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Need help with commute problem with operators

  1. Jan 18, 2014 #1
    Hi All,
    I just found this site and this is my first post here. I am working on getting my masters in polymer chemistry and started taking a class this semester which is pretty much all calculus and linear algebra and I just have a hard time with these subjects. I got a homework problem that I know how to do in theory, but in practice I just can't get it. The problem is

    1. [[∂][/2]/[∂x][2], [e][/-ax]-(1/[x][/n]]



    2. [AB]= 0 it commutes



    3. I know if (AB)ψ - (BA)ψ = 0 it commutes but I am having trouble mainly with the integration of the second term. I believe the first term goes to [ψ*[∂][/∂x]ψ-[∂][/∂x]ψψ*] + Cψψ* but I don't know where to go from there.

    Any help is appreciated. Thanks all!
     
  2. jcsd
  3. Jan 18, 2014 #2

    vela

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  4. Jan 18, 2014 #3
    Hi, sorry about that. I tried to follow the directions when I was typing but must have screwed up. Let me try again, the problem is

    [∂^2/∂x^2, e^-ax - (1/x^n)] I hope that makes more sense.
     
    Last edited: Jan 18, 2014
  5. Jan 18, 2014 #4
    [tex]
    ∂^2/∂x^2,

    ~e^-ax -1/x^n

    [/tex]
     
  6. Jan 18, 2014 #5
    alright, i'm getting better, the only thing wrong above there is that the ax is superscript along with the negative, but those are the two terms that I need to figure out if they commute. From the work I did they did not.
     
  7. Jan 18, 2014 #6

    strangerep

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    Chemguy251,

    Here is what I think you're asking, in corrected latex.
    $$
    \left[ \frac{\partial^2}{\partial x^2} ~,~ e^{-ax} - \frac{1}{x^n} \right] ~=~ ?
    $$(To see my original latex, click on the equation and you should get menu option to display the latex.)

    If the above is indeed your question, then... can you express the following
    $$
    [A^2, B + C]
    $$ in terms of simpler commutators? My A,B,C here are arbitrary noncommuting quantities.

    (Hint: use linearity of the commutator, and the Leibniz product rule.)

    Edit: BTW, you should be able to use the "Edit" button to add stuff to your recent posts if you wish. You don't always have to start a new post.
     
  8. Jan 19, 2014 #7
    Hi Strangerep, I have actually never done math like this before. This is a required course for my degree and there were no prerequisites for it. I took Calc and Calc II about 7 years ago but don't remember most of it so I am not sure what linearity of the commutator is... I googled the Leibniz product rule and I don't have a clue how to use that for this... The only thing I could think of doing with this would be to break it down to

    AA[B+c] - B+C[A^2] = ?
     
  9. Jan 19, 2014 #8

    strangerep

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    Have you studied any linear algebra? If not, you'll probably have to get a book like one of the Schaum Outlines and try to catch up. I would also ask the teachers what math is needed, and press them for more suggestions. There are clearly pre-requisites for this course. If they're not stated, that's an unsatisfactory situation, imho.

    But for purposes of just this exercise, you could look at linear operator, but I suspect most of the math language therein might be too advanced. For current purposes, if ##F## is a linear operator, then (e.g.,) ##F(x+y) = F(x) + F(y)##. The thing that's maybe confusing you is to understand that the commutator can be regarded as an operator with two arguments. To make this a little clearer, we could define a new symbolic operator "##Com##" via
    $$Com(A,B) ~:=~ AB - BA ~\equiv~ [A,B] ~.
    $$This is linear in both arguments, so, e.g.,
    $$Com(A,B+C) ~=~ Com(A,B) + Com(A,C) ~.$$
    It means that
    $$Com(AB,C) ~=~ A \, Com(B,C) + Com(A,C)B ~,
    $$(which can easily be proven in a couple of lines). Do you see the analogy with the Leibniz product rule for derivatives?

    Of course, nobody actually uses the symbol "##Com##". Instead, they write:
    $$[AB,C] ~=~ A[B,C] + [A,C]B ~.
    $$ Can you now apply the above to simplify your commutator down to sum of simpler commutators?
     
    Last edited: Jan 19, 2014
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