- #1
fay
Hello everyone,
I started reading the Nielsen and Chuang book on quantum computation and quantum informations. I got stuck by the last question of Problem 2.2. I got the other problems, but i can't see this one. I guess it's not really difficult, but as i am new in this field, some help will be nice :)
Suppose ##\rvert \psi \rangle ## is a pure state of a composite system with components ##A## and ##B##, such that: $$\vert \psi \rangle = \alpha \rvert \phi \rangle + \beta \rvert \gamma \rangle$$
Prove that:
$$ \textrm{Sch}(\psi) \geq \rvert \textrm{Sch}(\phi) - \textrm{Sch}(\gamma) \rvert$$
where ##\textrm{Sch}(x)## is the Schmidt number of the pure state labeled ##x##.
2. The attempt at a solution
Here is what i tried. Let's assume that ## \textrm{Sch}(\phi) > \textrm{Sch}(\gamma)##. If we write the Schmidt decomposition of ##\phi## and ##\gamma##:
$$
\rvert \phi \rangle = \sum_i \phi_i \rvert a_i^{\phi} \rangle \rvert b_i^{\phi} \rangle\\
\rvert \gamma \rangle = \sum_i \gamma_i \rvert a_i^{\gamma} \rangle \rvert b_i^{\gamma} \rangle
$$
we can then calculate the partial trace of ##\psi## regarding component ##A##:
$$
\rho \equiv tr_B(\rvert \psi \rangle \langle \psi \lvert) = |\alpha|^2 \rho_{\phi \phi} + |\beta|^2 \rho_{\gamma \gamma} + \alpha \bar{\beta} \rho_{\phi \gamma} + \bar{\alpha} \beta \rho_{\gamma \phi}
$$
where:
$$
\rho_{\phi \phi} \equiv tr_B(\rvert \phi \rangle \langle \phi \lvert) = \sum_i \phi_i^2 \rvert a_i^{\phi} \rangle \langle a_i^{\phi} \lvert\\
\rho_{\gamma \gamma} \equiv tr_B(\rvert \gamma \rangle \langle \gamma \lvert) = \sum_i \gamma_i^2 \rvert a_i^{\gamma} \rangle \langle a_i^{\gamma} \lvert\\
\rho_{\phi \gamma}\equiv tr_B(\rvert \phi \rangle \langle \gamma \lvert) = \sum_i \phi_i \rvert a_i^{\phi} \rangle \sum_j \gamma_j \langle b_j^{\gamma} | b_i^{\phi} \rangle \langle a_j^{\gamma} |\\
\rho_{\gamma \phi}\equiv tr_B(\rvert \gamma \rangle \langle \phi \lvert) = \sum_i \gamma_i \rvert a_i^{\gamma} \rangle \sum_j \phi_j \langle b_j^{\phi} | b_i^{\gamma} \rangle \langle a_j^{\phi} |
$$
it can be seen that ##Im(\rho_{\phi \gamma})## is a subspace of ##Im(\rho_{\phi})##, and similarly that ##Im(\rho_{\gamma \phi})## is a subspace of ##Im(\rho_{\gamma})##. We define ## P_{\gamma} ## the projector onto ##Im(\rho_{\gamma})## and
$$
P_{\gamma}^{\perp} = I - P_{\gamma}
$$
the projection onto the orthogonal complement of ##Im(\rho_{\gamma})##. As ##\textrm{Sch}(\gamma) < \textrm{Sch}(\psi)##, the subspace corresponding to ##P_{\gamma}^{\perp}## is not reduced to the zero vector.
We end-up with:
$$
\rho = P_{\gamma}( |\alpha|^2 \rho_{\phi \phi} + |\beta|^2 \rho_{\gamma \gamma} + \alpha \bar{\beta} \rho_{\phi \gamma} + \bar{\alpha} \beta \rho_{\gamma \phi} ) + P_{\gamma}^{\perp} ( |\alpha|^2 \rho_{\phi \phi} + \alpha \bar{\beta} \rho_{\phi \gamma} )
$$
As the subspaces corresponding to the two defined projectors are orthogonal:
$$
rank(\rho) = rank(P_{\gamma}( |\alpha|^2 \rho_{\phi \phi} + |\beta|^2 \rho_{\gamma \gamma} + \alpha \bar{\beta} \rho_{\phi \gamma} + \bar{\alpha} \beta \rho_{\gamma \phi} )) + rank(P_{\gamma}^{\perp} ( |\alpha|^2 \rho_{\phi \phi} + \alpha \bar{\beta} \rho_{\phi \gamma} )) \geq rank(P_{\gamma}^{\perp} ( |\alpha|^2 \rho_{\phi \phi} + \alpha \bar{\beta} \rho_{\phi \gamma} ))
$$
I was hopping to conclude by claiming that:
$$
rank(P_{\gamma}^{\perp} ( |\alpha|^2 \rho_{\phi \phi} + \alpha \bar{\beta} \rho_{\phi \gamma} )) \geq \textrm{Sch}(\phi) - \textrm{Sch}(\gamma)
$$
but it is not correct, as i find counter examples. Therefore i don't think that i am following the right track.
I started reading the Nielsen and Chuang book on quantum computation and quantum informations. I got stuck by the last question of Problem 2.2. I got the other problems, but i can't see this one. I guess it's not really difficult, but as i am new in this field, some help will be nice :)
Homework Statement
Suppose ##\rvert \psi \rangle ## is a pure state of a composite system with components ##A## and ##B##, such that: $$\vert \psi \rangle = \alpha \rvert \phi \rangle + \beta \rvert \gamma \rangle$$
Prove that:
$$ \textrm{Sch}(\psi) \geq \rvert \textrm{Sch}(\phi) - \textrm{Sch}(\gamma) \rvert$$
where ##\textrm{Sch}(x)## is the Schmidt number of the pure state labeled ##x##.
2. The attempt at a solution
Here is what i tried. Let's assume that ## \textrm{Sch}(\phi) > \textrm{Sch}(\gamma)##. If we write the Schmidt decomposition of ##\phi## and ##\gamma##:
$$
\rvert \phi \rangle = \sum_i \phi_i \rvert a_i^{\phi} \rangle \rvert b_i^{\phi} \rangle\\
\rvert \gamma \rangle = \sum_i \gamma_i \rvert a_i^{\gamma} \rangle \rvert b_i^{\gamma} \rangle
$$
we can then calculate the partial trace of ##\psi## regarding component ##A##:
$$
\rho \equiv tr_B(\rvert \psi \rangle \langle \psi \lvert) = |\alpha|^2 \rho_{\phi \phi} + |\beta|^2 \rho_{\gamma \gamma} + \alpha \bar{\beta} \rho_{\phi \gamma} + \bar{\alpha} \beta \rho_{\gamma \phi}
$$
where:
$$
\rho_{\phi \phi} \equiv tr_B(\rvert \phi \rangle \langle \phi \lvert) = \sum_i \phi_i^2 \rvert a_i^{\phi} \rangle \langle a_i^{\phi} \lvert\\
\rho_{\gamma \gamma} \equiv tr_B(\rvert \gamma \rangle \langle \gamma \lvert) = \sum_i \gamma_i^2 \rvert a_i^{\gamma} \rangle \langle a_i^{\gamma} \lvert\\
\rho_{\phi \gamma}\equiv tr_B(\rvert \phi \rangle \langle \gamma \lvert) = \sum_i \phi_i \rvert a_i^{\phi} \rangle \sum_j \gamma_j \langle b_j^{\gamma} | b_i^{\phi} \rangle \langle a_j^{\gamma} |\\
\rho_{\gamma \phi}\equiv tr_B(\rvert \gamma \rangle \langle \phi \lvert) = \sum_i \gamma_i \rvert a_i^{\gamma} \rangle \sum_j \phi_j \langle b_j^{\phi} | b_i^{\gamma} \rangle \langle a_j^{\phi} |
$$
it can be seen that ##Im(\rho_{\phi \gamma})## is a subspace of ##Im(\rho_{\phi})##, and similarly that ##Im(\rho_{\gamma \phi})## is a subspace of ##Im(\rho_{\gamma})##. We define ## P_{\gamma} ## the projector onto ##Im(\rho_{\gamma})## and
$$
P_{\gamma}^{\perp} = I - P_{\gamma}
$$
the projection onto the orthogonal complement of ##Im(\rho_{\gamma})##. As ##\textrm{Sch}(\gamma) < \textrm{Sch}(\psi)##, the subspace corresponding to ##P_{\gamma}^{\perp}## is not reduced to the zero vector.
We end-up with:
$$
\rho = P_{\gamma}( |\alpha|^2 \rho_{\phi \phi} + |\beta|^2 \rho_{\gamma \gamma} + \alpha \bar{\beta} \rho_{\phi \gamma} + \bar{\alpha} \beta \rho_{\gamma \phi} ) + P_{\gamma}^{\perp} ( |\alpha|^2 \rho_{\phi \phi} + \alpha \bar{\beta} \rho_{\phi \gamma} )
$$
As the subspaces corresponding to the two defined projectors are orthogonal:
$$
rank(\rho) = rank(P_{\gamma}( |\alpha|^2 \rho_{\phi \phi} + |\beta|^2 \rho_{\gamma \gamma} + \alpha \bar{\beta} \rho_{\phi \gamma} + \bar{\alpha} \beta \rho_{\gamma \phi} )) + rank(P_{\gamma}^{\perp} ( |\alpha|^2 \rho_{\phi \phi} + \alpha \bar{\beta} \rho_{\phi \gamma} )) \geq rank(P_{\gamma}^{\perp} ( |\alpha|^2 \rho_{\phi \phi} + \alpha \bar{\beta} \rho_{\phi \gamma} ))
$$
I was hopping to conclude by claiming that:
$$
rank(P_{\gamma}^{\perp} ( |\alpha|^2 \rho_{\phi \phi} + \alpha \bar{\beta} \rho_{\phi \gamma} )) \geq \textrm{Sch}(\phi) - \textrm{Sch}(\gamma)
$$
but it is not correct, as i find counter examples. Therefore i don't think that i am following the right track.