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Math Help - (Quantum) linear operators

  1. #1
    Junior Member froodles01's Avatar
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    (Quantum) linear operators

    Just starting third level Uni. stuff & am faced with linear operators from Quantum Mechanics & need a little help.
    OK, an operator, , is said to be linear if it satisfies the equation
    (α f1 + β f2) = α( f1) + β( f2)
    Fine

    but I have an equation I can't wrap my head around, maybe just rusty, a hint would be nice, though.

    1 = d/dx;
    2 =3 d/dx +3x^2;

     
    Find the new functions obtained by acting with each of these operators on
    (a) g(x, t) =3 d/dx +3x^2
    (b) h(x, t)=α sin(kx − ωt).

    Now
    1 g(x,t) = 6xt^3
    But not sure about how to get
    2 g(x,t) =

    how to get this middle bit, please . . . . .

    Answer is 18xt^3 + 9x^4 t^3
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  2. #2
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    Re: (Quantum) linear operators

    it appears you have not given a proper listing for g(x,t), as you just have the definiton of \hat{O}_2 repeated there.

    it is also unclear how "x" is supposed to operate on a function g(x,t).
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  3. #3
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    Re: (Quantum) linear operators

    Quote Originally Posted by froodles01 View Post
    Just starting third level Uni. stuff & am faced with linear operators from Quantum Mechanics & need a little help.
    OK, an operator, , is said to be linear if it satisfies the equation
    (α f1 + β f2) = α( f1) + β( f2)
    Fine

    but I have an equation I can't wrap my head around, maybe just rusty, a hint would be nice, though.

    1 = d/dx;

    Okay, that is "differentiate the function".

    2 =3 d/dx +3x^2;
    This says "differentiate the function, then add x^2 and multiply by 3.

     
    Find the new functions obtained by acting with each of these operators on
    (a) g(x, t) =3 d/dx +3x^2
    This is not a function. With that "d/dx" in it, it is another operator. Further, there is no "t" in it. In fact, this is the same as your definition of 2. I suspect you have copied the wrong thing.

    (b) h(x, t)=α sin(kx − ωt).

    Now
    1 g(x,t) = 6xt^3[/quote\
    This would be correct if g(x,t) were just 3x^2t^3 or that plus a constant.

    But not sure about how to get
    2 g(x,t) =
    Take your previous answer, which was apparently 6xt^3, add x^2, and multiply by 3: 3(6xt^3+ x^2)= 18xt^3+ 3x^2.


    how to get this middle bit, please . . . . .

    Answer is 18xt^3 + 9x^4 t^3
    I think you need to go back, read the problem again and see if you have not copied the first function incorrectly.
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  4. #4
    Junior Member froodles01's Avatar
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    Re: (Quantum) linear operators

    Hmmm. I have scanned the relevant bit in & attached it here.
    Aha! Yes, the devil's in the detail, of course & that's quite a big detail.
    Thank you for the heads up!

    (Quantum) linear operators-operator.jpegClick image for larger version. 

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    Last edited by froodles01; January 30th 2013 at 08:59 AM.
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