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4

No this is not possible. When people talk about the distribution of eigenvalues they mean the expected eigenvalues when the unitary matrices are sampled with respect to some measure. If you have a fixed unitary matrix $U$ then you can have absolutely any distribution of eigenvalues that lie on the unit circle in $\mathbb{C}$. For instance $\alpha I$ for any ...


1

I am creating an instance of your generic problem: If we consider $n=1$ then $Z$ Gate will become your $2^n \times 2^n$ "unknown" unitary matrix $U$. Then $\vert 0 \rangle$ and $\vert 1\rangle$ will be its eigenvectors and $1$ and $-1$ the corresponding eigenvalues. Now $\vert \phi\rangle$ is a vector which can be represented as a linear ...


0

Let us break down the problem into a simpler problem. Since you have n sets of quantum states, and every set is just a tensor product of the constituent states, each of them can be represented as a vector. For example, take ${|1_{1} \rangle, |2_{1} \rangle, ..., |k_{1} \rangle}$. If you actually evaluate the tensor product ${|1_{1} \rangle \otimes |2_{1} \...


0

Here's a solution that follows a slightly different idea, but provides the precise number of rows that need to be added to get a unitary matrix. Let $M$ be an arbitrary $n\times m$ matrix. As a first observation, note that the task of adding rows/columns can be reduced to that of adding rows to get $m$ orthonormal columns. Indeed, once this is done, we get ...


2

$U-I$ is a normal matrix so $||U-I||_{op}$ is its eigenvalue with the largest magnitude. The eigenvalue equation for this matrix is $$(U-I)|\psi\rangle=\lambda|\psi\rangle,$$ so $$|\lambda|^2=(\cos\phi-1)^2+\sin^2\phi=4\sin^2\frac{\phi}{2}\Rightarrow |\lambda|=2\left|\sin\frac{\phi}{2}\right|,$$ where $e^{i\phi}$ is some eigenvalue of $U$. Now, the ...


1

If I were you, I'd ignore the matrix $R$ and instead work with the matrix $Q$. They give you a conversion between vectors in the two different representations. First, I'm going to simplify things a bit by working with $$ \tilde Q=\left(\begin{array}{cc} e^{i\phi/2} & 0 \\ 0 & e^{-i\phi/2} \end{array}\right)Q. $$ You'll have to compensate for this in ...


2

Because for a general state $|\phi\rangle = \alpha |0\rangle + \beta |1\rangle$ with $\alpha,\beta \in \mathbb{C}$, we want to describe states with $\langle \phi|\phi\rangle = 1$. There are 4 degrees of freedom, two for the real and imaginary components of $\alpha$ and $\beta$ each. The normalization condition $\langle \phi|\phi\rangle = 1$ reduces that to 3 ...


4

This is because we are using a spherical coordinate system. You can think of this coordinate system as specifying a radius, then two angular coordinates $\theta$ and $\phi$. Pure single-qubit states have a Bloch vector with radius equal to unity, so they only need two more coordinates to fully specify the direction in which the vector points. When you look ...


1

Why is the amplitude simply so without including the $x \dot z$ factor? When you calculate the amplitude of the $|0\rangle^{\otimes n}$ state, you have $z = 0$ (the integer representation of the state you're looking at), so $x \dot z = 0$ for any $x$. Why is the resulting amplitude for the constant case is ±1 and 0 for the balanced case? For the constant ...


5

The terms $II$ and $ZZ$ do not uniquely specify the state $|11\rangle$ because you could equally have the state $|00\rangle$. Indeed, you should not include the identity term in your stabilizer. Thus, you need to add a second term, which could be either $-ZI$ or $-IZ$. Either way, you can easily see how to make a product $-ZI$ out of your stabilizers.


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