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*"we can calculate normalisation factor ##\Psi_0## of a wavefunction ##\Psi## if we integrate probability ##|\Psi|^2## over some volume and equate it to 1"*. Hence:

[itex]

\int\limits_{V} |\Psi|^2 \, \textrm{d}V= 1

[/itex]

Now how exactly do we integrate this? Please be specific, because in the post i linked to i got an anwser that the result of integration is

[itex]

\int\limits_{V} |\Psi|^2 \, \textrm{d}V = |\psi_0|^2 V

[/itex]

and i don't know how is this possible. Maybee my interpretation of this is wrong and this is why below i am supplying you with my interpretation.

**My interpretation:**

For the sake of clarity i will just choose some wave function for example ##\Psi = \Psi_0 \sin(\omega t - kx)##. I chose this as it is similar to an already known wave function of a sinusoidal wave ##A = A_0 \sin(\omega t - kx)## which i have been using allover wave physics. I don't know if i am allowed to choose the ##\Psi## like that because for now i don't know enough to know what i am alowed/not allowed to do in QM. If i understand this ##\Psi_0## in a vave function ##\Psi = \Psi_0 \sin(\omega t - kx)## is the normalisation factor i am seeking? (Please confirm this). So now i take an integral of the wavefunction and equate it to 1:

[itex]

\begin{split}

\int \limits^{}_{V} \left|\Psi \right|^2 \, \textrm{d} V &= 1\\

\int \limits^{}_{V} \big|\Psi_0 \sin (\omega t - kx) \big|^2 \, \textrm{d} V &= 1\\

&\dots

\end{split}

[/itex]

I get lost at the spot where i wrote down "##\dots##". I really don't know how to get ##|\psi_0|^2 V## as a result of integration.