# Relation/meaning between momentum and contours of constant equal phase of a wave function

Sometimes, mainly due to my limited knowledge of experimental modern physics, whenever I fancy and think about quantum physics, things appear really amusing and counter intuitive, and when if I don't get them resolved, I can't help but think about it all the time and waste a lot of precious time. As a part of it is this question. Apologies if my question appears too naive.

This question is about intuition behind a certain postulate/consideration and need not necessarily correspond to any physical reality. (So I expect some intuitive answers and not just like oh, it has been experimentally proven many times, at the moment i don't care about experiments until I acquire sufficient knowledge in theoretical side.

Question :

Consider a DeBroglie matter wave (I couldn't get any formula or equation of it from googling), what I am interested in is the equation of the Debroglie wave and explanation of all the physical parameters in it and reasons why it was formulated.

Then of course how it is said that $p = hk/{2\pi}$.

The Puzzle

Consider the quantum mechanical wave $$\psi(x,t) = e^{i(kx-\omega t)}$$(although not a valid one, I have considered it for simplicity)

Now consider the phase of $\psi(x,t)$ given as $\phi(x,t) = kx-\omega t$ and let $(x_{\theta},t_{\theta})$ be the set of points of constant phase, that is $\phi(x_{\theta},t_{\theta}) = \theta$. Observe that in a any local region, $d{x_{\theta}}/d{t_{\theta}} = \omega /k$. This apparently seems puzzling considering the way the DeBroglie momentum is proportional to $k$. Wonder is there any deeper meaning or intuition behind this? Appreciate your comments and suggestions.

## 1 Answer

The conditions that your de Broglie wave (also known as the plane wave) $\psi$ obey is simply $$i\hbar \frac{\partial}{\partial t }\psi = \omega \psi$$ and $$-i\nabla\cdot \psi = \vec k \cdot \psi$$ which says that the plane wave has the right energy $E=\hbar\omega$ and the right momentum $\vec p = \hbar \vec k$, respectively. In QM, these simple equations result from the simple quantum mechanical definitions of the operator of momentum and from Schrödinger's equation that is as simple as the first equation above when there is no potential energy etc. Usually we add potential energy terms or consider otherwise complicated Hamiltonians instead of just writing a number $\omega$ on the right hand side.

Your $dx_\theta / dt_\theta = \omega / k$ was calculated according to the recipe to calculate the so-called "phase velocity". Your resulbecause $\omega / k = E/p = 1/v$ in special relativity, assuming we include the total energy including $E_0=m_0c^2$ into $E$. So it's not $v$.

You may also choose the convention for $E$ that is just $p^2/2m$ for a small $p$, the non-relativistic formula subtracting $mc^2$. In that case, the phase velocity won't correspond to the actual speed of the particle, either, although it will be closer: the phase velocity will be $mv^2/2/mv = v/2$. But the "group velocity" (also the speed of small enough wave packets) given by $\partial \omega / \partial k$ will still coincide with the expected speed of the particle $v$ because the extra factor of two comes from differentiating the second power.

There's nothing wrong for the speed to be given by $\omega / k$ even though $k$ is directly and not indirectly proportional to the momentum. The different powers of $k$ are compensated by the fact that there's the extra $\omega$ in the formula which also depends on $v$. The non-relativistic result which gave us the phase velocity almost equal to the group velocity had $\omega=mv^2/2\hbar$ so aside from the factor of $1/2$, it was proportional to the second power of $v$, exactly enough to change $1/v$ to $v$.