# Tag Info

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The post measurement state is computed in the usual way: If you measure an observable with spectral decomposition $X = \sum_x x \ P_x$, $P$ being the projectors, the post measurement state if the outcome is $x$ is simply $\rho_x = \ P_x \ \rho \ P_x$ (up to normalization) To derive this in a neat way from the original postulate about pure states, you ...

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Even though you think of it as a single particle -- each of it's different properties like momentum, spin, etc (corresponding to each valid quantum number) sits in a Hilbert space of their own and the possible configurations of a particle sits in a tensor product of those Hilbert spaces. $$\mathcal{H_{particle}} = \mathcal{H_{momentum}} \otimes ... 0 "projection operators commute → they're the same" Are you sure he said this predicate ? or it is your own consequence? However, it is not true ! Consider two dimensional X-Projector And Y-Projector , they commute but they are not the same! 0 A complete set of eigenstates spans the whole space, not just the subspace the projection operators project on. In this set of eigenstates you also have a basis of the subspace belonging to the eigenvalue 0. 0 If you knew the maximum entropy S_{\text{max}} possible for a system then you know how many possible states there are because$$S_{\text{max}}=\sup_{p_n}\left\{-\sum_nkp_n\log p_n\right\}=k\log N,$$where k is Bolzmann's constant, and N is the number of states. There is a limit to the amount of entropy a volume of space can hold, and such a maximally ... 0 It's not exactly what you ask for, I guess, but there is a quite recent proposal by two very renowned physicists linking wormholes and entanglement: Cool horizons for entangled black holes by Maldacena and Susskind. A more accessible review of the idea can be found here: http://quantumfrontiers.com/2013/06/07/entanglement-wormholes/ Unfortunately, my ... 0 You could try something like this. Suppose you have two electrons in the singlet state. You can measure the electron's spin along three directions x,y,z. Regardless of the direction you measure the spin the probability of getting up or down is 1/2. If you measure the x spin of both electrons when you compare the results you find that they are opposite: if ... 1 To add to the answers above, an advantage to the quantum information theorist is that finite fields of ODD prime characteristic have nice properties, hence when using tools such as discrete phase spaces, qutrits can have properties that are otherwise hard to generalize to qubits. For example see arXiv:quant-ph/0602001. 2 The trace defined as you did in the initial equation in your question is well defined, i.e. independent from the basis when the basis is orthonormal. Otherwise that formula gives rise to a number which depends on the basis (if non-orthonormal) and does not has much interest in physics. If you want to use non-orthonormal bases, you should adopt a different ... 1 You can compute the trace of an endomorphism using any basis (including non-orthogonal ones). In Dirac notation, you show this by inserting the identity expressed in the new basis and re-arranging:$$\begin{align*} \sum\limits_{|s\rangle \in B} \langle s^*| \rho |s\rangle &= \sum\limits_{|s\rangle \in B} \langle s^*| \left( \sum\limits_{|t\rangle \in ...

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No, you can still only read 1 value out - $a$ and $b$ tell you the probability of either being read; unless either $a$ or $b$ are zero, the readout is probabilistic. [Caveat: I assume by 'read' we are talking about non-quantum results, i.e. not entangling with the qubit(s)] We have a single qubit in the state: $a|0\rangle + b|1\rangle$ The probabilities ...

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If you have these two states, when you take a measurement then the probability to obtain state $|0\rangle$ is $|a|^2$ and the probability of observing state $|1\rangle$ is $|b|^2$, thus its is required that $|a|^2 + |b|^2 = 1$. That means the system is definitely in one of these 2 states.

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Quantum Computation is testing the validity of QM. Which I guess is an interpretation to those people that think that QM is written in stone at the base of the universe. So it is testing interpretation, and thus may change it. Real physical attempts at Quantum Computation lean heavily on QM working perfectly. So these are then experiments in the validity of ...

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Every great physical theory has been proven wrong. That's of course not (yet or ever) true, but if history is any guide, physical theories fail. Newton's gravity is a prime example. Even Newton could see problems in its foundations, yet it is highly accurate and can be used to guide satellites and space probes. By questioning those foundations (and other ...

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In theory, there is one easy way to decide whether a quantum circuit can be realized in principle: Can we implement a universal gate set with the system? If yes, then we can implement any circuit, if no, then we can just implement circuits with the gates we know how to implement. So much for simple theory. The question however is much harder. First of all, ...

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Presently there is no known general argument to say wehther some qudit dimensions are better than others for implementing fault tolerant quantum computation schemes. (I know of no paper showing something like that.) However, it is true that sometimes you can gain something by using qudits. (Meaning that some particular codes work better if you use qudits.) ...

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This is a coupled linear differential equation. Just write it in a matrix form as $\dot{X} = M X$, where $X$ is a vector formed by $a$ and $b$, and $M$ is a matrix. The solution is similar to the one dimensional case, but you will get the exponential of the matrix $M$ in the solution. You can get the components of $\exp (M)$ by using the Taylor expansion of ...

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Using higher dimension (d>2) quantum systems, or qudits, indeed provides an advantage through greater control of the Hilbert space. In quantum key distribution (QKD) for example, qudits can enhance the average raw key rates as one can encode more bits per symbol. Even more, they improve the robustness or the noise tolerance of the QKD protocol -- Alice and ...

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There aren't really many books on quantum cryptography. The only one I am aware of is a book titled Applied Quantum Cryptography. The Nielsen and Chuang book has a few pages dedicated to quantum cryptography in chapter 12.6. However, I would recommend the following review papers on quantum cryptography as opposed to textbooks: Gisin et. al (2001) ...

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Closing the gap between lower and upper bounds for the tolerated errors in quantum key distribution protocols is a long standing problem in the field. In general, to give a lower bound on the key rate, one must provide a particular security proof of the protocol, but this proof may be suboptimal. For the BB84 protocol, the highest tolerable error rate I ...

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The Solovay-Kitaev algorithm is a known universal method that does what you want. Probably http://arxiv.org/abs/quant-ph/0505030 will be of interest for you.

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1) Basically, it seems what they implemented a version of the superdense coding theorem. What's that about? Well, the idea is the following: As always, Alice wants to send bits to Bob. Before they start, however, they already establish a connection by sharing a maximally entangled state $|\Phi\rangle=\frac{|00\rangle+|11\rangle}{\sqrt{2}}$. So, in ...

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The reduced matrix is defined as the partial trace of the density matrix. Be $A$, $B$ finite dimensional Hilbert spaces and be $T$ $\in$ $L(A \otimes B)$ (Linear operators on $A \otimes B$), then the partial trace of T is defined as $\rm{Tr}_B [T]$ in $L(A)$ is defined by \langle a | \rm{Tr}_B [T]| b \rangle = \sum_n \langle a | \langle n ...

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The reduced density matrix can be found by taking the trace over the subspaces of the Hilbert space that represent systems you're not interested in. For the Bell state the density matrix of the whole system is \tfrac{1}{2}(|00\rangle+|11\rangle)(\langle 00|+\langle 11|)\\ = \tfrac{1}{2} (|00\rangle\langle 00|+|00\rangle\langle 11|+|11\rangle\langle ...

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It's likely bad taste to answer your own question, however, here I am only pointing out the argument of another in the literature. I also won't accept this answer to allow others to respond. Surprisingly, my above guess that perfect copying is only limited by the probability of spontaneous emission... might actually be correct. Quoting from [Milonni, P. ...

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