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Let $H$ be a separable Hilbert space, $T(H)$ the set of trace-class operators, and $D(H)\subset T(H)$ the set of density operators (i.e., positive and having trace 1).

For any unit vector $\vert \alpha \rangle \in H$, consider the pure state $\vert \alpha \rangle\langle \alpha\vert \in D(H)$ represented in a complete orthonormal basis $\vert e_k \rangle $ as $\vert \alpha \rangle\langle \alpha\vert= \sum_{i=1,j=1} c_i c_j^* \vert e_i\rangle \langle e_j \vert$, where $c_i = \langle \alpha \vert e_i\rangle$. Is is possible to prove that this series converges under the trace norm, i.e., that $$\lim_{n\to \infty} \Big\Vert \vert \alpha \rangle \langle \alpha \vert - \sum_{i=1}^n \sum_{j=1}^n c_i c_j^* \vert e_i\rangle \langle e_j \vert \Big\Vert = 0,$$ where $\Vert \cdot \Vert$ is the trace norm?

If it does not, would it be possible to prove this under e.g., the operator norm topology?

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  • $\begingroup$ If $\rho$ is a density operator then what is $\langle \rho \vert e_i\rangle$? You can not have an inner product of an operator with a vector. The sum you wrote looks like a rank one operator. Is your $\rho$ supposed to be just $\rho x:=\langle x\vert c\rangle c$, where $c=\sum_{i=1}^\infty c_i e_i$ is a fixed vector? $\endgroup$
    – Conifold
    Sep 22, 2020 at 5:58
  • $\begingroup$ Apologies, that was a typo -- $\rho$ should be a pure state, and $c_i$ the inner product of $e_i$ with the corresponding vector. Fixed now. $\endgroup$
    – Artemy
    Sep 22, 2020 at 6:18
  • $\begingroup$ By "this series" do you mean the sequence $\sum_{i,j=1}^n c_i c_j^* \vert e_i\rangle \langle e_j \vert$ when $n\to\infty$, or do you really want the double series to converge? $\endgroup$
    – Conifold
    Sep 22, 2020 at 6:26
  • $\begingroup$ @Conifold I think the former is fine, I clarified the question. $\endgroup$
    – Artemy
    Sep 22, 2020 at 15:40

1 Answer 1

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A bit late but here is some work for posterity.

$$\lim_{n\to \infty} \Big\Vert \vert \alpha \rangle \langle \alpha \vert - \sum_{i=1}^n \sum_{j=1}^n c_i c_j^* \vert e_i\rangle \langle e_j \vert \Big\Vert = $$

$$ \lim_{n\to \infty} \Big\Vert \sum_{i=1}^\infty \sum_{j=1}^\infty c_i c_j^* \vert e_i\rangle \langle e_j \vert - \sum_{i=1}^n \sum_{j=1}^n c_i c_j^* \vert e_i\rangle \langle e_j \vert \Big\Vert = $$

$$ \lim_{n\to \infty} \Big\Vert \sum_{i=n+1}^\infty \sum_{j=n+1}^\infty c_i c_j^* \vert e_i\rangle \langle e_j \vert \Big\Vert \leq $$

(Triangle inequality for the trace-norm)

$$ \lim_{n\to \infty} \sum_{i=n+1}^\infty \sum_{j=n+1}^\infty |c_i c_j^*|\Big\Vert \vert e_i\rangle \langle e_j \vert \Big\Vert = $$ ( $\| |e_{i}\rangle \langle e_{j}| \|_{1} = 1$, check i!) $$ \lim_{n\to \infty} \sum_{i=n+1}^\infty \sum_{j=n+1}^\infty |c_i c_j^*| = 0 $$

If the values $|c_{j}| $ form a summable sequence then the last line is obviously true.

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