Momentum and position for free particle

Question

In the section of 'The free particle' in 'Introduction to quantum mechanics, second edition' by Griffiths page 65. He has the wave equation as $$\Psi(x,t) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty}\phi(k)e^{i(kx-\omega t)}dk$$ He then makes statements about if $\phi(k)$ has a large spread then the momentum is ill-defined. Why? Does $\phi(k)$ have such a strong influence on momentum? And how does a small spread in $\phi(k)$ correspond to a more well defined position?

Answer 1

We usually define $p=\hbar k$. In this post, I'll take $\hbar =1$ (see natural units) to simplify the notation. This means that $k=p$. This is not necessary, but IMHO this makes the arguments more transparent (besides the fact that natural units are indeed more natural once you get used to them). I usually recommend every learning physicist to try to use natural units whenever they can. If you don't like to, you can exchange $k\leftrightarrow p/\hbar$ in my answer.

In general, the expectation value of the momentum is given by $$ \langle k\rangle= \int_{-\infty}^{+\infty} \mathrm dk\ k\ |\phi(k)|^2 \tag{1} $$ which means that $|\phi(k)|^2$ has to decrease faster than $1/k^3$ as $k\to\infty$ for $\langle k\rangle$ to be well defined. In fact, we also want $\langle k^2\rangle$ to be well defined, which means we usually want $\phi(k)$ to be $\mathcal O(k^{-2})$ as $k\to\infty$.

If $\phi(k)$ has a large spread (it doesn't decrease fast enough), then the momentum is not well-defined, because the integrals that define momentum are divergent.

Why Does $\phi(k)$ have such a strong influence on momentum?

Well, $\phi(k)$ is the wave-function in momentum space, so all the information about the momentum of the system is contained in $\phi(k)$.

And how does a small spread in $\phi(k)$ correspond to a more well defined position?

Note that $\phi(k)$ is the Fourier Transform of $\Psi$, so that this statement is actually equivalent to the uncertainty principle of the Fourier Transform.

UPDATE

Here we prove that both expressions $(1)$ and $(2)$ are equivalent: $$ \langle k\rangle=-i\int_{-\infty}^{+\infty} \mathrm dx \ \Psi^*\partial_x\Psi \tag{2} $$

The proof is non-trivial, but in the end is just the differentiation property of the Fourier Transform. To prove the equivalence, we'll make use of the known fact $$ \int_{-\infty}^{+\infty} \mathrm dx\ \mathrm e^{iqx}=2\pi\delta(q) \tag{3} $$ for any $q\in\mathbb R$. Here, $\delta(q)$ is the Dirac delta function.

First, note that $$ \Psi=\frac{1}{\sqrt{2\pi}}\int \mathrm dk\ \phi(k)\;\mathrm e^{i(kx-\omega t)} \tag{4} $$ so that $$ \partial_x\Psi=\frac{i}{\sqrt{2\pi}}\int \mathrm dk\ k\;\phi(k)\;\mathrm e^{i(kx-\omega t)} \tag{5} $$

If we make the change of variable $k\to k_1$ in $(4)$ and $k\to k_2$ in $(5)$, and plug these into $(2)$ we get $$ \begin{align} \langle k\rangle&=\frac{1}{2\pi}\int \mathrm dx \int\mathrm dk_1\int\mathrm dk_2\ \overbrace{\phi^*(k_1)\;\mathrm e^{-i(k_1x-\omega t)}}^{\Psi^*}\ \overbrace{k_2\;\phi(k_2)\;\mathrm e^{i(k_2x-\omega t)}}^{\partial_x\Psi}=\\ &=\frac{1}{2\pi}\int\mathrm dk_1\int\mathrm dk_2\ k_2\ \phi^*(k_1)\;\phi(k_2)\int \mathrm dx\ \mathrm e^{i(k_2-k_1)x} \end{align} $$

Next, use $(3)$ to simplify the $x$ integral, where $q\equiv k_2-k_1$: $$ \begin{align} \langle k\rangle&=\int\mathrm dk_1\int\mathrm dk_2\ k_2\ \phi^*(k_1)\;\phi(k_2)\;\delta(k_2-k_1)=\\ &=\int\mathrm dk_1 k_1\ |\phi(k_1)|^2 \end{align} $$ which is just $(1)$ upon the change of variables $k_1\to k$. As you can see, both expressions are equivalent, which means that you can use whichever you like the most.