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?


1 Answer 1


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


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