At the introduction to quantum mechanic [phase](https://en.wikipedia.org/wiki/Phase_velocity) $v_p$ and [group](https://en.wikipedia.org/wiki/Group_velocity) $v_g$ velocities are often presented. I know how to derive $v_p$ and get equation:

$$
\scriptsize
v_p=\frac{\omega}{k}.
$$

What i dont know is how to explain a derivation of a group velocity $v_g$ to myself. Our professor did derive it, but i am having some difficulties with it.

****

1st he did a superposition of 2 waves with the same amplitude $s_0$:

$$
\scriptsize
\begin{split}
s &= s_0 \sin(\omega_1t-k_1x) + s_0 \sin(\omega_2t - k_2 x)\\
s &= s_0 \left[ \sin(\omega_1t-k_1x) + \sin(\omega_2t - k_2 x) \right]\\
s &= 2s_0 \left[ \sin\left(\frac{(\omega_1t-k_1x)+(\omega_2 t -k_2 x)}{2}\right) \cdot \cos\left(\frac{(\omega_1t-k_1x)-(\omega_2t - k_2 x)}{2}\right) \right]\\
s &= 2s_0 \left[ \sin\left(\frac{\omega_1 + \omega_2}{2} t - \frac{k_1 + k_2}{2} x\right) \cdot \cos\left(\frac{\omega_1 - \omega_2}{2} t - \frac{k_1 - k_2}{2} x\right) \right]\\
s &= 2s_0 \left[ \sin\left(\overline \omega t - \overline{k} x\right) \cdot \cos\left(\frac{\Delta \omega}{2} t - \frac{\Delta k}{2} x\right) \right]\\
\end{split}
$$

****

Here $\overline \omega$ is larger than $\Delta \omega$ and this is why:

 - $\scriptsize\sin \left(\overline{\omega}t - \overline k x\right)$ is a part which declares an *envelope* and
 - $\scriptsize\cos \left(\frac{\Delta{\omega}}{2}t - \frac{\Delta k}{2} x\right)$ is a part which declares *phases inside an envelope*.

![enter image description here][1]

****

Than professor takes only a part which declares an envelopa and says that phase of this part must be **constant** like this:

$$
\scriptsize\frac{\Delta{\omega}}{2}t - \frac{\Delta k}{2} x = const.
$$


**QUESTION:**
What does this mean? Does a constant phase mean to only look at one point which is allways at the same distance from $x$ axis? Please someone explain this a bit.

Well then he derives the group velocity easily from now on like this:

$$
\scriptsize
\begin{split}
\frac{\Delta{\omega}}{2}t - \frac{\Delta k}{2} x &= const.\\
\frac{\Delta k}{2} x &= \frac{\Delta{\omega}}{2}t -  const.\\
x &= \frac{\Delta{\omega}}{ \Delta k} t -  \frac{2}{\Delta k}const.\\
\end{split}
$$

If i partially diferentiate $x$ i finally get group velocity:

$$
\scriptsize
\begin{split}
v_g &= \frac{\partial x}{\partial t} \\
v_g&= \frac{\Delta{\omega}}{ \Delta k}\\
v_g&= \frac{\textrm d{\omega}}{ \textrm d k}
\end{split}
$$

  [1]: https://i.sstatic.net/GgLq5.png