6

First of all, note that the statement, as written, is wrong (or rather, it is correct only as long as the "$\equiv$" symbol is taken to mean "equal up to a phase"). An easy way to see it is by computing the determinant of $H=e^{i\pi H/2}$, which gives $-1=1$ (using $\det[\exp(A)]=\exp[\operatorname{Tr}(A)] $ for all $A$ and $\operatorname{Tr}(H)=0$). Now, ...


6

The Hadamard gate has close ties to the discrete Fourier transform. Consider the DFT for an $N$-level system: $$\vert j \rangle = \frac{1}{\sqrt{N}} \sum\limits_{k=0}^{N-1} e^{\frac{i2 \pi j k}{N}} \vert k \rangle.$$ For $N=2$ this is simply $$\vert j \rangle = \frac{1}{\sqrt{2}}\begin{bmatrix}1 & 1 \\ 1 & -1 \end{bmatrix} \, \vert k \rangle = H \...


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

If you look at the formula you want to prove term-by-term, you'll notice that the sum and the $(-1)^f(x)$ part is the same in both formulas; you just need to show that $$H^{\otimes n} |x\rangle = \frac{1}{\sqrt{2^n}} \left( \sum_{y=0}^{2^n-1} (-1)^{x \cdot y} |y\rangle \right )$$ You can either show this strictly by induction (similar to this question but ...


4

You did more than you needed to in part i. You effectively did part ii already. Because $$ g = \frac{u}{\sqrt{\det{u}}} $$ is in $SU(2)$, the phase you need to multiply by is $\frac{1}{\sqrt{\det{u}}}$. $$ \det H = -1/2-1/2 = -1 $$ So to get in $SU(2)$, you need to multiply $\sqrt{-1}=\pm i$. That is seen as the $e^{\pi i /2}$ you see as the prefactor ...


4

For questions like this, the conventional physics notation is easier to work with than the QIT gate notation. Define $\vec \sigma = (\sigma_1,\sigma_2,\sigma_3)$ to represent the three Pauli matrices $$\sigma_1 = X = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}, \;\;\; \sigma_2 = Y = \begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix}, \;\;\; \...


3

Hint: Consider the series expansion of the exponential in the case of matrices $M$ that satisfy $M^2=-1$ (in your case $M=i \frac{X+Z}{\sqrt(2)}$. You should find something akin to Euler's formula which renders the proof trivial.


3

The four Bell states are $$ |\Phi_{\pm}\rangle=(|00\rangle\pm|11\rangle)/\sqrt{2}\qquad |\Psi_{\pm}\rangle=(|01\rangle\pm|10\rangle)/\sqrt{2}. $$ So, let's consider what happens then we try and measure in the Bell basis, i.e. project onto one of these four states. If we started with the state $|00\rangle$, then we can write it as $$ |00\rangle=\frac{1}{\sqrt{...


3

Most probably because your input state was $|0\rangle$. The Hadamard gate has the property that $H^2=I$, thus acting on an arbitrary input $|\psi\rangle$ with the Hadamard gate twice in a row results in the output being identical to the input (that is, $|\psi\rangle$).


2

"Payload preperation gate" is just a fancy name for preparing the message with a more clear setting with the probability of 85.4% ∣0⟩ and 14.6% ∣1⟩ in this case. The chance of being a zero is less than that of one. A standard example of teleportation gives a 50% chance, in that case it is not a clear 0 or 1. The second H gate changes the relative phase of ...


2

You can easily check that $H^1=XHX$, which is another way to say that $H^1$ is the same as $H$ modulo swapping $|0\rangle$ and $|1\rangle$. This means that it is a different gate. On the other hand, it is "equivalent" in the sense that, given an arbitrary circuit given as a sequence of gates $\prod_k U_k$, if you swap all the $|0\rangle$s and $|1\rangle$s, ...


2

In general, given two matrices $A$ and $B$ of dimensions $n_1\times n_2$ and $m_1\times m_2$, respectively, their tensor product $A\otimes B$ can be represented using the Kronecker product as $$(A\otimes B)_{n_1 m_1,n_2m_2}=A_{n_1,n_2}B_{m_1, m_2}.$$ The indices on the left hand side are a standard way to enumerate the integers from $1$ to $n_1 m_1$ and from ...


2

What you are looking for is https://en.wikipedia.org/wiki/Kronecker_product Note that a column-vector can be considered as a matrix with the size $n \times 1$, so the Kronecker product rule also applies.


1

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,...


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.


1

In principle, yes, you can always do it. The Bloch representation can be generalised to arbitrary dimensions, and you can always parametrise states in it by their "angle coordinates". For example, you can write an arbitrary 3-modes pure state as $$|\psi\rangle=\cos\alpha|0\rangle + e^{i\theta}\sin\alpha\cos\beta|1\rangle+e^{i\phi}\sin\alpha\sin\beta|2\...


1

The state of single qubit can be described as a point on the Bloch sphere. All the allowed transformations of a single qubit can then be described as rotations on the Bloch sphere. Unfortunately, bigger quantum systems can no longer be described as fitting on a sphere like geometry. As a result, this idea of considering transformations as rotations does not ...


1

To do the step of the induction, you assume that you've already proven that $$H^{\otimes k} \left| 0 \right>^{\otimes k} = \frac{1}{\sqrt{2^k}} \sum_{i=0}^{2^k -1} \left| i \right>$$ (note the appropriate normalization factor). Now, you have to consider $H^{\otimes k+1} \left| 0 \right>^{\otimes k+1}$: $$H^{\otimes k+1} \left| 0 \right>^{\...


1

If you apply the same gate twice on a state $|\psi\rangle$ you will get out the state $|\psi\rangle$ if the hermitian of the gate is the same as the gate. In your case, you applied the Hadamard gate, $H$, twice on (presumably) the $|0\rangle$ state, and so you got $|0\rangle$ as your result. This happens because it is a requirement for quantum computing ...


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