Does Dirac notation apply to tensor product in tensor analysis?

[jk22]

Just a question : do we have in Dirac notation $$\langle u|A|u\rangle\langle u|B|u\rangle=\langle u|\langle u|A\otimes B|u\rangle |u\rangle$$ ?

 

[Geofleur]

That doesn't look right to me. If we translate the Dirac notation into tensor notation then $\langle u |$ corresponds to $u^1$, a linear functional and $|u\rangle$ corresponds to $\textbf{u}$, a vector. Then $\langle u | A | u \rangle$ is the same as $u^1(A\textbf{u})$, the functional $u^1$ acting on the vector $A\textbf{u}$. Similarly, $\langle u | B | u \rangle$ corresponds to $u^1(B\textbf{u})$. Then I guess we would write $u^1(A\textbf{u})u^1(B\textbf{u})$ in Dirac notation as $\langle u | \otimes \langle u | (A| u \rangle, B| u \rangle)$. The first $\langle u |$ would act on the $A | u \rangle$ and the second $\langle u |$ would act on the $B | u \rangle$.

 

[DEvens]

You have not defined what you mean by $\otimes$ in this case. As well, any kind of product of two operators is not necessarily going to be the same as the product of their results in brackets of this kind. How do the two operators act through each other? If they do.

 

[jk22]

I was thinkg of A and B be square matrices and $$\otimes$$ the kronecker product.

 

[kith]

Just a question : do we have in Dirac notation $$\langle u|A|u\rangle\langle u|B|u\rangle=\langle u|\langle u|A\otimes B|u\rangle |u\rangle$$ ?

So $A$ and $B$ act on the same space? Without context, the right-hand side looks like an unnecessary inflation of the state space but technically correct to me.

 

[kith]

Then I guess we would write $u^1(A\textbf{u})u^1(B\textbf{u})$ in Dirac notation as $\langle u | \otimes \langle u | (A| u \rangle, B| u \rangle)$. The first $\langle u |$ would act on the $A | u \rangle$ and the second $\langle u |$ would act on the $B | u \rangle$.

Your use of Dirac notation seems quite non-standard to me. I haven't seen it in QM texts. Instead of your $(\langle a| \otimes \langle b|) (|c\rangle, |d\rangle)$ I would write $(\langle a| \otimes \langle b|) (|c\rangle \otimes |d\rangle)$ which has the usual symmetry between bra and ket vectors.

 

[Geofleur]

Ah, I see. In tensor analysis, particularly in the modern sort where tensors are viewed as multilinear maps, that kind of notation is common. But the bra-ket notation is not used in that context, at least not in the books I have been reading. I'm just happy that what I wrote down actually does make sense!