According to Cohen-Tannoudji the wave-function for a one-dimensional free particle can be written as

$$ \psi (x,0)=\frac{1}{\sqrt{2 \pi}} \int g(k) e^{ikx} dk.$$

While $g(k)$ is not specified, there is a picture of its absolute value in the book and it looks Gaussian, with a peak at $k=k_0$. They state that we can write

$$g(k) = |g(k)| e^{i \alpha (k)},$$

where $\alpha(k)$ varies smoothly in the interval $\left[ k_0 - \frac{\Delta k}{2}, k_0 + \frac{\Delta k}{2} \right]$ over which $|g(k)|$ is appreciable. They go on to state that if $\Delta k$ is sufficiently small then expanding $\alpha(k)$ about $k_0$ allows us to approximate the wavefunction as

$$\psi (x,0)\approx \frac{e^{i \left[k_0 x + \alpha(k_0) \right]}}{\sqrt{2 \pi}} \int_{- \infty}^{\infty} |g(k)| e^{i(k-k_0)(x-x_0)} dk,$$


$$x_0=-\left[\frac{d \alpha}{dk} \right]_{k=k_0}.$$

The logic behind putting it in this format is to show that if $x$ varies appreciably from $x_0$ the large number of oscillations in the given k-interval will tend to make the integral quite small. However, if $x$ is close to $x_0$ $g(k)$ will not oscillate over the given interval and $\psi$ will be appreciable for that value of x. Cohen-Tannoudji then states that if $e^{i(k-k_0)(x-x_0)}$ "oscillates approximately once when $k$ traverses the domain $\Delta k$" then $\psi$ will be negligible. Mathematically, they state this condition as

$$\Delta k (x-x_0) \approx 1.$$

Cohen-Tannoudji then states, "If $\Delta x$ is the approximate width of the wave packet, we therefore have

$$\Delta k \Delta x \gtrsim 1."$$

My questions are

  1. Why is the condition of one oscillation $\Delta k (x-x_0) \approx 1,$ and not $\Delta k (x-x_0) \approx 2 \pi$?
  2. Even if I accept the previous condition, I don't understand the logic in making the jump to the statement $$\Delta k \Delta x \gtrsim 1."$$
  3. Anticipating a little further down the book, why is $g(k)$ considered the momentum-space wavefunction? Or, perhaps a better way of phrasing this is to ask why are the position and momentum basis related by a Fourier transform?

1 Answer 1


Regarding your third question. If you want to see the state vector $|\psi\rangle$ you considered explicitly, you must choose a basis to expand it. For example, the position eigenkets $|x\rangle$. Then you can expand your state kets with respect to $|x\rangle$. That is $|\psi\rangle=\int dx |x\rangle\langle x|\psi\rangle$. Where $\langle x|\psi\rangle=\psi(x)$ is just the wavefunction in real space.

Similarly, you can choose to expand with respect to eigenkets of momentum operator $p$. So you have:$$\psi(x)=\langle x|\psi\rangle=\int dp \langle x|p\rangle \langle p|\psi\rangle=\int dp \bar\psi(p)e^{ikx}$$

where $\bar\psi(p)=\langle p|\psi\rangle$ is just the wave function in momentum space.

Regarding your first and second question, my opinion is that don't treat the number 1 or $2\pi$ too seriously, since this argument itself is an approximation. It only want to show you that the wave package extension in real space $\Delta x$ times wave package extension in momentum space $\Delta k$ cannot be smaller than some value. Which is required by uncertainty principle.

Since$ |g(k)|$ is approximate zero outside $(k_0-\Delta k,k_0+\Delta k)$, we have:

$$\psi (x,0) \approx \frac{e^{i \left[k_0 x + \alpha(k_0) \right]}}{\sqrt{2 \pi}} \int_{ k-\Delta k}^{k+\Delta k} |g(k)| e^{i(k-k_0)(x-x_0)} dk$$

The main argument is that, if x is very near $x_0,\psi(x)$ cannot tends to zero, that is to say: given a $\Delta k$, $\Delta x$ cannot be arbitrarily small,(recall that $\Delta x$ is the extension of the wavefunction in real space, ie, the range that $\psi(x)$ is not approximate to zero), which is required by uncertainty principle.

In order to show this, let's consider a specific form of $|g(k)|=e^{(k-2)^2}$, the plot of it is shown below, which we can see in this case our $\Delta k\approx 2, k_0=2.$

enter image description here

We need to consider only the real part of the integral, because due to symmetry the imaginary part is zero.

Now consider the plot of the real part of the integrand, if we choose $x- x_0$ small enough, for example $x-x_0\approx1$, we can see that the integral can no way vanish in this case.

enter image description here

However, if we choose $x-x_0$ large enough, say, $x-x_0\approx 10$, the positive and negative will cancel each other and the integral will be very small.

enter image description here

So the range $\Delta x$ in which the wave function is appreciable should be greater than some specific value, it cannot be arbitraryly small.

Last to mention is that $\Delta x, \Delta k$ I described here is the half region that the wave function is appreciable. They are not $\sigma_x \sigma_k$ in the formula of uncertainty principle, usually the region that wave function is appreciable will be larger than the standard derivative and proportional to it.


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