# 11 Quantum Physics

To further understand, how quantum computers work, we take a look at the basics of quantum physics.

## 11.1 Wave function

The first concept we look into is the wave function of quantum mechanics. For this we will look at the experiment “particle in a well”. To keep the math simple, we assume the space to be 1-dimensional, so the position of the particle is confined to a line.

In this experiment, we have one particle and a potential, which is denoted by a function \(V: \mathbb{R} \rightarrow \mathbb{R}\). This function maps a position of a particle in \(\mathbb{R}\) to the energy which is needed to hold the particle in that position.

Classically a state of a system at time \(t\) is described by the position \(x(t) \in \mathbb{R}\) and the momentum \(p(t) \in \mathbb{R}\).

In the quantum world, we have a wave function \(\psi_t(x)\) with \(\psi_t: \mathbb{R} \rightarrow \mathbb{C}\) under \(t\in\mathbb{R}\), which takes the position of a particle as an input and outputs the amplitude of that particle, with \(t\) as the time.

To calculate the probability of a particle being in the interval \([a,b]\) at time \(t_0\), we can use the integral over the wave function: \[ \Pr[\text{Particle is in }[a,b]\text{ a time }t_0] = \int_a^b \lvert \psi_{t_0}(x) \rvert^2 dx \]

From this we can see that the integral \(\int \lvert \psi_{t_0} \rvert ^2 dx = 1\). The momentum is not needed for the wave function.

The inner product of two wave functions \(\mathbb{R} \rightarrow \mathbb{C}\) is given by \[ \braket{\psi|\phi} := \int \overline{\psi(x)} \cdot \phi(x) dx \]

The norm of a wave function is given by \[ \| \psi \|^2 :=\braket{\psi|\psi} = \int \lvert \psi(x) \rvert ^2 dx \]

In general, the wave function can have a different domain, e.g. \(\psi_t: \mathbb{R}^3 \rightarrow \mathbb{C}\) for a particle in 3D-space. Everything below works analogously in that case.

## 11.2 Energy / Hamiltonian

Given a particle in a state, this particle has some energy. Classically this energy is calculated from the potential and the momentum, with \[ \operatorname{Energy} := \underbrace{V(x)}_{\text{{potential energy}}} + \underbrace{\frac{p^2}{2m}}_{\text{kinetic energy}} \]

In the quantum world, we can calculate the potential energy using the wave function as follows: \[ \text{potential energy} := \int \underbrace{\lvert \psi(x) \rvert^2}_{\text{probability at pos. }x} \cdot V(x) dx \] The kinetic energy can be calculated as follows: \[ \text{kinetic energy}:= \int -\overline{\psi(x)} \cdot \frac{\hbar}{2m} \cdot \frac{\partial^2 \psi(x)}{\partial x^2} dx \] \(\hbar\) is the reduced Planck constant. We will not go further into reasoning for this definition.

From this we can calculate the energy in the quantum context: \[ \begin{aligned} \text{energy} :=& \int \overline{\psi(x)} \left(-\frac{\hbar}{2m} \frac{\partial^2 \psi(x)}{\partial x^2} + V(x)\psi(x)\right) dx\\ =&\int \overline{\psi(x)} (H\psi)(x) dx\\ =& \braket{\psi,H\psi} \end{aligned} \]

The operator \(H\) is called a Hamiltonian and is an operator that maps wave functions to wave functions, such that \(\braket{\psi,H\psi}\) is the energy. \(H\) maps \(\psi\) to \(H\psi\) which is defined as: \[ (H\psi)(x) = -\frac{\hbar}{2m} \frac{\partial^2 \psi(x)}{\partial x^2} + V(x)\psi(x) \tag{11.1}\]

Note that \(\hbar\) is the reduced plank constant and \(m\) is the mass. Also note that other physical systems (e.g. when there are more particles, or when there is not just a simple potential) may have a different definition of the Hamiltonian.

## 11.3 Schrödinger equation

### 11.3.1 Time-dependent Schrödinger equation

The time-dependent Schrödinger equation is denoted by:

\[ i\hbar \frac{\partial \psi_t(x)}{\partial t} = (H\psi_t)(x) \]

From this we see that the time development of \(\psi_t\) is determined by \(\psi_t\) at that moment (via \(H\psi_t\)).

### 11.3.2 Time-independent Schrödinger equation

For the time-independent Schrödinger equation, we try to find a wave function \(\psi: \mathbb{R} \rightarrow \mathbb{C}, \psi \neq 0\) and an energy \(E\) such that \[ H\psi = E\psi \] This means roughly that we try to find \(\psi\) that has the same energy everywhere. That is that we try to find a \(\psi\) with energy \(E\) (since the energy is \(\braket{\psi|H\psi}=\braket{\psi|E\psi}=E\braket{\psi|\psi}=1\)) and not a superposition of different energies.

Such a \(\psi\) is useful, since by the time-dependent Schrödinger equation, if \(H\psi = E \psi\), then \[ i\hbar \frac{\partial\psi_t(x)}{\partial t} = H\psi = E\psi_{t_0}(x) \]

Solving this differential equation gives us \(\psi\) with \[ \psi_t(x) = e^{-i E t/\hbar} \psi_{t_0}(x) \]

So all in all: If \(\psi_0\) is a solution of the time-independent Schrödinger equation, then \(\psi_t(x)= e ^{-i E t / \hbar} \psi_0(x)\) is the solution to the time-dependent Schrödinger equation with initial condition \(\psi_0\) (at time \(t=0\)).

Given any \(\psi_0\), we can try to rewrite \(\psi_0\) as \(\psi_0 = \sum_k a_k \psi^k\), where \(\psi^i, E_i\) are solutions of the time-independent Schrödinger equation and then the solution of the time-dependent Schrödinger equation is \[ \psi_t(x) = \sum_k a_k e^{-i E t /\hbar} \psi^k(x) \] Note that with \(\psi^k\), \(k\) is an index and not an exponent.

## 11.4 Infinite square well

Our goal is to solve the time-independent Schrödinger equation for the infinite square well. The infinite square well is defined to have a potential of \(V(x) = 0\) in the range \([0,1]\) and \(V(x) = \infty\) otherwise.

We now try to solve the time-independent Schrödinger equation \(H\psi= E\psi\), which for the Hamiltonian from Equation 11.1 becomes: \[ \begin{aligned} -\frac{\hbar}{2m} \frac{\partial^2 \psi(x)}{\partial x^2} + V(x)\psi(x) &= E\psi \end{aligned} \]

Since the potential of the particle is \(\infty\) outside of the range \([0,1]\), we know that \(\psi(x)=0\) outside of \([0,1]\), because there is no infinite energy. From this we also know that \(\psi(0)=0\) and \(\psi(1)=0\) by continuity.

Since either \(\psi(x)\) or \(V(x)\) are always 0 for any \(x\), we need to solve the term

\[ -\frac{\hbar}{2m} \frac{\partial^2 \psi(x)}{\partial x^2} = E\psi \] We assume for simplicity that \(m=1\) and \(\hbar = 1\) and we don’t specify units like meter, seconds, joule etc. All solutions to \(-\frac{1}{2} \frac{\partial^2\psi(x)}{\partial x^2} = E\psi\) have the form \[ \psi = A \sin(\gamma x) + B \cos(\gamma x) \] with \(E=\frac{\gamma^2}{2}\) for any \(A,B\in \mathbb{C}\) and \(\gamma \in \mathbb{R}\). Since \(\psi(0) = 0\), we know that \(B = 0\). Therefore \(\psi = A \sin(\gamma x)\). The scaling factor \(A\) has to fulfill \(A\neq 0\) since we only look for non-zero wave functions.

Since \(\psi(1) = 0\), we know that \(A \sin(\gamma) = 0\). Therefore \(\gamma\) needs to be a multiple \(k\) of \(\pi\) and also \(k>0\) because we want non-zero solutions \(\psi\neq 0\). So we set \(\gamma := (k+1)\pi\) for integers \(k\geq 0\). With \(E=\frac{\gamma^2}{2}\) we get: \[ E_k = \frac{(k+1)^2\pi^2}{2} \] and also that \[ \psi_k(x) = A\sin((k+1)\pi x) \]

For the rest of our lecture, we will set \(A\) to be \(A=\sqrt{2}\) because we are only interested in normalized wave functions.

So all in all: In the infinite square well the wave functions with no energy-superposition are \(\sin((k+1)\pi x)\) with \(E_k= \frac{(k+1)^2\pi^2}{2}\).

There is one important thing to note here: With the infinite square well, there are discrete energy levels, so for each energy level \(k\), we have an energy \(E_k=(k+1)^2\frac{\pi^2}{2}\) and no other energy levels (e.g \(\frac{1.4\pi^2}{2}\)) exist. \(E_k > 0\) also holds, so we can never have \(0\)-energy. This means that, different from the classical case, the particle can never be fully at rest. We will write \(\ket{k}:= \psi_k\).

We can now write each state as \[ \sum_{k\in\mathbb{N}} a_k \ket{m} \tag{11.2}\] If we ignore energies above a certain level, we get a system \(\ket{1},\ket{2},\ket{3},\dots,\ket{N}\), where any \(\psi\) can be written as \(\sum a_k \ket{k}\). That is useful if we do not want to think about infinite dimensional systems.