I am trying to show equality of two intermediate steps in the rearrangement of the Quantum Fourier transform definition, but I do not know how to rearrange the coefficients of a tensor product.
The text claims that $$ \frac{1}{2^{n/2}}\sum_{k_1=0}^{1}\sum_{k_2=0}^{1} \cdots \sum_{k_n=0}^{1}{e^{2\pi ij \left( \sum_{l=1}^n{k_l 2^{-l}} \right)} |k_1 \ldots k_n\rangle} = \frac{1}{2^{n/2}}\sum_{k_1=0}^{1}\sum_{k_2=0}^{1} \cdots \sum_{k_n=0}^{1}{ \bigotimes_{l=1}^n e^{2\pi ijk_l 2^{-l}} |k_l\rangle} $$
Isolating the parts that change leaves us with $$ e^{2\pi ij \left( \sum_{l=1}^n{k_l 2^{-l}} \right)} |k_1 \ldots k_n \rangle = \bigotimes_{l=1}^n e^{2\pi ijk_l 2^{-l}} |k_l \rangle. $$ If I were to look at a small case letting $ n = 3. $ I would get the following on the left hand side, $$ e^{2\pi i j \left( k_12^{-1} + k_22^{-2} + k_32^{-3} \right) } |k_1k_2k_3 \rangle, $$ and the following on the right hand side, $$ e^{2 \pi i jk_12^{-1}} |k_1\rangle \otimes e^{2 \pi i jk_22^{-2}}|k_2\rangle \otimes e^{2 \pi i jk_32^{-3}}|k_3\rangle. $$ Is there a rule that is similar to $$ a |k_1\rangle \otimes b |k_2\rangle = ab|k_1k_2\rangle$$ that will allow me to rewrite the RHS to be equal to the LHS as desired.
I would also like to ask for reference suggestions to strengthen my understanding of tensor product algebra as well.
Note: The \ket{} operator I typically use in LaTeX does not seem to be compatible here, is there a different solution besides using a | and \rangle?
I would suggest the notes on tensor products and quantum entanglement by Zwiebach : https://ocw.mit.edu/courses/physics/8-05-quantum-physics-ii-fall-2013/lecture-notes/MIT8_05F13_Chap_08.pdf . This is nice for physical application of quantum mechanics and shows the coherence between math and physics. This link could also be useful: http://hitoshi.berkeley.edu/221a/tensorproduct.pdf .
Now I will do my best answering the question. The tensor product of two states is just $ab|k_1 \rangle \otimes |k_2 \rangle$ in your notation (you can put $a,b$ out of the tensor product). Scalars can be taken out of the tensor product. Now we now that using a isomorphism for ket vectors with the standard basis we have $|k_1 \rangle \cong e_1$. Basically what you can see from this is that the tensor product of two vectors gives another vector using the Kronecker product for matrices (vector is a $(n \times 1)$-matrix. So we get $|k_1 \rangle \otimes |k_1 \rangle \cong e_1 \otimes e_1 = (10...0)^T \otimes (10...0)^T$
two vectors with $n$ entries. This gives a vector that looks like $(10...0)^T$, so the same form but now there are more zeros, namely $n^2$. You can simply multiply the number of entries of the objects you take the tensor product of. Now we can return to ket notation using the following isomorphism : $(10...0)^T \cong |k_1 k_1 \rangle$. Note that the Hilbert space itself is now a tensor product of Hilbert space i.e. $H=H_A \otimes H_B$ generally. So you just combine Hilbert spaces to get combined kets.
Your case has $|k_1 \rangle \otimes |k_2 \rangle \cong e_1 \otimes e_2 = (10...0)^T \otimes (01...0)^T = (0...1..0)^T$. The second vector has a 1 at the second entry and the last vector has a 1 at the $(n+1)$-th entry. We define now a isomorphism to return to ket notation using $(0...1..0)^T \cong |k_1 k_2 \rangle$. The last step means that we just use the label $k_1 k_2$ in the ket. Note that for $k_i,k_j$ with $i,j \in \{1,2,...,n\}$ there would be $n^2$ different kets (which you can label whatever you want, Pete is also valid for example if you like). The conclusion is that this is a valid thing to do what you said.
The last question: I also do not know how the write kets in another way here :). Hope this post helps, greets Dani.