1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Evaluating [C, D] commutator

  1. Oct 15, 2014 #1
    1. The problem statement, all variables and given/known data
    I am practicing problems from the textbook, but have no idea how to get to some of the solutions available in the back of the textbook....

    6-16. Evaluate the commutator [C,D] where C and D are given below:

    (e) C = d2/dx2, D = x
    (g) C = integral (x = 0 to infinite) dx, D = d/dx

    2. Relevant equations
    [C,D] = CD - DC

    3. The attempt at a solution
    e)
    CDf(x) = C(xf(x)) = d2xf(x)/dx2 = 0 + xd2f(x)/dx2
    DCf(x) = D(d2f(x)/dx2) =xd2f(x)/dx2

    [C,D] = xd2f(x)/dx2 - xd2f(x)/dx2 is what I got, but the solution, according to the textbook is:

    [C,D] = 2df(x)/dx...how do I get to here?

    g)
    CDf(x) = C(df(x)/dx) = integral (x=0 to infinite) df(x)/dx * dx = f(x) (integration from x = 0 to infinite)
    DCf(x) = D(integral f(x)) = d/dx integral f(x)*dx...which is what? I am lost here....

    Again, for this part, the solution is:
    [C,D]f(x) = -f(0)...could someone explain how this value was hinted?
     
  2. jcsd
  3. Oct 15, 2014 #2

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    (e) CDf = (xf)'' ... you made an error in the next two steps.
    (g) p1. cancel the dx's inside the integral: ##\int_a^b df = ?##
    p2. how does the definite integral result depend on x?
     
  4. Oct 15, 2014 #3

    RUber

    User Avatar
    Homework Helper

    For (e),
    ## (CD-DC)f(x)= \left[\frac{d^2}{dx^2}x - x\frac{d^2}{dx^2}\right] f(x) ##
    ## = \frac{d^2}{dx^2}xf(x) - x\frac{d^2}{dx^2}f(x) ##
    By product rule ## =\frac{d}{dx}(xf'(x)+f(x)) -xf''(x)##
     
  5. Oct 15, 2014 #4
    (e) d2xf(x)/dx2 =d/dx(d/dx xf(x)) = d/dx (f(x) + xdf(x)/dx) = df(x)/dx +df(x)/dx + xd2f(x)/dx2
    ...so:
    [C,D] = df(x)/dx +df(x)/dx + xd2f(x)/dx2 - xd2f(x)/dx2 = 2df(x)/dx---got it, thanks!
     
  6. Oct 15, 2014 #5
    By the way, for (g),

    As for (g)...so I guess I get integral (x = a to b) df = f (x= a to b) = b-a?

    Also, by definite integral....could you clarify as to what does "definite integral" mean? An integral without any infinity?
     
  7. Oct 15, 2014 #6

    RUber

    User Avatar
    Homework Helper

    For (g), where you have ##\frac{d}{dx} \int_0^\infty f(x) dx ##, you can think of a definite integral as a constant. Every f has an associated ##k_f=\int_0^\infty f(x) dx##. The fact that you are evaluating the interval at fixed points makes it definite. If the limit of integration were a function of x, you would have a different derivative of the integral.
     
  8. Oct 15, 2014 #7

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Well done - note: Leibnitx notation is somewhat eyewatering in text, it is usually clearer to write, say,
    CDf=(xf)'' = (f+xf')' = f' + (xf')' = 2f' + xf''.

    ... but this is where LaTeX comes into it's own.
    $$CDF = \frac{d^2}{dx^2}xf = \frac{d}{dx}\left(f + x\frac{df}{dx}\right)$$
     
  9. Oct 15, 2014 #8
    Okay! I have yet to use LaTex...which I have not yet figured out yet.

    Aside back to (g):

    is ABf(x) = f(x) - f(0) , for as you have so said, "integral df(x)/dx * dx" cancels out dx, making integral df(x)...which I have trouble thinking about.
    Also, for BAf(x)...d/dx integral f(x)*dx, dx is not cancelled out, so, I guess it would become something like f2(x)/2 over x=0 and x?

    I am sorry, but I am really confused when there are no equations within the integral, which I am used to.
     
  10. Oct 15, 2014 #9

    RUber

    User Avatar
    Homework Helper

    ##\int_0 ^\infty \frac {d}{dx} f(x) dx = \int_0 ^\infty f'(x) dx = \left. f(x) \right|_0^\infty ##
    Also you could look at
    ## \frac {d}{dx} \int_0 ^\infty f(x) dx = \lim_{x \to \infty } \frac {d}{dx} \int_0 ^x f(x) dx##
     
  11. Oct 15, 2014 #10

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Right ... but in your case the limits in the integral are for x, not f ... so how do you change variables in an integral?

    Of course you could treat the integral as an antiderivative ....
     
  12. Oct 15, 2014 #11

    RUber

    User Avatar
    Homework Helper

    What do you mean by?
    There are general rules for integrals.
    For example:
    ## \int_a^b f'(x) dx = f(a)-f(b) ##
    ##\frac{d}{dx} \int_a^x f(x) dx = f(x) ##
    You are not looking to evaluate f, you just need to know how the derivative and integral act upon eachother.
     
  13. Oct 15, 2014 #12

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    You don't have to memorize more rules - you can just follow the definitions...

    if ##g(x)=f'(x)##, then ##\int g(x)=f(x)+c## is the indefinite integral;
    then ##\int_a^b g(x)\;dx = f(b)-f(a)## is the definite integral.

    putting b=x in the definite integral would give f(x)-f(a) so I can do:
    ##\frac{d}{dx}\int_a^x g(x')\;dx' = \frac{d}{dx}\big(f(x)-f(a)\big) = f'(x) = g(x)##​

    Note: ##\int_a^x g(x)\;dx## is sometimes considered an abuse of notation.

    If you just treat the definite integral as a constant, you'd end up with $$\int_0^\infty \frac{d}{dx}f(x)\;dx - \frac{d}{dx}\int_0^\infty f(x)\;dx = [f(\infty)-f(0)] - 0$$... so unless you know that ##\lim_{x\to\infty}f(x)=## some specific value, you should consider writing the integral this way:
    $$\int_0^\infty f(x)\;dx = \lim_{x\to\infty}\int_0^xf(x')\;dx'$$ ... the limit notation is sort-of implied by the integration notation.

    You should already have done a lot of work on limits.
     
  14. Oct 15, 2014 #13
    GOTCHA--I finally remember--thank you very much. I figured out everything!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Evaluating [C, D] commutator
Loading...