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 ...


6

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


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

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

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

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

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

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

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

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 |+\...


4

These measreuements describe a non-projective measurement. We typically convert these into projective measurements by introducing ancilla qubits. In this case, define a unitary $U$ such that $$ U|0\rangle=\alpha|0\rangle+\sqrt{1-\alpha^2}|1\rangle. $$ Take the qubit that we want to measure, and introduce an ancilla in the state $|0\rangle$. Apply controlled-...


4

Yes, it is possible, same way as it is possible to measure a state with complex amplitudes in a basis with real amplitudes (say, a $|i\rangle$ state in the $[|0\rangle, |1\rangle]$ basis). Either way, the probability of measuring state $|\psi\rangle$ and getting measurement result corresponding to basis state $|a_i\rangle$ is defined as $P_i = |\langle a_i| \...


4

Write down the two reduced density matrices of the single qubits that you have access to. Apply the Helstrom measurement (there are several descriptions of this on the site already). The problem is that, in this case, the two reduced density matrices are the same. That means that you cannot tell them apart. More explicitly, $$ |\varphi_2\rangle=(I\otimes X)|...


4

You can calculate the probability of a given answer $\pm 1$ to each measurement by evaluating $$ \langle B|\frac{I+(-1)^{\eta_1}\vec{n}_1\cdot\vec{\sigma}}{2}\otimes\frac{I+(-1)^{\eta_2}\vec{n}_2\cdot\vec{\sigma}}{2}|B\rangle $$ Thus, the probability of equal measurement outcomes is $$ \langle B|\left(\frac{I+\vec{n}_1\cdot\vec{\sigma}}{2}\otimes\frac{I+\vec{...


4

In general, "operating on a state with an observable" does not have direct physical meaning (i.e. you cannot think of it as evolving the state doing something to it). What does have physical meaning, is applying a unitary operation to a state. Every unitary operator corresponds to a physical operation that you can (in principle) implement, transforming (...


4

How it works depends on the choice of quantum system used for computation. For any choice of quantum system, the common theme is that $\text{CNOT}$ does not collapse the wavefunction, i.e. force a choice between $\vert 0 \rangle$ and $\vert 1 \rangle$, while a measurement does. A simple example (oversimplified here) uses a non-linear Kerr medium to create ...


4

The question presupposes a misconception that the vector form of a state $|\psi\rangle$ exists independently of its density operator form $|\psi\rangle\langle\psi|$, which is often described as secondary. In reality, the density operator of a state is all that truly exists --- and even then, it only exists as statistical information. In fact, you can ...


4

Look like a lot of misunderstanding. If you measure a state $\alpha|0\rangle+\beta|1\rangle$ in computational basis, the state collapses either to $|0\rangle$ or $|1\rangle$. In Quantum Information we say that we measured state $|0\rangle$ or $|1\rangle$, or simply we measured $0$ or $1$. There is no chance to know $\alpha$ and $\beta$ from the measurement. ...


4

Now If I chose the standard basis $|0\rangle,|1\rangle$ what will be the result I will get? If I measure with respect of the projection operator $\langle 0|$ I get $\alpha$ and If I measure with respect of the projection operator $\langle 1|$ I get $\beta$. This is wrong. First of all, arguably the most natural kind of measurement in QM consists in choosing ...


4

Three outcomes amounts to more than one bit if the outcomes are all deterministic, and give you information about the original qubit. But suppose I have a coin (that is either heads or tails). I roll a dice, and if it comes 1 through 5, I tell you "H" or "T", depending on what the coin is. If it comes up 6, I tell you "6". There are three outcomes, but ...


3

You're correct that unitary gates like the Hadamard and CNOT do not perform any measurement. In fact, measurement isn't a unitary operation at all! There are separate measurements gates to measure the individual qubit states. The basis which the measurement gates will use is implementation dependent. However, in your specific circuit (N&C p. 188, Ex. 4....


3

Thanks for pointing this out. This is a bug that occurs when only a subset of the qubits are measured. It's being fixed. Until then, workarounds are: Use Aer instead of BasicAer (always the best thing to do when possible). Use LegacySimulators instead of BasicAer (this will give a deprecation warning). Install Qiskit from the master branch, where the issue ...


3

To answer this, contrast quantum teleportation with the swap gate. Ignoring everything in the middle, the effect of quantum teleportation is to get from state $|\phi00\rangle$ to $|00\phi\rangle$. This can obviously be accomplished with a simple application of the swap gate. So, why do we care about quantum teleportation? You'd never run the quantum ...


3

I assume that you mean to say that initially we are not aware of the quantum state of the qubit. The answer to your question is then, No. Say, the qubit was in some state $\alpha |0\rangle + \beta |1\rangle$. Now, if by any way we are able to prove that $\alpha =0$ (or $\beta =0$, for that matter), with a single shot experiment, then in a sense we are able ...


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