# Quantum Mechanics, a free particle prepared as a gaussian wavepacket

1. Apr 27, 2013

### Cogswell

1. The problem statement, all variables and given/known data
The problem is given in the attached image. I'm currently trying to work out question one.

2. Relevant equations

$$\phi (k) = \dfrac{1}{2 \pi} \int_{- \infty}^{ \infty} \Psi (x,0) e^{-ikx} dx$$

3. The attempt at a solution

Okay, so the first thing I did was to normalise it, but then I realised it was already normalised, so there's no need to do so there.

Using the above relevant equation:

$$\phi (k) = \dfrac{1}{2 \pi} \left( \dfrac{2a}{\pi} \right)^{\frac{1}{4}} \int_{- \infty}^{ \infty} e^{-ax^2} e^{ik'x} e^{-ikx} dx$$

$$\phi (k) = \dfrac{1}{2 \pi} \left( \dfrac{2a}{\pi} \right)^{\frac{1}{4}} \int_{- \infty}^{ \infty} e^{-ax^2 + ik'x -ikx} dx$$

$$\phi (k) = \dfrac{1}{2 \pi} \left( \dfrac{2a}{\pi} \right)^{\frac{1}{4}} \int_{- \infty}^{ \infty} e^{-[ax^2 + (ik - ik')x]} dx$$

Using the hint, $ax^2 + (ik - ik')x = u^2 - \dfrac{(ik - ik')^2}{4a}$ where $u = \sqrt{a} \left( x+\dfrac{(ik-ik')}{2a} \right)$
And therefore $du = \sqrt{a} dx$

So then the equation becomes:

$$\phi (k) = \dfrac{1}{2 \pi} \left( \dfrac{2a}{\pi} \right)^{\frac{1}{4}} \int_{x = - \infty}^{ x = \infty} exp - \left( u^2 - \dfrac{(ik - ik')^2}{4a} \right) dx$$

$$\phi (k) = \dfrac{1}{2 \pi} \left( \dfrac{2a}{\pi} \right)^{\frac{1}{4}} \int_{x = - \infty}^{ x = \infty} e^{-u^2} e^{\dfrac{(ik - ik')^2}{4a}} \dfrac{du}{\sqrt{a}}$$

$$\phi (k) = \dfrac{1}{2 \pi} \left( \dfrac{2a}{\pi} \right)^{\frac{1}{4}} \dfrac{1}{\sqrt{a}} e^{\dfrac{(ik - ik')^2}{4a}} \int_{x = - \infty}^{ x = \infty} e^{-u^2} du$$

Now how do I change the bounds of integration? If there were still from negative infinity to positive infinity, I'd get $\sqrt{ \pi}$ for the integral part, so is that right?

I feel like it's not from negative infinity to infinity though.

2. Apr 27, 2013

### phosgene

The bounds look correct, as your definition of u goes toward -∞ as x goes to -∞, and ∞ as x goes to ∞ (provided that $\sqrt{a}$ is taken to be positive, if it's not, the integral just becomes negative and the $\sqrt{a}$ in front of the integral will make it positive again).

Last edited: Apr 28, 2013
3. Apr 28, 2013

### Cogswell

Okay thanks.

I have done question 2 and this is what I got:

$$\Psi (x,t) = \dfrac{\dfrac{1}{2 \sqrt{2a}} \left( \dfrac{2a}{\pi} \right) ^{\frac{1}{4}} exp \left( \dfrac{(ix - \frac{2k'}{4a})^2}{4[\frac{1}{4a} + \frac{iht}{2m}]} \right)}{\sqrt{\frac{1}{4a} + \frac{iht}{2m}}}$$

where exp is the exponential function.
I asked a friend and he said he also got something this messy for the answer.

However, I run into trouble when I try to do the expectation values of $x$ and $x^2$ for question 3.
I remember my lecturer saying something about a way of solving this (if you know $<x>$ and $\sigma_x$ then you can work out $<x^2>$ without having to evaluate 2 more gaussian integrals.

So to work out $<x>$ I use:

$$<x> = \int^{\infty}_{- \infty} \Psi (x,t)^* (x) \Psi (x,t)dx$$

And of course I can extract out the constants so I get:

$$<x> = \left( \dfrac{\dfrac{1}{2 \sqrt{2a}} \left( \dfrac{2a}{\pi} \right) ^{\frac{1}{4}} }{\sqrt{\frac{1}{4a} + \frac{iht}{2m}}} \right)^2 \int^{\infty}_{- \infty} x \cdot exp \left( \dfrac{(ix - \frac{2k'}{4a})^2}{4[\frac{1}{4a} + \frac{iht}{2m}]} \right) exp \left( \dfrac{(-ix - \frac{2k'}{4a})^2}{4[\frac{1}{4a} - \frac{iht}{2m}]} \right)$$

$$<x> = \left( \dfrac{\dfrac{1}{2 \sqrt{2a}} \left( \dfrac{2a}{\pi} \right) ^{\frac{1}{4}} }{\sqrt{\frac{1}{4a} + \frac{iht}{2m}}} \right)^2 \int^{\infty}_{- \infty} x \cdot exp \left( \dfrac{(ix - \frac{2k'}{4a})^2}{4[\frac{1}{4a} + \frac{iht}{2m}]} + \dfrac{(-ix - \frac{2k'}{4a})^2}{4[\frac{1}{4a} - \frac{iht}{2m}]} \right)$$

I don't really know what to do now because I have imaginary's on the top and bottom of the fraction. It doesn't really seem like a gaussian integral.

Anyone know what I can do or if there's an easier/better way to do this (my lecturer said I can do all of question 3 on one page if I knew the shortcuts)? Doing the $<x^2>$ will be similar to this as well I presume.

4. Apr 29, 2013

### TheNutyProfesr

huh, how would I got about opening the attachment?

5. Apr 29, 2013

### Cogswell

I don't know, I'm sure I attached it before.

Here it is again... hopefully it works

#### Attached Files:

• ###### IMG_20130428_144247.jpg
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6. Apr 29, 2013

### phosgene

I think your lecturer was talking about finding $<x^2>$ with the relation σ=$\sqrt{<x^2> - <x>^2}$

Last edited: Apr 29, 2013
7. May 1, 2013

### TheNutyProfesr

Last edited: May 1, 2013
8. May 1, 2013

### Cogswell

Yea I've already solved it... I just have to transform this:

$$\left( \dfrac{\dfrac{1}{2 \sqrt{2a}} \left( \dfrac{2a}{\pi} \right) ^{\frac{1}{4}} }{\sqrt{\frac{1}{4a} + \frac{iht}{2m}}} \right)^2 \int^{\infty}_{- \infty} x \cdot exp \left( \dfrac{(ix - \frac{2k'}{4a})^2}{4[\frac{1}{4a} + \frac{iht}{2m}]} + \dfrac{(-ix - \frac{2k'}{4a})^2}{4[\frac{1}{4a} - \frac{iht}{2m}]} \right)$$

into the form $$\dfrac{1}{ \sigma \sqrt{2 \pi}} \int_{- \infty}^{\infty} e^{\frac{-(x- \mu)^2}{2 \sigma ^2}}$$

using a few algebra tricks and compare co-efficients. This is just the general equation for a normal distribution.

The function is already normalised and so $\mu$ is the expectation value of x and $\sigma_x ^2$ is the variance and I can use those 2 things to work out $<x^2>$