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The orthonormal basis $|j\rangle$ of the $d$ dimensional finite Hilbert space corresponds to a configuration space of equally spaced clockwise ordered $d$ points on a circle $S^1$ or equivalently, the vertices of a $d$-dimensional regular polygon. One may think of a point as a discrete location of a particle, then the shift operator $X$ shifts the particle ...


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The shift operator takes his name from the fact that it shifts the position of its input, as in, it sends $1\to2$, $2\to3$ etc, with the last computational basis element being sent back to the first one: $d\to 1$ (or the same thing starting with $0$, depending on notation). As per the "boost" operator $Z$, I have usually seen those referred to as "clock ...


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Annoyingly, the answer is "it depends on what you mean by $|\psi\rangle \otimes |\phi\rangle$." A tensor product of vectors from a collection of Hilbert spaces over the same field is simply a choice of one vector from each Hilbert space, with some equivalence relations modded out. (A tensor product of Hilbert spaces is more complicated.) Let $\...


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The idea of a tensor product is to link two Hilbert spaces together in a nice mathematical fashion so that we can work with the combined system. Normally, these two Hilbert spaces each consist of at least one qubit, and sometimes more. Let's say we have a qubit, which we label $a$, and a qubit which we label $b$. These qubits 'live' in the Hilbert spaces of $...


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$x \in \{0, 1\}^n$ means that in the sum we have all possible bitstrings with length $n$. Let's take $n = 3$: $$|\psi\rangle = \sum_{x \in \{0, 1\}^n} \alpha_x |x\rangle= \alpha_{000} |000\rangle + \alpha_{001} |001\rangle + \alpha_{010} |010\rangle + \\ + \alpha_{011} |011\rangle + \alpha_{100} |100\rangle + \alpha_{101} |101\rangle + \alpha_{110} |110\...


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The probability that the measurement outcome is 1 in the first qubit is $\sum|\alpha_x|^2$ for all $x$ whose first bit is 1.


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I'll comment on how you would obtain these quantities in a (generic, idealised) experimental scenario. To compute an expectation value $\langle \psi|O|\psi\rangle$ you need to be able to measure in the eigenbasis of $O$. Being observables, by definitions, Hermitian operators, you can always write $O$ as $O = \sum_k \lambda_k |u_k\rangle\!\langle u_k|$, ...


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Calculating the expectation value for a specific Hermitian operator This approach can be implemented with real Quantum Hardware and with a simulator. Every Hermitian operator can be decomposed in the sum of Pauli tensor product terms (Pauli terms) with real coefficients (see this thread [1]) $$H = a \cdot \sigma_z \otimes I + b \cdot\sigma_y \otimes \...


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If you assume that $e^{−2\pi z k} = 1$ then the sum is $ \sum_{z=0}^{m-1} \sum_{k=0}^{r-1} 1 = \sum_{z=0}^{m-1} r = mr, $ EDIT: However, at the same time $$ \begin{split} &\frac{1}{m\sqrt{r}}\sum_{z=0}^{m-1}\sum_{k=0}^{r-1}e^{-2\pi izk}e^{-2\pi i \frac{b}{r}k}|mk\rangle \\ =& \frac{1}{m\sqrt{r}}\sum_{z=0}^{m-1}\sum_{k=0}^{r-1}e^{-2\pi i \frac{b}{...


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If you know the quantum circuit for generating a particular state, starting from the all-zero state, it's easy enough to work out the stabilizers. You just start with stabilizers $K=III\ldots IZII\ldots I$, where you have one with a $Z$ on each qubit (i.e. the stabilizers of the all-zero state), and you just update them to $UKU^\dagger$. Particularly if you'...


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You need to watch out when you say 'stabilizer' or 'stabilizers' because there is a little bit of ambiguity in that terminology$^{1}$. The stabilizer $\mathcal{S}$ of a state $|\psi \rangle$ is the group of $n$-qubit Paulis of which $|\psi \rangle$ is a $+1$ eigenstate. That is, $|\psi \rangle$ is the shared $+1$ eigenspace of all these operators. We can ...


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