9

In simpler terms your question is: if noise/decoherence keeps entering the computation, how can a big computation possibly survive? The key concept you're missing is quantum error correction, which can pump noise/decoherence back out of the system. Of particular practical interest is the surface code.


9

Apart from the formal result about #P-hardness, there's something worth touching on, about the nature of strong simulation itself. I'll comment first on strong simulation, and then specifically on the quantum case. 1. Strong simulation even of classical randomised computation is hard Strong simulation is a very powerful concept — not only in the fact ...


8

We simply translate the binary result of a qubit measurement to our guess whether it's the first state or the second, calculate the probability of success for every possible measurement of the qubit, and then more find the maximum of a function of two variables (on the two-sphere). First, something that we won't really need, the precise description of the ...


7

An observable only needs to be Hermitian, and can have any real eigenvalues. They don't even need to be distinct eigenvalues: if there are repeated eigenvalues, we say that the eigenspace for that eigenvalue is degenerate. (In the case of observables on a qubit, having a repeated eigenvalue makes the observable rather uninteresting, because absolutely all ...


7

Yes. Remember that you require several properties of a projective measurement including $P_i^2=P_i$ for each projector, and $$ \sum_iP_i=\mathbb{I}. $$ The first of these show you that the $P_i$ have eigenvalues 0 and 1. Now take a $|\phi\rangle$ that is an eigenvector of eigenvalue 1 of a particular projector $P_i$. Use this in the identity relation: $$ \...


7

So, Bob is given the following state (also called the maximally-mixed state): $\rho = \frac{1}{2}|0\rangle\langle 0| + \frac{1}{2}|1\rangle\langle 1| = \begin{bmatrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{bmatrix}$ As you noticed, one nice feature of density matrices is they enable us to capture the uncertainty of an outcome of a measurement and ...


6

Such an example is given in Bennett et al., Quantum Nonlocality without Entanglement, Phys. Rev. A. 59, 1070 (1999).


5

So Alice sends Bob a qubit with the density matrix $$\rho = \frac{1}{2}|0\rangle\langle 0| + \frac{1}{2}|1\rangle\langle 1| = \begin{bmatrix} .5 & 0 \\ 0 & .5 \end{bmatrix}$$ as you said. (I've fixed the notation to make it a density matrix, what you wrote was in the structure of a state, but with non-normalized coeffecients. It is important to ...


5

With the given measurements, you cannot: there is no observable difference between many different states such as $|\pm\rangle=(|0\rangle\pm|1\rangle)/\sqrt{2}$. In order to determine what the state is completely, you need more measurements. If you're using projective measurements, you need two more. These would typically be projections onto the bases $$ |\...


5

You probably want to look at old posts about Simon's algorithm, such as the rather complete explanation I gave here, or talking more specifically about the number of times the algorithm has to be repeated. Yes, you have to repeat the algorithm several times to get different pieces of classical data, which you then process classically to get your final ...


5

Any set of commuting observables in any quantum state can be characterized by a joint classical distribution function describing the probabilities of its measurement outputs in that quantum state. Since you need a single observable and it is of course self commuting, the above is valid in your case. The obsevable in your case is: $$\sigma = \cos \phi \...


5

Given only one copy of such a state, it is not possible to determine it with any good probability. Reason being there is no way, in principle, to extract information from the system without making a measurement on the system. And when we go for a measurement, we project it to a basis element $i,j$ if we turn out to pick these by chance from the set $\{1,2,......


5

Cirq uses numpy's pseudo random number generator to pick measurement results, e.g. here is code from XmonStepper.simulate_measurement: def simulate_measurement(self, index: int) -> bool: [...] prob_one = np.sum(self._pool.map(_one_prob_per_shard, args)) result = bool(np.random.random() <= prob_one) [...] Cirq ...


5

No, no trick. You can’t prove it. One way to think about this is what if you had either the 1 state, or something that is arbitrarily close to the 1 state with a tiny amount of 0. You’re essentially asking if it’s possible to perfectly distinguish them. But if you could, you could copy them, and you’d have perfect cloning of a pair of non-orthogonal states, ...


5

You just need to do a bit more algebra: Note that $$ \sum_{i=0}^n (\overline{x_i+y_i})(x_i+y_i)=\langle x+y|x+y\rangle$$ and then you can distribute the right-hand side to get $$\langle x|x\rangle+\langle x|y\rangle+\langle y|x\rangle+\langle y|y\rangle.$$ Since $| x\rangle$ and $| y\rangle$ are normalized, we know that $\langle x|x\rangle=\langle y|y\...


5

If $P$ is (ortho)projector, that is $P^2=P=P^\dagger$, then we can define unitary $U = I - 2P$. You can verify $$ UU^\dagger = U^2 = (I-2P)(I-2P) = I-4P+4P = I $$ Now we can express $P=\frac{1}{2}(I-U), Q=\frac{1}{2}(I+U)$ and calculate $$ P\rho P + Q\rho Q = \frac{1}{4}(I-U)\rho(I-U) + \frac{1}{4}(I+U)\rho(I+U)= $$ $$ = \frac{1}{4}(\rho - U\rho - \rho U +...


5

$\text{Tr}(AB)$ is always real and non-negative if $A,B$ are positive semi-definite hermitian matrices. To see this note that $A = UDU^\dagger$, for some unitary $U$ and diagonal matrix $D$ with $d_{ii} \ge 0$. Then $\text{Tr}(AB) = \text{Tr}(UDU^\dagger B) = \text{Tr}(DU^\dagger B U).$ But $B^\prime = U^\dagger B U$ is also positive semi-definite and ...


4

Let's start from the state $$ |\Psi\rangle=\frac12\left(|0\rangle(|\psi\rangle|\phi\rangle+|\phi\rangle|\psi\rangle)+|1\rangle(|\psi\rangle|\phi\rangle-|\phi\rangle|\psi\rangle)\right). $$ There are a couple of ways to do the calculation. If you want to be formal, which typically leads to fewer mistakes, you identify the measurement operators on a single ...


4

Denote $| \psi_{i,j} \rangle = \frac{1}{\sqrt{2}} ( | i \rangle + | j \rangle) $ for fixed $i$ and $j$. Alice gives a randomly drawn one of this form. That is for each $0 \leq i<j \leq N$ she assigns a probability $p_{ij}$ and sends the corresponding state $| \psi_{ij} \rangle$ with that probability. When you say unknown of that form, I'll assume $p_{ij}=...


4

Less formally-stated than the other answers, but for beginners I like the intuitive method outlined by Prof. Vazirani in this video. Suppose you have a general two-qbit state: $|\psi\rangle = \begin{bmatrix} \alpha_{00} \\ \alpha_{01} \\ \alpha_{10} \\ \alpha_{11} \end{bmatrix} = \alpha_{00}|00\rangle + \alpha_{01}|01\rangle + \alpha_{10}|10\rangle + \...


4

Let's say you want to distinguish two states: $$|A\rangle = \cos \alpha |0\rangle + \sin \alpha |1\rangle \\ |B\rangle = -\sin \alpha |0\rangle + \cos \alpha |1\rangle$$ For your particular example $\cos \alpha = \frac {\sqrt{3}}{2}$ and $\sin \alpha = \frac{1}{2}$, so $\alpha = \frac{\pi}{6}$. These states are orthogonal and can be obtained from $|0\...


4

Suppose you have a state $\rho$, and a random process that changes this to a state $\rho_j$ with probability $p_j$. If you know what the value of $j$ is, your knowledge of the resulting state will be described by the corresponding $\rho_j$. If you have no information regarding $j$, your knowledge will be described by $$\sum_j ~ p_j ~ \rho_j$$ This is a ...


4

It is important to emphasise that a density matrix may not be absolute; it represents the state of somebody's knowledge of the system. To see this, consider 3 parties: Alice, Bob and Charlie. Alice prepares a qubit in either $|0\rangle$ or $|1\rangle$. Now, Alice knows which state she prepared (let's assume it's $|0\rangle$), so Alice's description of the ...


4

Yes. You already have the reasoning for why. Sometimes the basis is implied but not explicitly stated though. So caution when reading something like that.


4

measurements in every circuit can be postponed or never performed in a circuit while achieving the same functionality of the circuit That's correct. But if the circuit involves two parties, this process will introduce quantum operations between the two parties. It will require a quantum communication channel, so that the qubits can be shuttled back and ...


4

This phenomenon is sometimes known as a discretization of errors. It is a property of certain error correcting codes that allows it to work. It is described (somewhat briefly) in Section 10.2 of Nielsen and Chuang. Suppose that we have an arbitrary error that affects just one qubit, and suppose that we represent this error by a channel $\Phi$ mapping one ...


4

Note that measuring an observable is equivalent to projecting the quantum state into a particular eigenspace of the operator, and the measurement result tells you which eigenspace. So, in the case of measuring an observable on an eigenstate of that observable, you just project the state onto itself, and the outcome tells you the eigenvalue. So, (2)is ...


4

This question is actually entirely about the basics of measurement on a quantum system, and nothing to do with secret sharing. Let's state the measurement postulate of quantum mechanics as it applies to projective measurements: A measurement is described by a set of projectors $P_i$ satisfying $\sum_iP_i=1$. If a state $|\psi\rangle$ is being measured, ...


4

Without additional assumptions or context, there is no fundamental difference between an "$2^n$-dimensional qudit" and "$n$ qubits". Any "qudit system" over $2^n$ modes for some integer $n$ can be thought of as a system of $n$ qubits. Equivalently, an $n$-qubit system is nothing but a $2^n$-dimensional qudit system. The difference is in the fact that if you ...


4

The Hadamard gate is: $$\frac{1}{\sqrt 2} \left(|0\rangle \langle 0 | + |0\rangle\langle 1| + |1\rangle \langle 0| - |1\rangle \langle 1|\right)$$ And since $|+\rangle$ is $\frac{1}{\sqrt 2}\left(|0\rangle + |1\rangle \right)$, you can work out that $H(|+\rangle) = |0\rangle$ So, $$CNOT(H|+\rangle \otimes |+\rangle)$$ $$= CNOT(|0\rangle \otimes |+\...


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