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At the introduction to quantum mechanic phase $v_p$ and group $v_g$ 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.

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:

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

At the introduction to quantum mechanics, phase $v_p$ and group $v_g$ velocities are often presented. I know how to derive $v_p$ and get the equation:

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

What I don't 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.

First, 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 the part which declares an envelope and
• $\scriptsize\cos \left(\frac{\Delta{\omega}}{2}t - \frac{\Delta k}{2} x\right)$ is the part which declares phases inside an envelope.

Then, the professor takes only the part which declares an envelope and says that the 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 always at the same distance from the $x$ axis?

The professor then goes on to derive the group velocity 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 differentiate $x$, I finally get the group velocity:

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

At the introduction to quantum mechanic phase $v_p$ and group $v_g$ 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.

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:

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

At the introduction to quantum mechanic phase $v_p$ and group $v_g$ 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.

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:

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

At the introduction to quantum mechanic phase $v_p$ and group $v_g$ 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 $s_0$:

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

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.

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:

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

At the introduction to quantum mechanic phase $v_p$ and group $v_g$ 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.

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:

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