# Fourier Tranformation Identity

In our class we said for the following fourier transformation:

$$\phi(\vec r,t)=\int_{-\infty}^{\infty}d\omega\int d^3k \hat{\phi}(\vec k,\omega)e^{i(\vec k \vec r - \omega t)}$$

We said that $$\phi(\vec r,t)^*=\phi(\vec r,t)$$ and therefore it's fourier transform satisfies the condition $$\hat{\phi}^*(\vec k,\omega)=\hat{\phi}(-\vec k,-\omega).$$

So I have two questions about it:

1. How is $$\phi(\vec r,t)^*=\phi(\vec r,t)$$ when you have $$e^{i...}$$ and it's complex is $$e^{-i...}$$. How are they equal? + We don't know the equation of it's Fourier transform $$\hat{\phi}(\vec k,\omega)$$. It might contain complex numbers and therefore change when you consider it's conjugate.

2.How do we come up with this: $$\hat{\phi}^*(\vec k,\omega)=\hat{\phi}(-\vec k,-\omega)$$?

And if more info is needed. By solving the homogeneous wave equation one gets $$\omega = \pm ck$$ and you can write $$\hat{\phi}(\vec k,\omega)=2\omega f(\vec k,\omega)\delta(\omega^2- c^2k^2)$$

• I think Mathematics is better site for the question (?) Commented Jan 1, 2022 at 14:49

I think what they tried to say in class is that if $$\displaystyle \phi (\vec{r} ,t)$$ is a real function, $$\displaystyle \phi (\vec{r} ,t)^{*} =\phi (\vec{r} ,t)$$, then its Fourier transform satisfies $$\displaystyle \hat{\phi }(\vec{k} ,\omega )^{*} =\hat{\phi }( -\vec{k} ,-\omega )$$.

We can see this by the inverse transformation

$$\displaystyle \hat{\phi }(\vec{k} ,\omega ) =\int \frac{dt}{2\pi }\int \frac{d^{3} r}{( 2\pi )^{3}} \phi (\vec{r} ,t) e^{-i(\vec{k} \cdot \vec{r} -\omega t)}$$.

Taking the complex conjugate on both sides we get

$$\displaystyle \hat{\phi }(\vec{k} ,\omega )^{*} =\int \frac{dt}{2\pi }\int \frac{d^{3} r}{( 2\pi )^{3}} \phi (\vec{r} ,t) e^{+i(\vec{k} \cdot \vec{r} -\omega t)}$$

assuming that $$\phi (\vec{r} ,t)$$ is real. This can equivalently be written as

$$\displaystyle \hat{\phi }(\vec{k} ,\omega )^{*} =\int \frac{dt}{2\pi }\int \frac{d^{3} r}{( 2\pi )^{3}} \phi (\vec{r} ,t) e^{-i[( -\vec{k}) \cdot \vec{r} -( -\omega ) t]}$$

which by definition of $$\hat{\phi }$$ is $$\hat{\phi }( -\vec{k} ,-\omega )$$.

We can turn the argument around and say that if $$\hat{\phi }(\vec{k} ,\omega )^{*} =\hat{\phi }( -\vec{k} ,-\omega )$$ then $$\phi$$ is a real function, i.e $$\displaystyle \phi (\vec{r} ,t)^{*} =\phi (\vec{r} ,t)$$.

Taking the complex conjugate on both sides of

$$\displaystyle \phi (\vec{r} ,t) =\int d\omega \int d^{3} k\hat{\phi }(\vec{k} ,\omega ) e^{i(\vec{k} \cdot \vec{r} -\omega t)}$$

gives

$$\displaystyle \phi (\vec{r} ,t)^{*} =\int d\omega \int d^{3} k\hat{\phi }(\vec{k} ,\omega )^{*} e^{-i(\vec{k} \cdot \vec{r} -\omega t)} =\int d\omega \int d^{3} k\hat{\phi }( -\vec{k} ,-\omega ) e^{i[ -(\vec{k}) \cdot \vec{r} -( -\omega ) t]}$$.

Now we can change the integration variables to $$\displaystyle \omega '=-\omega ,\ \vec{k} '=-\vec{k}$$ with $$\displaystyle d\omega '=-d\omega ,\ d^{3} k'=-d^{3} k$$. Then we have that

$$\int d\omega \int d^{3} k=\int _{-\infty }^{\infty } d\omega \int _{-\infty }^{\infty } dk_{x}\int _{-\infty }^{\infty } dk_{y}\int _{-\infty }^{\infty } dk_{z}$$

$$=\int _{\infty }^{-\infty }( -d\omega ')\int _{\infty }^{-\infty }( -dk'_{x})\int _{\infty }^{-\infty }( -dk'_{y})\int _{\infty }^{-\infty }( -dk'_{z}) =\int _{-\infty }^{\infty } d\omega '\int _{-\infty }^{\infty } dk_{x} '\int _{-\infty }^{\infty } dk_{y} '\int _{-\infty }^{\infty } dk_{z} '$$

$$\displaystyle ={\textstyle \int d\omega '\int d^{3} k'}$$

So with our change of integration variables we can express $$\phi^*$$ as

$$\displaystyle \phi (\vec{r} ,t)^{*} =\int d\omega \int d^{3} k\hat{\phi }( -\vec{k} ,-\omega ) e^{i[ -(\vec{k}) \cdot \vec{r} -( -\omega ) t]} =\int d\omega '\int d^{3} k'\hat{\phi }(\vec{k} ',\omega ') e^{i(\vec{k} '\cdot \vec{r} -\omega 't)} =\phi (\vec{r} ,t)$$

• Ok, I understand my 2nd question but what about the first one? How is $\phi(\vec r, t)$ real when it contains the complex number in the power of $e$? Commented Jan 1, 2022 at 14:50
• As erikasan suggested, in your class you were probably imposing $\phi$ is real. Without context I can't tell why, though. Was it a QM class? What were you doing? Commented Jan 1, 2022 at 14:57
• @Feynman_00 It was electrodynamic class. We were simply dealing with the solution to the wave eq. And this was part of the solution. And I wanted to know the answer to those 2 questions. But I guess you can assume that $\phi(\vec r,t)$ is real Commented Jan 1, 2022 at 15:06