Localised wavepacket (basic question from Srednicki chap. 5)

Question

I'm currently working through Srednicki and I am confused by a couple of lines in chap. 5 (the LSZ reduction formula). He wants a wavepacket localised around $\mathbf{k = k}_1$ and $\mathbf{x} = 0$ at $t= 0$, giving it as

$$ \begin{align} a^{\dagger}_1 &\equiv \int d^3 k f_1(\mathbf{k}) a^{\dagger}(\mathbf{k}) \tag{5.6} \cr &\propto \int d^3 k \; d^3 x \; e^{-(\mathbf{k}-\mathbf{k}_1)^2/4\sigma^2} e^{i k x} (i \partial_0 \phi(x) - \omega \phi(x)) \tag{5.2+5.7} \end{align} $$

where $\phi(x)$ is an operator creating a state localised at $x$.

I don't understand how this is localised at $\mathbf{x} = 0$ when $t=0$. The same question is asked here but the one brief answer (saying to do a FT) sincerely leaves me none the wiser. It looks to me like integrating over $k$ gives us a Gaussian in $x$ centred at...$-i \mathbf{k}_1/2\sigma^2$?

Can anyone describe the exact steps I should be doing to see the localisation?

Answer 1

The basic idea underlying what's going on in that construction is the mathematical fact that the Fourier transform of a Gaussian centered around zero is also a Gaussian centered around zero. But what about a Gaussian centered around some other value? In the comments you wrote:

Why a Gaussian at zero would have its mode expansion dominated by low-frequency (near zero) modes, but one centred at 5 would not...is not intuitively clear to me.

That’s actually not true, a Gaussian centered at five would have the same frequency spectrum as the one centered at zero, but the phases would be different. Rather than writing a lot of formulas, I will provide a link to an awesome Fourier transform calculator from WolframAlpha that you can use to get a better feel for what the Fourier transforms of these functions look like.

Now, after getting a better feel for the math portion, here is the basic idea of what's going on in the quantum mechanical formulas in your question: $$a^{\dagger}(\mathbf{k}) $$ creates a state with a sharply peaked momentum, $$\int d^3 k f_1(\mathbf{k}) a^{\dagger}(\mathbf{k}) $$ creates a state with a Gaussian distribution of momenta centered around $\mathbf{k}_1$, assuming of course that $f_1$ is defined as such a Gaussian.

Answer 2

You did everything right. You're only lacking the correct interpretation.

The momentum-space wave packet is (Eq (5.9) in my version of Srednicki) $$ f_1(\mathbf k) \propto \exp(-(\mathbf k- \mathbf k_1)^2/4\sigma^2) .$$ You found the position-space wave packet $$ \tilde f_1(\mathbf x) \propto \exp(-\sigma^2\mathbf x^2-i\mathbf k_1\mathbf x)$$ by Fourier transforming, and you note that the maximum of the argument of the exponential is at imaginary value $\mathbf x=-i\mathbf k_1/2\sigma^2$. This is not a problem. Recall that the probability density of finding a particle at position $\mathbf x$ given a wave packet $\tilde f_1(\mathbf x)$ is $|\tilde f_1(\mathbf x)|^2$. That is $$|\tilde f_1(\mathbf x)|^2 \propto \exp(-2\sigma^2\mathbf x^2).$$ So the wave packet is sure enough centered at zero. The phase doesn't affect that.

In fact, as you saw, the phase vanishes too if $\mathbf k_1=0$. This reflects one of the basic properties of the Fourier transform, which is that shifting a function only changes its Fourier transform by a phase. As you anticipated in your comment, shifting a function cannot change its power spectrum (which is given by the magnitude of the Fourier transform).

Also note that this is related to one of the many neat properties of Gaussians. The wave packet in position-space is a function of a real variable (position is a real number). What would it mean to say the wave packet is centered at a complex value? In fact, you can show that none of the important properties of the Gaussian are affected by such a shift into the complex plane. For example, if your Gaussian is normalized to unity (as your wave packet should be), that is, if its integral over the real line is 1, then even if you shift the Gaussian into the complex plane its integral along the real line is still 1! You can check this by direct computation. Alternatively, note that the saddle point approximation for Gaussians is exact, and the Gaussian has no poles, so you can deform the contour off the real line to pass over the (complex) saddle point, and apply the saddle point "approximation" which only depends on the second derivative at the saddle point, that is, $\sigma$ for a Gaussian. You automatically get the same result.