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?

  • $\begingroup$ 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 '19 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 '19 at 17:28
  • $\begingroup$ Here is where I encountered with it : [link] : (physics.stackexchange.com/questions/276053/…) $\endgroup$ – Farshid Shateri Jan 31 '19 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 '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 $A$.

| cite | improve this answer | |
  • $\begingroup$ 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 '19 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 '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.

| cite | improve this answer | |
  • $\begingroup$ 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 '19 at 17:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.