Hello I am trying to find the fourier transform of a plane wave of the form $$\psi(x) = \frac {1}{\sqrt{2\pi \hbar}}\exp\left(\frac {i}{\hbar}p_0 x\right)$$ where $p_0$ is fixed and Real

I've worked through this far: $$(Ff)(p) = \frac {1}{\sqrt{2\pi \hbar}} \int_{-\infty}^\infty dx \frac {1}{\sqrt{2\pi \hbar}}\exp\left(\frac {i}{\hbar}p_0 x\right)\exp\left(\frac {-i}{\hbar}p x\right) = \frac {1}{{2\pi \hbar}} \int_{-\infty}^\infty dx \exp\left(\frac {i}{\hbar}x(p_0-p)\right)$$

Now I am a bit stuck on how to continue? I think I need to use the Dirac distribution somehow since it looks like: $$\frac {1}{{2\pi \hbar}} \int_{-\infty}^\infty dp \exp\left(\frac {i}{\hbar}p(x-x_0)\right) = \delta(x-x_0)$$

But I don't really know how that helps me solve the transform.

  • 1
    $\begingroup$ What you have is exactly your last formula, except with the roles of $x$ and $p$ swapped. $\endgroup$
    – Javier
    Commented Oct 31, 2020 at 15:12

1 Answer 1


You came very far and you are right about the Dirac-delta function So we use the following form of dirac-delta function. $$\int_{-\infty}^{\infty}e^{ik(x-x_0)}dk=2\pi\delta(x-x_0)$$

Proceeding from the last step:


putting $x=\hbar k$ $$\phi(p)=\frac{1}{2\pi\hbar}\int_{-\infty}^{\infty}\hbar e^{ik(p_0-p)}dk$$ $$\phi(p)=\frac{1}{2\pi}2\pi\delta(p-p_0)=\delta(p-p_0)$$

and that actually make sense because in position space you have a plane wave so that it's large uncertainity. In momentum space, you have a delta function and so the lowe uncertainity so that the product remain constant (Uncertainity principle) .

  • $\begingroup$ Ah I see thank you, I wasn't sure if I needed to adapt it in some other way for $\delta (p_0 - p)$ as opposed to $\delta (p - p_0)$. $\endgroup$
    – Allod
    Commented Oct 31, 2020 at 22:05

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