# Question about Gaussian wave package

## Homework Statement

At t=0 the particle is localized around $$x=x_0$$ that is: $$\langle x\rangle (0)=x_0$$, and the impulse is: $$\langle p\rangle (0)=\hbar k_0$$, we can describe the particle with the wave function:

$$\psi(x,0)=Ae^{-\frac{(x-x_0)^2}{2\sigma^2}+ik_0(x-x_0)}$$

Now, my question is: If I have localized particle around $$x=x_0$$ but my expected value of impulse at t=0 is zero, that means that my particle is at rest. Is then the wave package given by this wave function?:

$$\psi(x,0)=Ae^{-\frac{(x-x_0)^2}{2\sigma^2}$$ ?

dextercioby
Homework Helper
You know that the wavefunction is unique up to a complex factor of modulus =1, so

$$\psi(x,0)=Ae^{-\frac{(x-x_0)^2}{2\sigma^2}+ik_0(x-x_0)} = A e^{-\frac{(x-x_0)^2}{2\sigma^2}} e^{ik_0(x-x_0)} = A' e^{-\frac{(x-x_0)^2}{2\sigma^2}}$$

which is almost what you wrote.

Yeah, that phase vanishes when I look the probability, but where do I put the information about zero impulse expected value? That's what's confusing me.

I know that expected values are given like:

$$\langle x\rangle=\int_{-\infty}^\infty\psi^*x\psi dx$$ and the same for the p, but still I cannot see how that explicitly the expected value of impulse at t=0 determines that gaussian wave package (it gives him speed obviously)...

vela
Staff Emeritus
Homework Helper
Not really sure what you're asking. If all you have is <x>=x0 and <p>=p0, all you know is that

$$\langle x\rangle=\int_{-\infty}^\infty\psi^*x \psi \,dx = x_0$$

and

$$\langle p\rangle=\int_{-\infty}^\infty\psi^*\frac{\hbar}{i}\frac{d}{dx}\psi \,dx = p_0$$

The wave function

$$\psi(x,0)=Ae^{-\frac{(x-x_0)^2}{2\sigma^2}+ik_0(x-x_0)}$$

is one possible wave function that satisfies those conditions (where $p_0 = \hbar k_0$).

Well my actual question, which is kinda answered, and I found something in the books and so, is if I have
$$\langle x\rangle_0= x_0$$ and $$\langle p\rangle_0= 0$$ (at t=0), is my wave package described by:
$$\psi(x,0)=Ae^{-\frac{(x-x_0)^2}{2\sigma^2}}$$?

And I'm supposing that $$\sigma$$ is $$\Delta x$$ , right?

If I have two dimensions, the Gaussian is just product of one in x and one in y direction, right?

vela
Staff Emeritus
Homework Helper
$$\psi(x,0)=Ae^{-\frac{(x-x_0)^2}{2\sigma^2}+ik_0(x-x_0)}$$