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Taylor Series/Groups

  1. Dec 1, 2011 #1
    1. The problem statement, all variables and given/known data

    A translation operator T(a) coverts ψ(x) to ψ(x+a),

    T(a)ψ(x) = ψ(x+a)

    In terms of the (quantum mechanical) linear momentum operator p_x = -id/dx, show that T(a) = exp(iap_x), that is, p_x is the generator of translations. Hint. Expand ψ(x+a) as a Taylor series.

    2. Relevant equations

    Groups?

    3. The attempt at a solution

    I'm lost at the hint. To expand ψ(x+a) as a Taylor series, don't I need a point around which to expand it?
     
  2. jcsd
  3. Dec 1, 2011 #2

    HallsofIvy

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    Yes- a.
     
  4. Dec 1, 2011 #3
    Ok. so would the taylor series be:

    ψ(a+a)+ψ'(a+a)(x-a)+ψ''(a+a)*(x-a)^2/2!+... ? and maybe disregard the higher order terms O(3)? Or would it be

    ψ(a)+ψ'(a)*(x)+ψ''(a)*(x)^2/2!+... ? Either way, I'm not sure how I would prove that T(a) is the given exponential function..
     
  5. Dec 1, 2011 #4

    vela

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    You want to expand about x=x, i.e., ψ(x+a) = ψ(x)+....
     
  6. Dec 1, 2011 #5
    ok.. so i expanded that, and got ψ(x)+aψ'(x)+a^2*ψ''(x)/2!+ ...

    But the LHS gives me:

    e^(i*a*px)[ψ(x)]=e^(a*dψ/dx)=e^(-iak*ψ(x)), which is clearly not the left hand side. Am I interpreting the operator px wrong? This class is not a quantum dynamics class, so I'm having difficulty figuring out what i'm missing.
     
  7. Dec 2, 2011 #6

    vela

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    You can't pull ψ(x) into the exponent like that because ψ(x) isn't an eigenfunction of the operator [itex]\hat{p}_x[/itex].

    It's not be exactly obvious what the expression [itex]e^{ia\hat{p}_x}[/itex] means. How do you exponentiate an operator? The answer is that it's defined by the Taylor series for ex:
    [tex]e^{ia\hat{p}_x} \equiv 1+ia\hat{p}_x+\frac{(ia\hat{p}_x)^2}{2!}+\cdots[/tex]What do you get when you apply the righthand side to ψ(x)?
     
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