$\newcommand{\ud}{{#1^{\dagger}}} \newcommand{\bra}{\left\langle #1\right|} \newcommand{\ket}{\left| #1\right\rangle} \newcommand{\braket}{\left\langle #1 | #2\right\rangle} \newcommand{\braketo}{\left\langle #1 \right| #2 \left| #3\right\rangle} \newcommand{\ketbra}{\left| #1\right\rangle\left\langle #2\right|} \newcommand{\angle}{\left\langle #1\right\rangle}$

# Table of Contents

## The Hamiltonian

Similar to how position and momentum play a key role in expressing the total energy of a particle in classical physics, position and momentum are the primary observables in quantum mechanics. The quantum mechanical Hamiltonian operator for a particle moving in one dimension is: $$\hat{H}=\frac{\hat{p}_x^2}{2m}+V(\hat{x})$$

## Energy Eigenvalue Equation

There are few quantum mechanical problems that can be solved using abstract kets. It is generally more convenient to express quantum states as spatial functions known as wavefunctions, denoted as $\psi(x)$. This is a representation of the abstract quantum state, known as the position representation: $$\ket{\psi}\doteq\psi(x)$$ This representation uses the position eigenstates as the preferred basis. The energy eigenstates, written as wave functions, are denoted as $\ket{E_i}\doteq\phi_{E_i}(x)$. Thus, the energy eigenvalue equation becomes: $\hat{H}\phi_{E_i}(x)=E_i\phi_{E_i}(x)$. To solve this, we represent the operators in the Hamiltonian in the position representation. $$\hat{x}\doteq x,\quad\hat{p}\doteq -i\hbar\frac{d}{dx}$$ Thus, the eigenvalue equation becomes: $$\bigg(-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+V(x)\bigg)\phi_E(x)=E\phi_E(x)$$

## The Wavefunction

Unlike ket vectors, wavefunctions represent a continuous distribution of probabilities. The wavefunction is defined as the probability amplitude for the quantum state to be measured in the position eigenstate: $$\psi(x)=\braket{x}{\psi}$$ The probability distribution can be calculated by taking the absolute square of the probability amplitude. $$P(x)=|\braket{x}{\psi}|^2=|\psi(x)|^2$$ Probability of finding a particle in $a< x < b$: $$P_{a < x < b}=\int_a^b|\psi(x)|^2dx$$ Probability for finding a state with energy $E_n$: $$P_{E_n}=|\braket{E_n}{\psi}|^2=|\int_{-\infty}^\infty\phi^*_n(x)\psi(x)dx|^2=|c_n|^2$$ $$\phi_n(x)=\braket{x}{E_n}$$ where $\phi_n(x)$ is the energy eigenstate with energy $E_n$.

To find the expansion coefficient $c_n$, $$c_n=\int_{-\infty}^\infty\phi^*_n(x)\psi(x)dx$$ Note this is identical to the ket notation of $c_n=\braket{a_n}{\psi}$

Normalization condition: $$\int_{-\infty}^\infty P(x)dx=\int_{-\infty}^\infty|\psi(x)|^2dx=\int_{-\infty}^\infty\psi^*(x)\psi(x)dx=1$$

#### Bra-ket $\rightarrow$ Wavefunction

1. $\ket{\psi}\rightarrow\psi(x)$
2. $\bra{\psi}\rightarrow\psi^*(x)$
3. $\braket{}{}\rightarrow\int_{-\infty}^\infty dx$
4. $\hat{A}\rightarrow A(x)$

### Expectation Value

$$\angle{\hat{A}}=\braketo{\psi}{\hat{A}}{\psi}=\int_{-\infty}^\infty\psi^*(x)A\psi(x)dx$$ $$\angle{\hat{x}}=\braketo{\psi}{\hat{x}}{\psi}=\int_{-\infty}^\infty\psi^*(x)x\psi(x)dx$$ $$\angle{\hat{p}}=\braketo{\psi}{\hat{p}}{\psi}=\int_{-\infty}^\infty\psi^*(x)\big(-i\hbar\frac{d}{dx}\big)\psi(x)dx$$ $$\angle{H}=\sum_n |c_n|^2E_n$$

### Properties of Energy Eigenstates

1. Normalization: $$\braket{E_n}{E_n}=1\longleftrightarrow\int_{-\infty}^\infty|\phi_n(x)|^2dx=1$$
2. Orthogonality: $$\braket{E_n}{E_m}=\delta_{nm}\longleftrightarrow\int_{-\infty}^\infty\phi_n^*(x)\phi_m(x)dx=\delta_{nm}$$
3. Completenes: $$\ket{\psi}=\sum_n c_n\ket{E_n}\longleftrightarrow\psi(x)=\sum_n c_n\phi_n(x)$$

## The SE Recipe for Wavefunctions

1. Diagonalize $H$ to find energy eigenvalues and energy eigenstates. $$H\phi_n(x)=E_n\phi_n(x)$$
2. Write $\psi(0)$ as a superposition of energy eigenstates $$\psi(x,0)=\sum_n c_n\phi_n(x),\quad c_n=\int_{-\infty}^\infty\phi_n^*(x)\psi(x,0)dx$$
3. Multiply each term in the superposition (in the energy/momentum basis) by a phase, $e^{-iE_nt/\hbar}$, to get $\psi(x,t)$. $$\psi(x,t)=\sum_n c_n e^{-iE_nt/\hbar}\phi_n(x)$$
4. Calculate observable $\hat{A}$. $$\angle{A}=\int_{-\infty}^\infty\psi^*(x,t)A\psi(x,t)dx$$