# Is normalization consistent with Schrodinger's Equation?

Schrodinger's Equation does not set a limit on the size of wave functions but to normalize a wave function a limit must be set. How is this consistent physically and mathematically with Schrodinger's Equation.

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The title of this question is very confusing: Normalization of wave functions has absolutely nothing to do with renormalization. Renormalization can not be applied to wave functions. –  Dilaton Mar 29 '13 at 13:44

Schrodinger's equation is homogeneous -- so if $\phi_1,\phi_2,\cdots,\phi_n$ are solutions, $c_1\phi_1 + c_2\phi_2 + \cdots +c_n\phi_n$ is a solution.

More importantly, if $\phi$ is a solution, $A\phi$ is a solution as well. If $A$ is the normalization constant, we see that both non-normalized and normalized versions are valid solutions of Schrodinger's equation, making it consistent.

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Also the normalization is conserved by Schrodinger's equation, so once you normalize the wavefunction it stays normalized (of course I'm ignoring projective measurements, which are not described by the Schrodinger equation). –  Michael Brown Mar 29 '13 at 13:44

Start by considering some not necessarily normalised wavefunction that satisfies the general, not necessarily non-relativistic, schrodinger's equation: $$H\Psi=i\hbar\frac{\partial\Psi}{\partial t}$$

Now, consider some normalised wavefunction $\Psi_{\operatorname{corrected}}$. Clearly, $\Psi_{\operatorname{corrected}}=c\Psi$ for some constant $c$. Then, does this satisfy the generally Schrodinger's equation? $$H\Psi_{\operatorname{corrected}}=Hc\Psi=cH\Psi=ci\hbar\frac{\partial\Psi}{\partial t}=i\hbar\frac{\partial \left( c\Psi\right)}{\partial t}=i\hbar\frac{\partial\Psi_{\operatorname{corrected}}}{\partial t}{}{}$$

................................................................................Q.E.D............................................................................

It does satisfy schrodinger's generalised equation, (and thus, clearly, all its special cases).

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This answer is kind of pointless, it isn't helpful, you can just say "The SE is linear". –  Ron Maimon Aug 22 '13 at 22:15
@RonMaimon: I know, but it seemed that the OP couldn't understand that when some people wrote that in the comments. –  Dimensio1n0 Aug 23 '13 at 2:02
And the comments have disappeared... Weird. What's wrong with those comments? . –  Dimensio1n0 Aug 23 '13 at 5:09

MATHEMATICALLY, if a function $\phi$ is a solution of the SE: $$H\phi=E\phi$$ then $c\phi$ (c is a constant) is also a solution of it due to its linearity.

PHYSICALLY, however, we cannot choose any mathematical solution of the above SE equation to be the physical solution. Remember that we are working with Physics, not Mathematics. In Quantum Mechanics, wave function is probability amplitude and wave function squared is probability density. It's clear that we can find a particle in whole space with a probability of 1; or,in other work, we will definitely find out the particle in whole space. So, we must choose the constant $c$ so that the integral of the wave function squared over whole space must be 1 (normalization condition). The normalized solutions of the SE equation are really physical solutions, or wave functions.

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