
# Free Particle

## The Free Particle

For a free particle, $V(x)=0$ everywhere and the energy eigenvalue equation is $$\frac{d^2}{dx^2}\phi_E(x)=-\frac{2mE}{\hbar^2}\phi_E(x)=-k^2\phi_E(x)$$ Applying the Schrödinger time evolution, we get $$\psi_E(x,t)=Ae^{ik(x-\frac{\omega}{k}t)}+Be^{-ik(x+\frac{\omega}{k}t)}$$ We recognize the exponent as a function of position, with the negative sign indicating direction, so we simplify and rewrite the wavefunction as: $$\psi_E(x)=Ae^{ikx}$$

## Momentum Eigenstates

We recall that the momentum operator in the position representation is: $\hat{p}=-i\hbar\frac{d}{dx}$. Applying this to an eigenstate $\phi_k(x)$, we get $\hat{p}\phi_k(x)=\hbar k\phi_E(x)$. So, we have the momentum eigenvalue $p=\hbar k$ and the momentum eigenstate $\phi_p(x)=Ae^{ipx/\hbar}$.
For a free particle, where $V(x)=0$, momentum eigenstates are also energy eigenstates with energy $E=\frac{p^2}{2m}$. We can apply the normalization condition and the dirac-delta function to get $$A=\frac{1}{\sqrt{2\pi\hbar}}$$ $$\phi_p(x)=\frac{1}{2\pi\hbar}e^{ipx/\hbar}$$ To derive the general wavefunction, we apply the completenes theorem: $$\int_{-\infty}^\infty\ketbra{p}{p}dp=\mathbb{I}$$ \begin{align*} \psi(x)&=\braket{x}{\psi}\\ &=\braketo{x}{\bigg(\int_{-\infty}^\infty\ketbra{p}{p}dp\bigg)}{\psi}\\ &=\int_{-\infty}^\infty\braket{x}{p}\braket{p}{\psi}dp\\ &=\int_{-\infty}^\infty\phi_p(x)\psi(p)dp\\ &=\frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^\infty \psi(p)e^{ipx/\hbar}dp\quad\leftarrow\text{ Fourier Transform} \end{align*} The fourier transform allows for a conversion from the momentum space wavefunction to the position space wavefunction.
Conversely, the inverse fourier transform converts from the position space to the momentum space: $$\psi(p)=\frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^\infty \psi(x)e^{-ipx/\hbar}dx$$ Note:
• $\braket{x}{p}$ is the projection of the momentum eigenstate onto the position basis, which is the position representation of the momentum eigenstate wavefunction, $\phi_p(x)$.
• $\braket{p}{\psi}$ is the projection of the general state onto the momentum basis, $\psi(p)$.
* $\psi(x)$ and $\psi(p)$ are DIFFERENT functions.
$$\psi(x)=\frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^\infty \psi(p)e^{ipx/\hbar}dp\\\psi(p)=\frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^\infty \psi(x)e^{-ipx/\hbar}dx$$
In summary,
Position Space Momentum Space
Position Eigenstate $\ket{x_0}\doteq\delta(x-x_0)$ $\ket{x_0}\doteq\frac{1}{\sqrt{2\pi\hbar}}e^{-ip_0x/\hbar}$
Momentum Eigenstate $\ket{p_0}\doteq\frac{1}{\sqrt{2\pi\hbar}}e^{ip_0x/\hbar}$ $\ket{p_0}\doteq\delta(p-p_0)$
Position Operator $\hat{x}\doteq x$ $\hat{x}\doteq i\hbar\frac{d}{dp}$
Momentum Operator $\hat{p}\doteq -i\hbar\frac{d}{dx}$ $\hat{p}\doteq p$

## Scattering and Tunneling

For unbound states, the particle has enough energy to escape the potential well, $E>V_0$. We define the following potential energy function for the finite well: $$V(x)=\begin{cases} 0,&\text{outside}\\ -V_0,&\text{inside} \end{cases}$$ We define 2 wave vectors: $$k_1=\frac{\sqrt{2mE}}{\hbar}\text{ (outside well)},\quad k_2=\frac{\sqrt{2m(E+V_0)}}{\hbar}\text{ (inside well)}$$ $$\phi_E(x)=\begin{cases} Ae^{ik_1x}+Be^{-ik_1x},&x< -a\\ Ce^{ik_2x}+De^{-ik_2x},&-a < x< -a\\ Fe^{ik_1x}+Ge^{-ik_1x},&x> a\\ \end{cases}$$ Since the unbound states need not decay to $0$ at the endpoints, we cannot impose the normalization condition.
• Energy is not quantized.
• Scattering states have a continuous energy spectrum.
• We let $E$ be the initial condition.
Furthermore, we can treat the particle as coming from the left side from $-\infty$, so $G=0$ since no particles are coming from the right side. Imposing boundary conditions, we get: $$x=-a\begin{cases} \phi_E:Ae^{-ik_1x}+Be^{ik_1x}=Ce^{-ik_2x}+De^{ik_2x}\\ \frac{d\phi_E}{dx}:ik_1Ae^{-ik_1x}-ik_1Be^{ik_1x}=ik_2Ce^{-ik_2x}-ik_2De^{ik_2x} \end{cases}$$ $$x=a\begin{cases} \phi_E:Ce^{ik_2x}+De^{-ik_2x}=Fe^{ik_1x}\\ \frac{d\phi_E}{dx}:ik_2Ce^{ik_2x}-ik_2De^{-ik_2x}=ik_1Fe^{ik_1x} \end{cases}$$ Simplifying and eliminating, we get:

### Transmission Probability

$$T=\bigg|\frac{F}{A}\bigg|^2=\bigg(1+\frac{(k_1^2-k_2^2)^2}{4k_1^2k_2^2}\sin^2 2k_2a\bigg)^{-1}$$

### Reflection Probability

$$R=\bigg|\frac{B}{A}\bigg|^2=\bigg(1+\frac{4k_1^2k_2^2}{(k_1^2-k_2^2)^2\sin^2 2k_2a}\bigg)^{-1}$$

### Lossless Property

$$R+T=1$$