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Quantum mechanics operator relations. One dimensional problem involving position.

  1. Jan 26, 2013 #1
    1. The problem statement:

    Show that if the operator relation

    e^(ipa/ħ)xe^(-ipa/ħ) = x+a

    holds. The operator e^A is defined to the

    e^A= Ʃ(A^n)/n!
    n=0

    [Hint: Calculate e^(ipa/ħ)xe^(-ipa/ħ)f(p) where f(p)is any function of p, and use the representation x=iħd/dp]


    2. Relevant equations:

    I am not entirely sure but I think those presented in the problem statement are sufficient.


    3. The attempt at a solution:

    I calculated e^(ipa/ħ)xe^(-ipa/ħ)f(p) by replacing x=iħd/dp to get;

    e^(ipa/ħ)iħd/dp[e^(-ipa/ħ)f(p)]= e^(ipa/ħ){iħ[(-ia/ħ)e^(-ipa/ħ)f(p)+e^(-ipa/ħ)f'(p)]}
    = e^(ipa/ħ){ae^(-ipa/ħ)f(p)+e^(-ipa/ħ)f'(p)]}
    = af(p)+f'(p).
    ∴ af(p)+f'(p) = x + a.

    From this point I become entirely confused, I don't know how everything is suppose to tie into the power series.
     
  2. jcsd
  3. Jan 26, 2013 #2

    TSny

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    Hello RedMech. From the wording of the problem, I'm not sure what the question is. Could you restate what it is you are suppose to show?
     
    Last edited: Jan 26, 2013
  4. Jan 26, 2013 #3
  5. Jan 26, 2013 #4

    TSny

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    Well, not quite. You added the word "if" that threw me a bit. But, no problem. Your link cleared it up.

    This looks good except for a missing numerical factor for the part shown in red above.
    Your result will then have the form af(p)+cf'(p) where c is the missing factor. See what happens if you then rewrite f'(p) in terms of the x operator acting on f(p).

    You don't need to use the power series. I think the problem statement threw that in just to state the meaning of e^A.
     
  6. Jan 26, 2013 #5
    So af(p)+cf'(p), would be my final solution. I pay no attention to the power series?
     
  7. Jan 26, 2013 #6
    Not sure I know what you mean by this. The phrasing of the question has thrown me off so much that I don't know what I am looking for.
     
  8. Jan 26, 2013 #7

    TSny

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    No, that's not the final solution. You are asked to show that e^(ipa/ħ)xe^(-ipa/ħ) = x+a. So far you have shown that [e^(ipa/ħ)xe^(-ipa/ħ)]f(p) = af(p)+cf'(p) for any function f(p) and where c is the factor you dropped (you'll need to go back and find the value of c). You still need to show that e^(ipa/ħ)xe^(-ipa/ħ) is equivalent to x+a. But, you are almost there. You just need to rewrite the last term of af(p)+cf'(p) in terms of the operator x.

    You will not need the power series.
     
  9. Jan 26, 2013 #8
    Okay, I'm with you now. Let me see what I can do and I'll come back with my result.
     
  10. Jan 26, 2013 #9
    C=iħ. Giving us af(p) + iħd/dpf(p).

    Leading to
    [e^(ipa/ħ)xe^(-ipa/ħ)]f(p)= [a + iħd/dp]f(p).

    ∴e^(ipa/ħ)xe^(-ipa/ħ)=a + iħd/dp as required. Does this seem right?
     
  11. Jan 26, 2013 #10

    TSny

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    Yes, I think that's right.
     
  12. Jan 26, 2013 #11
    Thank you so much. This makes sense. I see why I needed to bring in the f(p) function. My careless differentiation mistake and that power series confused me.
     
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