From 1871fd820a97c2d11a23a2b8ea47f4949065afd7 Mon Sep 17 00:00:00 2001 From: Zheyuan Wu <60459821+Trance-0@users.noreply.github.com> Date: Mon, 7 Jul 2025 01:31:48 -0500 Subject: [PATCH] updates --- content/Math401/Math401_P1.md | 64 ++++++++++++++++++++++++--- content/Math401/Math401_P1_1.md | 2 +- content/Math401/Math401_T6.md | 2 + public/Math401/Superdense_coding.png | Bin 0 -> 12815 bytes 4 files changed, 60 insertions(+), 8 deletions(-) create mode 100644 public/Math401/Superdense_coding.png diff --git a/content/Math401/Math401_P1.md b/content/Math401/Math401_P1.md index 1f70b18..caa5372 100644 --- a/content/Math401/Math401_P1.md +++ b/content/Math401/Math401_P1.md @@ -92,11 +92,47 @@ Let $\epsilon>0$. Then for any subset $A\subset S^n$, given the area $\sigma_n(A The above two inequalities is not proved in the Book _High-dimensional probability_. -To continue prove the theorem, we use sub-Gaussian concentration of sphere $\sqrt{n}S^n$. +To continue prove the theorem, we use sub-Gaussian concentration *(Chapter 3 of _High-dimensional probability_ by Roman Vershynin)* of sphere $\sqrt{n}S^n$. -This will leads to some constant $C>0$ such that +This will leads to some constant $C>0$ such that the following lemma holds: +#### The "Blow-up" lemma +Let $A$ be a subset of sphere $\sqrt{n}S^n$, and $\sigma$ denotes the normalized area of $A$. Then if $\sigma\geq \frac{1}{2}$, then for every $t\geq 0$, + +$$ +\sigma(A_t)\geq 1-2\exp(-ct^2) +$$ + +where $A_t=\{x\in S^n: \operatorname{dist}(x,A)\leq t\}$ and $c$ is some positive constant. + +#### Proof of the Levy's concentration theorem + +Proof: + +Without loss of generality, we can assume that $\eta=1$. Let $M$ denotes the median of $f(X)$. + +So $\operatorname{Pr}[|f(X)\leq M|]\geq \frac{1}{2}$, and $\operatorname{Pr}[|f(X)\geq M|]\geq \frac{1}{2}$. + +Consider the sub-level set $A\coloneqq \{x\in \sqrt{n}S^n: |f(x)|\leq M\}$. + +Since $\operatorname{Pr}[X\in A]\geq \frac{1}{2}$, by the blow-up lemma, we have + +$$ +\operatorname{Pr}[X\in A_t]\geq 1-2\exp(-ct^2) +$$ + +And since + +$$ +\operatorname{Pr}[X\in A_t]\leq \operatorname{Pr}[f(X)\leq M+t] +$$ + +Combining the above two inequalities, we have + +$$ +\operatorname{Pr}[f(X)\leq M+t]\geq 1-2\exp(-ct^2) +$$ > The Levy's lemma can also be found in _Metric Structures for Riemannian and Non-Riemannian Spaces_ by M. Gromov. $3\frac{1}{2}.19$ The Levy concentration theory. @@ -122,10 +158,10 @@ Hardcore computing may generates the bound but M. Gromov did not make the detail Choose a random pure state $\sigma=|\psi\rangle\langle\psi|$ from $A'\otimes A$. -The expected value of the entropy of entanglement is kown and satisfies a concentration inequality. +The expected value of the entropy of entanglement is known and satisfies a concentration inequality. $$ -\mathbb{E}[H(\psi_A)] \leq \log_2(d_A)-\frac{1}{2\ln(2)}\frac{d_A}{d_B} +\mathbb{E}[H(\psi_A)] \geq \log_2(d_A)-\frac{1}{2\ln(2)}\frac{d_A}{d_B} $$ From the Levy's lemma, we have @@ -133,15 +169,29 @@ From the Levy's lemma, we have If we define $\beta=\frac{d_A}{\log_2(d_B)}$, then we have $$ -\operatorname{Pr}[H(\psi_A) \geq \log_2(d_A)-\alpha-\beta] \leq \exp\left(-\frac{(d_Ad_B-1)C\alpha^2}{(\log_2(d_A))^2}\right) +\operatorname{Pr}[H(\psi_A) < \log_2(d_A)-\alpha-\beta] \leq \exp\left(-\frac{(d_Ad_B-1)C\alpha^2}{(\log_2(d_A))^2}\right) $$ -where $C$ is a small constatnt and $d_B\geq d_A\geq 3$. - +where $C$ is a small constant and $d_B\geq d_A\geq 3$. #### ebits and qbits ### Superdense coding of quantum states +It is a procedure defined as follows: + +Suppose $A$ and $B$ share a Bell state $|\Phi^+\rangle=\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle)$, where $A$ holds the first part and $B$ holds the second part. + +$A$ wish to send 2 classical bits to $B$. + +$A$ performs one of four Pauli unitaries on the combined state of entangled qubits $\otimes$ one qubit. Then $A$ sends the resulting one qubit to $B$. + +This operation extends the initial one entangled qubit to a system of one of four orthogonal Bell states. + +$B$ performs a measurement on the combined state of the one qubit and the entangled qubits he holds. + +$B$ decodes the result and obtains the 2 classical bits sent by $A$. + + ### Consequences for mixed state entanglement measures #### Quantum mutual information diff --git a/content/Math401/Math401_P1_1.md b/content/Math401/Math401_P1_1.md index b622cb2..4f9019a 100644 --- a/content/Math401/Math401_P1_1.md +++ b/content/Math401/Math401_P1_1.md @@ -12,4 +12,4 @@ Practically speaking: - Entanglement and non-orthogonality ## MM space -- + diff --git a/content/Math401/Math401_T6.md b/content/Math401/Math401_T6.md index 3ab27d8..4059db0 100644 --- a/content/Math401/Math401_T6.md +++ b/content/Math401/Math401_T6.md @@ -582,6 +582,8 @@ $B$ performs a measurement on the combined state of the one qubit and the entang $B$ decodes the result and obtains the 2 classical bits sent by $A$. +![Superdense coding](https://notenextra.trance-0.com/Math401/Superdense_coding.png) + ## Section 4: Quantum automorphisms and dynamics Section ignored. \ No newline at end of file diff --git a/public/Math401/Superdense_coding.png b/public/Math401/Superdense_coding.png new file mode 100644 index 0000000000000000000000000000000000000000..00dde4d806579c70bb820ae08c0a5e9b85e0e001 GIT binary patch literal 12815 zcmdUWWl)>Z)-IGnixi5JV#T3Q+_ktnEl^woEuP{MC=`ObTX853MS{Brcemi~&X=Au z^ZmOs_vgKtNmk^2*WOF_T6;eUR#TCEgYgan2?^!+x}PsoqtG$sUvoK>DtQ-ZOQz~mvZqU|GB4N6^11Mc4pt}oTU=b+ zeha&;H`Is+RGdmuN=k}M;{SEhVM-EcvdkF6!+f8hYzp_&s`5P7VWDs%W>e1l8g4yS zFX%(DB_bu&^_j`M?_))k*)VWuAz7{?`jRDuyl2fErEWZY_j+>UC4pn7le4xbp5vIF zPP5G$uqR*cJp||GCcd|5J^2YBE!B0&Xx?{NNx4-V^qtem_JeeT^7h5gCjyRR6rD(x zt-Rq!K&wcyQ>6F6_-*-UUvE<)1X7O|-XFEF89ozV7LeN52z3jWg|mhwbN}=dwn@8z zE2DY8m5*^}z{PcB!T`hon<#b^>pX?q9Zfdr=~8GqLMsQ88EzhvPEgjQ4rZ9z&(HeY6;et`jSS{M%q>%eL5jd9h}WPHSaz$b>)8c8HoJ#bG633m^ci}vPxZk{-F zE;v6uT^r4kjAPLq*Z)xQbXg8oud_2S^g5LW$xFQ?;M}auLkEb80G*udGO>s^UI;O0 zH}#o`w3@iC^)&I@QsoFO=j}Ud6R_$hne|4f|IuOs6}5S6B?Alf>dFKG)GSd1%UOe@J3H9M#Jio<^4r^0f@o;g8v!kfo`-0khmI~L9uRv>Fx9|~t?qftSuu3A7 z9S2PR`m!EnmcPasf9#tj!nQ2Gu=+KxKa zwsmR$UW%NUWj_QH#UMARGMYeorn%4=emVG#G&m5o{Hx^ciP3egt|lVd^ur|r7c)Y^ zj_g-X19IVFgU~P{UnPO=TKD6L$F=q$(QgGe-jYcTZdkUDLqZ`C2(x-$lwMRQPvv$kV=OSS`#V<5FP$ntuqmDxK)+*a2^SkNEa7Dy; z=|mfNOh|u85+U`}&YXyF@1i33Uy>FWZ@Yq(tGs_-w0o>-@vNZrM=SPTuMWN|-`Hnm zP<>;O^SVSg+KKSP+VQb!&0XZl6&w22(tmSCm$lpIs*%{)`TXam>Gu(4y($_dE8NrN z2HkPMeG7i_+42qI2T|Z_Ld}7R+X?ss6s|4mQBv>Dhl?Ae*7w#?u)D~h81HCSfS6cz znx-yxw#otDxo3Ez_3=1QOGX|sQ*p#h(;{9?eb?$)GTlDNl1zhH*~T!()J4E!a~^Mm z>pvWAWkVqTOy&@Na=wQbFBlKeu4rEIM~xX{A?>TQJXztSuN*2%KNf__iH>H;cA>jZ z7K(r7s~{KaYnK8Mwhd(1iyDI#u1VCSMX+%DDKP-lDt*lo&VTM2OmuKStE=IqSPOpr zO|4Jj%V5DziSkmkD~yQ2Xh!z5$`7!obIADP(poHqy5d~H1-FeOYLe)ZQfJW#gsq)( z5kaT$4p7{cV%MO4oG_Qw=zDBRQKerPghP2sv?peDAy{AIByu8aE1yaQwFQZIu~o^t z27!ha)6Y}A(XNo)`5NYymX^QYXNL6~a@^e9jBA!>l?MjI#=xtD7Z~m0@k4aDiP=$( zud2IQET!(LSc-+y?kM!gJtSgEKI&ElNCsi6=P3|^nDCjTS~N0WwF-YKDj>oY$-`gv z6`>Dkr@=m7FX6M^S)C*yA(4}ic){~+l0ET=QB+ivmfn(|ogN=eUiR5Il`d(DR5Yio zL3zD{fb&RA8>Qzsyv3wD*HCNf4*hm-%Rb6-yX&S_i@fDs&ZYLl<$kCZeS*LUmf3N2 zwOc`D-7PXyg-@>WO+Tr7wJxPfQ)@6&`s2`N+{7#nfa~w?$T_EO5)917{0s;K{!yQQ zV-xheZu}=*Gjy3Ix1%_0fYEq1qGD%%?w~puGcnHwAbUOSo&0xuYyQb8wCNp(sk!7A zuxRp9@`JLJzMn)7`PR2Kodq}Htk{fC)uic(y-PfW>u6WAl4Sf%Vb=xm=9AxHOs31Y zLX{xd>hS{Id=&AerbW5M=D2up3R-lCf4zFv_4k(yD(j)}Q-pU7s`b_T&zdIaqm$t4 zWcjXLEHY2lbA~(POav+Co+y^%Z<;M*RvZ}*loOI9SWbVa271@57DRP#AU)lDK5 zsGlFFrU$qT!vUvN4S@A|gc9E-qV%jXBj*&n2xLCN`YmJOTG8!LwH4a*sMf!{|zAv!+C2sobrnvp=G#Y@#n{DQ;p_O~$Cu{o8)BN@Rh1;GueHi@oj)aSK{ zG6B1Oh%C<^?bBSf8V11mm{}bAOnF65k3&bo78=8}>^5?(I zw%+B(^zLoN95gb#n=@m3mXPlneGZNYE6>y8>u{(ech=mT^?`Dd$<7206)m6257vyL z{jPCMjyA9($*JuXcaBXUagszTo2D&Aa`+6mw6Nu#wae(*mfTi3DE+R3pCtS=ji zR9e5V>776eqtN=&=D=a6<@Ml>#&bS+Mmg6!k7T$@(n&1hZ#8!G`HQwk5TV$QwVNH* zzgAPJI5RV={=YVM4MmHAxIyZClKM&0{SNB8FM3|0hW$L+n(43Cbz#lmmq@!dxJj(9 zI+*W{C2QVM?5>io!eUAu69hmR1w!~zuHaqwsY&zF`<9UX`gpTY0*4ut9w}5TXh(Q9 zyx0(j1W9^u4IGLr>6EN(W=Jv@HI>6qluE@a-Ahai;xP{3N$(3mUm|3eO6!=a1jO?2 zO6rG51Y*h)v-dC?L-+}uxw=r9%n7LN_G8uozqgjIpRnl3xTu&NUGthDrqDE0VG&fOCAQ*73?* zxI{C%SP>)iGnw<#k;)P`NfZRV!e92n%fcX`X5Cca+gLYmQR9{#&ERH z6A{kSQ?`FHdpq+xYK?pRWtc2| z1=X zl^e<}UC1&c|I;XCn&ZAo(jbBXjvbIaXT4HxwavB*klAQ+EE9#UKyVKb7?so=X^Ks-=_gF$eY79ATa`^1@AB-W8U~L+i zFq^);5r*PaDiukDeD&|xJFEm9QSloP4gyIgCG=>OaagBq(J4b%Qeg9YU%i_#3b==0 z$lCMj#=vG45VWM<=emV(J=RMls!e|Yx~X76GD=hU;9g8V+{qt{C-|!H_2U1K2Hyg>v*-{b`b?o zukd(p<^o)beLrm~oW-Rj$7=mjz^pB=&9qibDWMh5j^nqjY27=U*CP8(%DcObNwc*O zXbPqcnb95fVNgl)V*5_MFBP?bqdVbf{tB-Je14{|7$u0n_;$9x4;mMMPZlcLmQjk! 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