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While finding the wave function of a free particle say $\psi(x)$, my textbook says that this is nonphysical because the wave velocity is not same as particle velocity of the free-particle. I don't understand why this is a reasonable objection? Since we have established beforehand said that $\psi(x)$ is not a physical wave, then why are we bothered to have particle and wave to stick together when they move. One possible explanation I came up with is that after some time, separation between wave and particle will be so great that the particle will be at some zero probability zone of wave (because wave is moving slowly), but then I feel that why the wave function will have such a shape to begin with, as it can have very large spread so that no matter how far particle goes, it remain still in its domain?

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It is true that the wave function $\psi(x)$ is already not an observable, hence the statement about the group velocity is not needed to "falsify" the possible claims for $\psi(x)$ to be physical. However, given a state $|\psi\rangle \in \mathcal{H}$ there might still be a temptation to identify its wave function with a possible particle behaviour in terms of expectation values: this is a tricky subject and must be handled with care; as an example comes the group velocity issue. Notice that this applies as well in classical wave theory, no need to move to quantum mechanics.

Let us assume that a measurement of the position of the particle in the state $|\psi\rangle$ were possible (let us leave behind all the issues related to the position operator and its unfair properties): a first measure gives $x_1$, a second measure gives $x_2$ and so on and so forth. After infinite measures you will end up with a distribution of positions with expectation value $\langle x\rangle$. Given the classical interpretation, this would represent somehow where the particle could most likely have been during our infinite measurement procedure. Taking derivatives with respect to the time one obtains the mean velocity $\frac{d\langle x\rangle}{dt}$.

However, this has very little to do with the actual behaviour of the wave function itself. Given a state $|\psi\rangle$ and its wave function expressed in Fourier transform $$ \langle x|\psi\rangle = \psi(x) = \int \textrm{d}k\,\tilde{c}(k)\textrm{e}^{-i(kx-\omega t)} $$ the velocity of each single component $\tilde{c}(k)$ is defined as $v(k(\omega))=\frac{dk}{d\omega}$. This is usually referred to as phase velocity and for general functions has nothing to do with the aforementioned mean velocity as taken from the expectation value. If you still want to identify the classical particle (and the classical velocity) with the expectation values of the measures you perform, you have to conclude that they are different from each single group phase component as part (in an infinite sum) of the actual wave function itself.

separation between wave and particle will be so great

There is no "separation" between particles and waves. Particles are particles (whatever that means) and their states are represented by an element in the Hilbert space, whose projection onto the $x$ basis gives the wave function. As such, wave functions can be interpreted as some sort of distributions of the states onto the positions, but their plane wave components do not represent single particles themselves.

I feel that why the wave function will have such a shape to begin with, as it can have very large spread

It is true that the dispersion of a wave packet may increase as time goes by, therefore even starting with a well-localised packet (somewhere in space) after some time the shape will be spread out and the probabilities (after performing infinite measures) of finding the particle anywhere in the space are comparable.

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