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I saw an awesome derivation of Schrodinger's equation on Wikipedia. Part of it relies on:

Since time-evolution must preserve the norm of the wave-function, it follows that $U(t', t)$ must be a member of the unitary group of operators acting on wave-functions, where $\Psi(t') = U(t', t)\Psi(t)$

So, the question is: Why should time evolution preserve the norm of a wave-function?

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The norm of the wavefunction gives the probability that the particle still exists at all. So the norm of the wavefunction changing implies that particles can either be created or destroyed.

In ordinary, non-relativistic quantum mechanics, particle number is a conserved quantity. Particles can neither be created or destroyed, and so the norm of the wavefunction stays constant.

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  • $\begingroup$ When it is said that unitary operators preserve norm of a function, do they preserve $$\int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} |\Psi(x, y, z, t)|^2 dx dy dz$$ or $$|\Psi(x, y, z, t)|$$ Because I think you are talking about the first, while norm means the second $\endgroup$ – PhyEnthusiast Mar 4 '18 at 9:20
  • $\begingroup$ @PhyEnthusiast The norm is not $\int\left|\Psi\right|d^3x$, but $\sqrt{\int\left|\Psi\right|^2d^3x}$. So conserving the first is the same as conserving the norm. $\endgroup$ – Chris Mar 4 '18 at 9:31
  • $\begingroup$ But $\sqrt{|\Psi|^2 d^3 x}$ is not the same as $|\Psi|$, is it? So, which is called the norm $\endgroup$ – PhyEnthusiast Mar 4 '18 at 9:41
  • $\begingroup$ They preserve the first one. The norm-squared is your first expression, and the square root of the first expression is called the norm. $\endgroup$ – pianyon Mar 4 '18 at 9:42
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    $\begingroup$ @PhyEnthusiast The issue is that there are infinitely many different ways to define a 'norm'. Chris is talking about the physically meaningful one. $\endgroup$ – knzhou Mar 4 '18 at 10:11
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If you look for a probability interpretation of QM, you get a continuity equation of the form $$\partial_{t}\rho+ \vec{\nabla}\cdot\vec{j}=0,$$ where $$\rho\left(\vec{x},t\right)\equiv\psi^*\left(\vec{x},t\right)\psi\left(\vec{x},t\right)$$ and $$\vec{j}\equiv\frac{\hbar}{2im}\left(\psi^*\vec{\nabla}\psi - \psi\vec{\nabla}\psi^*\right).$$ Integrating the continuity equation over a finite volume $V$ and using the divergence theorem, you can find $$\frac{\mathrm{d}}{\mathrm{d}t}\int_V \mathrm{d}^3x\psi^*\psi=-\frac{\hbar}{2im}\int_\sigma \mathrm{d}\sigma\left(\psi^*\partial_n\psi-\psi\partial_n \psi^*\right),$$ where $\sigma\equiv\partial V$ is the smooth boundary surface of $V$ and $\partial_n\left(\cdot\right)$ indicates the normal derivative (directional derivative taken in the direction normal to the surface). If you let the volume $V$ tend to the whole space, $\mathbb{R}^3$, since your wavefunction is in the Hilbert space $\mathscr{L}^2\left(\mathbb{R}^3\right)$, last equation becomes $$\frac{\mathrm{d}}{\mathrm{d}t}\int_V \mathrm{d}^3x\psi^*\psi\equiv\frac{\mathrm{d}}{\mathrm{d}t}\left|\psi\left(\vec{x},t\right)\right|^2=0,$$ so you can see that the preservation in time of the norm of the wavefunction is required in order to get a probability interpretation of the theory.

Note that, at this level, you're not allowing decays, for example.

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Schrodinger's equation evolved in order to solve the problems that arose for small dimensions: photoelectric effect, spectra of hydrogen modeled by the Bohr model, black body radiation, etc. It is a mathematical model, to start with successful in predicting the decay spectra.

I evolved into a theory, quantum mechanics, which instead of determining positions at time t, gives the probability of finding a particle at time t. The probability given by the complex conjugate squared of the wavefunction given by the Schrodinger equation to start with.

Probability should always be bounded by 0 and 1, by construction.

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