# Questions tagged [quantum-state]

Quantum systems can mathematically be described by their 'quantum state'. When the system is closed/isolated, the state is 'pure' and can be written as a sum (i.e. 'superposition') of basis vectors. When the system is a subsystem of an open system, the state is instead usually 'mixed' and cannot be written as a pure state, so has to be written as a density matrix. Consider using the density-matrix tag when relevant

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### What are the constraints for the coefficents of the basis states in quantum computing?

$\newcommand{\ket}{\left|#1\right>}$ It's known that the Kolmogorov axioms characterise a probability distribution: Probability of an event is a non-negative real number. The sum of all ...
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### What is the “quantum phase” of a quantum state?

On this page IMBQ docs, until the sentence '..and since the global phase of a quantum state is not detectable..' I follow everything. However 'quantum phase' is introduced without any explaination? ...
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### Unitary Transformations for Schmidt Decomposition

$\newcommand{\ket}{|#1\rangle}$ Suppose a pure state $\ket{\psi}$ has a Schmidt decomposition given by $\ket{\psi^{SD}}$, which can be obtained via the diagonalization of the reduced density matrix ...
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### How to perform a plot histogram for a circuit?

I have created a circuit and I don't know how to plot a histogram. I tried to plot a histogram but it gives me output for 0000 case only, how to get to know the probability for all of the cases. The ...
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### What is the best way to extend a state $\rho_S$ to a tensor product of spaces ${\cal H}_S\otimes{\cal H}_A?$

Let $\Phi_S$ be an operator acting on a space $\mathcal H_S$. If we introduce an ancilla $A$, the total space becomes $\mathcal H_S\otimes \mathcal H_A$ and I can naturally extend the operator $\Phi_S$...
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### On existence of orthonormal basis for each subsystem in Separable state [closed]

A separable state in $\mathcal{H}_{a}\otimes\mathcal{H}_{b}$ is given by $$\rho_{s}=\sum_{\alpha,\beta}p(\alpha,\beta)|\alpha\rangle\!\langle\alpha|\otimes|\beta\rangle\!\langle\beta|.$$ Now, my ...
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### What are the conditions under which an unknown quantum state is learnable with arbitrary precision?

Assume that we have an unknown quantum state and we need to learn that unknown state with arbitrary precision. Under what conditions can we learn the unknown state with arbitrary precision? One ...
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### Restrictions on Quantum Gates and Pure Partially Traced Output States

Suppose we have a quantum circuit that contains an arbitrary number of quantum gates and takes as an input more than a single qubit, say three. What are the restrictions on the quantum gates and the ...
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### How do I find the state of each qubit at the end of the circuit?

I have this Quantum Fourier Transform (QFT) and I want to know how to find the final state of each qubit if q0, q1, ...
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### Lower bounds on the number of measurements outcomes required for quantum state tomography

It seems that in order to reconstruct a quantum state, a large number of measurements is typically used. Are there any known theoretical lower bounds on the number of measurements required to ...
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### What is the eigenvalue distribution of arbitrary unitary matrices?

I had a question regarding the nature of the eigenvalue distribution of unitary matrices. Searching for the answer I found that the unitary matrices which are sampled randomly have a defined ...
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### Distinguishing $\frac{| 0 \rangle + e^{i\theta} |1 \rangle}{\sqrt{2}}$ from $| 0 \rangle/|1 \rangle$ with probability $1/2 + \epsilon$

I am given one copy of one of two quantum states - $\frac{| 0 \rangle + e^{i\theta} | 1 \rangle}{\sqrt{2}}$, for some unknown fixed $\theta$. One of $| 0 \rangle/|1 \rangle$ - don't know which one, ...
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### The expectation of a measurement of qubit 2 after qubit 1 has been measured

In section 1.2.4 (page 13) of these lecture notes http://users.cms.caltech.edu/~vidick/teaching/fsmp/fsmp.pdf, it says \begin{aligned}\left\langle\psi\left|X_{1}^{0} Z_{2} X_{1}^{0}\right| \psi\right\...
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### Why does $(XZ\otimes I)|\Phi^+\rangle$ equal the Bell state $|\Psi^-\rangle$?

I'm slightly confused by the solution provided below by a suggested solution online to convert |$\phi^+$⟩ to |$\psi^-$⟩. I tried doing the operation XZ but I got $\frac{1}{\sqrt2}$(|10⟩-|01⟩) instead ...