Wavefunction of a free particle has carrier and envelope parts

[Kashmir]

If

$\psi(x, t)=\left(\frac{1}{2 \pi \alpha^{2}}\right)^{1 / 4} \frac{1}{\sqrt{\gamma}} e^{i p_{0}\left(x-p_{0} t / 2 m\right) / \hbar} e^{-\left(x-p_{0} t / m\right)^{2} / 4 \alpha^{2} \gamma}$where
* $\gamma=1+\frac{i t} {\tau}$( a complex number)

* $\tau=\frac{m h}{2 \beta^{2}}$McIntyre says
" ... this wave packet has a carrier wave part that is characterized by $p_{0}$ and propagates at the phase velocity $p_{0} / 2 \mathrm{~m}$, and an envelope part that is characterized by the momentum width $\beta$ (through the $\alpha$ parameter) and propagates at the group velocity $p_{0} / \mathrm{m}$. As we expected, the envelope is a Gaussian function"

I'm not able to understand how the wavefunction has a carrier and an envelope part.

Can anyone help me with this.

 

[DrClaude]

One exponential is complex, so it represents "only" a complex phase factor, dependent on position (and time). When calculating the probability density, i.e., $| \psi|^2$, it will simply cancel itself out.

The other exponential is real and forms a Gaussian. It is the envelope that defines what $| \psi|^2$ will look like as a function of position.

 

[vanhees71]

Be careful! The 2nd exp-factor is not real since $\gamma$ is complex.

 

[Kashmir]

Be careful! The 2nd exp-factor is not real since $\gamma$ is complex.

Exactly.that's why I am not understanding it.

 

[vanhees71]

Just decompose the argument of the 2nd exp-factor in real and imaginary part. It's just a bit of algebra.

 

[Kashmir]

Just decompose the argument of the 2nd exp-factor in real and imaginary part. It's just a bit of algebra.

The coefficient is also complex ie $1/\sqrt{(\gamma)}$ is complex ,both the exponentials are complex so we've three factors having real and imaginary parts and upon multiplication they'll give many more real terms. And it's confusing to see the envelope and carrier wave.

 

[anuttarasammyak]

\gamma=(1+\frac{t^2}{\tau^2})^{1/2}e^{i\eta}
where
\eta=\arctan \frac{t}{\tau}
So
\gamma^{-1/2}=(1+\frac{t^2}{\tau^2})^{-1/4}e^{-i\eta/2}and
\gamma^{-1}=\frac{1}{1+\frac{t^2}{\tau^2}}-i \frac{\frac{t}{\tau}}{1+\frac{t^2}{\tau^2}}
Try these substitutions in your formula to get
\psi = A e^B e^{iC}
where A, B and C are real.

 

[Kashmir]

\gamma=(1+\frac{t^2}{\tau^2})^{1/2}e^{i\eta}
where
\eta=\arctan \frac{t}{\tau}
So
\gamma^{-1/2}=(1+\frac{t^2}{\tau^2})^{-1/4}e^{-i\eta/2}and
\gamma^{-1}=\frac{1}{1+\frac{t^2}{\tau^2}}-i \frac{\frac{t}{\tau}}{1+\frac{t^2}{\tau^2}}
Try these substitutions in your formula to get
\psi = A e^B e^{iC}
where A, B and C are real.

Thank you. I'm working on it.

 

[Meir Achuz]

Draw a picture of a bell shaped curve. Then draw a lot of sine curves inside.
The bell curve moves with the group velocity and the sine waves inside move at the faster phase velocity,