In the Schrödinger equation, at any given time $t$ we should jointly add another sub equation, like $$||\psi_t(x)|| = 1$$ where $\psi_t(x) = \Psi(x,t)$, and then try to solve the two equations simultaneously. Why not? I know it does not yield, but I am always baffled, who is doing the normalization of the wave function? Observers, the system, the measuring process, God?

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    $\begingroup$ this is a physics question and answer site, God is metaphysics. There is no WHO here. There are data/observations and mathematical formulae that fit the data/observations: that is what the Schrodinger equation and the normalization impozed are, a mathematical model. $\endgroup$ – anna v Jan 3 '15 at 14:31
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    $\begingroup$ The normalisation is input by hand (by physicists), and is done so to conserve total probability. Otherwise, if you had say a state in some linear superposition, you could end up predicting a probability greater than one for some observation. $\endgroup$ – Akoben Jan 3 '15 at 14:39
  • $\begingroup$ @annav : I don't mean literally. Anyway thats the interpretation of a true experimental physicst! $\endgroup$ – Rajesh Dachiraju Jan 3 '15 at 14:39
  • $\begingroup$ I think that this happens because the evolution operator has norm 1. $\endgroup$ – nicoguaro Jul 18 '17 at 20:26

Nobody is "doing the normalization".

Normalization is not even necessary. We often normalize for convenience, since that means that the Born rule for $\lvert \psi \rangle$ being the state $\lvert \phi \rangle$ reads

$$ P(\psi,\phi) = \lvert \langle\psi\vert\phi\rangle \rvert ^2$$

which is certainly easier to recall/write than

$$ P(\psi,\phi) = \frac{\lvert \langle\psi\vert\phi\rangle \rvert ^2}{\lvert \langle\phi\vert\phi\rangle \rvert \lvert \langle\psi\vert\psi\rangle \rvert }$$

but nothing in the formalism forces normalisation. The basic principle says that states are rays in the Hilbert space, so that $\lvert \psi \rangle$ and $c\lvert \psi \rangle$ represent the same state for all $c \in \mathbb{C}$, and are, for all purposes, fully equivalent representants of the same state. (This, by the way, means that if we want a space where every element corresponds to a distinct quantum state, we should look at the projective Hilbert space instead)

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    $\begingroup$ I would say the normalisation is input by hand in order to conserve total probability $\endgroup$ – Akoben Jan 3 '15 at 14:36
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    $\begingroup$ I would just add that the Schrodinger equation as it is ensures that the norm of the wavefunction is constant. One can see it explicitly if we write the solution as $\ |\psi \rangle_t = e^{-iHt/\hbar} |\psi \rangle_0 $: the "time evolution operator" is unitary, so the inner product $\langle \psi |\psi \rangle_t = \langle \psi|\psi \rangle_0$ $\endgroup$ – SuperCiocia Jan 3 '15 at 14:40
  • $\begingroup$ The solution of the linear Schrödinger equation of quantum mechanics preserves the $L^2$ norm of the function. Therefore it is sufficient that the norm is one at the initial time, and then it is one at any time. As already said, the choice of norm one is a convention due to the probabilistic interpretation; nothing forbids to choose 3 as the wavefunction "normalization". $\endgroup$ – yuggib Jan 3 '15 at 14:40
  • $\begingroup$ @SuperCiocia and yuggib : I tend to disagree with you. Example see the free particle wave like the one in this question, read all the comments by Phoenix there. physics.stackexchange.com/q/156355/540 $\endgroup$ – Rajesh Dachiraju Jan 3 '15 at 14:44
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    $\begingroup$ @RajeshD this is not a square integrable solution of the Schrödinger equation, and therefore it is not an acceptable wavefunction. Linear Schrödinger equations (with self-adjoint generators, as usual in quantum mechanics) admit a unique, norm-preserving, global solution in the physical Hilbert space $\endgroup$ – yuggib Jan 3 '15 at 14:47

Suppose $\psi$ satisfies the (dimensionless) time-dependent Schrödinger equation: $$ i\frac{\partial\psi}{\partial t}=-\frac{\partial^2\psi}{\partial x^2}+V(x)\psi. $$ It will also satisfy the conjugate equation: $$ -i\frac{\partial\psi^*}{\partial t}=-\frac{\partial^2\psi^*}{\partial x^2}+V(x)\psi^* $$ Now consider how the normalisation changes over time: $$\begin{align} \frac{\partial}{\partial t}\int\psi^*\psi dx &= \int\left(\psi\frac{\partial\psi^*}{\partial t}+\frac{\partial\psi}{\partial t}\psi^*\right)dx \\ &= \int\left(\psi\frac{1}{-i}\left(-\frac{\partial^2\psi^*}{\partial x^2}+V(x)\psi^*\right)+\psi^*\frac{1}{i}\left(-\frac{\partial^2\psi}{\partial x^2}+V(x)\psi\right)\right)dx\\ &=0 \end{align}$$ In the last step we use integration by parts with the assumption that everything goes to zero on the boundary of our domain of integration. (Or use the fact that the momentum operator is Hermitian.)

So if you start with a normalised wave function it stays normalised.

The answer to your question is: Schrödinger.


I think it's a very good question. As a specific case for example for the $\psi$ of a particle, we say that $\int||\psi_t(x)|| = 1$, and what does it mean? it means that we have a particle. it means it can be found in a time in some space. and how do we say that?

I think it's just a logical reasoning and it's according to what we have persevered of nature from the beginning up to now that: if we have a particle it is (must BE) in some space-time. So the probability of finding it in all space and time (universe under which we do experiment) should be equal to 1.


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