

• You are write. But, I think I had a typo. By 1, I mean identity operator not a 1 bra in second subsystem – Farshid Shateri Jan 31 '19 at 17:44
• @FarshidShateri yes, that is how I interpreted it. It maps $|b\rangle$ to itself, not to a number as a bra would have done – doetoe Jan 31 '19 at 17:56

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.

• 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. – Farshid Shateri Jan 31 '19 at 17:15