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    Griffiths Problems 02.12

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    Ema Maharani
    This document discusses using raising and lowering operators to express position (x) and momentum (p) in terms of the harmonic oscillator energy eigenstates. It shows that the expectation values of x and p are zero, and calculates the mean squares of x and p. The uncertainty principle for the harmonic oscillator is derived. Finally, it shows that the kinetic energy is half the total energy as expected.

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    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd

    Griffiths Problems 02.12

    Uploaded by

    Ema Maharani
    34 views3 pages
    This document discusses using raising and lowering operators to express position (x) and momentum (p) in terms of the harmonic oscillator energy eigenstates. It shows that the expectation values of x and p are zero, and calculates the mean squares of x and p. The uncertainty principle for the harmonic oscillator is derived. Finally, it shows that the kinetic energy is half the total energy as expected.

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    This document discusses using raising and lowering operators to express position (x) and momentum (p) in terms of the harmonic oscillator energy eigenstates. It shows that the expectation values of x and p are zero, and calculates the mean squares of x and p. The uncertainty principle for the harmonic oscillator is derived. Finally, it shows that the kinetic energy is half the total energy as expected.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    34 views3 pages

    Griffiths Problems 02.12

    Uploaded by

    Ema Maharani
    This document discusses using raising and lowering operators to express position (x) and momentum (p) in terms of the harmonic oscillator energy eigenstates. It shows that the expectation values of x and p are zero, and calculates the mean squares of x and p. The uncertainty principle for the harmonic oscillator is derived. Finally, it shows that the kinetic energy is half the total energy as expected.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    of 3

    HARMONIC OSCILLATOR - RAISING AND LOWERING

    OPERATOR CALCULATIONS

    Link to: physicspages home page.


    To leave a comment or report an error, please use the auxiliary blog.
    Post date: 17 Jul 2012.
    Reference: Griffiths, David J. (2005), Introduction to Quantum Mechan-
    ics, 2nd Edition; Pearson Education - Problem 2.12.
    Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press.
    Section 7.4, Exercise 7.4.2.
    In the study of the harmonic oscillator, we can express x and p in terms
    of the raising and lowering operators:
    r
    h̄
    x = (a+ + a− ) (1)
    2mω
    r
    h̄mω
    p = i (a+ − a− ) (2)
    2
    We now have
    r
    h̄
    Z
    hxi = ψn∗ (a+ + a− )ψn dx (3)
    2mω
    = 0 (4)

    The reason this is zero is that, as we saw when working out the normaliza-
    tion of the stationary states,

    √
    a+ ψn = n + 1ψn+1 (5)
    √
    a− ψn = nψn−1 (6)

    a+ a− ψn = nψn (7)
    a− a+ ψn = (n + 1) ψn (8)

    and since the wave functions are orthogonal, we get


    Z Z
    ψn∗ ψn+1 dx = ψn∗ ψn−1 dx = 0 (9)
    1
    HARMONIC OSCILLATOR - RAISING AND LOWERING OPERATOR CALCULATIONS 2

    Similarly:
    r
    h̄mω
    Z
    hpi = i ψn∗ (a+ − a− )ψn dx (10)
    2
    = 0 (11)

    for the same reason.


    For the mean squares:
     Z
    h̄
    2
    hx i = ψn∗ (a+ + a− )(a+ + a− )ψn dx (12)
    2mω
     Z
    h̄
    = ψn∗ (a+ a− + a− a+ )ψn dx (13)
    2mω
     
    h̄
    = (2n + 1) (14)
    2mω
     
    h̄ 1
    = n+ (15)
    mω 2
    In going from the first to the second line, we’ve thrown out terms where we
    integrate two orthogonal functions. For example,
    Z Z
    ψn∗ a+ a+ ψn dx ψn∗
    p
    = (n + 1) (n + 2)ψn+2 dx (16)
    = 0 (17)
    We have used the relations above and the fact that ψn is normalized to
    get the third line.
    Similarly:

    h̄mω
    Z
    2
    hp i = − ψn∗ (−a+ a− − a− a+ )ψn dx (18)
    2 
    1
    = h̄mω n + (19)
    2
    The uncertainty principle then becomes
    q
    σp σx = hx2 i hp2 i (20)
     
    1
    = h̄ n + (21)
    2
    HARMONIC OSCILLATOR - RAISING AND LOWERING OPERATOR CALCULATIONS 3

    and the kinetic energy is

    hp2 i 1
     
    1
    hT i = = h̄ω n + (22)
    2m 2 2
    which is half the total energy, as it should be.
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