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| pure-state [May 23, 2026 at 20:31] – Ivan Janevski | pure-state [June 13, 2026 at 03:13] (current) – external edit 127.0.0.1 |
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| # Pure state | # Pure state |
| **Pure state** is a state that completely encodes the information about a quantum system. If a state is pure it can be represented by a state vector $\lvert\psi\rangle$ in Hilbert space $\mathcal{H}$. If this information leaks into the environment the state is called "mixed". The state vector alone is no longer sufficient to describe the quantum system and is upgraded $\rho$. This is why the environment is a major enemy of quantum computing. If we consider the Hilbert space that combines both the environment quantum system we're looking at $\mathcal{H}\otimes\mathcal{H}_\text{Env}$ | **Pure state** is a quantum state that is completely described by a single state vector $\lvert\psi\rangle$. It carries the maximum possible information about a quantum system. When information about the system leaks into the environment the state becomes [[mixed-state|mixed]] — this process is called decoherence, and it is one of the primary challenges in quantum computing. |
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| | Classical bits can be `0` or `1`, encoded as voltages such as $0\,\text{V}$ and $5\,\text{V}$. These voltages are not perfectly precise — they fluctuate due to thermal noise, electromagnetic interference, and current leakage — but digital logic is designed to be resilient to such fluctuations. A pure qubit state $\lvert\psi\rangle = a\lvert 0\rangle + b\lvert 1\rangle$ is far more fragile. Any interaction with the environment perturbs the complex amplitudes $a$ and $b$, entangling the qubit with environmental degrees of freedom and converting the pure state into a mixed one. |
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| | ## Density matrix representation |
| | Any pure state $\lvert\psi\rangle$ can be equivalently represented as a [[density-matrix|density matrix]] $\rho = \lvert\psi\rangle\langle\psi\rvert$. Pure states are exactly the states whose density matrix satisfies $\text{tr}(\rho^2) = 1$. On the [[bloch-sphere]], pure states correspond to points on the surface of the unit sphere (the pure states are the ones "on" the sphere, while mixed states are "inside" the sphere). |
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| | $$\rho = \lvert\psi\rangle\langle\psi\rvert, \qquad \text{tr}(\rho^2) = 1$$ |
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| Consider a classical bit: | |
