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Operators in Quantum Mechanics

  1. Sep 23, 2010 #1
    1. The problem statement, all variables and given/known data
    Find the operator for position [tex]x[/tex] if the operator for momentum p is taken to be [tex]\left(\hbar/2m\right)^{1/2}\left(A + B\right)[/tex], with [tex]\left[A,B\right] = 1[/tex] and all other commutators zero.

    2. Relevant equations
    Canonical commutation relation
    [tex]\left [ \hat{ x }, \hat{ p } \right ] = \hat{x} \hat{p} - \hat{p} \hat{x} = i \hbar[/tex]


    3. The attempt at a solution
    Using [tex]c = \left(\hbar/2m\right)^{1/2}[/tex]
    [tex]\hat{x} \hat{p} f - \hat{p} \hat{x} f = i \hbar[/tex]
    [tex]\hat{x} c \left(\hat{A} + \hat{B}\right) f - c \left(\hat{A} + \hat{B}\right) \hat{x} f = i \hbar[/tex]
    [tex]c \hat{x} \left(\hat{A} + \hat{B}\right) f - c \left(\hat{A} + \hat{B}\right) \hat{x} f = i \hbar[/tex]
    [tex]\hat{x} \hat{A} f + \hat{x} \hat{B} f - \hat{A} \hat{x} f - \hat{B} \hat{x} f = i \hbar / c[/tex]
    [tex]\hat{x} \hat{A} f - c \hat{A} \hat{x} f + \hat{x} \hat{B} f - \hat{B} \hat{x} f = i \hbar / c[/tex]
    [tex]\left [ \hat{x}, \hat{A} \right ] + \left [ \hat{x}, \hat{B} \right ] = i \hbar / c[/tex]
    "all other commutators zero"
    [tex]0 + 0 = i \hbar / c[/tex]

    :confused:

    Problem 1.2 from http://www.oup.com/uk/orc/bin/9780199274987/" [Broken].
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 23, 2010 #2
    If [A,B]=1, then what will be the commutator of A with A+B? Can you go from there?

    P.S. Your "f" on the left hand side is unnecessary.
     
  4. Sep 23, 2010 #3
    If [A, B] = 1 then [A, A + B] = 1

    [A, A + B] = A(A + B) - (A + B)A = AA + AB - AA - BA = AB - BA = [A, B] = 1

    We have p = c(A + B) and [x, p] = iħ.

    xp - px = iħ
    cx(A + B) - c(A + B)x = iħ
    x(A + B) - (A + B)x = iħ/c | multiply on A from the left
    Ax(A + B) - A(A + B)x = iħ/c A | A(A + B) = 1 + (A + B)A
    Ax(A + B) - (1 + (A + B)A)x = iħ/c A
    Ax(A + B) - x - (A + B)Ax = iħ/c A
    [Ax, A + B] - x = iħ/c A
    x = -iħ/c A

    Checking: [x, p] = xp - px = (-iħ/c A)(c(A + B)) - (c(A + B))(-iħ/c A) = -iħ A(A + B) + iħ (A + B) A = -ih

    Ouch! Like always accurate within a sign... :cry:

    Anyway, arkajad, many thanks for guiding me. I didn't have enough sleep last night. Maybe tomorrow I'll find the missing sign...
     
    Last edited: Sep 23, 2010
  5. Sep 23, 2010 #4
    Get a sleep and you will have it. But, perhaps, first prove for yourself a very useful identity: for any number a, and any operators A,B,C we have that

    [A,B+C]=[A,B]+[A,C]
    [A+B,C]=[A,C]+[B,C]
    [aA,B]=a[A,B]
    [A,aB]=a[A,B]

    To save some work use [A,B]=-[B,A] as many times as useful. Once you are done - remember these rules, as they will be handy in the future.
     
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