5
$\begingroup$

$\newcommand{\ket}[1]{\lvert#1\rangle}$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 $$ \frac{1}{2^{n/2}}\sum_{k_1=0}^{1}\sum_{k_2=0}^{1} \cdots \sum_{k_n=0}^{1}e^{2\pi ij \left( \sum_{l=1}^n{k_l 2^{-l}} \right)} \ket{k_1 \ldots k_n} = \frac{1}{2^{n/2}}\sum_{k_1=0}^{1}\sum_{k_2=0}^{1} \cdots \sum_{k_n=0}^{1}{ \bigotimes_{l=1}^n e^{2\pi ijk_l 2^{-l}} \ket{k_l}} $$

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.

$\endgroup$
  • 2
    $\begingroup$ see edit for how to use the ket command. Also, the answer to the question is essentially yes, that is how tensor products work. Indeed, the tensor product operation is (multi)linear by definition, as also pointed out in the wiki: en.wikipedia.org/wiki/… $\endgroup$ – glS Aug 14 '18 at 20:37
3
$\begingroup$

Expanding and generalising from Jitendra's answer: the key observation in this case is that you must use how scalar factors behave over tensor products. Specifically, $$ a (U \otimes V) = (aU) \otimes V = U \otimes (aV), $$ or more generally $$ a_1 a_2 \cdots a_k \,(U_1 \otimes U_2 \otimes \cdots \otimes U_k) = (a_1 U_1) \otimes (a_2 U_2) \otimes \cdots \otimes (a_k U_k). $$ Let's consider the left-hand side of the expression which you isolated, $$ \exp\bigl(2\pi ij \sum_{\ell=1}^n{k_\ell 2^{-\ell}} \bigr) |k_1 \cdots k_n \rangle \;: $$ we may use the fact that $\exp(a+b+\cdots+h) = \exp(a) \exp(b) \cdots \exp(h)$ to re-express this as $$ = \Bigl[\, \prod_{\ell=1}^n \exp\bigl(2\pi ij k_\ell \big/ 2^{\ell} \bigr) \Bigr] |k_1 \cdots k_n \rangle \;. $$ We next use the fact that $\lvert k_1 k_2 \cdots k_n \rangle$ is short-hand for a tensor product of operators (state-vectors to be specific): $$ = \Bigl[\, \prod_{\ell=1}^n \exp\bigl(2\pi ij k_\ell \big/ 2^{\ell} \bigr) \Bigr] \Bigl[\, \bigotimes_{\ell=1}^n |k_\ell \rangle \Bigr] \;. $$ Now we use the way that scalars interact with tensor products: $$ = \bigotimes_{\ell=1}^n \Bigl[ \exp\bigl(2\pi ij k_\ell \big/ 2^{\ell} \bigr) |k_\ell \rangle \Bigr] \;, $$ which was what we wanted to show.

$\endgroup$
1
$\begingroup$

Isn't $$ e^{2\pi i j \left( k_12^{-1} + k_22^{-2} + k_32^{-3} \right) } |k_1k_2k_3 \rangle, $$ same as $$ e^{2 \pi i jk_12^{-1}} |k_1\rangle \otimes e^{2 \pi i jk_22^{-2}}|k_2\rangle \otimes e^{2 \pi i jk_32^{-3}}|k_3\rangle. $$

The kets gets the kronecker product according to $ a |k_1\rangle \otimes b |k_2\rangle = ab|k_1k_2\rangle$ and the exponentials adds up according to $e^{x+y}=e^x.e^y$. Just keep in mind that this identity of exponential holds only if $x$ and $y$ commutes. In your case the power to which exponentials are raised are just scalar numbers and thus commutes.

$\endgroup$

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.