# It is possible to express the convolution between two quantum functions $\psi_{1}(x)$ and $\psi_{2}(x)$ in terms of a inner product?

The question of the title is due to the following methodology. Let us consider two arbitray quantum functions $$\psi_{1}(x)$$ and $$\psi_{2}(x)$$, such that the convolution between them is

$$\left\lbrace \psi_{1}(x) \ast \psi_{2}(x) \right\rbrace (y) =\int_{-\infty}^{+\infty}dx ~\psi_{1}(x) \psi_{2}(y-x). \tag{1}$$

Now, take into account the definition for the inner product between two quantum functions:

$$\left(\phi_{1}, \phi_{2} \right)=\int_{-\infty}^{+\infty} dx~\phi_{1}^{\ast}(x) \phi_{2}(x) \tag{2};$$

then, by comparing Eqs. (1) and (2) we can associate $$\psi_{1}(x)$$ with a real valued function, such that $$\psi_{1}^{\ast}(x)=\psi_{1}(x)$$ (this could be the case for a real Gaussian function, for example $$\psi_{1}(x)=C_{1}e^{-C_{2}x^2}$$, being $$C_{1}$$ and $$C_{2}$$ real constants). Therefore, with this in mind, and considering Eq. (2), we can write the convolution of Eq. (1) as

$$\left\lbrace \psi_{1}(x) \ast \psi_{2}(x) \right\rbrace (y) =(\psi_{1}(x), \psi_{2}(y-x)) \tag{3}$$

being $$y$$ an arbitrary displacement in the function $$\psi_{2}(x)$$. The Eq. (3) is specially relevant for me since I have the probability of an event as proportional to the squared modulus of the convolution between the two quantum functions $$\psi_{1}(x)$$ and $$\psi_{2}(x)$$, that is,

$$\mathcal{P}_{\text{event}}=\frac{1}{\pi^{2}}\left|\int_{-\infty}^{+\infty}\psi_{1}(x) \psi_{2}(y-x)\right|^{2}. \tag{4}$$

Then, I could write Eq. (4) as an inner product, that is,

$$\mathcal{P}_{\text{event}}=\frac{1}{\pi^{2}} \left| (\psi_{1}(x),\psi_{2}(y-x))\right|^{2},. \tag{5}$$

With this in mind, we could use the Schwarz innequality: $$\left|(\psi_{1},\psi_{2}) \right|^2 \leq (\psi_{1},\psi_{1})^2 (\psi_{2},\psi_{2})^2$$ to find the maximum probability of the event as follows. Since the functions $$\psi_{1}(x)$$ and $$\psi_{2}(y-x)$$ are normalized: $$\left(\psi_{1}, \psi_{1}\right)=1$$, $$\left(\psi_{2}, \psi_{2}\right)=1$$, we deduce through Schwarz inequality that $$\left|(\psi_{1},\psi_{2}) \right|^2\leq 1$$. Therefore, the maximum value of $$\mathcal{P}_{\text{event}}$$ happens when $$\left|(\psi_{1},\psi_{2}) \right|^2=1$$; hence, using this in Eq. (5), we have

$$\text{Max}\left[ \mathcal{P}_{\text{event}}\right]=\frac{1}{\pi^2}. \tag{6}$$

Then, it is valid to express the convolution of Eq. (1) as an inner product when $$\psi_{1}(x)$$ is a real valued function?

You don't even need $$\psi_1$$ to be real valued right? Your probability appears to depend on two functions as $$$$\mathcal{P}_{\mathrm{event}}[\psi_0, \psi_1] = \frac{1}{\pi^2} |(\psi_0, \psi_1)|^2.$$$$ The fact that the first argument can be defined by $$$$\psi_0(x) = \psi_2^*(y - x)$$$$ for some $$\psi_2$$ and $$y$$ is probably important for physical interpretations. But to bound it, everything that you're saying works.