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If $\mid\phi\rangle=a_0\mid 00\rangle + a_1\mid 01\rangle +a_2\mid 10\rangle +a_3\mid 11\rangle$ and $P_0=\mid 0\rangle\langle 0\mid \otimes I$, how do we show that $\langle\phi \mid P_0\mid \phi\rangle =\langle\phi \mid a_0\mid 00\rangle+\langle \phi\mid a_1\mid 01\rangle =|a_0|^2+|a_1|^2$

I get the following if we only consider the first quibit (assuming that's what $\otimes I$ does,.... I'm still not sure on that either) :

$(a_0 \langle 0\mid +a_1\langle 0\mid)(\mid 0\rangle\langle0\mid)(a_0\mid0\rangle + a_1\mid0\rangle)=$

$(a_0 \langle 0\mid +a_1\langle 0\mid)(a_0\mid 0\rangle\langle0\mid0\rangle+a_1\mid 0\rangle\langle0\mid0\rangle$)$ = a_0^2+2a_0a_1+a_1^2$

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I changed "$<$" and "$>$" to "$\langle$" and "$\rangle$". Seconds later, rschwieb changed "$|$" to "$\mid$". The difference between "$|$" and "$\mid$" is this: "$a|b$" versus "$a\mid b$". –  Michael Hardy Aug 11 '12 at 19:30

2 Answers 2

up vote 2 down vote accepted

Here's the complete calculation:

First, use linearity to get the individual terms: $$\begin{aligned} \langle\phi\rvert P_0\lvert\phi\rangle &= \left(a_0^*\langle00\rvert + a_1^*\langle01\rvert + a_2^*\langle10\rvert + a_3^*\langle11\rvert\right) P_0 \left(a_0\lvert00\rangle + a_1\lvert01\rangle + a_2 \lvert10\rangle+a_3\lvert11\rangle\right)\\ &= a_0^*a_0\langle00\rvert P_0\vert00\rangle + a_0^*a_1\langle00\rvert P_0\vert01\rangle + \dots + a_3^*a_3\langle11\rvert P_0\vert11\rangle \end{aligned}$$ Note that $\lvert ij\rangle = \lvert i\rangle\otimes\lvert j\rangle$. Thus we have $$\begin{aligned} \langle ij\rvert P_0 \lvert kl\rangle &= \langle ij\rvert\left(\lvert0\rangle\langle0\rvert\otimes I\right)\lvert kl\rangle\\ &= \langle i\rvert0\rangle\langle0\rvert k\rangle\cdot \langle j\rvert I\lvert l\rangle\\ &= \delta_{i0}\delta_{k0}\delta_{jl} \end{aligned}$$ That is, if all factors are tensor products, you just evaluate every term separately and get the tensor product of the results (the tensor product of numbers is just the normal product of numbers). This is a general rule, so for example $(A\otimes B)(C\otimes D)=(AC)\otimes(BD)$, or $(A\otimes B)(\lvert\phi\rangle\otimes\lvert\psi\rangle) = (A\lvert\phi)\otimes (B\lvert\psi\rangle)$. Now it's just a matter of inserting that in the first equation and noticing that the only non-zero terms are those with $i=k=0$ and $j=l$, and the corresponding terms have prefactors $a_0^*a_0=\left|a_0\right|^2$ and $a_1^*a_1=\left|a_1\right|^2$. Thus you finally get $$ \langle\phi\rvert P_0\lvert\phi\rangle = \left|a_0\right|^2 + \left|a_1\right|^2$$

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Hey, thanks. An answer that doesn't actually assume I know what I'm asking. –  Squirtle Aug 11 '12 at 21:23
Will you kindly explain the step $\langle ij|(|0\rangle \langle 0|\otimes I)kl \rangle= \langle i|0\rangle \langle 0|k \rangle \langle j|I|l \rangle$? –  user12290 Oct 23 '12 at 20:13
@user12290: Sorry, I didn't see your comment until now (as you might have noticed on my activity, I've not been here for quite some time). The point is that for tensor products, $(A\otimes B)(C \otimes D) = (AC) \otimes (BD)$. Thus we actually get $\left<i\middle|0\right>\left<0\middle|k\right> \otimes \left<i\middle|I\middle|l\right>$. But both factors are just complex numbers, therefore the tensor product reduces to the ordinary product. –  celtschk May 26 '13 at 13:29

The projector $|0\rangle\langle0|$ projects onto the $|0\rangle$ state of the first factor. Thus it leaves the first and second terms invariant and annihilates the third and fourth terms (assuming that $|0\rangle$ and $|1\rangle$ form an orthonormal basis of the first factor). Then multiplying by $\langle\phi|$ and using the orthonormality of $|0\rangle$ and $|1\rangle$ in both factors yields the result.

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Really my question is: What does |0><0|⊗I mean? How is it read? Like (|0><0|)⊗I... or what? Even the most simple example, <x|(|0><0|⊗I) |x> means nothing to me (where I assume |x> is a two-quibit. –  Squirtle Aug 11 '12 at 19:30
@dustanalysis: Then why didn't you ask that? –  joriki Aug 11 '12 at 19:30
Okay, my bad. So..... –  Squirtle Aug 11 '12 at 19:34

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