• dingo_d

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?:

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

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)...

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?

If you're assuming the wave function has the form

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

then the answer to your first question is yes. You just set k0 to 0.

And yes, Δx=σ, which you can easily verify by finding <x2>.

And yes, the product of single-variable Gaussians will gives you a multi-variate Gaussian, which, without knowing more about the problems, I'd guess is what you want. It's generally possible though for there to be an xy cross term which rotates the Gaussian in the xy plane.