Movement of a wave packet of a free particle

Jillds

In my course there's a chapter with the mathematical explanation to find the real expression and localisation of a free particle with the superposited wave function. The same is used to explain the movemement of a wave packet (which is a free particle). I've worked out almost all the math behind every step, but the last.

$\psi (x,t) = e^{ik_{0}x-iw(k_{0})t} \Big( \frac{\pi}{\alpha+i\beta t} \Big)^{\frac{1}{2}} e^{-\big[\frac{(x-v_g t)^2}{4(\alpha +i\beta t)}\big]}$

Next we take the square of the absolute of the wave function. I know the first exponent would equal to 1 as a square, as it is phasefactor. And if I calculate the rest as a square I have.

$|\psi (x,t)|^2 = \frac{\pi}{\alpha+i\beta t} e^{-\big[\frac{(x-v_g t)^2}{2(\alpha +i\beta t)}\big]}$

My course, however, has a different result, and I haven't got a clue what my professor did to get that result:

$|\psi (x,t)|^2 = \Big( \frac{\pi^2}{\alpha^2+\beta^2 t^2} \Big)^{\frac{1}{2}} e^{-\alpha \big[\frac{(x-v_g t)^2}{2(\alpha^2 +\beta^2 t^2)}\big]}$

How does one get $\alpha^2 +\beta^2 t^2$ ?

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jtbell

Mentor
When you take the complex conjugate of $\psi$ (in order to find $|\psi|^2 = \psi^*\psi$), you have to change i to -i everywhere.

• Jillds

Jillds

Thank you,

so, if I understand you correctly I have to do

$\big( \frac{\pi}{\alpha + i\beta t} \cdot \frac{\pi}{alpha - i\beta t}\big)^{\frac{1}{2}} = \big( \frac{\pi^2}{\alpha^2 - i^2\beta^2 t^2} \big)^{\frac{1}{2}} = \big( \frac{\pi^2}{\alpha^2 + \beta^2 t^2} \big)^{\frac{1}{2}}$

But how does that help with the power of the exponent?

Jillds

Ok, I worked it out for the exponent as follows

$e^{-[\frac{(x-v_g t)^2}{4(\alpha + i\beta t)}]} \cdot e^{-[\frac{(x-v_g t)^2}{4(\alpha - i\beta t)}]}$
By then adding the powers of the exponents and working out the numenators of the fractions I get the desired result.

Thanks!

• vanhees71

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