One-shot assisted concentration of coherence
Min-Hsiu Hsieh
arxiv:1804.06554 with Madhav Krishnan V & Eric Chitambar
Resource theory of coherence
Free Entities
Expensive Entities
\mathcal{I} = \left\{ \rho = \sum\limits_i p_i | i \rangle \langle i | \right\}
\rho\neq \sum\limits_i p_i | i \rangle \langle i |
Incoherent operations
Coherence creating operations
Fixed basis
\{ | i \rangle \}
| i \rangle \langle i |
| i \rangle \langle i |
Definitions
- Set of Incoherent states
\mathcal{I}
\delta = \sum\limits_i p_i | i \rangle\langle i |
- Set of Incoherent operations
\Lambda_{\mathcal{I}} : \mathcal{I} \rightarrow \mathcal{I}
\mathcal{O}
Measure of coherence
- Relative entropy of coherence
C_r(\rho)
C_r(\rho) = \min\limits_{\delta\in \mathcal{I}} S( \rho \| \delta)
= S(\Delta(\rho)) - S(\rho),^1
\text{where } \Delta \text{ dephases } \rho \text{ (deletes off-diagonal terms)}
1.Winter, A., & Yang, D. : PRL 116.12, 120404, (2016)
Maximally coherent state
| \Phi_M \rangle = \sum\limits_{i =1 }^M \frac{1}{\sqrt{M}} | i \rangle
Assisted distillation of coherence
Alice
Bob
\left(| \psi \rangle^{AB} \right)^{\otimes n}
\Lambda \text{ (unrestricted)}
\Lambda_{\mathcal{I}}
One-way classical communication
\Phi_2^{\otimes m}
Why assisted?
- The distributed scenario models how to localize coherence on an inaccessible remote target without a quantum channel.
- One-way communication is easier to model because there is no feedback to consider.
- When Bob's state is a qubit and the shared state is pure, then 2-way communication offers no advantage over 1-way communication.
- This can be relavent to some biological systems where coherence is important, open-destination quantum metrology where coherence improves precision. Or any protocol that requires a highly coherent initial state.
Assisted distillation of coherence
| Optimal rate |
S(\Delta(\rho^B))^1
n \rightarrow \infty
n = 1
1. Chitambar, E., et al. Physical review letters 116.7 (2016): 070402.
?
Method of the proof
- The assisted distillation task can be broken down into two parts
- Alice performs POVM on her part of the shared state and sends the result to Bob.
\psi^{AB}
\{ P^A_i\}
2. Bob performs an Incoherent operation on his part of the state based on Alice's outcome.
\Lambda_i
\max\limits_{\lbrace P^A_i \rbrace_i}\max\limits_{M \in \mathbb{N}} \left\lbrace \log_2 M : \max\limits_{\lbrace \Lambda^{B}_i \rbrace_i} F^{2}\left( \sum\limits_{i}p_i \Lambda_i^{B}(\rho_{i}^{B}), \Phi_{M}^{B^{\prime}} \right) \geq 1 - \epsilon \right\rbrace
\max\limits_{\lbrace p_i, \psi^{B}_i \rbrace_i}\max\limits_{M \in \mathbb{N}} \left\lbrace \log_2 M : \max\limits_{\lbrace \Lambda^{B}_i \rbrace_i} F^{2}\left( \sum\limits_{i}p_i \Lambda_i^{B}(\psi_{i}^{B}), \Phi_{M}^{B^{\prime}} \right) \geq 1 - \epsilon \right\rbrace
\text{where, }\sum\limits_i p_i \psi_i = \rho^B = \text{Tr}_A(\psi^{AB})
p_i \rho_i^B = \text{Tr}_A((P_i^A \otimes \mathbb{I}^B ) \psi^{AB})
\rho_i^B
\rho_i^B
- The best strategy is for Alice to create an optimal ensemble on Bob's side and for Bob to optimally distill this ensemble
- We need to calculate the optimal concentration rate for an ensemble of pure states
\mathfrak{E} = \{ p_i, \psi_i \}
Pure state concentration
C_c(\psi, \epsilon) := \max\limits_{M \in \mathbb{N}} \left\lbrace \log_2 M : \max\limits_{\Lambda \in \mathcal{O}} F^2(\Lambda(\psi), \Phi_{M}) \geq 1- \epsilon \right\rbrace
\max\limits_{\overline{\psi} \in b_*(\psi, \epsilon)} S_{min}(\Delta(\overline{\psi})) - \delta \leq C_{c}(\psi, \epsilon) \leq \max\limits_{\overline{\psi} \in b^{\prime}_*(\psi, 2\epsilon)} S_{min}(\Delta(\overline{\psi}))
S_{min}(\rho) := - \log_2(\lambda_{max}(\rho))
b^{\prime}_*(\rho, \epsilon) = \{ \overline{\psi} \text{ s.t. } \text{Tr}(\overline{\psi}) \leq 1, F(\rho, \overline{\psi}) \geq 1 - \epsilon \}
b_*(\rho, \epsilon) = \{ \overline{\psi} \in b^{\prime}_*(\rho, \epsilon) \text{ s.t. } \text{Tr}(\overline{\psi}) = 1 \}
\text{ If } \Delta(\Phi_M) \succ \Delta(\psi) :
Proof Sketch direct
\psi
\Phi_M
\Lambda_{\mathcal{I}}
=S_{min}(\Delta(\psi))
\log_2M
M
Winter, A., & Yang, D. : PRL 116.12, 120404, (2016)
\cdot \psi
\cdot \Lambda(
\cdot \Phi_{\overline{M}}
\cdot\overline{\psi}
\epsilon
\epsilon^{\prime}
\text{with } \epsilon^{\prime} \text{ error, optimal}
\text{rate } M^*, \text{ will be max }
\text{over}
\text{ball}
\text{around } \psi
\log_2M^* \geq \max\limits_{\overline{\psi} \in b_*(\psi, \epsilon)}S_{min}(\Delta(\overline{\psi}))
\psi
)
\Lambda
\Lambda
Proof Sketch: Converse
\text{Starting with an operator identity :}
\Phi_M \delta \Phi_M = \frac{1}{M}\Phi_M
\delta \in \mathcal{I}
\delta
\bullet \text{ Assume the optimal rate with error } \epsilon \text{ is } M, \text{ acheived by } \Lambda
\log_2M \leq \min\limits_{\gamma \in \mathcal{I}} \left\{ -\log_2\text{Tr}(\overline{\psi}\gamma) \right\}
\overline{\psi} \text{ depends on } \Lambda, \Phi_M, \psi
\overline{\psi}
\cdot \psi
\cdot \overline{\psi}
= \max\limits_i \langle i | \overline{\psi} | i \rangle
= S_{min}(\Delta(\overline{\psi}))
2\epsilon
\log_2 M \leq \max\limits_{\overline{\psi} \in b_*^{\prime}(\psi, 2\epsilon)} S_{min}(\Delta(\overline{\psi}))
Ensemble concentration
C_c(\mathfrak{E}, \epsilon) := \max\limits_{M \in \mathbb{N}} \left\lbrace \log_2M : \max\limits_{ \lbrace\Lambda_{i}\rbrace_i} F^{2}\left(\sum\limits_{i}p_i\Lambda_i(\psi_i), \Phi_{M} \right) \geq 1 - \epsilon \right\rbrace
\bullet \text{ Let }\mathfrak{E} = \{ p_i, \psi_i \} \text{ be a pure state ensemble}
\bullet \text{ We define the optimal concentration rate for } \mathfrak{E} \text{ as}
\max\limits_{\overline{\mathfrak{E}} \in b(\mathfrak{E}, \epsilon)}F_{min}^{\Delta}(\overline{\mathfrak{E}}) - \delta \leq C_{c}(\mathfrak{E}, \epsilon) \leq \max\limits_{\overline{\mathfrak{E}} \in b(\mathfrak{E}, 2\epsilon)} F_{min}^{\Delta}(\overline{\mathfrak{E}})
\text{where, } F^{\Delta}_{min}(\mathfrak{E}) = \min\limits_{\psi_i \in \mathfrak{E}} S_{min}(\Delta(\psi_i))
Asymptotic limit
\bullet \text{ The coherence of assistance is defined as,}
D_a(\rho^B) = \max\limits_{\mathfrak{E}_{\rho}=\lbrace p_i, \psi_i \rbrace_i} \sum\limits_i p_i S(\Delta(\psi_i)) = D_c^{A | B}(\psi^{AB}) = S(\Delta(\rho^B))^1
Chitambar, E., et al. Physical review letters 116.7 (2016): 070402.
\bullet \text{ We define the one-shot coherence of assistance as,}
C_a(\rho, \epsilon) = \max\limits_{\mathfrak{E}_{\rho}} C_{c}(\mathfrak{E}_{\rho}, \epsilon)
\lim\limits_{\epsilon \rightarrow 0}\lim\limits_{n \rightarrow \infty}\frac{1}{n} C_a(\rho^{\otimes n}, \epsilon) = \lim\limits_{n \rightarrow \infty}\frac{1}{n}D_a(\rho^{\otimes n})
Conclusions
- We derived bounds on the pure state concentration of coherence
- We generalize this bound to an ensemble of pure states and use this to find the one-shot coherence concentration.
- We recover the correct asymptotic behaviour.
- Our converse proof works for entanglement also and we are trying to generalize to arbitrary resource.
- Similar questions are open for when 2-way communication is allowed
Thank you!
One Shot Assisted Concentration of Coherence (10)
By madhav_krishnan
One Shot Assisted Concentration of Coherence (10)
Seminar USyd
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