Expectation Value for Gaussian Wave Packet.

[atomicpedals]

Homework Statement

Calculate <x> for the Gaussian wave packet $\psi(x)=Ne^{-(x-x₀)/2k²}$

Homework Equations

$\left\langle x \right\rangle = \int dx x|\psi(x)|²$

The Attempt at a Solution

So I've been reviewing for the up-coming midterm and I've had the painful realization that I'm basically killing myself with basic issues; in this case solving the integral.

$\left\langle x \right\rangle = \int dx x|\psi(x)|²<br /> = \int dx x|Ne^{-(x-x₀)/2k²}|²<br /> =N²\int dx x|e^{-(x-x₀)/2k²}|²$

It's at this point that I know there a u substitution to be made (at least I'm pretty sure that's the best way to solve the integral), but I'm not exactly sure for which value? Is there a better way to solve this integral?

Thanks for any pointers!

 

[G01]

Let $u=x-x_o$.

That should give you two terms that both proportional to standard Gaussian integrals. Consider the symmetry of the integrand and the solution shouldn't be hard to come by.

 

[atomicpedals]

Ah, ok. Since the other expectation values are next on the block, and I can screw up any given integral is the following mathematically "legal"?

$=N²\intdxx|e^-(x-x0)/2k²|²$
$=N²\intdux|e^-u/2k²|\intdu|e^-u/2k²|$
$=N²x\intdu|e^-u/2k²|\intdu|e^-u/2k²|$

 

[atomicpedals]

Am I way off or does this come out to be <x>=x0?

 

[G01]

Am I way off or does this come out to be <x>=x0?

That's the answer.

Your Latex is not formatted correctly, so just to make sure that we are on the same page, this is the type of integral we are considering:

$$<x^n>=\int dx x^n e^{\frac{-2(x-x_o)^2}{2k^2}}$$

Correct? Quote my post to see how I coded the above equation.

 

[atomicpedals]

Yep, that's the one! (sorry about that, Latex and I have a sordid history of being hit or miss)

 

[G01]

Oops, I actually forgot a square in the numerator. Check that integral again and make sure it's still right.

 

[atomicpedals]

There should also be a negative sign, using numerical syntax:

int(x^n*exp[(-(x-x0)^2)/(k^2))]dx).

 

[G01]

There should also be a negative sign, using numerical syntax:

int(x^n*exp[(-(x-x0)^2)/(k^2))]dx).

Ooops. Yes, of course! Typo#2 corrected.


Now, do you still need advice on finding the other expectation values?

 

[atomicpedals]

The basic technique is still the same correct? <x^2>=<psi|x^2|psi> or <p>=<psi|p|psi>

 

[G01]

The basic technique is still the same correct? <x^2>=<psi|x^2|psi> or <p>=<psi|p|psi>

Yes, the basic idea is the same. Try them out. If you get stuck, feel free to come back and ask for some advice.