The wave function is defined as $$ \Psi(x,t) $$
To get the probability, they squared it with a modulus bracket $$ |\Psi(x,t)|^2 $$ Because amplitude can also be -ve but the probability cannot be. My question is, What is the actual point of both mod and squaring together?
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weegee
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Keep in mind $$ |\Psi(x,t)|^2 = \Psi(x,t)^*\Psi(x,t)$$
Where $*$ indicates complex conjugacy. So the modulus is guaranteed to be real by this definition of the inner product.
The modulus indicates the probability density, the chance of finding the particle between $x$ and $x+dx$.
TurlocTheRed
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2Why is there an integral there? The absolute value of $\Psi$ is just $\sqrt{\Psi^*\Psi}$ – J_P Jun 11 '19 at 22:13
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Bas habit. Fixed. Thanks! – TurlocTheRed Jun 11 '19 at 22:15
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That's right. An integral arises if one wants to calculate the probability that a particle is located in a set $\Omega$ at time $t$. $P=\int_{\Omega}|\Psi(x,t)|^2dx$ – Peter Melech Jun 11 '19 at 22:20
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This is a good point. Even better would be to start with bilinear form $\Psi(x,t)\bar\Psi(x',t')$. – user Jun 11 '19 at 22:31
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This is physics. $\Psi$ can have complex values. So its square need not be positive.
In quantum mechanics the wave function must be complex. We (so far) have not found how to do it with only real values.
GEdgar
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I never knew that it could also be a complex number. Thanks for clarification – weegee Jun 11 '19 at 21:59
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Isn't $|\Psi(x,t)|^2=\Psi(x,t)\overline{\Psi(x,t)}\geq 0$ i.e. in particular real and can be seen as a probability density as explained e.g. here? – Peter Melech Jun 11 '19 at 22:06