23 lines
3.4 KiB
Markdown
23 lines
3.4 KiB
Markdown
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# Qubit
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Qubit is a [quantum computing](quantum.md) equivalent of a [bit](bit.md). While bits in classical computers can have one of two state -- either 0 or 1 -- a qubit can additionally have infinitely many states "in between" 0 and 1 (so called [superposition](superposition.md)). Physically qubits can be realized thanks to quantum states of particles, e.g. the polarization of a photon or the spin of a photon. Qubits are processed with [quantum gates](quantum_gate.md).
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**Whenever we measure a qubit, we get either 1 or 0**, just like with a normal bit. However during quantum computations the internal state of a qubit is more complex. This state determines the **probabilities** of measuring either 1 or 0. When the measurement is performed (which is basically any observation of its state), the qubit state collapses into one of those two states.
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Now we will be dealing with so called **pure states** -- these are the states that can be expressed by the following representation. We will get to the more complex (mixed) states later.
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The state of a qubit can be written as
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*A * |0> + B * |1>*
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where *A* and *B* are [complex numbers](complex_number.md) such that *A^2 + B^2 = 1*, *|0>* is a vector [0, 1] and *|1>* is a vector [1, 0]. *A^2* gives the probability of measuring the qubit in the state 0, *B^2* gives the probability of measuring 1.
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The vectors *|0>* and *|1>* use so called bra-ket notation and represent a vector basis of a two dimensional state. So the qubit space is a point in a space with two axes, but since *A* and *B* are complex, the whole space is four dimensional (there are 4 variables: *A* real, *A* imaginary, *B* real and *B* imaginary). However, since *A + B* must be equal to 1 ([normalized](normalization.md)), the point cannot be anywhere in this space. Using logic^TM we can figure out that the final state of a qubit really IS a point in two dimensions: a point on a sphere (Bloch sphere). A point of the sphere can be specified with two coordinates: *phase* ([yaw](yaw.md), 0 to 2 [pi](pi.md), can be computed from many repeated measurements) and *p* ([pitch](pitch.md), says he probability of measuring 1). It holds that:
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*A = sqrt(1 - p)*
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*B = e^(i * phase) * sqrt(p)*
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The sphere has the state |0> at the top (north pole) and |1> at the bottom (south pole); these are the only points a normal bit can occupy. The equator is an area of states where the probability of measuring 0 and 1 are equal (above the equator gives a higher probability to 0, below the equator to 1).
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Now a qubit may actually be in a more complex state than the pure states we've been dealing with until now. Pure states can be expressed with the state vector described above. Such a state is achieved when we start with a qubit of known value, e.g. if we cool down the qubit, we know it has the value *|0>*, and transforming this state with quantum gates keep the state pure. However there are also so called **mixed states** which are more complex and appear e.g. when the qubit may have randomly been modified by an external event, or if we start with a qubit of unknown state. Imagine if we e.g. start with a qubit that we known is either *|0>* or *|1>*. In such case we have to consider all those states separately. A mixed state is composed of multiple pure states. Mixed states can be expressed with so called **density matrices**, an alternative state representation which is able to encode these states.
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