Who is doing the normalization of wave function in the time evolution of wave function?

Question

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?

Answer 1

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)

Answer 2

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.

Answer 3

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.