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
$$R+T=1$$