URGENT HELP NEEDED - Last Modern Physics I exam tomorrow

I've done reasonably well in the class so far. However, I have not done well since starting modern quantum mechanics. The last several problem sets have covered quantum mechanics, and I failed Problem Set 6. I've never failed anything in college. The last exam in Modern Physics I (no comprehensive final) is tomorrow. I know this thread is a bit lengthy, and I'd be willing to paypal some $$$ if someone wants to really help me out. If I'm having this much trouble in the beginning of modern quantum mechanics, I've deeply worried about Modern Physics II next semester.

Let's just focus on the integration, particularly the wave function in momentum space integrals.

Re: URGENT HELP NEEDED - Last Modern Physics I exam tomorrow

Well the reason you split them up is because is because the absolute value. Just like if
[tex] f(x) = |x| [/tex]

If you integrated that from -2 to 2 you would have to split it up into two different integrals. When x < 0 you have to integrate y = -x and when x > 0 you have to integrate y = x. so
[tex]

Re: URGENT HELP NEEDED - Last Modern Physics I exam tomorrow

Yes. Look of the graph of |x| compared to just x. It might help you visualize it. Also graph [tex] e^{-x} [/tex] and [tex] e^{-|x|} [/tex] and see the difference. Especially in integrating the function.

Re: URGENT HELP NEEDED - Last Modern Physics I exam tomorrow

What's your major/degree?

I shouldn't have changed the integration limits. I did that on a few ones. lol.

Yeah, I saw that in the book. If you integrate an odd function over symmetric limits, you get a zero value. I assume the oddness of the function is based on the exponent of x. x^1 and x^17 are odd, but x^2 is even.

So that immediately we notice that this function is anti-symmetric when n is odd. In such case, for every point +x there is a point of equal and opposite value at -x. Thus the integral over all space = 0

For n=2 this function is symmetric and so you can evaluate the integral as twice the integral from [itex]x= 0[/itex] to [itex]x = +\infty[/itex], which is a well-defined standard integral. I can't decipher how the TA solution got there, but I get the same answer.

Re: URGENT HELP NEEDED - Last Modern Physics I exam tomorrow

That's correct. Odd functions over even intervals (- some value to + some value) will give 0. Try integrating x on [-1,1] -- it's the same concept. That is also really nifty for evaluating higher dimension integrals.

The idea will come in handy when dealing with different quantum states.

Re: URGENT HELP NEEDED - Last Modern Physics I exam tomorrow

Wow. I never would have figured out what square to complete! The second term is simply a constant. Why is the first term integral equal to one? I'm not familiar with the trick of the Fourier transform of a Gaussian. It might be in m QM book, but I'll have to look.

Re: URGENT HELP NEEDED - Last Modern Physics I exam tomorrow

You can observe the why the first integral is equal to one by a transform of variables, from x to say y
[tex]y = x - \frac{ip}{\hbar \alpha}[/tex]
[tex]\rm{d}y = \rm{d}x[/tex]

So it the solution to the integral is just a standard integral with the argument that the complex term does not change your limits of integration.

Gaussian distributions are very important in quantum mechanics, maths and optics because they are Fourier transforms of themselves (in other words, your answer should certainly have the following form)

[tex]\psi(p) = A\exp\left[-\beta p^2\right][/tex]

where [itex]A[/itex] and [itex]\beta[/itex] are some constants.

Re: URGENT HELP NEEDED - Last Modern Physics I exam tomorrow

Sorry, bad communication on my part. The "trick" is to re-write the equation by "completing the square" in the integrand exponential; just so that you get something you can integrate easily.