Find Wave Function given <x>, sigmax, <p>

[friedymeister]

Homework Statement

Come up with a wave function Psi[x] that satisfies the given known values:
<x>=-1
sigma x = 1
<p> = h bar

Homework Equations

The Attempt at a Solution

So far I have this equation, which satisfies <x>, <p>, but not sigma x.
1/[Pi]^(1/4) E^(i (x + 2)) E^(-(1/2) (x + 1)^2)

 

[bdforbes]

What does sigma x refer to?

 

[friedymeister]

Sigma x is the standard deviation of x.

 

[genneth]

Remember that the standard deviation of a gaussian is controlled by dividing the x^2 term by sigma^2.

 

[yaychemistry]

Try looking at the below wikipedia article on Gaussian or Normal distributions for a good way to create a wavepacket with a given uncertainty ($$\sigma_x$$) and position $$\langle x\rangle$$.

This wikipedia link should give an adequate definition:
http://en.wikipedia.org/wiki/Normal_distribution

 

[friedymeister]

For some reason, I just can't find the function.

I don't know how I could make a wave function that could possibly have a standard distribution of 1 and an expected value of -1, graphically.

 

[yaychemistry]

A Gaussian wavefunction has the form
$$A\exp\left(-\frac{\left(x - x_0\right)^2}{2\sigma^2}\right)$$
where A is the normalization, $$x_0$$ is the center of the Gaussian, and $$\sigma$$ is the standard deviation.

Try calculating
$$\sigma = \sqrt{\langle x^2\rangle - \langle x\rangle^2}$$ to convince yourself of this property of Gaussians and then see if you can put the momentum part into the total wavefunction.

 

[friedymeister]

I have tried this, the only problem is that the function must also have the momentum, -h bar, which requires another exponential term, exp(-i x).

Also, sigma^2 would be 1, and that would have the form A * exp^(-i x)*exp^((-(x+1)^2)/2) where A = 1/ pi^(1/4)

Because of the first exponential function, exp^(-i x), the standard deviation becomes 1/rad(2).

How can I get around this?

 

[yaychemistry]

I'm unsure how you're getting the standard deviation to be $$1/\sqrt{2}$$ because of the complex exponential part. I'm assuming rad(2) is the square root.

Don't forget that to calculate the expectation value you use the complex conjugate on the left, e.g.
$$\langle x \rangle = \int \psi^{*}\left(x\right)x\psi\left(x\right)\,dx$$.
So the complex exponential times its complex conjugate should just give 1 inside the integral and not affect the expectation value of $$x$$ and $$x^2$$.

One more thing to consider, when you multiply two Gaussians together you get another Gaussian:

$$A\exp\left(-\frac{\left(x-x_0\right)^2}{2\sigma^2}\right) * A\exp\left(-\frac{\left(x-x_0\right^2}{2\sigma^2}\right) = A^2\exp\left(-\frac{\left(x - x_0\right)^2}{\sigma^2}\right)$$
This about this relates the standard deviation of $$\psi\left(x\right)$$ to $$\psi^{*}\left(x\right)\psi\left(x\right)$$.

 

[friedymeister]

Yeah, that's how I've been getting <x>.

It must the be exp^(i x) that's causing the problems. The x in the exponent must be messing something up.

How can I get around this?

 

[yaychemistry]

I'm pretty sure the complex exponential should cancel in the calculation of the x and x^2 expectation values.

Remember what I mentioned about the product of Gaussians. This means that while the standard deviation of $$\psi\left(x\right)$$ might be $$\sigma$$ the standard deviation of $$|\psi\left(x\right)|^2$$ will be $$\sigma/\sqrt{2}$$...

 

[friedymeister]

Well, sigma x comes out to 1/rad(2) for the sqrt((integral from -inf to +inf of Psi* x^2 Psi) - (integral from -inf to +inf of Psi* x Psi)^2).

How would I change the wave function to make sigma x = 1 then?

 

[yaychemistry]

Try changing the value of $$sigma$$ in the Gaussian part of the wavefunction and see how that changes what you calculate $$\sigma_x$$. See if can figure it out own your own exactly what you need to change then.
Also, remember the correct normalization of a Gaussian involves $$\sigma$$ as well. It should be in the wikipedia article.

 

[friedymeister]

Oh that's awesome. Thanks, I changed 1/2 to 1/4 and then renormalized the wave function, and everything works perfectly.

Thanks a lot.

 

[yaychemistry]

Glad I could help :)