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--- qubit [May 14, 2026 at 11:38] – external edit 127.0.0.1
+++ qubit [May 25, 2026 at 13:16] (current) – Ivan Janevski
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 A classical bit can be `0` or `1`. There are no probabilities involved at all, but we can assign them anyway. If a bit is `1`, then it's `1` with 100% probability and `0` with 0% probability. Conversely, if a bit is `0`, then it's `1` with 0% probability and `0` with 100% probability.
-A qubit works in a very similar way. It also has two states: $\lvert 0\rangle$ and $\lvert 1\rangle$, except the total probability is distributed between those two. A qubit is often in both states at the same time: a small amount in state $\lvert 0\rangle$ and a small amount in state $\lvert 1\rangle$. A qubit  $\lvert\psi\rangle$ is written in the following form, using what is known as [[dirac-notation|Dirac notation]]:
+A qubit works in a similar way. It also has two states: $\lvert 0\rangle$ and $\lvert 1\rangle$, except the total probability is distributed between those two. A qubit is often in both states at the same time: a small amount in state $\lvert 0\rangle$ and a small amount in state $\lvert 1\rangle$. A qubit  $\lvert\psi\rangle$ is written in the following form, using what is known as [[dirac-notation|Dirac notation]]:
 $$\lvert\psi\rangle = a\lvert 0\rangle + b\lvert 1\rangle$$
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 $$|a|^2 + |b|^2 = 1$$
-## States
-### Basis state
-### Pure state
-A qubit is in a pure state when it's in one of the basis states with 100% probability. When we say 100% probability we mean a complex probability amplitude $1 + i0$. Conversely, when we say 0% probability we mean a complex probability amplitude $0 + i0$.
-Because there are only two basis states, and a pure state is when a qubit is in one of the two basis states, there are also only two pure states:
-$$ \lvert\psi\rangle = \underbrace{\left(1 + i0\right)}_{a}\cdot \lvert 0\rangle + \underbrace{\left(0 + i0\right)}_{b}\cdot\lvert 1\rangle$$
-$$ \lvert\psi\rangle = \underbrace{\left(0 + i0\right)}_{a}\cdot\lvert 0\rangle + \underbrace{\left(1 + i0\right)}_{b}\cdot\lvert 1\rangle$$
-### Mixed state
-A qubit is in a mixed state when total probability distributed among states.
-$$ \lvert\psi\rangle = \underbrace{\left(\frac{1}{2} + i0\right)}_{a}\cdot \lvert 0\rangle + \underbrace{\left(\frac{1}{2} + i0\right)}_{b}\cdot \lvert 1\rangle$$
-$$ \lvert\psi\rangle = \left(\frac{1}{\sqrt{2}} + i 0 \right)\cdot\lvert 0\rangle + \left(\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}}\right)\cdot \lvert 1\rangle$$
-When a qubit is measured the state collapses to a pure state. In both of these examples, the probabilities are equally distributed. Measuring these qubits will yield either state with 50% probability.
-### Invalid state
-A qubit is in an invalid state when the square magnitutdes of its probability amplitudes don't add up to $1$. This is not allowed by quantum mechanics so a qubit will never find itself in an invalid state.
-An example is a qubit that's in both states with 100% probability, like $\lvert\psi\rangle = \lvert 0\rangle + \lvert 1 \rangle$. Expanding the notation to see the structure more clearly:
-$$ \lvert\psi\rangle = \left(1 + 0i\right)\lvert 0\rangle + \left(1 + 0i\right)\lvert 1\rangle$$
-When we sum the square magnitudes we get the result $2$
-$$ |1 + 0i|^2 + |1 + 0i|^2 = 1^2 + 1^2 = 2$$
-The interpretation of this result is that the qubit exists with a total probability of 200%, which doesn't make sense. The total probability distribution among all states must always add up to 100%.
-## Evolution
-The equation that governs quantum mechanics is the Shrodinger equation. In the most general case it's written in the following form.
-$$i\hbar\frac{\partial}{\partial t}\lvert \psi\rangle = H\lvert\psi\rangle$$
-The ket $\lvert \psi\rangle$ is a quantum state. Mathematically, it's a vector in a Hilbert space. The operator $H$ is called a Hamiltonian. Mathematically, it's a linear Hermitian opeartor acting on vectors in that space. The Hamiltonian represents the total energy of the system (kinetic energy + potential energy).
-In quantum mechanics, there is a postulate that the system should be fully described its state $\lvert\psi\rangle$. This is not a law of nature. Rather it's our responsibility to make that so, by choosing a Hilbert space that appropriately captures all possible states. Often it's possible to combine Hilbert spaces with a tensor product to make a more general Hilbert space that fits our needs.
-When the Hilbert space is chosen, one identifies a Hamiltonian that captures the energies of the system. This is done either by looking (e.g. the spherical symmetry in the Hydrogen atom) or just by constructing one that satisfies being a linear Hermitian operator. The Hamiltonian is then plugged in the Schrodinger equation as solved for state  $\lvert\psi\rangle$.
-### Hydrogen atom
-Let's take a particle like an electron inside a hydrogen atom. A postulate in quantum mechanics is that an electron should be fully described by its state $\lvert\psi\rangle$. We need a Hilbert space that captures all possible states of an electron. But first, what is an electron state?
-An electron is an object in 3D space. Like with qubit, its probability amplitudes are complex numbers. Therefore a state should be something that associates a complex number $\mathbb C$ with every point in space $\mathbb R^3$. So we'll define the electron state as a function $\lvert\psi\rangle: \mathbb R^3\mapsto\mathbb C$.
-We also need the electron to exist *somewhere* in the universe. According to the Born rule, the probability of finding an electron at a point $(x, y, z)$ is proportional to the square magnitude of the complex number associated with that point. Therefore, if we integrate square magnitude of the function across all of 3D space we should get $1$.
-So, to capture all possible electron states, we choose the Hilbert space to be a space of all square-integrable functions. This is compactly written as $L^2(\mathbb R^3)$.
-$$ \lvert\psi\rangle\in \left\{ f:\mathbb R^3\mapsto\mathbb C \text{ such that } \int_{\mathbb R^3} |f(x)|^2 \mathrm dx = 1\right \} \quad \text{ same as } \quad \lvert\psi\rangle\in L^2(\mathbb R^3)$$
-When it comes to the Hamiltonian, we need a Hermitian linear operator acting on vectors in our Hilbert space. It's out of the scope of this section, but it can be proven that if we take the classical equation for energy $E = \frac{p^2}{2m} + V$, and promote $p\mapsto -i\hbar\nabla^2$ and $V\mapsto V$ to operators in Hilbert space we get:
-$$H = -\frac{\hbar^2}{2m}\nabla^2 + V$$
-The Schrodinger equation becomes:
-$$-i\hbar\frac{\partial}{\partial t}\lvert\psi\rangle = \left(-\frac{\hbar^2}{2m}\nabla^2 + V\right)\lvert\psi\rangle$$
-If we let the potential $V$ be the Coulomb potential $V = e^2/4\pi\varepsilon_0r$ created by the presence of a proton, solving the Schrodinger equation yields the equation for electron orbitals in 3D space.
-### Qubit
-Let's turn back to the qubit. A postulate in quantum mechanics is that a qubit should be fully described by its state $\lvert\psi\rangle$. We need a Hilbert space that captures all possible states of a qubit. But what is a qubit state?
-Unlike the electron, qubit is quite limited in its degrees of freedom: instead of being distributed across all of 3D space, there are only two places the probability amplitudes can go to. We can do something similar to an electron and define a qubit state to be a function $\lvert\psi\rangle: \{0, 1\}\mapsto\mathbb C$. But this is isomorphic to the column vector $\mathbb C^2$. So rather, it's better if we define the state to be a column vector $\lvert\psi\rangle\in\mathbb C^2$. This lets us take advantage of linear algebra.
-So, to capture all qubit states, we choose the Hilbert space to be the 2-dimensional complex vector space $\mathbb C^2$.
-$$ \lvert \psi\rangle\in \left\{\begin{pmatrix}a \\ b\end{pmatrix}\text{ where } a, b\in \mathbb C \text{ and } |a|^2 + |b|^2 = 1 \right \} \text{ same as } \lvert\psi\rangle\in\mathbb C^2$$
-Now we need a Hamiltonian. The Hamiltonian is a bit odd. It encodes the energy of the system -- but what is the kinetic and potential energy of a qubit? The truth is, qubit doesn't have those. Kinetic and potential energies arise when a system has spatial degrees of freedom. A qubit has two internal degrees of freedom instead. A Hamiltonian can still be defined, just not in terms of kinetic and potential energies.
-Mathematically, Hamiltonian is a linear hermitian operator over Hilbert space. Our Hilbert space is $\mathbb C^2$ so linear operators are complex 2x2 matrices. To satisfy being a Hamiltonian, we just need our 2x2 matrix to be Hermitian. A Hermitian matrix is one where if you do a transpose and apply complex conjugation to all its elements it stays the same. A general 2x2 hermitian matrix for some real numbers $x,y,z,w\in\mathbb R$ is:
-$$H = \begin{pmatrix}x & z - iw \\ z + iw & y \end{pmatrix}$$
-When we plug this in the Schrodinger equation we get the following expression:;
-$$i\hbar\frac{\partial}{\partial t}\begin{pmatrix}a(t)\\b(t)\end{pmatrix} = \begin{pmatrix}x & z - iw \\ z + iw & y \end{pmatrix}\cdot\begin{pmatrix}a(t)\\b(t)\end{pmatrix}$$
-This is a first order differential equation with respect to time, so when we solve it for the column vector (a, b):
-$$\begin{pmatrix}a(t)\\b(t)\end{pmatrix} = e^{iH(t)/\hbar}\cdot\begin{pmatrix}a(0)\\b(0)\end{pmatrix}$$
-Notice $H$ is in an exponential function $e^x$, yet $H$ is a 2x2 matrix. How do we define an exponential function of a matrix? We can do it by defining $e^x$ via its Maclaurin series $e^x = 1 + x + x^2 / 2 + \cdots$.
-$$e^H = 1 + H + \frac{H^2}{2} + \frac{H^2}{6} + \cdots$$
-## Visualization
-A classical bit only has two states, `0` and `1`. Therefore you can easily picture a classical bit as an on-off switch.
-A qubit is a little more complex. However we've seen it consists of two probability amplitudes $\alpha$ and $\beta$, both of which are complex numbers. According to the Schrodinger equation their phases evolve in opposite directions (both at the same angular frequency), but their amplitudes do not change.
-Therefore, you can picture a qubit like an analog clock with two hands: One hand spinning clockwise and the other hand spinning counter-clockwise. They're both spinning at the same speed, just in the opposite direction. One is shorter and the other one is longer. The area of the circle they sweep represents the probability. Area of a circle is proportional to the radius squared $\pi r^2$. This matches the Born rule that that the probability of collapsing a qubit in a given state is given by magnitude squared $P(0) = |\alpha|^2$ and $P(1) = |\beta|^2$.