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Gaussian Wavepacket Momentum Squared

  1. Mar 23, 2012 #1
    1. The problem statement, all variables and given/known data

    I will not post a specific problem, but rather, I would like to ask a general question. Say I am given a Gaussian wavepacket (function psi(x,t) ) and asked to find the expectation value for x-squared and momentum squared. Now, x-squared is rather straightforward integration since the operator for x-squared does not actually change (act on) the wave wave function. However, when it comes to momentum squared, it can quickly get nasty and easy to make mistakes in calculations due the double derivative and complicated Gaussian integrals involved.

    Is there any shortcut one can take to solve such problems (finding momentum squared) in simpler manner? Maybe along the lines of V=0 ,thus, simplifying the Schroedinger Equation and expressing the p^2(psi) part?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 23, 2012 #2
    If you have something like
    [tex] \psi (p) = \frac{1}{(2 \pi \hbar)^{1/2}} \int dx \, \phi(x) e^{-ipx}, [/tex]
    it is legitimate to use
    [tex] \psi^* (p) = \frac{1}{(2 \pi \hbar)^{1/2}} \int dx \, \phi(x)^* e^{ipx}, [/tex]
    In other words, you can take the complex conjugate of an integral before integrating (at least the mathematicians tell me so) because it's just an infinite sum, so it just involves summing an infinite number of complex conjugates, which is the same as performing the original sum and conjugating the result. I'm not sure if this helps, but if you're trying to find
    [tex] \psi^* \psi = \lvert \psi \lvert^2 [/tex]
    it should.
     
    Last edited: Mar 24, 2012
  4. Mar 24, 2012 #3
    Here's one thing. For
    [tex] \psi(x,t)= \frac{1}{ \sqrt{2 \pi \hbar}} \int dp \, \phi(p)e^{i(kx- \omega t)},[/tex]
    [tex] \psi(x,t)^*= \frac{1}{ \sqrt{2 \pi \hbar}} \int dp \, \phi(p)^*e^{-i(kx- \omega t)}[/tex]
    [tex] \big< \psi \big\lvert \,p^2 \big\lvert \psi \big> = \frac{1}{2 \pi \hbar} \int \phi(p) \,p^2 \phi^*(p) dp[/tex]
    Makes life a lot easier. Now, a trick that saved my tail when I was preparing for my qualifying exam, and one that EVERYONE should know about, is this:
    [tex] \tag{1} - \bigg< \psi \bigg\lvert \,\frac{ \hbar^2}{2m}\frac{d^2}{dx^2} \bigg\lvert \psi \bigg> = \frac{ \hbar^2}{2m} \int^{ \infty}_{ -\infty} \bigg\lvert \frac{d\, \psi(x)}{dx} \bigg\lvert^2 dx [/tex]
    Or, in three dimensions,
    [tex] \tag{2} - \bigg< \psi \bigg\lvert \,\frac{ \hbar^2}{2m} \Delta \bigg\lvert \psi \bigg> = \frac{ \hbar^2}{2m} \int \bigg( \boldsymbol{ \nabla} \, \psi^*( \mathbf{x})\bigg) \cdot \bigg( \boldsymbol{ \nabla} \, \psi( \mathbf{x})\bigg) d^3 \mathbf{x} [/tex]
    Where [tex] \Delta = \boldsymbol{ \nabla} \cdot \boldsymbol{ \nabla}= \nabla^2[/tex]
    Either way, you can see that using the trick in Eq.(1) or (2) allows you to really only take one derivative, which, as you know, will save you a huge mess. Especially if you're dealing with the variational principle. You can prove either of these relations using integration by parts, but I can guarantee you they are 100% accurate (see Zettili, Quantum Mechanics, Concepts and Applications).
     
  5. Mar 24, 2012 #4
    I like the equation (1). So, essentially, I will still need to take space derivatives. However, one of them will be of the bra part. Thus, I could just quickly reverse the signs of i's on the derivative of the ket and roll with the integral, is that correct?
     
  6. Mar 24, 2012 #5
    Well, your bra should just be the Hermitian conjugate of your ket, right? I'm not sure what your level of quantum is, so I'm not really sure if I'm dumbing things down or making them complicated.
     
  7. Mar 24, 2012 #6
    Yes, it is. I meant the same thing. First semester undergrad, but we are on the same page.
     
  8. Mar 24, 2012 #7
    Ok, great. You just want to take the derivative of the ket (not the conjugate), don't sweat the bra. If you were just doing this problem straight out, you'd do
    [tex] \big< \psi \big\lvert \, \hat p^2 \big\lvert \psi \big> = \int dx \, \psi^*(x) \, \hat p^2 \, \psi(x) = -\hbar^2 \int dx \, \psi^*(x) \, \frac{d^2 \psi(x)}{dx^2} [/tex]
    But with the new trick, you just do
    [tex] \hbar^2 \int dx \, \bigg\lvert \frac{d \psi(x)}{dx}\bigg\lvert [/tex]
    That's it- that's the formula for momentum expectation. Just take [tex]\frac{d \psi(x)}{dx}[/tex] square the magnitude, integrate over x, and multiply by hbar. Don't worry about the conjugate, just psi. Keep your signs the way they are above. Why were you going to take the derivative of the conjugate, for separation of variables?
     
  9. Mar 24, 2012 #8
    Please discard my comment about taking derivative of the bra :). I realize from your last post that it will not be needed. Will the formula (1) work for any general case and not just a free particle given by a gaussian packet?
     
  10. Mar 24, 2012 #9
    Yes! That's the amazing thing about it! Take the particle in a box, or infinite square well, with
    [tex]V=0,~~0<x<a; V= \infty,~~x<0,~x>a[/tex]
    Now plug
    [tex] \psi(x)= \sin\bigg(\frac{ \pi x}{a}\bigg)[/tex]
    into both methods. With straight-forward way, you'd get
    [tex] -\hbar^2 \int^a_0 dx\, \sin\bigg(\frac{ \pi x}{a}\bigg) \bigg( -\frac{\pi^2}{a^2} \sin\bigg(\frac{ \pi x}{a}\bigg)\bigg) = \frac{ \hbar^2 \pi^2}{2a}[/tex]
    With the new way, you'd have
    [tex] \hbar^2 \int^a_0 dx\, \bigg( \frac{ \pi}{a} \cos\bigg(\frac{ \pi x}{a} \bigg) \bigg)^2 = \frac{ \hbar^2 \pi^2}{2a}[/tex]
    It's awesome, and will save you a HUGE pain in the *** later on, believe me. Not only is it convenient, but if you're not sure if you're function has a continuous first derivative, it will keep you from making mistakes. If you have a function that is discontinuous at some value of x, just use this trick and you will avoid nasty mistakes like getting a negative kinetic energy term. This is an invaluable trick for undergrad/early grad QM, and I've never met another person familiar with it.
     
  11. Mar 24, 2012 #10
    Grrrrrr .... I wish I knew this about a month ago when I had all kinds of expectations of momentum squared to calculate. :)
     
  12. Mar 24, 2012 #11
    God, do I know it. Do yourself a favor (I tell this to everybody) and buy a copy of Zettili's book. I've been in grad school for over two years, but I've had this book since the end of my undergrad. It is absolutely shredded. I still use it all the time, even though I never do non-relativistic QM anymore. It's just so good, it will show you tricks and explain things in ways that make every part of physics easier. When you get into perturbation theory and scattering, you MUST own this book. If you follow Griffiths or almost any other book, you will waste your time. With Zettili, You will learn things in half the time, and understand them better than your classmates. This book should be your undergrad quantum bible.
     
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