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Homework Help: HELP NEEDED - Last Modern Physics I exam tomorrow

  1. Apr 27, 2010 #1
    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. :frown:

    Let's just focus on the integration, particularly the wave function in momentum space integrals.
    Problem Set 6: #2, 5, 8, 16

    Problems

    G2-2.jpg

    G2-5.jpg

    G2-8.jpg

    G2-16.jpg

    My Pitiful Work

    ps6a.jpg

    ps6b.jpg

    Solutions Posted by TA

    ps6corr1.jpg

    ps6corr2.jpg

    Problem Set 7: #5, 6, 14

    Problems

    G3-56.jpg

    G3-14.jpg

    My Pitiful Work

    ps7a.jpg

    ps7b.jpg

    Solutions Posted by TA

    ps7corr.jpg
     
    Last edited: Apr 27, 2010
  2. jcsd
  3. Apr 27, 2010 #2
    Re: URGENT HELP NEEDED - Last Modern Physics I exam tomorrow

    Line 2 comes directly from the question. All she did was square both sides of the given expression and and solve for [tex] v^2 - {v_0}^2 [/tex].

    For the third line:

    [tex]v_g = -{\lambda}^2\frac{dv}{d\lambda} [/tex]

    [tex] v^2 - {v_0}^2 = \frac{c^2}{{\lambda}^2} [/tex]

    Take the derivative:

    [tex] 2v\frac{dv}{d\lambda} - 0 = \frac{-2c^2}{{\lambda}^3} \rightarrow \frac{dv}{d\lambda} = -\frac{c^2}{v{\lambda}^3} [/tex]

    Plug that into

    [tex]v_g = -{\lambda}^2\frac{dv}{d\lambda} [/tex]

    And then plug the original equation for [tex] \lambda [/tex] into it.
     
  4. Apr 27, 2010 #3
    Re: URGENT HELP NEEDED - Last Modern Physics I exam tomorrow

    Zach, thanks. I understand everything except the derivative. Why is the zero there?

    I'm a bit disappointed I couldn't even pull the first correction line out of my butt. :rolleyes:
     
  5. Apr 27, 2010 #4
    Re: URGENT HELP NEEDED - Last Modern Physics I exam tomorrow

    Because [tex] v_0 [/tex] is just a constant. I tried to figure some of the other problems out but failed miserably :grumpy:.
     
  6. Apr 27, 2010 #5
    Re: URGENT HELP NEEDED - Last Modern Physics I exam tomorrow

    Ah. Of course it's a constant.

    Do those momentum-space wave function or expectation-value integrals make sense to you?
     
  7. Apr 27, 2010 #6
    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]

    \int_{-2}^{2} |x| dx = \int_{-2}^{0} -x dx + \int_{0}^{2} x dx [/tex]


    I'm not sure if this is what you were asking though.
     
  8. Apr 27, 2010 #7
    Re: URGENT HELP NEEDED - Last Modern Physics I exam tomorrow

    Ah. Because it's the absolute value of x?
     
  9. Apr 27, 2010 #8
    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.
     
  10. Apr 27, 2010 #9
    Re: URGENT HELP NEEDED - Last Modern Physics I exam tomorrow

    Of course, they give very different function values. Dammit. I forget these important details and it completely screws me up.

    Now, what about #14 and 16? I think #5 might be a lost cause for me. lol.
     
  11. Apr 27, 2010 #10
    Re: URGENT HELP NEEDED - Last Modern Physics I exam tomorrow

    I am terrible at Quantum lol. But on 16 why did you change the limits from -inf -> inf into
    0 -> inf?

    Also do you get the odd function thing?

    I must go i'll try to solve over the night.
     
    Last edited: Apr 27, 2010
  12. Apr 27, 2010 #11
    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.
     
  13. Apr 28, 2010 #12
    Re: URGENT HELP NEEDED - Last Modern Physics I exam tomorrow

    For question 14, I think you're correct just need the trick for the Fourier transform of a Gaussian...

    So you have the Fourier transform on your wavepacket:

    [tex]\mathcal{F}\left[ \psi(x) \right]= \left( \frac{\alpha}{\pi}\right)^{\frac{1}{4}}\frac{1}{\sqrt{2\pi \hbar}} \int^\infty_{-\infty} \exp\left[-\frac{\alpha}{2}x^2 + i \frac{p}{\hbar}x \right]\,\rm{d}x[/tex]

    We can re-write the exponential term by using a complete the squares method ie..

    [tex]-\frac{\alpha}{2}x^2 + i \frac{p}{h}x = -\frac{\alpha}{2}\left(x - \frac{ip}{\hbar\alpha} \right)^2 + \frac{p^2}{\hbar^2 \alpha^2}[/tex]

    And the other crucial point for evaluating this integral is

    [tex]\sqrt{\frac{\alpha}{2\pi}} \int^\infty_{-\infty} \exp\left[ -\frac{\alpha}{2}(x - \frac{ip}{\hbar\alpha})^2\right] \,\rm{d}x = 1[/tex]

    Just noticed that my integral slightly differs from your TA solution, the trick is fine just check whether I missed a symbol or not.

    Hope this helps, and good luck :wink:
     
  14. Apr 28, 2010 #13
    Re: URGENT HELP NEEDED - Last Modern Physics I exam tomorrow

    For question 16, I started with

    [tex]\left\langle x^n \right\rangle = \int^{\infty}_{-\infty}\psi^\ast(x) x^n \psi(x) \rm{d}x[/tex]

    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.
     
  15. Apr 28, 2010 #14
    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.
     
  16. Apr 28, 2010 #15
    Re: URGENT HELP NEEDED - Last Modern Physics I exam tomorrow

    Thanks for the help, guys. I'll have to look it over when I have some free time here at work.
     
  17. Apr 28, 2010 #16
    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.
     
  18. Apr 28, 2010 #17
    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.

    [tex]\sqrt{\frac{\alpha}{2\pi}} \int^\infty_{-\infty}\exp\left[ -\frac{\alpha}{2}y^2\right] \rm{\, d}y = \sqrt{\frac{\alpha}{2\pi}} \sqrt{\frac{2\pi}{\alpha}}[/tex]

    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.
     
  19. Apr 28, 2010 #18
    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.
     
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