Wave functions in position and momentum space
Contents:
# Wave functions in position space
Let’s talk about the 1D case only:
$$ \hat x|x'\rangle = x'|x'\rangle, \qquad \langle x''|x'\rangle = \delta(x' - x'') $$
(normalized). We’re talking continuous spaces here:
$$ |\alpha\rangle = \int dx'|x'\rangle\langle x'|\alpha\rangle $$
and $|\langle x'|\alpha\rangle|^2dx'$ $\to$ the prob. of the particle being found in the interval $x'$ to $x' + dx'$.
We recall that $\langle x'|\alpha\rangle$ is the wave function of the state $|\alpha\rangle$:
$$ \boxed{\langle x'|\alpha\rangle = \psi_\alpha(x')} $$
and we know the prob. is $|\psi_\alpha(x')|^2dx'$, which is what we have above too.
# Inner products
Consider the inner product $\langle\beta|\alpha\rangle$. From $|\alpha\rangle = \int dx'|x'\rangle\langle x'|\alpha\rangle$,
$$ \langle\beta|\alpha\rangle = \int dx'\langle\beta|x'\rangle\langle x'|\alpha\rangle $$
$$ \boxed{\langle\beta|\alpha\rangle = \int dx'\psi_\beta^{\ast}(x')\psi_\alpha(x')} $$
$\to$ represents the prob. amplitude of the state $|\alpha\rangle$ to be found in $|\beta\rangle$.
# Operators
Consider also the application of an operator $f(\hat x)$:
$$ f(\hat x)|\alpha\rangle = \int dx'f(\hat x)|x'\rangle\langle x'|\alpha\rangle = \int dx'f(x')|x'\rangle\langle x'|\alpha\rangle $$
and
$$ \langle\beta|f(\hat x)|\alpha\rangle = \int dx'\langle\beta|x'\rangle f(x')\langle x'|\alpha\rangle $$
$$ \boxed{\langle\beta|f(\hat x)|\alpha\rangle = \int dx'\psi_\beta^{\ast}(x')f(x')\psi_\alpha(x')} $$
↳ only true if $f$ is a function of the observable $\hat x$. Why?
In general,
$$ \langle\beta|\hat A|\alpha\rangle = \int dx'\langle\beta|\hat A|x'\rangle\langle x'|\alpha\rangle $$
We can introduce the identity by the completeness of the $|x''\rangle$:
$$ 1 = \int dx''|x''\rangle\langle x''| $$
$$ \begin{aligned} \langle\beta|\hat A|\alpha\rangle &= \int dx'\int dx''\langle\beta|x''\rangle\langle x''|\hat A|x'\rangle\langle x'|\alpha\rangle \\ &= \int dx'\int dx''\psi_\beta^{\ast}(x'')\langle x''|\hat A|x'\rangle\psi_\alpha(x') \end{aligned} $$
If $\hat A = f(\hat x)$, say $\hat x^2$, then
$$ \hat x^2|x'\rangle = x'^2|x'\rangle \implies \langle x''|\hat x^2|x'\rangle = x'^2\delta(x' - x'') $$
(since $\langle x''|x'\rangle = \delta(x' - x'')$). The delta kills one of the integrals:
$$ \langle\beta|\hat x^2|\alpha\rangle = \int dx'\langle\beta|x'\rangle x'^2\langle x'|\alpha\rangle = \int dx'\psi_\beta^{\ast}(x')x'^2\psi_\alpha(x') $$
and in general, if $\hat A = f(\hat x)$,
$$ \boxed{\langle\beta|f(\hat x)|\alpha\rangle = \int dx'\psi_\beta^{\ast}(x')f(x')\psi_\alpha(x')} $$
$f(\hat x)$ on the left is an operator, the $f(x')$ inside is not an operator (just a function). That’s the why: for a function of $\hat x$, the matrix $\langle x''|\hat A|x'\rangle$ is diagonal. For a general $\hat A$ (like $\hat p$) it isn’t.
# Momentum in the position basis
Let’s apply an infinitesimal translation $\mathcal J(\Delta x)$ to an arbitrary state $|\alpha\rangle$:
$$ \begin{aligned} \left(1 - \frac{ip_x\Delta x}{\hbar}\right)|\alpha\rangle &= \int dx'\mathcal J(\Delta x)|x'\rangle\langle x'|\alpha\rangle \\ &= \int dx'|x' + \Delta x\rangle\langle x'|\alpha\rangle \end{aligned} $$
Introduce a dummy $y = x' + \Delta x$, so $dy = dx'$ and $x' = y - \Delta x$:
$$ \begin{aligned} \int dy|y\rangle\langle y - \Delta x|\alpha\rangle &= \int dx'|x'\rangle\langle x' - \Delta x|\alpha\rangle \\ &= \int dx'|x'\rangle\left(\langle x'|\alpha\rangle - \Delta x\frac{\partial}{\partial x'}\langle x'|\alpha\rangle\right) \end{aligned} $$
through a first order Taylor expansion:
$$ \langle x' - \Delta x|\alpha\rangle = \langle x'|\alpha\rangle - \frac{\partial}{\partial x'}\langle x'|\alpha\rangle\cdot\Delta x $$
And on the other side,
$$ \left(1 - \frac{ip_x\Delta x}{\hbar}\right)|\alpha\rangle = \int dx'|x'\rangle\langle x'|\alpha\rangle - \frac{i\Delta x}{\hbar}p_x|\alpha\rangle $$
Comparing these two, we get
$$ \boxed{p_x|\alpha\rangle = \int dx'|x'\rangle\left(-i\hbar\frac{\partial}{\partial x'}\langle x'|\alpha\rangle\right)} $$
i.e. $p_x = -i\hbar\frac{\partial}{\partial x}$ in the position basis. To see it as a wave function, hit it with $\langle x''|$:
$$ \langle x''|\hat p|\alpha\rangle = \int dx'\underbrace{\langle x''|x'\rangle}_{\delta(x'' - x')}\left(-i\hbar\frac{\partial}{\partial x'}\langle x'|\alpha\rangle\right) $$
$$ \boxed{\langle x'|\hat p|\alpha\rangle = -i\hbar\frac{\partial}{\partial x'}\langle x'|\alpha\rangle} $$
$\to$ momentum in the position basis.
# Momentum space wave function
In the $p$ basis, we can span the entire space too:
$$ 1 = \int dp'|p'\rangle\langle p'|, \qquad |\alpha\rangle = \int dp'|p'\rangle\langle p'|\alpha\rangle $$
$$ \hat p|p'\rangle = p'|p'\rangle, \qquad \langle p'|p''\rangle = \delta(p' - p'') $$
(orthonormal).
$$ \boxed{\langle p'|\alpha\rangle = \phi_\alpha(p')} $$
$\to$ called the momentum space wave function. If $|\alpha\rangle$ is normalized, $\langle\alpha|\alpha\rangle = 1$
$$ \implies \langle\alpha|\alpha\rangle = \int dp'\langle\alpha|p'\rangle\langle p'|\alpha\rangle = 1 $$
$$ \implies \int dp'|\langle p'|\alpha\rangle|^2 = 1 \implies \int dp'|\phi_\alpha(p')|^2 = 1 $$
The interpretation of $\langle p'|\alpha\rangle$: the prob. that a measurement of $p$ gives an eigenvalue $p'$ within a narrow interval $dp'$ is $|\langle p'|\alpha\rangle|^2dp'$.
# The momentum eigenfunction
Recall that
$$ \langle x'|\hat p|\alpha\rangle = -i\hbar\frac{\partial}{\partial x'}\langle x'|\alpha\rangle $$
Let $|\alpha\rangle = |p'\rangle$:
$$ \langle x'|\hat p|p'\rangle = -i\hbar\frac{\partial}{\partial x'}\langle x'|p'\rangle $$
but $\hat p|p'\rangle = p'|p'\rangle$
$$ \implies \boxed{p'\langle x'|p'\rangle = -i\hbar\frac{\partial}{\partial x'}\langle x'|p'\rangle} $$
$$ \implies \boxed{\langle x'|p'\rangle = N\exp\left(\frac{ip'x'}{\hbar}\right)} $$
$\to$ note that this is a function of 2 variables. It can be regarded as the wave function for the momentum eigenstate $|p'\rangle$ $\implies$ the momentum eigenfunction.
To find $N$, note:
$$ |x''\rangle = \int dp'|p'\rangle\langle p'|x''\rangle \implies \langle x'|x''\rangle = \int dp'\langle x'|p'\rangle\langle p'|x''\rangle $$
and since $\langle x'|x''\rangle = \delta(x' - x'')$ (normalized),
$$ \boxed{\int dp'\langle x'|p'\rangle\langle p'|x''\rangle = \delta(x' - x'')} $$
With $\langle x'|p'\rangle = N\exp\left(\frac{ip'x'}{\hbar}\right)$ and $\langle p'|x''\rangle = N^{\ast}\exp\left(-\frac{ip'x''}{\hbar}\right)$,
$$ \begin{aligned} \delta(x' - x'') &= |N|^2\int dp'\exp\left(\frac{ip'(x' - x'')}{\hbar}\right) \\ &= 2\pi\hbar|N|^2\delta(x' - x'') \end{aligned} $$
(using $\int dk\ e^{ik(x' - x'')} = 2\pi\delta(x' - x'')$ with $k = p'/\hbar$).
$$ \implies \boxed{N = \frac{1}{\sqrt{2\pi\hbar}}} $$
$\to$ this is the normalization constant (choosing it real and positive).
$$ \implies \boxed{\langle x'|p'\rangle = \frac{1}{\sqrt{2\pi\hbar}}\exp\left(\frac{ip'x'}{\hbar}\right)} $$
# Switching between the two
We know $|\alpha\rangle = \int dp'|p'\rangle\langle p'|\alpha\rangle$, so
$$ \langle x'|\alpha\rangle = \int dp'\langle x'|p'\rangle\langle p'|\alpha\rangle, \qquad \langle p'|\alpha\rangle = \int dx'\langle p'|x'\rangle\langle x'|\alpha\rangle $$
$$ \boxed{ \begin{aligned} \psi_\alpha(x') &= \frac{1}{\sqrt{2\pi\hbar}}\int dp'\exp\left(\frac{ip'x'}{\hbar}\right)\phi_\alpha(p') \\ \phi_\alpha(p') &= \frac{1}{\sqrt{2\pi\hbar}}\int dx'\exp\left(-\frac{ip'x'}{\hbar}\right)\psi_\alpha(x') \end{aligned} } $$
↳ the wave functions in the position and momentum basis, written in terms of each other (they’re Fourier transforms of each other).
# Gaussian wave packets
Wave function in the position basis:
$$ \boxed{\langle x'|\alpha\rangle = \frac{1}{\pi^{1/4}\sqrt d}\exp\left(ikx' - \frac{x'^2}{2d^2}\right)} $$
(a plane wave $e^{ikx'}$ under a Gaussian envelope of width $d$).
Recall $|\alpha\rangle = \int dx'|x'\rangle\langle x'|\alpha\rangle$, so
$$ \langle p'|\alpha\rangle = \int dx'\langle p'|x'\rangle\langle x'|\alpha\rangle, \qquad \langle p'|x'\rangle = \frac{1}{\sqrt{2\pi\hbar}}\exp\left(-\frac{ip'x'}{\hbar}\right) $$
$$ \langle p'|\alpha\rangle = \int_{-\infty}^{\infty}dx'\frac{1}{\sqrt{2\pi\hbar}}\frac{1}{\pi^{1/4}\sqrt d}\exp\left(-\frac{ip'x'}{\hbar} + ikx' - \frac{x'^2}{2d^2}\right) $$
$$ \boxed{\langle p'|\alpha\rangle = \sqrt{\frac{d}{\hbar\sqrt\pi}}\exp\left(-\frac{(p' - \hbar k)^2d^2}{2\hbar^2}\right)} $$
$\to$ also a Gaussian, centred at $p' = \hbar k$.
# Uncertainty of the Gaussian
To obtain $\langle(\Delta x)^2\rangle = \langle x^2\rangle - \langle x\rangle^2$ we need these two.
$$ \langle x\rangle = \int_{-\infty}^{\infty}dx'\langle\alpha|x'\rangle x'\langle x'|\alpha\rangle $$
with (writing $C = \frac{1}{\pi^{1/4}\sqrt d}$)
$$ \langle\alpha|x'\rangle = C\exp\left(-ikx' - \frac{x'^2}{2d^2}\right) \implies \langle\alpha|x'\rangle\langle x'|\alpha\rangle = C^2\exp\left(-\frac{x'^2}{d^2}\right) $$
$\to$ a Gaussian (the phases cancel).
$$ \implies \langle x\rangle = \int_{-\infty}^{\infty}C^2\exp\left(-\frac{x'^2}{d^2}\right)x'dx' = 0 $$
$\to$ by symmetry (odd integrand). And
$$ \langle x^2\rangle = \int dx'\langle\alpha|\hat x^2|x'\rangle\langle x'|\alpha\rangle = \int dx'x'^2\langle\alpha|x'\rangle\langle x'|\alpha\rangle $$
$\to$ remember this trick.
$$ = \frac{1}{\sqrt\pi d}\int_{-\infty}^{\infty}dx'x'^2\exp\left(-\frac{x'^2}{d^2}\right) $$
$$ \boxed{\langle x^2\rangle = \frac{d^2}{2}} \implies \langle(\Delta x)^2\rangle = \langle x^2\rangle - \langle x\rangle^2 = \frac{d^2}{2} $$
Similarly (doing the same thing with the momentum space Gaussian, $\langle p\rangle = \hbar k$),
$$ \langle(\Delta p)^2\rangle = \langle p^2\rangle - \langle p\rangle^2 = \frac{\hbar^2}{2d^2} + \hbar^2k^2 - \hbar^2k^2 = \frac{\hbar^2}{2d^2} $$
$$ \therefore \boxed{\langle(\Delta x)^2\rangle\langle(\Delta p)^2\rangle = \frac{\hbar^2}{4}} $$
$\to$ exactly!
↳ this is an equality, and hence the Gaussian is the minimum uncertainty wave packet.