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1/1.2 Quantum 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1/1.4 Quantum 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1/1.5 Quantum parallelism.html":[".","/","1)^{f(0)}|0\\rangle+(","1)^{f(1)}|1\\rangle]\\frac{1}{\\sqrt{2}}(|0\\rangl","1)^{f(x)}\\frac{1}{\\sqrt{2}}(|0\\rangl","1.5","1}|x\\rangl","1}|x\\rangle|f(x)\\rangl","1}|x\\rangle|f(x)\\rangle中。然而，提取这个非局域信息并不容易。","2}}",":","\\frac{1}{2^{n/2}}\\sum_{x=0}^{2^n","\\rangle。","\\sum_{x=0}^{2^{n}","deutsch解释地更具体，deutsch指出量子计算机的潜能在于量子并行性(quantum","deutsch问题的解决方案告诉我们，如何巧妙利用那些编码在量子非局域关联中的有效信息是设计量子算法的关键。","f(0)=f(1)（称作balanc","f(0)\\neq","f(1)（称作constant","f(1)，也还是需要做两次计算，耗时48小时。","f(x)\\rangl","f(x)\\rangle)=|x\\rangle(","feynman的想法在1985年被david","function和const","function）那么我们总会得到|","function）那么我们总会得到|+\\rangle，","function，但量子计算机只需要一次计算就可以完成任务。","parallel","parallelism)。","parallelism)。考虑一个黑盒，可以将一个比特x映射到另一个比特f(x)上，但我们不知道内部发生了什么，也许计算过程很复杂，因为它耗时24小时。一共有四种可能的函数映射关系f(x)，我们想要知道这个黑盒在计算什么。如果要计算f(0)和f(1)的结果，则总共要消耗两倍时间，即48小时。如果我们退一步，只想验证f(0)=f(1)还是f(0)\\neq","parallelism）。函数再复杂都只需要计算一次，这就是量子并行性的威力。这种大规模并行性也是shor设计因数分解算法的出发点。","quantum","qubit上的函数的全局性质感兴趣，此函数具有2^n个可能的输入，如f(|0100...\\rangle)。要得到函数f(x)完整的映射表，需要执行2^n次f的计算。然而这个计算对于n\\gg1的情况完全不可行，比如n=100时需要10^{30}次计算。但对于一个执行幺正变换","qubit的幺正变换：","qubit转换到2","u_f","u_f:","u_f:\\frac{1}{\\sqrt{2}}(|0\\rangle+|1\\rangle)\\frac{1}{\\sqrt{2}}(|0\\rangl","u_f:|x\\rangle|0\\rangle\\rightarrow|x\\rangle|f(x)\\rangl","x_0的函数值f(y_0)了。在这种情况下，量子计算机并不比经典计算机更强大。","|1\\oplu","|1\\rangle).","|1\\rangle)\\rightarrow","|1\\rangle)\\rightarrow\\frac{1}{\\sqrt{2}}[(","|1\\rangle)，那么映射","|\\psi\\rangle\\equiv\\left[\\frac{1}{\\sqrt{2}}(|0\\rangle+|1\\rangle)\\right]^{n}=\\frac{1}{2^{n","|x\\rangle\\frac{1}{\\sqrt{2}}(|0\\rangl","|x\\rangle\\frac{1}{\\sqrt{2}}(|f(x)\\rangl","|x\\rangle|y\\rangle\\rightarrow|x\\rangle|y\\oplu","|x\\rangle也改为另一个叠加态：","于是我们将f(x)分离到了前面系数中（与x有关的相位项）。","其中\\oplus代表“异或”。此操作表示，若f(x)=1就翻转y，若f(x)=0则不做任何操作。deutsch问的问题是，能不能只执行一次操作就验证f(0)与f(1)到底相同还是不同？","函数f的全局信息都包含在了这个态中，只要我们想个办法提取出来即可。","只执行一次上述变换，产生了状态","就这样，我们解决了deutsch的问题，而且发现了经典计算机和量子计算机的区别：经典计算机必须运行两次黑盒计算才能区分balanc","我们再将第1个qubit","最后，将第一个比特投影到一组基向量|\\pm\\rangle=\\frac{1}{\\sqrt{2}}(|0\\rangle\\pm|1\\rangle)上。显然，如果：","正如前面提到的，量子信息是编码在物理系统不同部分之间的非局域关联（纠缠）中的，比如上述f(x)的信息包含在了量子计算机的输入寄存器和输出寄存器的纠缠\\frac{1}{2^{n/2}}\\sum_{x=0}^{2^n","比如，假如我要去测量输入寄存器，它就会塌缩到某一个本征态|x_0\\rangle，这个本征态是从2^n个可能的态中随机选择的。这个过程会得到状态|x_0\\rangle|f(x_0)\\rangle。我们可以继续测量输出寄存器，但只能得到测量结果f(x_0)。但是由于两个寄存器之间的纠缠被第一次测量破坏，我们就没有机会再通过测量去确定任何y_0\\neq","现在假设我们对作用在n","现在看看量子计算机能不能只用一次计算（耗时24小时）就能完成任务。尽管映射f(x)可能是不可逆的，但量子计算过程是幺正的且一定是可逆的。考虑一个从2","由于黑盒是一台量子计算机，我们可以选取输入态为|0\\rangle和|1\\rangle的叠加态：","的量子计算机来说，我们可以选取存储在输入寄存器中的初态为：","若选取第2个qubit为叠加态|y\\rangle=\\frac{1}{\\sqrt{2}}(|0\\rangl","这个量子计算呈现出了“大规模量子并行性”（massiv","这是因为量子计算机操作的不是单个态而是叠加态，因此可以一次获得映射函数的“全局”信息，这个信息同时依赖于f(0)和f(1)。在这个意义上我们称它具有量子并行性(quantum"],"Chapter 1/1.6 A new classification of 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1/1.7 What about 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1/1.8 Quantum error-correcting codes.html":["&","&+b\\frac{1}{2^{3/2}}(|000\\rangl","&0\\end{pmatrix}+i\\varepsilon_y\\begin{pmatrix}","&0\\end{pmatrix}+i\\varepsilon_z\\begin{pmatrix}","&0\\end{pmatrix}|0\\rangle=|1\\rangle，该bit变成|0\\rangle\\rightarrow|0\\rangle+i\\varepsilon_x|1\\rangle。其余情况以此类推。","&0\\end{pmatrix}。","&0\\end{pmatrix}，phas","&0\\end{pmatrix}，使得整体状态出现一个小误差","&=","&\\hat{b}(|000\\rangl","&\\hat{b}(|000\\rangle+|111\\rangle)=(|111\\rangle+|000\\rangle),\\\\","0","1","1)(|000\\rangl","1.8","1\\end{pmatrix}.","1\\end{pmatrix}，或表示这两个flip同时发生的算符\\begin{pmatrix}","1。因此只要再任意选取另一对做一次两两比较就能找出到底哪一组的phase与其他两组不同，然后对那个组里某一个qubit做幺正的相位变换即可翻转符号，完成纠错。","\\\\0","\\\\1","\\\\i","\\begin{align}","\\end{align}","\\hat{a}|\\phi\\rangle\\equiv\\hat{a}[(|000\\rangle+|111\\rangle)(|000\\rangl","\\underset{算符(y\\oplu","a\\frac{1}{2^{3/2}}(|000\\rangle+|111\\rangle)(|000\\rangle+|111\\rangle)(|000\\rangle+|111\\rangle)\\\\","a|0\\rangle+b|1\\rangle\\rightarrow","a|100\\rangle+b|011\\rangle，测量(y\\oplu","a|\\bar{0}\\rangle+b|\\bar{1}\\rangl","a|\\bar{0}\\rangle+b|\\bar{1}\\rangle=a|000\\rangle+b|111\\rangle.","bit","bit测量值就构成了我们用来诊断误差是否产生了的症状（syndrome）。下面举两例说明。","bit测量算符(y\\oplu","bit算符(y\\oplu","b|\\bar{1}\\rangle，量子信息就被破坏了，而且由于用了三个qubit，phas","code","code）是pet","correct","eror，我们将在不改变叠加态a|\\bar{0}\\rangle+b|\\bar{1}\\rangle所包含的量子信息的情况下纠正这1个bit","error","errors还存在。因为一旦某一个qubit发生phas","error。","error发生的可能性变大为原来的3倍。不要怕，我们已经通过添加冗余的qubits解决了bit","error！","error（这会在编码数据中引起bit","error（这会在编码数据中引起phas","error）。","error）发生的可能性约为|ε|^4。","error）；","error，对于编码以前的本征态貌似只需要测量|\\bar{0}\\rangle或|\\bar{1}\\rangle其中一个即可。但是对于编码以后的叠加态a|\\bar{0}\\rangle+b|\\bar{1}\\rangle应该如何测量呢？","error，然后对单个qubit做适当的幺正变换纠错。","error，现在我们来添加冗余的phases来解决phas","error，还能克服phas","error，这个误差改变了那一组里|000\\rangle和|111\\rangle的相对符号：","error，那么a|\\bar{0}\\rangle+b|\\bar{1}\\rangle就变成了a|\\bar{0}\\rangl","flip","flip有没有发生。在大部分情况下我们都能将系统投影回正确的初态，但有小概率o(|\\varepsilon|^2)某一个bit会出现较大误差：bit","flip有没有发生，用6","flip的误差。","flip的那一组进行纠错。这就需要一个6","flip算符\\begin{pmatrix}","flip纠错。","flip表示为\\begin{pmatrix}","flip，或phas","flip，或两者同时出现。从测量结果推测，我们可以知道哪个qubit发生了bit","flip，或只有某一组发生了phas","flip，或哪一组发生了phas","flip，或某两组发生了phas","flip，或每一个qubit的幺正变换发生小误差。下面这段讲的是如果某两个qubits发生bit","flip，量子纠错就无能为力了。","observables）来诊断误差——因此在我们试图纠错时也可能进一步破坏数据。然而我们将会证明，尽管恢复过程有偶发的误差，它还是很有效的。","phase","quantum","qubits中每一个bit的幺正变换都发生了小误差。最一般的单个qubit幺正变换（不考虑物理上不相关的全局相位）","qubits态会塌缩到对应的本征态，比如|\\bar{0}\\rangle，导致系数a,b中蕴含的量子信息丢失。","qubits来编码：","qubits编码更多，不仅能克服singl","qubits编码的冗余度相比3","qubits表示|\\bar{0}\\rangle和|\\bar{1}\\rangle。由于每一组内部具有3","qubits里的每个qubit都出现幺正误差，然后我们测量错误症状，即用2","qubits，一共3组同样的量子态构成的9","qubit冗余，可以用前面提到的方法进行每组内部的singl","qubit可观测量\\hat{a}有区别。","qubit可观测量\\hat{b}与前面讨论的6","qubit可观测量（比如可以同时翻转一个组里3个qubits的可观测量\\hat{b}，它就能通过本征值来区分|000\\rangle+|111\\rangle和|000\\rangl","qubit可观测量，例如用“将所有6个bit翻转”这个可观测量\\hat{a}（该观测量并不会破坏量子信息，即相对相位，而只会改变整体相位）：","qubit态|x,y,z\\rangle=|000\\rangle\\text{or}|111\\rangle，我们定义一个2","qubit或6","qubit的“共同可观测量”（collect","qubit算符\\hat{a}测量phas","shor提出的，这个发现迎来了发展迅速的新学科——量子纠错码理论。","u=1+o(\\varepsilon)=1+i\\varepsilon_x\\begin{pmatrix}","z)=(0,0)本征态}{(a|000\\rangle+b|111\\rangle)}+","z)=(0,1)本征态}{(a\\varepsilon|100\\rangle+b\\varepsilon|011\\rangle)}。此时测量(y\\oplu","z)=(0,1)，而(01)_{2进制}=(1)_{10进制}告诉我们要把第1位qubit翻转过来，这样就完成了纠错。仅有第二或第三位翻转同样适用。","z)有很大概率(p=1","z)测量bit","z)的值会告诉我们具体是哪一位。","z)的测量值正好是被翻转的那个qubit位置的二进制表示。这个2","z)，其中异或操作\\oplus也可以等价定义为“相加后取模2”，我们发现(y\\oplu","z,x\\oplu","|000\\rangl","|000\\rangle)]=","|000\\rangle+|111\\rangle\\rightarrow|000\\rangl","|000\\rangle\\rightarrow|000\\rangle+\\varepsilon|100\\rangle,\\\\","|0\\rangle\\rightarrow|\\bar{0}\\rangle\\equiv\\frac{1}{2^{3/2}}(|000\\rangle+|111\\rangle)(|000\\rangle+|111\\rangle)(|000\\rangle+|111\\rangle),\\\\","|0\\rangle\\rightarrow|\\bar{0}\\rangle\\equiv|000\\rangle,\\\\","|111\\rangle)(|000\\rangl","|111\\rangle),","|111\\rangle).","|111\\rangle)=(","|111\\rangle)]=\\hat{a}[(|111\\rangle+|000\\rangle)(|111\\rangl","|111\\rangle,\\\\","|111\\rangle\\rightarrow|000\\rangle+|111\\rangle.","|111\\rangle\\rightarrow|111\\rangle+\\varepsilon|011\\rangle.","|111\\rangle）：","|1\\rangle\\rightarrow|\\bar{1}\\rangle\\equiv\\frac{1}{2^{3/2}}(|000\\rangl","|1\\rangle\\rightarrow|\\bar{1}\\rangle\\equiv|111\\rangle.","|\\phi\\rangl","|\\varepsilon|^2)得到本征值(0,0)，同时也投影到正确的本征态上，消除了误差；但也有小概率(p=|\\varepsilon|^2)得到本征值(0,1)，即上一个例子中的态，此时根据本征值只需翻转第一位bit即可完成纠错恢复到正确的态。","上述机制不但可以进行纠错，而且也避免了量子退相干。无论发生了哪种误差都能恢复到正确的量子态，我们避免了量子态与环境发生任何纠缠。","上述量子纠错机制利用了误差的局域性假设来以非局域性方式编码量子信息——量子信息存储在数个qubits之间的关联中。仅仅测量9个qubits中的任何1个qubit是无法区分|\\bar{0}\\rangle和|\\bar{1}\\rangle的，因为不管编码的信息是什么，我们总是有同样的1/2概率得到|0\\rangle或|1\\rangle。","且允许9","但我们没必要局限在单个qubit测量，我也可以一次同时对两个qubit测量，同时测量就足够探测到是否有翻转误差了。对3","但是这种双相误差（doubl","使得这一组的相对相位与其他两组不同。可以通过组与组两两比较（而不是直接测量一个组内部的相对相位而破坏编码信息）找出发生phase","假如我们制备了初态：","即用每组3","只有第1点困难phase","可以展开到o(\\varepsilon)为：","后续的课程中我们将进一步讨论量子纠错机制的几个方面：","和经典纠错一样，存在“好”的量子纠错机制，允许我们得到任意高的可靠性，只要每个bit的错误率足够小。","因此只要|ε|足够小，那么通过编码是可以提高量子信息可靠性的。","在两个不同的组里发生phase","在两组中的每个组里都有两位bit","如果不是某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1/1.9 Quantum 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物理事实，因为经典的图灵机并不是一个恰当的能描述真实物理世界中运行的计算模型。","如果量子计算机最终成为实用设备，那么可能需要构造量子硬件的新想法。","存储（storage）：能存储qubit足够长时间，来完成有意义的计算。","它表示从输入\\{a,","实际上，退相干并不是我们面临的唯一问题。即使我们能把量子计算机与环境完全隔绝，也不能保证量子计算机无误差地运行。量子计算机执行的量子逻辑门一次只对少数qubits做幺正变换，比如这个幺正变换是4\\times4的幺正矩阵，作用在2","对于每一个bit的幺正变换都发生小误差的情况|000\\rangle\\rightarrow|000\\rangle+\\varepsilon_1|100\\rangle+\\varepsilon_2|010\\rangle+\\varepsilon_3|001\\rangle也同样处理，会有小概率o(|\\varepsilon_i|^2)投影到三个错误态（某一位翻转了）其中之一，然后测量(y\\oplu","对于经典计算机，现代数字电路已经足够可靠。它们通过能量耗散的形式来获得高精度的计算过程。我们可以想象一个作用在1","对于量子信息，出错的地方会更多。除了bit","少数服从多数投票机制中，我们需要测量每个bit来探测并纠正错误，但我们测量qubits总是会干扰它们编码的量子信息。","就这样，我们解决了deutsch的问题，而且发现了经典计算机和量子计算机的区别：经典计算机必须运行两次黑盒计算才能区分balanc","尽管你桌上的台式电脑不是量子计算机，但它仍旧是个很厉害的设备：原理上，它能够执行任何想得到的（conceivable）计算。实际上有一些计算是无法完成的——要么时间太长，要么内存溢出。但如果你有一台内存无穷大的电脑，且愿意等待足够长时间，那么任何能够称作计算的工作都能由你手头这台小小的计算机完成。在这个意义上，我们称它是一台“通用计算机”（univers","当然，如果我们只测量单个qubit，比如测量第一个qubit得到|0\\rangle，那么3","当然，量子纠错也是个耗散过程，因为关于误差的信息从量子系统中被抹去了。关于误差的信息记录在了测量结果中，一旦测量结果被擦除，它就以热量形式耗散掉了。","必须不断收集和存储气体分子的信息。如果demon的记忆容量有限，那么它就不能无限制地降低气体温度，因为当demon的记忆容量用完后，它必须清除记忆信息才能存储新的粒子信息；而擦除过程需要消耗能量，这些能量抵消了产生温度差（或降温）带来的好处。当然，这个解释会招来质疑：在记忆内存没用完之前，或demon不去擦除记忆，或者我们在擦除前研究整个系统的热力学，难道热二定律就可以不成立了吗？","总结一下量子纠错机制：","我也不能确定。但是也许仅仅提到以下事实就足以让人印象深刻：目前没有哪个经典计算机能够精确预测哪怕只是中等数量qubits（100个的数量级）量子计算机的行为。这说明少量的qubits就有很大的潜力等待我们去挖掘。","我们不测量数据就能测量出误差的种类，因此不会获取（从而干扰）编码好的量子信息。","我们之前假设，执行“纠错恢复”过程本身是无瑕疵的。但是误差诊断机制很复杂——要测量2","我们再将第1个qubit","我们可以用激光将离子冷却到远低于\\nu的温度，从而使每个振动模式有极大可能占据其量子力学基态。","我们可以避免这个经典计算机能力的困境吗？反过来思考一下，我们只要求模拟能给所有可能的输出结果分配一个概率分布，但是中间过程不必完整模拟n","我们可能永远不会看到半死半活的猫，但是也许我们可以制备和储存一只半死半活的编码猫。","我们已经看到，以下三个相互契合的因素相结合，使得这一主题亦可赛艇。","我们很荣幸能见证在实验室里对量子信息进行相干操纵时代的到来。","我们无法对自己手上未知的量子态进行拷贝（量子不可克隆原理），但我们可以让另一个人制备好这些拷贝（他是可以做到的，因为他明确地知道这个态是什么），只要不告诉我们这个态就可以了。","我们注意到，量子算法是一种概率算法(probabilist","我们的核心思想就是一句话：","我们还可以打开频率为\\omega的脉冲旋转磁场（其中\\omega是自旋向上和向下的能级差），从而引起自旋的拉比振荡。","或编码成叠加态：","所有这些在数十年前就已经被化学家们熟知。但仅在去年，gershenfeld和chuang，以及cory，fahmy和havel独立指出，nmr提供了一种实用的量子计算实现方式。但由于以下几个原因，这个想法并不显然。","新方案使用核磁共振（nmr）技术。","既然量子信息有这么多奇特性质，我们就会期待量子理论可以更新我们对计算的认知。peter","时间”不确定关系限制。相比典型的振动能级差\\nu来说，激光光子能量的不确定性很小，因此每个激光光子持续的时间应该比\\nu^{","是2bit输入1bit输出，通过输出无法唯一确定输入。由于这个逻辑门会擦除1bit信息（对所有可能的输入a,b做平均意义下），于是根据landauer’","显然，量子计算硬件还处于起步阶段。","最后三个pauli矩阵可以看成bit","最后，将第一个比特投影到一组基向量|\\pm\\rangle=\\frac{1}{\\sqrt{2}}(|0\\rangle\\pm|1\\rangle)上。显然，如果：","最近，nmr方法已用于制备3","最重要的是，nmr系统非常热。自旋的典型温度（例如室温）可能比|0\\rangle和|1\\rangle之间的能级差大一百万倍。这意味着计算机的量子状态（单个分子的自旋）噪声非常大——受到强烈的随机热涨落的影响，这种噪声会掩盖量子信息。此外，我们实际上不是在单个分子上执行处理","本文章翻译于caltech教授john","本课程的主题可以从不同角度切入，但这门课程会采用理论物理学家的视角。","正如之前所说，量子计算机所处理的量子信息的本质属性是系统各组成部分的非局域关联——大部分量子信息处于系统各部分的非局域关联中。如果一次只观测某个子系统，只能获得微乎其微的信息。","正如前面提到的，量子信息是编码在物理系统不同部分之间的非局域关联（纠缠）中的，比如上述f(x)的信息包含在了量子计算机的输入寄存器和输出寄存器的纠缠\\frac{1}{2^{n/2}}\\sum_{x=0}^{2^n","此处n是子系统熵的最大可能取值，对应于该子系统不携带任何信息。于是，对于n很大的情况，如果我们只分开测量每个子系统的话，只能得到指数级小量的信息。","此外，不确定性原理要求，两个非对易物理量无法同时具有确定值。因此测量物理量a势必会影响紧接着的对b（与a不对易）的测量结果。所以，获取一个物理系统的信息的操作会不可避免干扰这个系统的状态，此限制在经典物理里并无对应。","此外，自旋具有偶极","每个离子的量子态是由基态|g\\rangle（或表示成|0\\rangle）和某个长寿命的亚稳激发态|e\\rangle（或表示成|1\\rangle）的线性组合。这两个能级的相干叠加态：a|g\\rangle+be^{i\\omega","比如，假如我要去测量输入寄存器，它就会塌缩到某一个本征态|x_0\\rangle，这个本征态是从2^n个可能的态中随机选择的。这个过程会得到状态|x_0\\rangle|f(x_0)\\rangle。我们可以继续测量输出寄存器，但只能得到测量结果f(x_0)。但是由于两个寄存器之间的纠缠被第一次测量破坏，我们就没有机会再通过测量去确定任何y_0\\neq","注.","然而shor发现，量子计算机竟然可以在多项式时间o[(\\ln","然而我们并不能用同样地方法对量子计算机进行降温。和环境接触做降温能够提升经典信息的可靠性，但这么做会完全破坏量子信息，正所谓甲之蜜糖乙之砒霜。更广泛地来说，误差累积对于经典可逆计算同样是个问题。要避免误差的累积，我们必须丢弃关于误差的信息，即进行纠错，而扔掉信息总是个耗散过程（landauer'","然而，bennett(1973)发现任意计算都能用可逆操作实现，因此原理上任意计算都可以没有耗散，即不会消耗能量。例如，我们可以构造上述nand门的可逆版本——toffoli门（右上图）——toffoli门可以保留所有输入信息：","环境也许偶尔会扰动单个qubit，等效于“测量”了它，但是单个qubit携带的信息几乎为零，因此扰动单个qubit几乎不会破坏量子信息。量子纠错机制的基础，就在于非局域的信息不受局域干扰影响这个事实。","现在假设9","现在假设在症状测量中存在以下情况，则量子纠错将会失效：","现在假设我们对作用在n","现在假设某一组发生了phase","现在可以开始讨论量子计算了。我们制备具有n个qubits的标准初态|0\\rangle|0\\rangl","现在想象一个频率为(\\omega","现在我们假设某一个bit发生了bit","现在我们想对量子计算机进行纠错，而量子纠错会面临许多困难：","现在看看量子计算机能不能只用一次计算（耗时24小时）就能完成任务。尽管映射f(x)可能是不可逆的，但量子计算过程是幺正的且一定是可逆的。考虑一个从2","现在，qubits由特定分子中的某些核自旋携带。在施加恒定磁场的情况下，每个自旋可以与外磁场对齐|\\uparrow\\rangle=|0\\rangle或反对齐|\\downarrow\\rangle=|1\\rangle。","用经典计算机进行质因数分解，大家公认需要\\log(n)的超越多项式(superpolynomial)时间（虽然还没被严格证明），即当$n$增长时，最差情况需要的时间比\\log(n)的任何次幂都大。已知的最好算法(numb","由于相移取决于两个光子的极化，这是一个非平凡的2","由于离子之间的库伦排斥，离子之间是充分隔开的，因此它们能被脉冲激光单个地激发。如果一个激光光子的频率调成激发频率\\omega，且聚焦在第n个离子上，那么|0\\rangle和|1\\rangle之间就激发出了拉比振荡（rabi","由于黑盒是一台量子计算机，我们可以选取输入态为|0\\rangle和|1\\rangle的叠加态：","的量子计算机来说，我们可以选取存储在输入寄存器中的初态为：","目前最大的运算量是几百个工作站算一个月，将130位数分解成65位数。由此推算，分解400位数需要$10^{10}$年，相当于宇宙年龄！因此，即便科技大幅进步，质因数分解400位数仍旧是个遥不可及的梦想。","离子阱量子计算机最大的缺陷之一是，它本质上是一台运行很慢的设备。它的运行速度最终受到“能量","精度（precision）：要让设备可靠运行，量子逻辑门的执行必须高精度。","经典信息可以完美拷贝，而量子信息的复制受制于不可克隆定理，不能高保真地拷贝。","经典的复杂性理论研究的是困难和容易问题的区分标准。通常来说，“困难”和“容易”是由需要多少时间或存储容量决定的。但如何定义有意义的区分标准，使得它不依赖于我们使用的硬件呢？真正的用来区分简单和困难计算问题的标准应该是普适的，它不以使用高性能还是老破旧的计算机而改变。","经过适当的脉冲时间后，第n个离子的激发态|e\\rangle_n将旋转为|g\\rangle_n，而cm谐振子则从其基态|0\\rangle_{cm}跃迁至其第一激发态|1\\rangle_{cm}（产生了一个cm“声子”）。","结论很惊奇，原理上做任何计算都可以不消耗能量！但现实中，每个逻辑门器件都会消耗远比$kt\\ln2$大得多的能量。因此工程上讨论landauer’","网页版浏览请点击：https://chenhuang94.github.io/john_preskill_quantum","考虑将1qubit编码成3qubits：","自旋需要很长时间才会驰豫或退相干，因此qubits的保留时长比较理想。","至此，我们已经克服了上一小节提出的量子纠错的后3个困难。我们可以不破坏叠加态的信息而进行测量(第3点)，量子测量可以将“有小误差的状态”投影到没有误差的态，或者投影到翻转了一个qubit的具有较大离散误差的态（第2点）。第4点被巧妙避开了，因为a|\\bar{0}\\rangle+b|\\bar{1}\\rangle并不是通过克隆得到的，它与克隆操作得到的三重拷贝态(a|0\\rangle+b|1\\rangle)^3不是一回事。","若两组的phase一样，则特征值为+1，反之则如上为","若要更高的正确率，只需要再多加几位重复位即可，即使用n","若选取第2个qubit为叠加态|y\\rangle=\\frac{1}{\\sqrt{2}}(|0\\rangl","要了解量子纠错原理，最佳方法也许是回过头去看shor的原始纠错码——量子纠错是经典3bit重复位机制的推广。但要注意，对于量子编码，我们要能够不直接测量编码信息就达到纠错的目的（因为直接测量会破坏量子信息）。","要提取有用信息则需要一个3","要操作量子计算机，不仅要能可靠地存储量子信息，还要能处理它。我们将会展示使用量子逻辑门来处理量子信息。","要读出离子的态很容易，只要通过测量把它投影到\\{|g\\rangle,|e\\rangle\\}基矢即可。准备一束激光，它能将离子的基态|g\\rangle激发到一个短寿命的激发态|e'\\rangle。当该激光照射离子时，每一个处于|0\\rangle的qubit不断地吸收、释放光子，发出荧光（fluoresces）。而处于态|1\\rangle的qubit保持不被激发的状态。","解决maxwell’","许多复杂性理论一般将算法的复杂性分为多项式时间(polynomi","译者注：9","译者注：rabi","译者注：上段讨论的是离子（单个qubit）的量子态。频率为\\omega的激光能把单个离子从基态打到激发态。","译者注：上段讨论的是集体振动（简正模）的量子态，即“声子”的量子态。频率为\\nu的激光能把质心模cm","译者注：为什么环境对猫测量选用的基矢一定要是\\{|\\text{dead}\\rangle,|\\text{alive}\\rangle\\}而不是一加一减两个叠加态？我猜想是这两个“能量”本征态的本征值相差很大造成的，即死猫和活猫的“能量”相差很大。真的是这样吗，如何定量证明？","译者注：举个例子，假设|0\\rangle\\equiv(1,0)，|1\\rangle\\equiv(0,1)，那么bit","译者注：在上述三种量子纠错机制中，我们都假设了要么只有某一个qubit发生bit","译者注：此处不理解。","译者注：此处不理解。“每个振动模式”是不是指，除了cm","译者注：此处不理解。探测是否发生了phase","译者注：此处不理解，为什么是等概率的？","译者注：此处不理解，需要举例。暂时不知道如何将前面的例子用在这里进行说明。","译者注：注意此处并不是在量子纠错（不破坏量子信息）。而是直接探测量子信息，测量完会破坏量子信息。因此，此处的3","误差是局域的，而量子信息是非局域的。前面讨论中我们心照不宣地做了一个很重要的隐含假设：在足够好的近似下，不同qubit的误差之间不存在关联，即假设“同时造成2个qubits产生误差”的事件的发生概率远比“只导致1个qubit错误”的事件的概率要小的多。这在物理上当然是存疑的——我们很容易就能构想一种同时导致两个qubits发生误差的过程。如果这种关联误差很普遍，上述量子纠错机制就不可靠了。","误差被数字化。尽管量子信息的误差可能是小量，我们还是可以将有误差的量子态投影到无误差量子态，或投影到一个离散的误差态集合中的某个态上，后者很容易通过幺正变换纠错。","读出（readout）：能高效可靠测量qubits。","课程网址：http://theory.caltech.edu/~preskill/ph219/index.html#lectur","质因数分解是一个很难对付的(intractable)计算问题：答案一旦找到的话很容易验证，只要相乘就可以了；但难点就在于答案很难找到。如果p和q是很大的质数，那么他们的乘积n=pq很容易计算，所需要的基本的位操作(bit","质因数分解问题，从复杂性理论（complex","过去15年量子力学理论最重要的发展之一，是我们越来越有底气来回答schrodinger的疑惑。猫态|\\text{cat}\\rangle理论上可以存在，但是我们很少能见到，因为它极端不稳定。schrodinger的这只猫从来没有真正于环境完美隔绝，因为猫不会脱离周围环境而存在（空气，地面，阳光照射过来的光子等）。假设有人制备了薛定谔猫态|\\text{cat}\\rangle，那么编码在|\\text{dead}\\rangle和|\\text{alive}\\rangle叠加态中的量子信息会瞬间转移到猫和环境的纠缠中消散而去，永远无法再获得。这是因为，环境和猫的相互作用，本质上是环境不断对猫进行测量，将猫投影到|\\text{dead}\\rangle或|\\text{alive}\\rangle中的一个本征态，这个过程叫作“（量子）退相干”（decoherence）。","这个想法是将几个中性原子捕获在一个小的高精度光学腔中。","这个操作擦除了原先存储在第n个离子里的1","这个量子计算呈现出了“大规模量子并行性”（massiv","这意味着第m个离子(m\\neq","这是因为量子计算机操作的不是单个态而是叠加态，因此可以一次获得映射函数的“全局”信息，这个信息同时依赖于f(0)和f(1)。在这个意义上我们称它具有量子并行性(quantum","通过对脉冲时间的调制，我们可以对单个自旋上执行所需的幺正变换（就像在对ion","逻辑门操控（gates）：能操纵单个qubit的量子态，且能引起qubits之间受控的相互作用，即能执行量子逻辑门。","逻辑门通常是不可逆的，比如nand门","那么a|000\\rangle+b|111\\rangle\\rightarrow","量子信息发轫于1980年代，于1990年代全面开花。我们将会在后面讨论：量子信息的压缩，量子系统能够编码的经典信息的上下界，通过含噪声的量子信道传输的可靠量子信息的上下界。","量子信息是连续的，因此小扰动会把a|0\\rangle+b|1\\rangle的系数a和b改变小量，这些小量会随时间不断累积。相反，经典纠错纠正的是比较大的误差（bit","量子信息这次存储在了原子的内部状态中。","量子复杂性（quantum","量子硬件是可以建造出的。","量子计算机可以解决难题。","量子误差可以被校正。使用合适的编码方法，我们可以保护复杂的量子系统免受退相干的破坏。","量子问题有量子的解决办法。第一个量子纠错码（quantum","隔绝（isolation）：与环境有效隔绝，最小化退相干误差。","需要说明的是，尽管物理原理不同，量子计算机可以计算的，经典计算机也都能计算。经典计算机可以储存和旋转向量，也可以模拟量子测量过程（投影到正交基向量）。因此，经典计算机能够以任意精度模拟量子计算机。用哪种计算机其实不是重点，因为什么问题是“可计算的”对两者来说是相同的。","顺着shor的思路，我们对单个qubit用9","频率\\nu最低的振动模是质心模（cent","（n\\gg1）个qubits的量子系统，它被分为三个子系统，每个子系统各自有n个qubits，分别叫做子系统(1)，(2)和(3)。我们随机选取一个3n","（但仍然需要更精确地表述那些量子计算机比传统计算机具有更大的优势的问题。）","）角度看，是一类有趣的难解问题，即无法在（输入大小的）多项式时间内解决的问题，在上例中输入大小即为\\log","，而是在在包含10^{23}个“计算机”的宏观系综上，我们从该设备读出的信号实际上是该系综的平均值。但是由于量子测量的随机性，量子算法是概率性的。因此，系综平均并不等价于在单个设备上运行计算；系综平均可能会掩盖真实的结果。"],"pipeline":["stopWordFilter","stemmer"]},"store":{"./":{"url":"./","title":"Introduction","keywords":"","body":"Introduction\n本文章翻译于Caltech教授John Preskill的Course Information for  Physics 219/Computer Science 219 Quantum Computation (Formerly Physics 229)课程讲义，我对其中某些内容做了一定程度的精简，并在必要的地方加了一点自己的注释方便理解。\n网页版浏览请点击：https://chenhuang94.github.io/John_Preskill_Quantum-Information-and-Computation/\n课程网址：http://theory.caltech.edu/~preskill/ph219/index.html#lecture\n"},"Chapter 1/Chapter 1.html":{"url":"Chapter 1/Chapter 1.html","title":"Chapter 1. Introduction and Overview","keywords":"","body":"本课程的主题可以从不同角度切入，但这门课程会采用理论物理学家的视角。\n"},"Chapter 1/1.1 Physics of information.html":{"url":"Chapter 1/1.1 Physics of information.html","title":"1.1 Physics of information","keywords":"","body":"1.1 Physics of information\n信息，是编码在物理系统里的；计算，是基于物理器件实现的。因此，信息论可以和物理过程联系起来。目前为止，人们对使用和操纵信息的物理约束有几个里程碑式的理解：    \n\nLandauer’s principle. Landauer(1961)指出信息擦除是个耗散过程(dissipative process)。他的理由是擦除总是伴随相空间的缩减，因此不可逆。例如，用一个放有单粒子的中间有隔板的盒子来存储1bit信息。信息擦除过程指的是我们瞬间抽掉中间的隔板，然后在最右侧墙壁插入隔板（不管之前粒子在左半边还是右半边），接着用隔板把粒子缓慢推往左半边，完成信息擦除。这个过程中，单粒子气体减小了\\Delta S=k\\ln2的熵。进一步地，如果该过程是在温度T下的恒温过程，那么外界对气体做功W=kT\\ln2，这些能量耗散到热源。因此，擦除信息必然需要消耗能量。\n\nReversible computation. 逻辑门通常是不可逆的，比如NAND门\n\r\n  (a,b)\\rightarrow\\neg(a\\wedge b)\r\n  \n是2bit输入1bit输出，通过输出无法唯一确定输入。由于这个逻辑门会擦除1bit信息（对所有可能的输入a,b做平均意义下），于是根据Landauer’s principle的要求，要运行这个逻辑门，外界至少做功W=kT\\ln2。因此，如果供电电量有限，计算时长会存在一个理论上界。左下图是NAND门示意：\n             \n然而，Bennett(1973)发现任意计算都能用可逆操作实现，因此原理上任意计算都可以没有耗散，即不会消耗能量。例如，我们可以构造上述NAND门的可逆版本——Toffoli门（右上图）——Toffoli门可以保留所有输入信息：\n\r\n  (a,b,c)\\rightarrow(a,b,c\\oplus a\\wedge b)\r\n  \n它表示从输入\\{a, b, c\\}映射到输出\\{a,b,c\\ \\text{XOR}\\ (a\\ \\text{AND}\\ b)\\}。它是可逆的3bit输入门，只有前两位都是1才翻转第三位。Toffoli门中c=1的那些行对应的输出output的第三列就对应NAND门。因此我们能够将一个不可逆计算NAND门转换成可逆计算Toffoli门，且原理上能够不产生能量损耗。当然，同时也产生了冗余的信息“垃圾”。有人会说，这个所谓的可逆（无耗散）计算只不过推迟了能量消耗，因为擦除这些冗余信息需要额外能量。对此，Bennett指出一个可逆计算机能够运行到计算结束，输出结果的拷贝（拷贝是恒等映射，因此是可逆计算），然后将之前的计算过程反向，恢复到初态。这样就不需要额外能量来擦除这些冗余”垃圾“信息了。\n结论很惊奇，原理上做任何计算都可以不消耗能量！但现实中，每个逻辑门器件都会消耗远比$kT\\ln2$大得多的能量。因此工程上讨论Landauer’s limit没太大价值。然而，当运算器件越来越小时，计算消耗的能量如果散发不出去，可能会让器件损坏，此时研究可逆计算就变得重要了。\n\nMaxwell’s demon. 上述Landauer和Bennett的洞见引导了Bennett在1982年协调了麦克斯韦妖和热力学第二定律的矛盾。麦克斯韦想象了一盒气体，中间的隔板把它分为A，B两部分空间。隔板上有个demon操纵的小阀门，demon可以观察靠近阀门的气体分子，仅允许速度快的分子从A经过阀门进入B，而让速度慢的分子从B进入A。于是，A不断降温，B不断升温，且这个过程中几乎没有能量损耗（我们假设操纵阀门不需要能量，或者小到可忽略不计）。热量就这样从低温热源流向高温热源而不伴随任何能量损耗，显然违背了热力学第二定律。这就是著名的Maxwell’s paradox。\n解决Maxwell’s paradox的关键是要理解demon 必须不断收集和存储气体分子的信息。如果demon的记忆容量有限，那么它就不能无限制地降低气体温度，因为当demon的记忆容量用完后，它必须清除记忆信息才能存储新的粒子信息；而擦除过程需要消耗能量，这些能量抵消了产生温度差（或降温）带来的好处。当然，这个解释会招来质疑：在记忆内存没用完之前，或demon不去擦除记忆，或者我们在擦除前研究整个系统的热力学，难道热二定律就可以不成立了吗？\n为了回应这个质疑，我们就要在记忆擦除之前，定义与存储的信息相对应的信息熵。Leo Szilard(1929)在研究Maxwell demon中定义了1bit信息的概念（\"bit\"的名称是后来Tukey命名的），并且认为获取1bit信息等价于熵增加\\Delta S=k\\ln2。显然Szilard并没有完全抓住Landauer’s principle的核心要点，因为后者说的是信息擦除必须要付出代价（熵or能量），而不是信息获取。但不管怎样，Szilard仍旧是研究信息的物理原理的一位先驱。\n\n\n以上例子表明，物理学与信息论的学科交叉催生出了许多很有趣的结果，吸引了一大批物理学家和计算机科学家。\n"},"Chapter 1/1.2 Quantum information.html":{"url":"Chapter 1/1.2 Quantum information.html","title":"1.2 Quantum information","keywords":"","body":"1.2 Quantum information\n我们的核心思想就是一句话：\n\r\n\\textbf{“Information is physical.\"}\r\n\n从根本上说，世界是由量子物理支配的，那么量子物理是如何揭示信息的本质的呢？\n在量子理论早期，人们就已经认识到信息的经典认知需要修改了。比如，（由量子原理支配的）盖革计数器记录的触发(clicks )被认为是真正随机（truly random）的泊松过程；然而经典决定性(deterministic)力学并没有真正随机性之说（即便复杂一点的混沌系统也只是在效果上和随机系统相差无几，但两者在本质上并不相同）。\n此外，不确定性原理要求，两个非对易物理量无法同时具有确定值。因此测量物理量A势必会影响紧接着的对B（与A不对易）的测量结果。所以，获取一个物理系统的信息的操作会不可避免干扰这个系统的状态，此限制在经典物理里并无对应。\n“获得信息”与“产生干扰”的取舍与量子随机性（quantum randomness）有关。正是因为测量结果包含了随机因子（绝对相位），所以我们无法从测量结果推断初始态的所有信息，即初始的相位。\n“信息的获取会对状态产生干扰”也和量子与经典信息的另一个区别有关，即量子不可克隆原理(the no-cloning principle: quantum information cannot be copied with perfect fidelity) (Wootters&Zurek1982, Dieks1982)。假设我们的确可以完美克隆一个量子态，我们就能测量这个复制态的物理量，而不用干扰原来的态，不可克隆原理否定了这种可能性。相反，经典信息就可以完美复制，这也是我们能使用U盘拷贝，或做硬盘备份的原因。\n上述量子信息的性质很重要，但是量子信息与经典信息真正本质的区别还是John Bell(1964)指出的，量子力学的预测不可能用局域隐变量理论（local hidden variable theory）来复现。Bell提出量子信息被编码在物理体系不同部分的非局域关联（nonlocal correlations）中，而这种关联在经典物理里找不到对应。我们将会在后面仔细讨论Bell的定理。\n量子信息发轫于1980年代，于1990年代全面开花。我们将会在后面讨论：量子信息的压缩，量子系统能够编码的经典信息的上下界，通过含噪声的量子信道传输的可靠量子信息的上下界。\n"},"Chapter 1/1.3 Efficient quantum algorithms.html":{"url":"Chapter 1/1.3 Efficient quantum algorithms.html","title":"1.3 Efficient quantum algorithms","keywords":"","body":"1.3 Efficient quantum algorithms\n既然量子信息有这么多奇特性质，我们就会期待量子理论可以更新我们对计算的认知。Peter Shor (an AT&T computer scientist and a former Caltech undergraduate)在1994年给了一个量子计算机高效地完成质因数分解的算法，堪比平地一声雷。Shor证明了，至少在原理上，量子计算机可以高效地进行质因数分解。\n质因数分解是一个很难对付的(intractable)计算问题：答案一旦找到的话很容易验证，只要相乘就可以了；但难点就在于答案很难找到。如果p和q是很大的质数，那么他们的乘积n=pq很容易计算，所需要的基本的位操作(bit operations)数大约是\\log_2p\\cdot\\log_2q。但是，给定n，要找到p和q就很难了。\n用经典计算机进行质因数分解，大家公认需要\\log(n)的超越多项式(superpolynomial)时间（虽然还没被严格证明），即当$n$增长时，最差情况需要的时间比\\log(n)的任何次幂都大。已知的最好算法(number field sieve算法)需要耗时约\n\r\n\\exp[(64/9)^{1/3}(\\ln n)^{1/3}(\\ln \\ln n)^{2/3}],\r\n\n目前最大的运算量是几百个工作站算一个月，将130位数分解成65位数。由此推算，分解400位数需要$10^{10}$年，相当于宇宙年龄！因此，即便科技大幅进步，质因数分解400位数仍旧是个遥不可及的梦想。\n质因数分解问题，从复杂性理论（complexity theory ）角度看，是一类有趣的难解问题，即无法在（输入大小的）多项式时间内解决的问题，在上例中输入大小即为\\log n。从实践角度看，它也是很有实际应用价值的一类问题，因为质因数分解的难度就是现代公钥加密技术的基础，比如RSA加密技术。\n然而Shor发现，量子计算机竟然可以在多项式时间O[(\\ln n)^3]内完成质因数分解计算！因此，假设我们已经有了量子计算机，它能在一个月内质因数分解一个130位的数，那么它能在3年内就质因数分解一个400位的数。问题越难（位数越多），量子计算机的优势就越明显。在量子计算机面前，现有的RSA技术将无法保密。\nShor的结果激发了我（本书中“我”皆代表作者Preskill，译者注）对量子信息的兴趣，如果不是Shor，我觉得我也不会教这门课。思考Shor的结果对于复杂性理论、量子理论和技术应用是很有意思的一件事。\n"},"Chapter 1/1.4 Quantum complexity.html":{"url":"Chapter 1/1.4 Quantum complexity.html","title":"1.4 Quantum complexity","keywords":"","body":"1.4 Quantum complexity\n在Shor之前，已经有人对量子计算跃跃欲试。“量子系统可以用来执行计算”这个想法分别由Paul Benioff和Richard Feynman在1982年独立提出。这个想法随着微型电路的小型化而自然产生，因为到很小尺度，量子效应必然显著，Benioff就基于这个想法。但是Feynman的出发点很不一样，要了解Feynman的洞见，先得说说量子信息和量子计算的数学描述。\n传统比特(bit)最小的不可分单元用0和1两种状态表示，而量子比特(qubit)用归一正交基|0\\rangle 和|1\\rangle两个态张成的二维复空间表示。一个归一化的量子比特表示为|\\psi\\rangle=a|0\\rangle\t+b|1\\rangle，其中|a|^2+|b|^2=1，且a,b\\in\\mathbf{C}。我们可以测量|\\psi\\rangle到两个基的投影大小。测量结果是不确定的：测量结果是|0\\rangle态的概率是|a|^2，|1\\rangle态的概率是|b|^2。\nN个qubits构成的量子态可以用2^N维希尔伯特空间的向量表示，比如|01110010 · · ·1001\\rangle。如果每一个2^N维正交归一基被编号为|x\\rangle，那么我们就可以定义这个空间里的一般状态\\sum_{x=0}^{2^N-1}a_x|x\\rangle，其中\\sum_x|a_x|^2=1。如果我们在每一个qubit位上都用投影到\\{|0\\rangle,|1\\rangle\\}正交基的测量方法来测量N个qubits，那么得到|x\\rangle态的概率为|a_x|^2。\n现在可以开始讨论量子计算了。我们制备具有N个qubits的标准初态|0\\rangle|0\\rangle · · · |0\\rangle，或表示成|x=0\\rangle。然后对这个初态做幺正变换U（U可分解成标准化量子逻辑门的乘积，每个标准量子门每次只对几个qubits做幺正变换）。幺正变换U完成后，测量所有qubits在基向量\\{|0\\rangle,|1\\rangle\\}上的投影，这个测量值就是计算的输出结果，所以最终的输出结果是可以打印在纸上的经典信息。\n我们注意到，量子算法是一种概率算法(probabilistic algorithm)，即由于量子测量过程的随机性，同样的程序可能得到不同的输出，最终得到的是所有可能输出的一个分布函数。实际上，Shor的因数分解算法也只能保证大概率能得到正确的质因数；当然这也可接受，因为很容易验证结果是否正确。\n需要说明的是，尽管物理原理不同，量子计算机可以计算的，经典计算机也都能计算。经典计算机可以储存和旋转向量，也可以模拟量子测量过程（投影到正交基向量）。因此，经典计算机能够以任意精度模拟量子计算机。用哪种计算机其实不是重点，因为什么问题是“可计算的”对两者来说是相同的。\n但用经典计算机模拟量子计算机也有局限，需要考虑到这种模拟的时长。比如模拟N=100qubits的量子计算机，表示任意一个一般的状态|\\psi\\rangle就需要用到2^{100}\\approx10^{30}个复数，这个存储能力已经远超任何经典计算机了，而且对10^{30}维向量进行旋转操作也远超任何经典计算机的计算能力。相比之下，用N=100经典比特表示的任意一个一般状态，尽管它同样可以表示2^{100}个可能的状态，但只需要用长度为N的二进制字符串描述即可。这样的话，纵使经典计算机可以模拟量子计算机，当N变大的时候，用经典模拟量子也会无能为力，因为希尔伯特空间是在太大了！Feynman正是基于此才考虑到，量子计算机或许可以执行经典计算机不可能完成的任务（当然，量子计算机也可以轻而易举地模拟它自身！）。Shor的结果似乎印证了这个观点。\n我们可以避免这个经典计算机能力的困境吗？反过来思考一下，我们只要求模拟能给所有可能的输出结果分配一个概率分布，但是中间过程不必完整模拟N-qubit的量子态演化过程。因此，我们可以设计一种经典算法的概率化版本(probabilistic classical algorithm)：输入并不唯一决定输出，而是输出结果服从与量子计算结果相同的概率分布。这是一种局域的模拟（local simulation）：每一步中，每个qubit有确定值，但处理这个qubit的量子逻辑门依据（伪）随机数选定所有可能的操作方式之一。这种模拟方式大大简化了完全跟踪量子态演化的方式。\n但Bell不等式用数学定理不可辩驳地指出，经典算法的概率化模拟不可能实现：不存在局域概率化（local probabilistic algorithm）算法可以复现量子力学的结果！（译者注：量子关联是非局域的，即隐变量理论是不正确的。）因此，尽管没有已知的证明，但是用经典计算机模拟量子计算机似乎是很困难的。\n为了理解为何量子信息的数学描述如此复杂，我们可以想象一个3N （n\\gg1）个qubits的量子系统，它被分为三个子系统，每个子系统各自有N个qubits，分别叫做子系统(1)，(2)和(3)。我们随机选取一个3N qubits的量子态，然后把这三个子系统分开，把(1)发送到Santa Barbara，把(3)发送到San Diego，把(2)留在Pasadena。现在我们想通过测量来尽可能多地找出系统的信息。为了简单起见，假设我们手上有无数份原系统的拷贝，数量足够多，让我们可以测量任何想知道的可观测量注。除了一项限制条款：我们每次测量只能作用于其中一个子系统，不允许有跨越子系统边界的集体测量。于是，对于一个3N qubit典型的量子系统来说，我们的测量几乎得不到任何关于该量子态的有用信息。几乎所有用来区分不同量子态的信息都在子系统(1)、(2)、(3)测量结果的非局域关联（nonlocal correlations）中。这些关联就是Bell发现的用物理来描述量子信息所必须的非局域关联性。\n信息量可以用熵来表征：熵越大表示信息越少。Don Page发现，如果我们随机从3N qubits中选择一个态，那么总会发现它每个子系统的熵非常接近\n\r\nS\\simeq N-2^{-(N+1)},\r\n\n此处N是子系统熵的最大可能取值，对应于该子系统不携带任何信息。于是，对于N很大的情况，如果我们只分开测量每个子系统的话，只能得到指数级小量的信息。\n也就是说，如果我们不考虑San Diego，Pasadena和Santa Barbara的测量结果之间是如何关联的话，我们的测量只能提供很少的信息。此处，“对于关联的测量”表示集体测量（collective measurement），即使它实际上可以是由不同的人对同一份拷贝的不同子系统各自做测量，然后打电话互相通知测量结果。通过测量关联，我们可以知道更多；原理上，我们可以完全重建出这个量子态。\n任何一个关于3N qubits量子态的描述都必须把这些非局域的关联表征出来，然而这是非常复杂的一件事。这也是为何用经典模拟一个大的量子系统需要巨量计算资源的原因。当系统的不同部分之间存在这种非局域的关联时，我们称这些不同部分彼此“纠缠”（entangled），它表示我们不能通过简单地把系统分开单独研究每个部分，来获得整体的状态信息。\n\n注. 我们无法对自己手上未知的量子态进行拷贝（量子不可克隆原理），但我们可以让另一个人制备好这些拷贝（他是可以做到的，因为他明确地知道这个态是什么），只要不告诉我们这个态就可以了。 ↩\n\n"},"Chapter 1/1.5 Quantum parallelism.html":{"url":"Chapter 1/1.5 Quantum parallelism.html","title":"1.5 Quantum parallelism","keywords":"","body":"1.5 Quantum parallelism\nFeynman的想法在1985年被David Deutsch解释地更具体，Deutsch指出量子计算机的潜能在于量子并行性(quantum parallelism)。考虑一个黑盒，可以将一个比特x映射到另一个比特f(x)上，但我们不知道内部发生了什么，也许计算过程很复杂，因为它耗时24小时。一共有四种可能的函数映射关系f(x)，我们想要知道这个黑盒在计算什么。如果要计算f(0)和f(1)的结果，则总共要消耗两倍时间，即48小时。如果我们退一步，只想验证f(0)=f(1)还是f(0)\\neq f(1)，也还是需要做两次计算，耗时48小时。\n现在看看量子计算机能不能只用一次计算（耗时24小时）就能完成任务。尽管映射f(x)可能是不可逆的，但量子计算过程是幺正的且一定是可逆的。考虑一个从2 qubit转换到2 qubit的幺正变换：\n\r\nU_f : |x\\rangle|y\\rangle\\rightarrow|x\\rangle|y\\oplus f(x)\\rangle .\r\n\n其中\\oplus代表“异或”。此操作表示，若f(x)=1就翻转y，若f(x)=0则不做任何操作。Deutsch问的问题是，能不能只执行一次操作就验证f(0)与f(1)到底相同还是不同？\n由于黑盒是一台量子计算机，我们可以选取输入态为|0\\rangle和|1\\rangle的叠加态：\n\n若选取第2个qubit为叠加态|y\\rangle=\\frac{1}{\\sqrt{2}}(|0\\rangle-|1\\rangle)，那么映射\n\r\n  U_f: |x\\rangle\\frac{1}{\\sqrt{2}}(|0\\rangle-|1\\rangle)\\rightarrow |x\\rangle\\frac{1}{\\sqrt{2}}(|f(x)\\rangle-|1\\oplus f(x)\\rangle)=|x\\rangle(-1)^{f(x)}\\frac{1}{\\sqrt{2}}(|0\\rangle-|1\\rangle).\r\n  \n于是我们将f(x)分离到了前面系数中（与x有关的相位项）。\n\n我们再将第1个qubit |x\\rangle也改为另一个叠加态：\n\r\n  U_f:\\frac{1}{\\sqrt{2}}(|0\\rangle+|1\\rangle)\\frac{1}{\\sqrt{2}}(|0\\rangle-|1\\rangle)\\rightarrow\\frac{1}{\\sqrt{2}}[(-1)^{f(0)}|0\\rangle+(-1)^{f(1)}|1\\rangle]\\frac{1}{\\sqrt{2}}(|0\\rangle-|1\\rangle).\r\n  \n\n最后，将第一个比特投影到一组基向量|\\pm\\rangle=\\frac{1}{\\sqrt{2}}(|0\\rangle\\pm|1\\rangle)上。显然，如果：\n\nf(0)=f(1)（称作balanced function）那么我们总会得到|+\\rangle，\nf(0)\\neq f(1)（称作constant function）那么我们总会得到|-\\rangle。\n\n\n\n就这样，我们解决了Deutsch的问题，而且发现了经典计算机和量子计算机的区别：经典计算机必须运行两次黑盒计算才能区分balanced function和constant function，但量子计算机只需要一次计算就可以完成任务。\n这是因为量子计算机操作的不是单个态而是叠加态，因此可以一次获得映射函数的“全局”信息，这个信息同时依赖于f(0)和f(1)。在这个意义上我们称它具有量子并行性(quantum parallelism)。\n现在假设我们对作用在N-qubit上的函数的全局性质感兴趣，此函数具有2^N个可能的输入，如f(|0100...\\rangle)。要得到函数f(x)完整的映射表，需要执行2^N次f的计算。然而这个计算对于N\\gg1的情况完全不可行，比如N=100时需要10^{30}次计算。但对于一个执行幺正变换\n\r\nU_f:|x\\rangle|0\\rangle\\rightarrow|x\\rangle|f(x)\\rangle\r\n\n的量子计算机来说，我们可以选取存储在输入寄存器中的初态为：\n\r\n|\\psi\\rangle\\equiv\\left[\\frac{1}{\\sqrt{2}}(|0\\rangle+|1\\rangle)\\right]^{N}=\\frac{1}{2^{N / 2}} \\sum_{x=0}^{2^{N}-1}|x\\rangle\r\n\n只执行一次上述变换，产生了状态\n\r\n\\frac{1}{2^{N/2}}\\sum_{x=0}^{2^N-1}|x\\rangle|f(x)\\rangle\r\n\n函数f的全局信息都包含在了这个态中，只要我们想个办法提取出来即可。\n这个量子计算呈现出了“大规模量子并行性”（massive quantum parallelism）。函数再复杂都只需要计算一次，这就是量子并行性的威力。这种大规模并行性也是Shor设计因数分解算法的出发点。\n正如前面提到的，量子信息是编码在物理系统不同部分之间的非局域关联（纠缠）中的，比如上述f(x)的信息包含在了量子计算机的输入寄存器和输出寄存器的纠缠\\frac{1}{2^{N/2}}\\sum_{x=0}^{2^N-1}|x\\rangle|f(x)\\rangle中。然而，提取这个非局域信息并不容易。\n比如，假如我要去测量输入寄存器，它就会塌缩到某一个本征态|x_0\\rangle，这个本征态是从2^N个可能的态中随机选择的。这个过程会得到状态|x_0\\rangle|f(x_0)\\rangle。我们可以继续测量输出寄存器，但只能得到测量结果f(x_0)。但是由于两个寄存器之间的纠缠被第一次测量破坏，我们就没有机会再通过测量去确定任何y_0\\neq x_0的函数值f(y_0)了。在这种情况下，量子计算机并不比经典计算机更强大。\nDeutsch问题的解决方案告诉我们，如何巧妙利用那些编码在量子非局域关联中的有效信息是设计量子算法的关键。\n"},"Chapter 1/1.6 A new classification of complexity.html":{"url":"Chapter 1/1.6 A new classification of complexity.html","title":"1.6 A new classification of complexity","keywords":"","body":"1.6 A new classification of complexity\n尽管你桌上的台式电脑不是量子计算机，但它仍旧是个很厉害的设备：原理上，它能够执行任何想得到的（conceivable）计算。实际上有一些计算是无法完成的——要么时间太长，要么内存溢出。但如果你有一台内存无穷大的电脑，且愿意等待足够长时间，那么任何能够称作计算的工作都能由你手头这台小小的计算机完成。在这个意义上，我们称它是一台“通用计算机”（universal computer）。\n经典的复杂性理论研究的是困难和容易问题的区分标准。通常来说，“困难”和“容易”是由需要多少时间或存储容量决定的。但如何定义有意义的区分标准，使得它不依赖于我们使用的硬件呢？真正的用来区分简单和困难计算问题的标准应该是普适的，它不以使用高性能还是老破旧的计算机而改变。\n许多复杂性理论一般将算法的复杂性分为多项式时间(polynomial time)和指数时间(exponential time)。任意算法A，都可以定义复杂度T_A(N)，N是用bit表示的算法输入值的长度。T_A(N)表示对任意N-bit输入所需的最长时间步，即最坏情况下的最长时间。比如，若算法A是质因数分解算法，那么T_A(N)就是最坏情况下分解一个N-bit数所需的时间步。多项式时间指的是T_A(N)\\leq\\text{Poly}(N)，即解决问题所需的时间随N的增长不会快于N的某个幂次。\n如果不是多项式时间，则统称为指数时间。当然也有超多项式(superpolynomial)时间O(N^{\\log N})，随N的增长速度远比指数时间小，但也被误称为指数时间。这种是一种合理的区分问题难易的标准。但使用这种区分标准真正的原因是它不依赖于硬件，即不依赖于我们使用普通电脑还是超算。这种区分标准的普适性来自于计算机科学：不管在什么样的计算机上，如果你的运行时间是多项式时间，那么用我的计算机运行同样的算法也需要多项式时间；即使运算不了，我也可以仿真你的计算机每一个操作步骤，而这种仿真也只需要运行多项式时间，综合效果还是多项式时间。同样地，你的计算机也可以仿真我的，于是我们就能对哪个算法是多项式时间达成一致了。（当然，需要排除掉那些“不现实”的计算机，比如有无穷多个结点的并行计算机）。\n如果量子计算机也能被这样仿真，像1.4节中所说的那样，那么量子计算机就只是技术上的进步了，却无法让经典计算机的多项式时间这个数学事实改变（量子计算机也只是多项式时间）。然而，Shor算法揭示了（并没有被证明），不可能用多项式时间来仿真一台量子计算机！这就改变了一切！对于经典的通用计算机，复杂性理论30余年来证明的这些定理，作为数学事实还是成立的。但复杂性理论也许不能作为真实的物理事实，因为经典的图灵机并不是一个恰当的能描述真实物理世界中运行的计算模型。\n假如区分复杂性的量子标准和经典标准不一样（仅是猜测但未证实），那么这个结论会动摇计算机科学的基础。长远来看，它还会对技术进步产生深远影响。但它对物理学的重要性在哪儿？\n我也不能确定。但是也许仅仅提到以下事实就足以让人印象深刻：目前没有哪个经典计算机能够精确预测哪怕只是中等数量qubits（100个的数量级）量子计算机的行为。这说明少量的qubits就有很大的潜力等待我们去挖掘。\n"},"Chapter 1/1.7 What about errors.html":{"url":"Chapter 1/1.7 What about errors.html","title":"1.7 What about errors?","keywords":"","body":"1.7 What about errors?\n与Shor质因数分解算法同样重要的是，量子信息还有一个重要特性是量子纠错（quantum error correction）。只有有了这个能力，量子计算的实用化才能看到曙光。\n正如之前所说，量子计算机所处理的量子信息的本质属性是系统各组成部分的非局域关联——大部分量子信息处于系统各部分的非局域关联中。如果一次只观测某个子系统，只能获得微乎其微的信息。\n不幸的是，在现实中这种非局域关联非常脆弱，很容易被破坏。原因在于量子系统很难与环境（一个更大的系统）完全隔绝，一旦系统和环境相互作用产生了纠缠，我们最初在系统内部编码的量子信息就被转移到系统和环境的关联中去了。在那种情况下，我们无法仅通过观察系统本身得到有用信息。实际上，这个有用信息不可逆转地丢失了。对一个宏观系统来说，即使系统和环境相互作用很微弱，这个不可逆转移也会发生地很快。\nErwin Schrodinger曾经揶揄量子力学主流解释的拥护者，说在他们的理论中，叠加态会导致一只又死又活的猫存在，这就是那只著名的“薛定谔的猫”：|\\text{cat}\\rangle=\\frac{1}{\\sqrt{2}}(|\\text{dead}\\rangle+|\\text{alive}\\rangle)。Schrodinger认为这个态是量子力学白玉之微瑕，因为现实中从没人看到过一只半死半活的猫，只有非死即活。\n过去15年量子力学理论最重要的发展之一，是我们越来越有底气来回答Schrodinger的疑惑。猫态|\\text{cat}\\rangle理论上可以存在，但是我们很少能见到，因为它极端不稳定。Schrodinger的这只猫从来没有真正于环境完美隔绝，因为猫不会脱离周围环境而存在（空气，地面，阳光照射过来的光子等）。假设有人制备了薛定谔猫态|\\text{cat}\\rangle，那么编码在|\\text{dead}\\rangle和|\\text{alive}\\rangle叠加态中的量子信息会瞬间转移到猫和环境的纠缠中消散而去，永远无法再获得。这是因为，环境和猫的相互作用，本质上是环境不断对猫进行测量，将猫投影到|\\text{dead}\\rangle或|\\text{alive}\\rangle中的一个本征态，这个过程叫作“（量子）退相干”（decoherence）。\n\n译者注：为什么环境对猫测量选用的基矢一定要是\\{|\\text{dead}\\rangle,|\\text{alive}\\rangle\\}而不是一加一减两个叠加态？我猜想是这两个“能量”本征态的本征值相差很大造成的，即死猫和活猫的“能量”相差很大。真的是这样吗，如何定量证明？\n\n同样道理，要执行复杂的量子计算，不可避免要搭建规模比较大的量子系统（尽管也许没有猫那么大），但同样不幸的是，这么大的量子系统不可能与环境完全隔离，因此量子系统的叠加态总会退相干，使得量子计算机失效。换句话说，量子计算机和环境的接触（退相干）会在量子信息中引入误差。要让量子计算机正常工作，我们必须避免或纠正这个误差。\n实际上，退相干并不是我们面临的唯一问题。即使我们能把量子计算机与环境完全隔绝，也不能保证量子计算机无误差地运行。量子计算机执行的量子逻辑门一次只对少数qubits做幺正变换，比如这个幺正变换是4\\times4的幺正矩阵，作用在2 qubits上。当然，这些幺正矩阵构成了一个连续统（continuum）。理想情况是对2 qubits做幺正变换U_0，但实际上该变换会有误差，于是真实的变换是U=U_0(1+O(\\varepsilon))，即U会和U_0相差O(\\varepsilon)。在做了O(1/\\varepsilon)次逻辑门变换之后，这些误差将累积到无法忽略，导致计算失败。经典计算机也同样有这个误差累积的问题，但经典计算机处理的是离散逻辑（discrete logic），小误差并不会带来很大的问题。\n对于经典计算机，现代数字电路已经足够可靠。它们通过能量耗散的形式来获得高精度的计算过程。我们可以想象一个作用在1 bit上的经典逻辑门，表现形式为处于双势阱中某一侧的小球。该逻辑门可以把小球推一下，让它越过中间的势垒跑到另一侧。当然，这个逻辑门可能不会完美运转；它也许会稍微用力过头，或者稍微用力不足。随着时间推移，这些瑕疵会逐渐累积，导致误差。\n为了改进，我们在每次逻辑门作用后，对这个bit进行（等效的）降温。这是一个耗散过程，把热量耗散到环境中，同时也压缩小球的相空间，使得它接近势能的局部极小值。于是，计算过程产生的小误差最终仅仅是加热了环境，并没有牺牲计算准确度，影响计算结果。\n然而我们并不能用同样地方法对量子计算机进行降温。和环境接触做降温能够提升经典信息的可靠性，但这么做会完全破坏量子信息，正所谓甲之蜜糖乙之砒霜。更广泛地来说，误差累积对于经典可逆计算同样是个问题。要避免误差的累积，我们必须丢弃关于误差的信息，即进行纠错，而扔掉信息总是个耗散过程（Landauer's principle）。\n不要急，经典纠错已经有解决方案了。比如将1bit复制为3bit重复的值(0)\\rightarrow(000)，这样即使有一位出错变为(100)，也可以通过投票机制（majority voting scheme，少数服从多数）来纠错。当然这也只能以一定概率保证不出错。假设每一个bit出错的概率为p，则两个bit出错（翻转）的情况有三种，概率为3p^2(1-p)，三个同时出错的概率为p^3，因此，投票机制失败的概率为3p^2(1-p)+p^3=3p^2-2p^3。由于不使用投票机制的话（只用1bit）出错概率为p，所以如果让纠错方案更好，必须有3p^2-2p^3，即p。\n若要更高的正确率，只需要再多加几位重复位即可，即使用N-bit重复位（尽管这样并不是效率最高的）。假设每一个bit独立同分布，其不出错的概率都是P=1/2+\\varepsilon，由中心极限定理，当位数N\\rightarrow\\infty时，这些bit的平均值服从宽度为1/\\sqrt{N}的高斯分布，此时投票机制出错的概率为P_{error}\\sim e^{-N\\varepsilon^2}。因此，对任意\\varepsilon>0，只要重复位足够多（冗余度N足够大），我们就可以保证任意小的出错率。即使是\\varepsilon也不是大问题，只需要明确投票机制给出的永远是相反的答案即可。唯一有问题的是\\varepsilon=0情况，此时一串Nbit的字符串完全随机，无法提供有用信息。\n在1950年代，John Von Neumann展示了一台具有噪声部件的经典计算机能够可靠工作，只要施加足够多的冗余度即可。他指出，如有必要，我们可以对每一个逻辑门计算许多次，然后接受主流多数的输出结果。(Von Neumann was especially interested in how his brain was able to function so well, in spite of the unreliability of neurons. He was pleased to explain why he was so smart.) \n现在我们想对量子计算机进行纠错，而量子纠错会面临许多困难：\n\nPhase errors. 对于量子信息，出错的地方会更多。除了bit-flip errors，即qubit错误翻转|0\\rangle\\rightarrow|1\\rangle,|1\\rangle\\rightarrow|0\\rangle外，还会有phase errors，即|0\\rangle\\rightarrow|0\\rangle,|1\\rangle\\rightarrow-|1\\rangle。这个误差是很致命的，比如它会把态\\frac{1}{\\sqrt{2}}(|0\\rangle+|1\\rangle)变成其正交态\\frac{1}{\\sqrt{2}}(|0\\rangle-|1\\rangle)。经典的编码方式并不能避免phase errors。\nSmall errors. 量子信息是连续的，因此小扰动会把a|0\\rangle+b|1\\rangle的系数a和b改变小量，这些小量会随时间不断累积。相反，经典纠错纠正的是比较大的误差（bit flip）。\nMeasurement causes disturbance. 少数服从多数投票机制中，我们需要测量每个bit来探测并纠正错误，但我们测量qubits总是会干扰它们编码的量子信息。\nNo cloning. 经典信息可以完美拷贝，而量子信息的复制受制于不可克隆定理，不能高保真地拷贝。\n\n"},"Chapter 1/1.8 Quantum error-correcting codes.html":{"url":"Chapter 1/1.8 Quantum error-correcting codes.html","title":"1.8 Quantum error-correcting codes","keywords":"","body":"1.8 Quantum error-correcting codes\n量子问题有量子的解决办法。第一个量子纠错码（quantum error-correcting code）是Peter Shor提出的，这个发现迎来了发展迅速的新学科——量子纠错码理论。\n要了解量子纠错原理，最佳方法也许是回过头去看Shor的原始纠错码——量子纠错是经典3bit重复位机制的推广。但要注意，对于量子编码，我们要能够不直接测量编码信息就达到纠错的目的（因为直接测量会破坏量子信息）。\n考虑将1qubit编码成3qubits：\n\r\n|0\\rangle\\rightarrow|\\bar{0}\\rangle\\equiv|000\\rangle,\\\\\r\n|1\\rangle\\rightarrow|\\bar{1}\\rangle\\equiv|111\\rangle.\r\n\n或编码成叠加态：\n\r\na|0\\rangle+b|1\\rangle\\rightarrow a|\\bar{0}\\rangle+b|\\bar{1}\\rangle=a|000\\rangle+b|111\\rangle.\r\n\n\n现在我们假设某一个bit发生了bit flip eror，我们将在不改变叠加态a|\\bar{0}\\rangle+b|\\bar{1}\\rangle所包含的量子信息的情况下纠正这1个bit flip的误差。\n当然，如果我们只测量单个qubit，比如测量第一个qubit得到|0\\rangle，那么3 qubits态会塌缩到对应的本征态，比如|\\bar{0}\\rangle，导致系数a,b中蕴含的量子信息丢失。\n但我们没必要局限在单个qubit测量，我也可以一次同时对两个qubit测量，同时测量就足够探测到是否有翻转误差了。对3 qubit态|x,y,z\\rangle=|000\\rangle\\text{or}|111\\rangle，我们定义一个2 bit测量算符(y\\oplus z,x\\oplus z)，其中异或操作\\oplus也可以等价定义为“相加后取模2”，我们发现(y\\oplus z,x\\oplus z)的测量值正好是被翻转的那个qubit位置的二进制表示。这个2 bit测量值就构成了我们用来诊断误差是否产生了的症状（syndrome）。下面举两例说明。\n\n如果仅有第一位bit翻转，得到a|000\\rangle+b|111\\rangle\\rightarrow a|100\\rangle+b|011\\rangle，测量(y\\oplus z,x\\oplus z)=(0,1)，而(01)_{2进制}=(1)_{10进制}告诉我们要把第1位qubit翻转过来，这样就完成了纠错。仅有第二或第三位翻转同样适用。\n\n如果不是某一个bit翻转，而是某一个bit的幺正变换发生了小误差U=1+O(\\varepsilon)=1+i\\varepsilon\\begin{pmatrix} 0 & 1 \\\\1 &0\\end{pmatrix}，使得整体状态出现一个小误差\n\r\n  |000\\rangle\\rightarrow|000\\rangle+\\varepsilon|100\\rangle,\\\\\r\n  |111\\rangle\\rightarrow|111\\rangle+\\varepsilon|011\\rangle.\r\n  \n\n\n那么a|000\\rangle+b|111\\rangle\\rightarrow \\underset{算符(y\\oplus z,x\\oplus z)=(0,0)本征态}{(a|000\\rangle+b|111\\rangle)}+ \\underset{算符(y\\oplus z,x\\oplus z)=(0,1)本征态}{(a\\varepsilon|100\\rangle+b\\varepsilon|011\\rangle)}。此时测量(y\\oplus z,x\\oplus z)有很大概率(p=1-|\\varepsilon|^2)得到本征值(0,0)，同时也投影到正确的本征态上，消除了误差；但也有小概率(p=|\\varepsilon|^2)得到本征值(0,1)，即上一个例子中的态，此时根据本征值只需翻转第一位bit即可完成纠错恢复到正确的态。\n  对于每一个bit的幺正变换都发生小误差的情况|000\\rangle\\rightarrow|000\\rangle+\\varepsilon_1|100\\rangle+\\varepsilon_2|010\\rangle+\\varepsilon_3|001\\rangle也同样处理，会有小概率O(|\\varepsilon_i|^2)投影到三个错误态（某一位翻转了）其中之一，然后测量(y\\oplus z,x\\oplus z)的值会告诉我们具体是哪一位。\n至此，我们已经克服了上一小节提出的量子纠错的后3个困难。我们可以不破坏叠加态的信息而进行测量(第3点)，量子测量可以将“有小误差的状态”投影到没有误差的态，或者投影到翻转了一个qubit的具有较大离散误差的态（第2点）。第4点被巧妙避开了，因为a|\\bar{0}\\rangle+b|\\bar{1}\\rangle并不是通过克隆得到的，它与克隆操作得到的三重拷贝态(a|0\\rangle+b|1\\rangle)^3不是一回事。\n只有第1点困难phase errors还存在。因为一旦某一个qubit发生phase error，那么a|\\bar{0}\\rangle+b|\\bar{1}\\rangle就变成了a|\\bar{0}\\rangle-b|\\bar{1}\\rangle，量子信息就被破坏了，而且由于用了三个qubit，phase error发生的可能性变大为原来的3倍。不要怕，我们已经通过添加冗余的qubits解决了bit-flip error，现在我们来添加冗余的phases来解决phase-flip error！\n顺着Shor的思路，我们对单个qubit用9 qubits来编码：\n\r\n|0\\rangle\\rightarrow|\\bar{0}\\rangle\\equiv\\frac{1}{2^{3/2}}(|000\\rangle+|111\\rangle)(|000\\rangle+|111\\rangle)(|000\\rangle+|111\\rangle),\\\\\r\n|1\\rangle\\rightarrow|\\bar{1}\\rangle\\equiv\\frac{1}{2^{3/2}}(|000\\rangle-|111\\rangle)(|000\\rangle-|111\\rangle)(|000\\rangle-|111\\rangle).\r\n\n即用每组3 qubits，一共3组同样的量子态构成的9 qubits表示|\\bar{0}\\rangle和|\\bar{1}\\rangle。由于每一组内部具有3 qubit冗余，可以用前面提到的方法进行每组内部的single bit-flip纠错。\n\n译者注：9 qubits编码的冗余度相比3 qubits编码更多，不仅能克服single bit-flip error，还能克服phase flip error。\n\n\n现在假设某一组发生了phase flip error，这个误差改变了那一组里|000\\rangle和|111\\rangle的相对符号：\n\r\n|000\\rangle+|111\\rangle\\rightarrow|000\\rangle-|111\\rangle,\\\\\r\n|000\\rangle-|111\\rangle\\rightarrow|000\\rangle+|111\\rangle.\r\n\n使得这一组的相对相位与其他两组不同。可以通过组与组两两比较（而不是直接测量一个组内部的相对相位而破坏编码信息）找出发生phase flip的那一组进行纠错。这就需要一个6-qubit可观测量，例如用“将所有6个bit翻转”这个可观测量\\hat{A}（该观测量并不会破坏量子信息，即相对相位，而只会改变整体相位）：\n\r\n\\hat{A}|\\phi\\rangle\\equiv\\hat{A}[(|000\\rangle+|111\\rangle)(|000\\rangle-|111\\rangle)]=\\hat{A}[(|111\\rangle+|000\\rangle)(|111\\rangle-|000\\rangle)]=-|\\phi\\rangle\r\n\n若两组的phase一样，则特征值为+1，反之则如上为-1。因此只要再任意选取另一对做一次两两比较就能找出到底哪一组的phase与其他两组不同，然后对那个组里某一个qubit做幺正的相位变换即可翻转符号，完成纠错。\n\n译者注：此处不理解。探测是否发生了phase flip error，对于编码以前的本征态貌似只需要测量|\\bar{0}\\rangle或|\\bar{1}\\rangle其中一个即可。但是对于编码以后的叠加态a|\\bar{0}\\rangle+b|\\bar{1}\\rangle应该如何测量呢？\n\n\n现在假设9 qubits中每一个bit的幺正变换都发生了小误差。最一般的单个qubit幺正变换（不考虑物理上不相关的全局相位） 可以展开到O(\\varepsilon)为：\n\r\nU=1+O(\\varepsilon)=1+i\\varepsilon_x\\begin{pmatrix} 0 & 1 \\\\1 &0\\end{pmatrix}+i\\varepsilon_y\\begin{pmatrix} 0 & -i \\\\i &0\\end{pmatrix}+i\\varepsilon_z\\begin{pmatrix} 1 & 0 \\\\0 &-1\\end{pmatrix}.\r\n\n最后三个Pauli矩阵可以看成bit flip算符\\begin{pmatrix} 0 & 1 \\\\1 &0\\end{pmatrix}，phase flip算符\\begin{pmatrix} 1 & 0 \\\\0 &-1\\end{pmatrix}，或表示这两个flip同时发生的算符\\begin{pmatrix} 0 & -i \\\\i &0\\end{pmatrix}。\n\n译者注：举个例子，假设|0\\rangle\\equiv(1,0)，|1\\rangle\\equiv(0,1)，那么bit flip表示为\\begin{pmatrix} 0 & 1 \\\\1 &0\\end{pmatrix}|0\\rangle=|1\\rangle，该bit变成|0\\rangle\\rightarrow|0\\rangle+i\\varepsilon_x|1\\rangle。其余情况以此类推。\n\n假如我们制备了初态：\n\r\n\\begin{align}\r\na|\\bar{0}\\rangle+b|\\bar{1}\\rangle &= a\\frac{1}{2^{3/2}}(|000\\rangle+|111\\rangle)(|000\\rangle+|111\\rangle)(|000\\rangle+|111\\rangle)\\\\\r\n&+b\\frac{1}{2^{3/2}}(|000\\rangle-|111\\rangle)(|000\\rangle-|111\\rangle)(|000\\rangle-|111\\rangle),\r\n\\end{align}\r\n\n且允许9 qubits里的每个qubit都出现幺正误差，然后我们测量错误症状，即用2 bit算符(y\\oplus z,x\\oplus z)测量bit-flip有没有发生，用6-qubit算符\\hat{A}测量phase-flip有没有发生。在大部分情况下我们都能将系统投影回正确的初态，但有小概率O(|\\varepsilon|^2)某一个bit会出现较大误差：bit flip，或phase flip，或两者同时出现。从测量结果推测，我们可以知道哪个qubit发生了bit flip，或哪一组发生了phase error，然后对单个qubit做适当的幺正变换纠错。\n\n译者注：在上述三种量子纠错机制中，我们都假设了要么只有某一个qubit发生bit flip，或只有某一组发生了phase flip，或每一个qubit的幺正变换发生小误差。下面这段讲的是如果某两个qubits发生bit flip，或某两组发生了phase flip，量子纠错就无能为力了。\n\n\n现在假设在症状测量中存在以下情况，则量子纠错将会失效：\n\n在两组中的每个组里都有两位bit flip error（这会在编码数据中引起phase error）；\n在两个不同的组里发生phase error（这会在编码数据中引起bit flip error）。 但是这种双相误差（double phase error）发生的可能性约为|ε|^4。 因此只要|ε|足够小，那么通过编码是可以提高量子信息可靠性的。\n\n\n译者注：此处不理解，需要举例。暂时不知道如何将前面的例子用在这里进行说明。\n\n上述机制不但可以进行纠错，而且也避免了量子退相干。无论发生了哪种误差都能恢复到正确的量子态，我们避免了量子态与环境发生任何纠缠。\n当然，量子纠错也是个耗散过程，因为关于误差的信息从量子系统中被抹去了。关于误差的信息记录在了测量结果中，一旦测量结果被擦除，它就以热量形式耗散掉了。\n后续的课程中我们将进一步讨论量子纠错机制的几个方面：\n\n和经典纠错一样，存在“好”的量子纠错机制，允许我们得到任意高的可靠性，只要每个bit的错误率足够小。\n我们之前假设，执行“纠错恢复”过程本身是无瑕疵的。但是误差诊断机制很复杂——要测量2 qubit或6 qubit的“共同可观测量”（collective observables）来诊断误差——因此在我们试图纠错时也可能进一步破坏数据。然而我们将会证明，尽管恢复过程有偶发的误差，它还是很有效的。\n要操作量子计算机，不仅要能可靠地存储量子信息，还要能处理它。我们将会展示使用量子逻辑门来处理量子信息。\n\n\n总结一下量子纠错机制：\n\n误差被数字化。尽管量子信息的误差可能是小量，我们还是可以将有误差的量子态投影到无误差量子态，或投影到一个离散的误差态集合中的某个态上，后者很容易通过幺正变换纠错。\n我们不测量数据就能测量出误差的种类，因此不会获取（从而干扰）编码好的量子信息。\n误差是局域的，而量子信息是非局域的。前面讨论中我们心照不宣地做了一个很重要的隐含假设：在足够好的近似下，不同qubit的误差之间不存在关联，即假设“同时造成2个qubits产生误差”的事件的发生概率远比“只导致1个qubit错误”的事件的概率要小的多。这在物理上当然是存疑的——我们很容易就能构想一种同时导致两个qubits发生误差的过程。如果这种关联误差很普遍，上述量子纠错机制就不可靠了。\n\n上述量子纠错机制利用了误差的局域性假设来以非局域性方式编码量子信息——量子信息存储在数个qubits之间的关联中。仅仅测量9个qubits中的任何1个qubit是无法区分|\\bar{0}\\rangle和|\\bar{1}\\rangle的，因为不管编码的信息是什么，我们总是有同样的1/2概率得到|0\\rangle或|1\\rangle。\n\n译者注：此处不理解，为什么是等概率的？\n\n要提取有用信息则需要一个3-qubit可观测量（比如可以同时翻转一个组里3个qubits的可观测量\\hat{B}，它就能通过本征值来区分|000\\rangle+|111\\rangle和|000\\rangle-|111\\rangle）：\n\r\n\\begin{align}\r\n&\\hat{B}(|000\\rangle+|111\\rangle)=(|111\\rangle+|000\\rangle),\\\\\r\n&\\hat{B}(|000\\rangle-|111\\rangle)=(-1)(|000\\rangle-|111\\rangle).\r\n\\end{align}\r\n\n\n译者注：注意此处并不是在量子纠错（不破坏量子信息）。而是直接探测量子信息，测量完会破坏量子信息。因此，此处的3-qubit可观测量\\hat{B}与前面讨论的6-qubit可观测量\\hat{A}有区别。\n\n环境也许偶尔会扰动单个qubit，等效于“测量”了它，但是单个qubit携带的信息几乎为零，因此扰动单个qubit几乎不会破坏量子信息。量子纠错机制的基础，就在于非局域的信息不受局域干扰影响这个事实。\n"},"Chapter 1/1.9 Quantum hardware.html":{"url":"Chapter 1/1.9 Quantum hardware.html","title":"1.9 Quantum hardware","keywords":"","body":"1.9 Quantum hardware\n量子复杂性（quantum complexity）和量子纠错（quantum error correction）伴随着前赴后继的实验工作，这些实验专注于处理相干的量子信息。我们会简要介绍这些工作。\n为了建造一台量子计算机，我们要能操作qubits。满足这一点的硬件要求是：\n\n存储（Storage）：能存储qubit足够长时间，来完成有意义的计算。\n隔绝（Isolation）：与环境有效隔绝，最小化退相干误差。\n读出（Readout）：能高效可靠测量qubits。\n逻辑门操控（Gates）：能操纵单个qubit的量子态，且能引起qubits之间受控的相互作用，即能执行量子逻辑门。\n精度（Precision）：要让设备可靠运行，量子逻辑门的执行必须高精度。\n\n下面讲量子计算机的实现方式，Ion Trap，Cavity QED，NMR。\n1.9.1 Ion Trap\nIgnacio Cirac和Peter Zoller提出了一种实现这些目标的可能方法，并由国家标准技术研究所（NIST）的Dave Wineland实验组以及其他实验组跟踪研究。 在此方案中，用一个linear Paul trap中囚禁的单个离子代表一个qubit。\n每个离子的量子态是由基态|g\\rangle（或表示成|0\\rangle）和某个长寿命的亚稳激发态|e\\rangle（或表示成|1\\rangle）的线性组合。这两个能级的相干叠加态：a|g\\rangle+be^{i\\omega t}|e\\rangle能够存活与该激发态相当的时间（尽管这个叠加态的相对相位会由于能级差而振荡）。这些离子与外界分隔地足够好，因此最主要的退相干机制是自发衰减（spontaneous decay）。\n要读出离子的态很容易，只要通过测量把它投影到\\{|g\\rangle,|e\\rangle\\}基矢即可。准备一束激光，它能将离子的基态|g\\rangle激发到一个短寿命的激发态|e'\\rangle。当该激光照射离子时，每一个处于|0\\rangle的qubit不断地吸收、释放光子，发出荧光（fluoresces）。而处于态|1\\rangle的qubit保持不被激发的状态。\n由于离子之间的库伦排斥，离子之间是充分隔开的，因此它们能被脉冲激光单个地激发。如果一个激光光子的频率调成激发频率\\omega，且聚焦在第n个离子上，那么|0\\rangle和|1\\rangle之间就激发出了拉比振荡（Rabi oscillations）。通过适当调整激光脉冲时间并选择合适的相位，我们能施加任意的单个qubit幺正变换。特别地，激光脉冲作用在|0\\rangle上可以制备出|0\\rangle和|1\\rangle的任意线性组合。\n\n译者注：Rabi oscillations简单解释：对于一个二能级系统，施加一个电磁波，当该电磁波频率为某恰当值时，原子会不断在E1能级和E2能级间跃迁，这种现象称之为拉比振荡。拉比振荡是个随机过程，我们无法断定任意时刻原子的能级，只能做统计学分析。设对在同一时刻进行测量n次，其中原子位于E1的次数为n1，则可得原子在该时刻处于E1的概率为P=n1/n，再对P进行不同时刻分析，发现P呈现周期性变化，其波形为余弦函数\\cos\\omega t，该\\omega随外界条件改变，称为拉比振荡频率。\n译者注：上段讨论的是离子（单个qubit）的量子态。频率为\\omega的激光能把单个离子从基态打到激发态。\n\n但是设计和搭建量子计算硬件最困难的部分在于，如何让两个量子位相互作用。在离子阱中，由于离子之间的库仑排斥，量子位之间会产生相互作用。由于库仑排斥，整个离子系统存在一系列振动的耦合简正模（normal modes）。当离子吸收或发射一个激光光子，这个离子的质心会反冲。 但是如果激光调整地适当，那么当单个离子吸收或发射时，某些简正模会是多个离子作为一个整体反冲，这就是莫斯鲍尔效应（Mossbauer effect）。\n频率\\nu最低的振动模是质心模（center-of-mass (cm) mode）。在该模式下，这些离子在简谐势阱中以步调一致（lockstep）的方式振荡。 我们可以用激光将离子冷却到远低于\\nu的温度，从而使每个振动模式有极大可能占据其量子力学基态。 \n\n译者注：此处不理解。“每个振动模式”是不是指，除了cm mode还有其他normal modes，每种mode都有基态，对应一个声子激发。实际上，cm mode对应的声子可以有1个，2个……，其他normal  mode也会有1,2……个声子。这样理解是否正确？\n译者注：上段讨论的是集体振动（简正模）的量子态，即“声子”的量子态。频率为\\nu的激光能把质心模cm model对应的声子从基态打到激发态。\n\n现在想象一个频率为(\\omega-\\nu)，其中\\omega>\\nu，的激光照射在第n个离子上。 经过适当的脉冲时间后，第n个离子的激发态|e\\rangle_n将旋转为|g\\rangle_n，而cm谐振子则从其基态|0\\rangle_{cm}跃迁至其第一激发态|1\\rangle_{cm}（产生了一个cm“声子”）。 但是，直积态|g\\rangle_n|0\\rangle_{cm}不会发生共振跃迁到任何其他态上，因此不受脉冲影响。 于是，该激光脉冲导致了以下幺正变换：\n\r\n\\begin{array}{l}\r\n|g\\rangle_{n}|0\\rangle_{c m} \\rightarrow|g\\rangle_{n}|0\\rangle_{c m} \\\\\r\n|e\\rangle_{n}|0\\rangle_{c m} \\rightarrow-i|g\\rangle_{n}|1\\rangle_{c m}\r\n\\end{array}\r\n\n这个操作擦除了原先存储在第n个离子里的1 bit信息，并且将那个1 bit信息储存在了所有离子的集体状态中。\n\n译者注：此处不理解。\n(1) 为什么是频率(\\omega-\\nu)？\n(2)为什么移除的信息量是1 bit？\n(3)为什么对直积态的操作是幺正变换？\n\n这意味着第m个离子(m\\neq n)的运动状态会受到第n个离子内部状态的影响。在这个意义下，我们成功地引发了离子之间的相互作用。要完成量子逻辑门，我们应该将量子信息从cm声子转移回某个离子的内部状态中。这个过程应该设计成，在逻辑门施加结束后，cm mode总是会返回到它的基态|0\\rangle_{cm}。比如，Cirac和Zoller揭示了量子异或（XOR，or controlled not）门|x, y\\rangle \\rightarrow|x, y \\oplus x\\rangle可以通过对离子阱施加5次激光脉冲完成。对于单个束缚的离子，单个声子的有条件激发（conditional excitation）已经被NIST实验组实现了。\n离子阱量子计算机最大的缺陷之一是，它本质上是一台运行很慢的设备。它的运行速度最终受到“能量-时间”不确定关系限制。相比典型的振动能级差\\nu来说，激光光子能量的不确定性很小，因此每个激光光子持续的时间应该比\\nu^{-1}要长。实际上，\\nu的数量级在100kHz。\n1.9.2 Cavity QED\n在加州理工学院的Jim Kimble研究小组正在寻求另一种硬件设计（由Pellizzari，Gardiner，Cirac和Zoller提出）。 这个想法是将几个中性原子捕获在一个小的高精度光学腔中。 量子信息这次存储在了原子的内部状态中。 但这里的原子之间有相互作用，是因为它们都耦合到腔体中电磁场的简正模，而不是离子阱中的与振动模（声子）耦合。 同样，通过脉冲激光驱动跃迁，我们可以在一个原子中诱导一个跃迁，该跃迁取决于另一个原子的内部状态。\n另一种可能性是不以离子内部状态而是以光子极化状态来存储量子比特。 于是，可以将束缚原子当做让一个光子与另一个光子相互作用的中介（而不是使用光子将一个原子耦合到另一个原子）。 两年前在 Kimble小组的“flying qubit”实验中，他们演示了双光子量子门的操作，其中一个光子的圆极化影响了另一个光子的相位：\n\r\n\\begin{aligned}\r\n|L\\rangle_{1}|L\\rangle_{2} & \\rightarrow|L\\rangle_{1}|L\\rangle_{2} \\\\\r\n|L\\rangle_{1}|R\\rangle_{2} & \\rightarrow|L\\rangle_{1}|R\\rangle_{2} \\\\\r\n|R\\rangle_{1}|L\\rangle_{2} & \\rightarrow|R\\rangle_{1}|L\\rangle_{2} \\\\\r\n|R\\rangle_{1}|R\\rangle_{2} & \\rightarrow e^{i \\Delta}|R\\rangle_{1}|R\\rangle_{2}\r\n\\end{aligned}\r\n\n其中|L\\rangle,|R\\rangle表示具有左和右圆极化的光子状态。为了实现这种相互作用，一个光子存储在腔中，其中|L\\rangle极化不耦合到原子，但是|R\\rangle极化和原子有很强的耦合。第二个光子穿过腔体，并且和第一个光子类似，某一个极化与原子相互作用更强。仅当两个光子均具有|R\\rangle极化时，第二个光子波包才会获得特定的相移e^{i\\Delta}。 由于相移取决于两个光子的极化，这是一个非平凡的2 qubit量子门。\n1.9.3 NMR\n在过去的一年中，出现了第三种（黑马）硬件方案，它越过了ion trap和Cavity QED，在相干量子处理领域处于领先地位。 新方案使用核磁共振（NMR）技术。 现在，qubits由特定分子中的某些核自旋携带。在施加恒定磁场的情况下，每个自旋可以与外磁场对齐|\\uparrow\\rangle=|0\\rangle或反对齐|\\downarrow\\rangle=|1\\rangle。 自旋需要很长时间才会驰豫或退相干，因此qubits的保留时长比较理想。\n我们还可以打开频率为\\omega的脉冲旋转磁场（其中\\omega是自旋向上和向下的能级差），从而引起自旋的拉比振荡。 通过对脉冲时间的调制，我们可以对单个自旋上执行所需的幺正变换（就像在对ion trap的讨论中一样）。 分子中的所有自旋都暴露于旋转磁场，但只有那些能产生共振的自旋才会响应。\n此外，自旋具有偶极-偶极相互作用，可以利用这种耦合来执行逻辑门。|\\uparrow\\rangle和|\\downarrow\\rangle之间的能级差实际上取决于相邻自旋的状态。 因此，驱动脉冲是否达到能够翻转自旋的共振，取决于另一个自旋的状态。\n所有这些在数十年前就已经被化学家们熟知。但仅在去年，Gershenfeld和Chuang，以及Cory，Fahmy和Havel独立指出，NMR提供了一种实用的量子计算实现方式。但由于以下几个原因，这个想法并不显然。\n最重要的是，NMR系统非常热。自旋的典型温度（例如室温）可能比|0\\rangle和|1\\rangle之间的能级差大一百万倍。这意味着计算机的量子状态（单个分子的自旋）噪声非常大——受到强烈的随机热涨落的影响，这种噪声会掩盖量子信息。此外，我们实际上不是在单个分子上执行处理\n，而是在在包含10^{23}个“计算机”的宏观系综上，我们从该设备读出的信号实际上是该系综的平均值。但是由于量子测量的随机性，量子算法是概率性的。因此，系综平均并不等价于在单个设备上运行计算；系综平均可能会掩盖真实的结果。\nGershenfeld和Chuang，还有Cory，Fahmy和Havel解释了如何克服这些困难。 他们描述了如何通过对热力学系综执行适当的操作，从而制备、操纵和监视“等效的纯态”（effective pure states）。他们的想法是，在检测信号时，用做平均的方式把那些分子的涨落特性抹去，以便仅仅测量相干特性。 他们还指出，某些量子算法（包括Shor的因式分解算法）可以转换成确定性的形式（这样至少大部分量子计算机都会给出相同的答案）；然后再对许多计算取平均值，就不会影响正确结果了。\n最近，NMR方法已用于制备3 qubits的最大纠缠态（maximally entangled state），这是以前从未实现过的。\n显然，量子计算硬件还处于起步阶段。 在尝试进行雄心勃勃的计算之前，我们需要将现有硬件放大多个数量级（包括存储的qubits数量和可以施加的逻辑门数量）。在NMR方法中，存在一项原理上特别严格的限制，因为相干信号与背景（噪声）的比值随每个分子的自旋数量增加而呈指数下降。 在实践中，以超过10个的数量级来执行NMR量子计算将非常具有挑战性。\n如果量子计算机最终成为实用设备，那么可能需要构造量子硬件的新想法。\n"},"Chapter 1/1.10 Summary.html":{"url":"Chapter 1/1.10 Summary.html","title":"1.10 Summary","keywords":"","body":"1.10 Summary\n到此，我们结束了对量子计算的介绍性概述。 我们已经看到，以下三个相互契合的因素相结合，使得这一主题亦可赛艇。\n\n量子计算机可以解决难题。 似乎一种新的复杂性分类已经建立起来了，这种分类建立在物理学的基本定律之上，而不是传统的复杂性理论上。 （但仍然需要更精确地表述那些量子计算机比传统计算机具有更大的优势的问题。）\n\n量子误差可以被校正。使用合适的编码方法，我们可以保护复杂的量子系统免受退相干的破坏。 我们可能永远不会看到半死半活的猫，但是也许我们可以制备和储存一只半死半活的编码猫。\n\n量子硬件是可以建造出的。 我们很荣幸能见证在实验室里对量子信息进行相干操纵时代的到来。\n\n\n在此课程中，我们的目标是加深对以上三点的理解。\n"}}}