6

If the simulator is saying that state 00 occurs 75% of the time then the simulator has a bug. Reordering measurements can't make certain outcomes more likely in that way. It would violate the no communication theorem.


5

It appears that you have some confusion regarding the basic notions of density operators and "dimension". Why $d^2$ dimensions "are required to describe" a density matrix isn't the right question to ask; density matrices are $d^2$ dimensional objects in the same sense that vectors in $\Bbb R^3$ are 3-dimensional (i.e., the cardinality of any basis set of $\...


5

You specifically ask about qubits, so I'll keep it to that. Imagine you have a state $$ |\psi\rangle=\sum_{x\in\{0,1\}^n}a_x|x\rangle. $$ You can choose to look at each qubit. I'll take the first qubit for the sake of simplicity. We have that $$ |\psi\rangle=|0\rangle\sum_{y\in\{0,1\}^{n-1}}a_{0y}|y\rangle+|1\rangle\sum_{y\in\{0,1\}^{n-1}}a_{1y}|y\rangle $$ ...


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

Note that $H | 0 \rangle = | + \rangle = \frac{1}{\sqrt{2}}(| 0 \rangle + | 1 \rangle)$. So, after the two Hadamard gates the state will be $$ | 0 \rangle \otimes H| 0 \rangle \otimes H| 0 \rangle = \frac{1}{2} | 0 \rangle \otimes (| 0 \rangle + | 1 \rangle)\otimes (| 0 \rangle + | 1 \rangle) = $$ $$ = \frac{1}{2} (| 000 \rangle + | 001 \rangle + | 010 \...


3

Given an arbitrary state $|\psi\rangle$, if it is expressed in the computational basis as $|\psi\rangle=\sum_k c_k |k\rangle$, then it will give the $k$-th result (when measuring in the computational basis) with probability $|c_k|^2$. Note that here by "computational basis" I simply mean the measurement basis under consideration. If you consider another ...


3

Preliminary I would like to rewrite the equation that you have in a slightly different manner. Since a density matrix can be written as a matrix, we can also write it down as a linear combination of elements from a basis for the space of density matrices. We can use essentially any basis to do this, but some are preferred: most notably, the Pauli basis. For ...


3

Let’s say that Alice, Bob, and Charlie hold random bits, which are either all or all (so, they’re classically correlated). If all three of them get together, they can see that their bits are correlated, and the same is true if only two of them are together. If Alice, Bob, and Charlie all have either the bits $\{1_A, 1_B, 1_C\}$ or the bits $\{0_A, 0_B, 0_C\}...


3

Yes you are absolutely right with your understanding but here is the game: All the Qubits are starting with the |0> i.e. value 0 until the Hadamard gate is applied. The Hadamard gates just makes the probability of the Qubit being 0 or 1 to be 50-50 i.e $\frac{1}{\sqrt{2}}|\mathbf{0}> + \frac{1}{\sqrt{2}}|\mathbf{1}>$ Now in the second example you ...


3

In theory, there is no difference between this and measuring one qubit and then the other. One way of thinking about this is from the perspective of inertial frames of reference in special relativity - for two separated parties, there is no global definition of simultaneous. Different observers can observe different orderings of space-like separated events. ...


3

I am not following all of the calculations in your post (for one thing because your first displayed equation does not mention $C_1$), but I know why the goal of the exercise is true. In fact it is true for any code at all, not just a CSS code. If $\mathcal{C} \subseteq \mathcal{H}_\text{qubit}^{\otimes n}$ is any code, then you get an equivalent code $\...


3

There isn't. A density matrix encodes all the knowledge available about a state, therefore if two states are described by the same density matrix, they are indistinguishable. Ket vectors differing by only a global phase have always the same density matrix, and represent the same physical state.


3

We can begin this question by comparing two small cases, a qubit with a "rebit", the latter being a qubit confined to real amplitudes. The set of pure state of a rebit makes a great circle in the Bloch sphere, say a vertical circle. The corresponding measure on the probability interval is given by projecting uniform measure on the circle onto the $z$ ...


3

This is a neat idea. However, having individual overlaps being small isn't sufficient in a quantum system. For example, imagine the set of overlaps $$ \langle V_1|V_N\rangle=0,\qquad \langle V_1|V_n\rangle=\langle V_N|V_n\rangle=\epsilon \quad \forall n=2,\ldots,N-1, \langle V_n|V_m\rangle=0 \quad \forall n,m=2,\ldots,N-2 $$ Remember that these are supposed ...


2

Let's define the kets, ket0 = {{1},{0}};ket1 = {{0},{1}}; This function produces input from a string, f[x_?StringQ] := ToExpression[StringJoin["ket",x]]; This function produces the diagonal matrix corresponding to a string "000", matrixFunc[x_]:= KroneckerProduct@@ f/@ StringPartition[x,1] . ConjugateTranspose[KroneckerProduct@@ f/@ StringPartition[x,...


2

The idea behind pseudo-random number generators is to generate random numbers using the software, which is convenient and cost-effective. The pseudo-random number generators are one-way functions so if you know the seed you can predict their subsequent values but you cannot determine the sequence of values if you do not know the seed. Quantum randomness is ...


2

Depends on what pseudorandom generator Python is using; if the pseudorandom generator is cryptographically secure, then it is computationally impossible to distinguish whether a bit string is generated by the pseudorandom generator or is truly random. On the other hand, "truly random" quantum generator may fail classical randomness tests, because real-world ...


2

Consider an arbitrary vector $\boldsymbol v\equiv(v_i)_{i=1}^N\in\mathbb C^{nm}$ of length $n m$ (if you care only about qubits, just fix $n=m=2$). For $\boldsymbol v$ to have an $(n,m)$ partition means that we can write it as $\boldsymbol v=\boldsymbol u\otimes \boldsymbol w$ for some $\boldsymbol u\in\mathbb C^n$ and $\boldsymbol w\in\mathbb C^m$. If this ...


2

In the line immediately before Step 2, you have like terms that can be combined. Combine these, then renormalize the vector. Recognize that your misstep lies in the properties of the amplitude. Generally, for some $a, b$ components and $x, y$ separation: $$ |a + bi|^2 \neq |x + yi|^2 + |(a - x) + (b - y)i|^2 $$ In this case, the like terms aren't combined,...


2

Yes, the overall density matrix shared between Alice and Bob is $|\psi\rangle\langle\psi|$. To get the desnity matrix of either Alice or Bob, you should calculate $$ \text{Tr}_B|\psi\rangle\langle\psi|\qquad\text{Tr}_A|\psi\rangle\langle\psi| $$ respectively. However, in this particular case, the calculation is much simply. Let $|\phi\rangle$ be the Bell ...


2

The speed up is not expressed by the exponentially many basis states that a quantum system can be in. The speedup comes from being able to recombine the probability amplitudes associated to the these basis states so that a measurement can output the correct result of the computation with sufficiently high probability. If the exponentially many basis ...


2

$N$ qubits have $2^N$ basis states, and $2^N$ probabilities (to be in each of basis states). The question "does the number of probabilities express the exponential speed-up ?" does not make sense to me.


2

Yes, you can write a diagonally polarized state as a linear superposition of horizontally and vertically polarized states ("mixture" isn't the right term though; it's still a pure state). For instance, the $45^{\circ}$ diagonally polarized state $\lvert \nearrow \rangle$ may be expressed as $$\lvert \nearrow \rangle = \frac{1}{\sqrt 2} \lvert \rightarrow \...


1

Sometimes, mainly in popular articles and books, it is stated that $n$ qubits can be in all $2^n$ possible zero-one combinations simultaneously (the superposition of these combinations) and any calculation on this register is done with all these combinations. This is used as an explanation for speed-up brings by a quantum computer. However, this is ...


1

I hinted in the comments of your previous question that the answers to the above questions have a shared basis for an answer, and since you're already asking about that too I guess you have found that: the Bloch sphere. The Bloch sphere is a method of visualizing the state of a qubit. However, to really understand this visualization, you really need to use ...


1

The ion states are the different energy states of the ion. These different energy states can be denoted with the total spin of the ion and the phonon levels of the ion. If we look at a Helium atom which has two electrons the total spin can be S = 0 if the electrons are in the following spin state: $$ \frac{1}{\sqrt{2}}(\uparrow \downarrow-\downarrow \...


1

I tried to prepare and measure this Bell state on IBM Q. There is a low probablity, around few percent, that you measure states $|01\rangle$ or $|10\rangle$. But this is caused by a thermal noise and spontaneous break of the entangled state.


1

Please be careful with your notation; don't get confused with the number of qubits that are in $|0_{Eve}\rangle$ (I'm not necessarily saying that you are - just mentioning this as a precaution). Since the state $|\psi\rangle = \frac{1}{\sqrt{N}}\sum_{j \in 0,1,2....N-1}|j\rangle$ that Alice wants to send is a $n$-qubit state with $N=2^{n}$, Alice needs to ...


1

It does exist. It's basically the controlled-not gate generalised to higher dimensional systems. The important thing to realise is that means that the bit that Bob ends up with will be highly entangled with what Eve has, and that will have significant observable consequences. As part of a cryptographic protocol, for exampe, Bob could detect that Eve is ...


1

There is no long math, and you need not online calculators. When you simply omit $|+\rangle$ state of the first qubit after the measurement, the resulting 2-qubit state $|\Phi\rangle$ is unnormalized. You just need to factor out the state $|-\rangle$ of the first qubit and normalize the remaining state of the second qubit before obtaining the final answer.


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