Computer Science Stack Exchange is a question and answer site for students, researchers and practitioners of computer science. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I can not understand the following equality $$\langle ij|(|0\rangle \langle 0|\otimes I)kl \rangle= \langle i|0\rangle \langle 0|k \rangle \langle j|I|l \rangle?$$

Also to estimate phase $\phi$ in Nielsen & Chuang book, I can not understand why $(|0 \rangle + e^{2\pi i 2^{t-1}\phi} |1 \rangle)(|0 \rangle + e^{2\pi i2^{t-2}\phi }|1 \rangle)\cdots (|0 \rangle + e^{2\pi i 2^{0}\phi} |1 \rangle)= \displaystyle\sum_{k=0}^{2^t-1}e^{2\pi i \phi^k} |k\rangle$.

Will you kindly help me?

share|cite|improve this question

migrated from Oct 27 '12 at 6:24

This question came from our site for theoretical computer scientists and researchers in related fields.

Isn't there an error in the powers? should be $|0 \rangle + e^{2\pi i 2^{t-1}\phi} |1 \rangle$ instead, right? – Ran G. Oct 27 '12 at 6:49
Yes, there was a mistake. – user12290 Oct 30 '12 at 19:53
I changed it, verify that it is correct (you can edit if there's still an error) – Ran G. Oct 31 '12 at 1:33
up vote 10 down vote accepted

A tensor product of operations, $I\otimes J$ say, acts on each subsystem separately: if $\phi$ and $\psi$ are states and $I$ and $J$ are operators then $$(I\otimes J)(\phi\otimes \psi) = (I\phi) \otimes (J\psi)$$ In bra-ket notation the state $\phi\otimes \psi$ can be denoted $|\phi\rangle|\psi\rangle$. In your first equation, the $\langle i|0\rangle\langle 0|k\rangle$ factor and $\langle j|I|l\rangle$ factor just separate in this way.

The algebra behind the second equation is basically: $$(1+x^{2^0})\dots(1+x^{2^{t-1}})=1+x+x^2+x^2+\dots+x^{2^t-1}$$ except that the "1" is replaced by $| 0\rangle$, and the $x$ is replaced by $\exp(2\pi i \phi |1\rangle)$ (clash of my notation: $\phi$ is now a number). The only difference is that the multiplication is really a tensor product, and with bosons $|1\rangle\otimes |1\rangle=|2\rangle$.

share|cite|improve this answer
+1: good answer. One caveat: you can get into trouble by assuming that with bosons, |1⟩⊗|1⟩=|2⟩, and naively manipulating equations. Bosons are a little more complicated than that. – Peter Shor Oct 24 '12 at 13:09

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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