Problem 2.2 in Nielsen & Chuang - Properties of the Schmidt number

[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 :)

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.

 

[mighty2000]

Any help would be appreciated.Hello,

I am a fellow scientist and I would be happy to help you with this problem. First, let's define the Schmidt decomposition for a pure state $\rvert \psi \rangle$ of a composite system with components $A$ and $B$:

$$
\rvert \psi \rangle = \sum_i \lambda_i \rvert a_i \rangle \rvert b_i \rangle
$$

where $\lambda_i$ are the Schmidt coefficients and $\rvert a_i \rangle$ and $\rvert b_i \rangle$ are orthonormal bases for components $A$ and $B$, respectively. The Schmidt number of a pure state is defined as the number of non-zero Schmidt coefficients, so for our state $\rvert \psi \rangle$, we have $\textrm{Sch}(\psi) = k$, where $k$ is the number of non-zero Schmidt coefficients.

Now, let's look at the Schmidt decomposition for the states $\rvert \phi \rangle$ and $\rvert \gamma \rangle$ given in the problem:

$$
\rvert \phi \rangle = \sum_i \phi_i \rvert a_i^{\phi} \rangle \rvert b_i^{\phi} \rangle\\
\rvert \gamma \rangle = \sum_i \gamma_i \rvert a_i^{\gamma} \rangle \rvert b_i^{\gamma} \rangle
$$

where $\phi_i$ and $\gamma_i$ are the Schmidt coefficients and $\rvert a_i^{\phi} \rangle$, $\rvert b_i^{\phi} \rangle$, $\rvert a_i^{\gamma} \rangle$, and $\rvert b_i^{\gamma} \rangle$ are orthonormal bases for components $A$ and $B$, respectively. We can see that the Schmidt numbers for these states are $\textrm{Sch}(\phi) = n$ and $\textrm{Sch}(\gamma) = m$, where $n$ and $m$ are the number of non-zero Schmidt coefficients for $\rvert \phi \rangle$ and $\rvert \gamma \rangle$, respectively.

Now, to prove the inequality, we will use the Cauchy-S