13  Quantum Physics

To further understand how quantum computers work, we take a look at the basics of quantum physics. We will omit or simplify a lot of physics details and strive to give a first impression only.

13.1 Quantum state

As an example, we consider a single atom with one electron.

At any time, the electron can be found in a superposition between (infinitely) many positions, leading to it being in a “fuzzy probability cloud”. Certain superpositions are special¹; we call them orbitals. A photon can then be in superposition between different orbitals (i.e., a linear combination of the different “probability clouds”)².

We can assign each orbital a number, e.g. \(0, 1, 2, \dots\). In each orbital the electron has a specific energy level \(E_0, E_1, E_2, \dots\). If the electron is in a particular orbital, it is said to be in the corresponding quantum state \(\ket{0}, \ket{1}, \ket{2}, \dots\).

Using this notation, we can describe the quantum state of the electron with an infinite one-dimensional vector:

\[ \begin{pmatrix} \alpha_0 \\ \alpha_1 \\ \alpha_2 \\ \vdots \end{pmatrix} = \alpha_0 \ket{0} + \alpha_1 \ket{1} + \alpha_2 \ket{2} + \dots. \]

The coefficients \(\alpha_i\) are complex numbers that represent the probability amplitude of the electron being in the corresponding quantum state \(\ket{i}\). With probability \(|\alpha_i|^2\) the electron is in the quantum state \(\ket{i}\) (if we measure).

13.2 Time evolution of a quantum state

We consider an electron in the quantum state \(\ket{n}\) with the energy level \(E_n\). After a time \(t\) the electron is in the quantum state \[ e^{-i E_n t / \hbar} \ket{n}, \] where \(\hbar = 1.054571817\ldots \cdot 10^{-34} \text{Js}\) is the so-called reduced plank constant. In the following we use \(\hbar = 1\) and omit the units for simplicity.

This means that if the electron is in some orbital, and we wait, then the electron is still in the same orbital. But the phase will have changed.

Example: Quantum state evolution

We assume an electron to be in the quantum state \[ \ket{+} = \frac{1}{\sqrt{2}} \ket{0} + \frac{1}{\sqrt{2}} \ket{1}. \] Furthermore, we set \(E_0 = 0\) and \(E_1 = 1\).

If we wait the time \(t = \pi\), the quantum state evolves to \[ \begin{aligned} &\quad \: \frac{1}{\sqrt{2}} e^{-i E_0 t / \hbar} \ket{0} + \frac{1}{\sqrt{2}} e^{-i E_1 t / \hbar} \ket{1} \\ &= \frac{1}{\sqrt{2}} e^{-i 0 \pi / 1} \ket{0} + \frac{1}{\sqrt{2}} e^{-i 1 \pi / 1} \ket{1} \\ &= \frac{1}{\sqrt{2}} e^0 \ket{0} + \frac{1}{\sqrt{2}} e^{-i \pi} \ket{1} \\ &= \frac{1}{\sqrt{2}} \ket{0} - \frac{1}{\sqrt{2}} \ket{1} \\ &= \ket{-}. \end{aligned} \]

13.3 Energy / Hamiltonian

From now on, we assume that there are only finitely many orbitals \(N\). (This is not true but allows us to avoid mathematical subtleties.)

So far, we know that the quantum state \(\ket{\psi} = \alpha_0 \ket{0} + ... \alpha_N \ket{N}\) has probability \(|\alpha_i|^2\) of being in quantum state \(\ket{i}\), which has energy level \(E_i\).

Therefore, the total energy \(E_\text{total}\) of the electron can be calculated as \[ E_\text{total} = |\alpha_0|^2 \cdot E_0 + \dots + |\alpha_N|^2 \cdot E_N = \sum_{i=0}^N |\alpha_i|^2 \cdot E_i. \tag{1} \]

There is also another method to calculate the energy of the electron. For this we need the Hamiltonian:

Definition 13.1 (Hamiltonian, diagonal case) For an electron with the basis quantum states \(\ket{0}, \ket{1}, \dots, \ket{N}\) and the corresponding energy levels \(E_0, E_1, \dots, E_N\) the Hamiltonian is defined by

\[ H \coloneqq \begin{pmatrix} E_0 & 0 & \dots & 0 \\ 0 & E_1 & \dots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \dots & E_N \end{pmatrix} = \operatorname{diag}(E_0, E_1, \dots, E_N). \]

With this Hamiltonian \(H\), the total energy \(E_\text{total}\) can be calculated as \[ E_\text{total} = \braket{\psi | H | \psi}. \] \(\braket{\psi | H | \psi}\) is the inner product of \(\ket{\psi}\) and \(H \ket{\psi}\).

This follows from \[ \begin{aligned} E_\text{total} &= \sum_{k=0}^N |\alpha_k|^2 E_k = \sum_{k=0}^N \overline{\alpha} E_k \alpha \\ &= \sum_{k=0}^N \sum_{l=0}^N \overline{\alpha_k} E_{kl} \alpha_l = \sum_{k=0}^N \overline{\alpha_k} \Biggl( \sum_{l=0}^N E_{kl} \alpha_l \Biggr) \\ &= \sum_{k=0}^N \overline{\alpha_k} \Bigl( H \ket{\psi} \Bigr)_k \\ &= \braket{\psi | H | \psi}. \end{aligned} \]

Why is the concept of a Hamiltonian useful? Why don’t we just use the formula \((1)\)? The reason is that is general, our basis states \(\ket{0}, \ket{1}, \dots\) will not always be pure energy states. Then \((1)\) does not apply. But we can always define some matrix \(H\) that describes the total energy:

Definition 13.2 (Hamiltonian) The total energy of an \(N\)-dimensional quantum system is described by matrix \(H \in \mathbb{C}^{N \times N}\), the Hamiltonian \(H\) is always Hermitian (meaning \(H^\dagger\) = H). The total energy of any quantum state \(\ket{\psi}\) is then \(\braket{\psi | H | \psi}\),

What Hamiltonian corresponds to a specific quantum state depends on the specific physical system.

13.4 Schrödinger equation

If \(H(t)\) is the Hamiltonian of our system at time \(t\), and \(\psi(t)\) is the quantum state at time \(t\), then the following holds: \[ \frac{d \psi(t)}{d t} = \frac{i}{\hbar} H(t) \psi(t). \] This is the time-dependent Schrödinger equation.

From now on we assume that \(H(t)\) is constant, i.e. \(H(t) = H\) for all \(t\).

For the next step, we need a new mathematical operation: The exponentiation of a matrix.

Definition 13.3 (Matrix exponentiation) For a diagonal matrix \(D\), the exponentiation is defined as: \[ e^D \coloneqq \begin{pmatrix} e^{D_{11}} & 0 & \dots & 0 \\ 0 & e^{D_{22}} & \dots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \dots & e^{D_{nn}} \end{pmatrix}. \]

Let \(A\) be a diagonalizable matrix, then there exists a unitary matrix \(U\) and a diagonal matrix \(D\) such that \(A = U^\dagger D U\). In this case the exponentiation of \(A\) is defined as \[ e^A = e^{U^\dagger D U} \coloneqq U^\dagger e^D U. \]

The eigenvectors of \(H\) are called “energy eigenstates”. They have a very simple time evolution:

If \(\psi(0)\) is an energy eigenstate of \(H\) with the total energy \(E\) as eigenvalue, meaning \(H \psi(0) = E \psi(0)\), then the solution of the time-dependent Schrödinger equation is \[ \psi(t) = e^{-i E t / \hbar} \psi(0). \]

We can check the plausibility of this result by substituting it into the time-dependent Schrödinger equation: \[ \begin{aligned} \frac{\partial \psi(t)}{\partial t} &= \frac{\partial}{\partial t} \biggl( e^{-i \frac{E}{\hbar} t} \psi(0) \biggr) = \frac{\partial}{\partial t} \biggl( e^{-i \frac{E}{\hbar} t} \biggr) \psi(0) \\ &= \biggl( -i\frac{E}{\hbar} \biggr) e^{-i \frac{E}{\hbar} t} \psi(0) = -\frac{i}{\hbar} e^{-i \frac{E}{\hbar} t} \biggl( E \psi(0) \biggr) \\ &= -\frac{i}{\hbar} e^{-i \frac{E}{\hbar} t} \biggl( H \psi(0) \biggr) = -\frac{i}{\hbar} H \biggl( e^{-i \frac{E}{\hbar} t} \psi(0) \biggr) \\ &= -\frac{i}{\hbar} H \psi(t). \end{aligned} \]

This shows that \(\psi\) is indeed a solution. We omit the proof that it is unique.

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1. Roughly speaking, they are eigenvectors of the infinitely-dimensional Hamiltonian of the electron. We explain the concept of Hamiltonians in Section 13.3.

2. For mathematical reasons beyond the scope of this exposition, any valid “probability cloud” equals a superposition of many orbitals.