# Tensor Product in Quantum Computation

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?

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

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 $|1\rangle\otimes |1\rangle=|2\rangle$.

• +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. Oct 24 '12 at 13:09