I have read many times in several articles (such as https://arxiv.org/abs/1609.05033) that the entanglement of formation EoF puts a lower bound on entanglement dimensionality $d$ (i.e., the Schmidt number):

$$\log_2d\geq E_{\rm{oF}}.$$

However, I was not able to find any proof of this fact. How would one go about proving it?


1 Answer 1


For a pure state $\psi_{AB}$ with Schmidt rank $r$ the maximum of $$E_{oF}(\psi_{AB})=H({\rm tr}_B(|\psi_{AB}\rangle\langle\psi_{AB}|))=-\sum_{k=1}^r\lambda_k\log_2\lambda_k$$ is known to equal $\log_2(r)$, achieved on a maximally entangled state (i.e. $\lambda_k=\frac1r$). For mixed states $E_{oF}(\rho_{AB}):=\min \sum_k\lambda_kE_{oF}(\psi_{AB}^k)$ where the minimum is taken over all pure state decompositions (and this minimum is achieved somewhere!) meaning the above bound carries over: $$ \sum_k\lambda_kE_{oF}(\psi_{AB}^k)\leq \Big(\sum_k\lambda_k\Big)\log_2(\max_k r_k)=\log_2(\max_k r_k)\leq\log_2(\min\{\dim\mathcal H_A,\dim\mathcal H_B\}) $$ For further reading refer, e.g., to Chapter in the book of Khatri & Wilde.


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