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Math Help - Qm 1

  1. #1
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    Qm 1

    Alright, got lots and lots of Quantum questions. A large variety.

    Note everything in bold = operator

    1.) These are all for a simple harmonic oscillator...

    a.) Prove the expression of \bold{a^\dagger} from \bold{a} in the equation below (1.1) using <\phi|\bold{A^\dagger}|\psi> = <\psi|\bold{A}|\phi>^*

    (1.1) \bold{a} = \sqrt{m\omega/2\hbar}\bold{x} + i\bold{p}/\sqrt{2m\hbar\omega}

    (1.1 continued) \bold{a^\dagger} = \sqrt{m\omega/2\hbar}\bold{x} - i\bold{p}/\sqrt{2m\hbar\omega}

    b.) Write \bold{H} in terms of \bold{x} and \bold{p} and prove the expression of H in terms of \bold{a} and \bold{a^\dagger}.

    c.) Find:

    [\bold{a^\dagger},\bold{a}] = ?
    [\bold{H},\bold{a^\dagger}] = ?
    [\bold{H},\bold{a}] = ?

    d.) If |U_n> is the eigenstate of \bold{H} w/ eigenvalue E_n, ie., \bold{H}|U_n> = E_n|U_n>, prove |U_m> = \bold{a^\dagger} | U_n> is also an eigenstate of H w/ a different eigenvalue. Determine the eigenvalue that corresponds to |U_m>.

    ------------------------------------------------

    We know \bold{H}|U_n = E_n|U_n> is the time indept. Schrod. equation, where |U_n> is the eigenstate of \bold{H}.

    We'll denote \bold{a}|U_n> to be U_m.

    I will prove that \bold{H}|\bold{a}|U_N> = (E_n - \hbar\omega)\bold{a}|U_n>. In order to do so, I conveniently have to know what [\bold{H},\bold{a}] = ? (this is a question in part c...).

    So my work:

    [\bold{H},\bold{a}] = [\hbar\omega\bold{a^\dagger}\bold{a} + \frac{1}{2}\hbar\omega, \bold{a}]

    We can use [\bold{A},\bold{C}] + [\bold{B},\bold{C}] = [\bold{A} + \bold{B}, \bold{C}]

    = [\hbar\omega\bold{a^\dagger},\bold{a}] + [\mbox{this ends up being 0 since we have a constant}]

    = \hbar\omega[\bold{a^\dagger}\bold{a},\bold{a}]

    = \hbar\omega\left(\bold{a^\dagger}[\bold{a},\bold{a}] + [\bold{a^\dagger},\bold{a}]\bold{a}\right)

    We know [\bold{a},\bold{a}] = 0, so we're left with [\bold{a^\dagger},\bold{a}]. Apparently this is equal to -1 but I'm not sure how, and I have to show why..

    So any way, this would be equal to -\hbar\omega\bold{a}.

    So we know \bold{H}\bold{a} - \bold{a}\bold{H} = -\hbar\omega\bold{a}

    We have \bold{a}\bold{H} - \hbar\omega\bold{a})|U_n>

    = \bold{a}\bold{H}|U_n> - \hbar\omega\bold{a}|U_n>

    = E_n\bold{a}|U_n> - \hbar\omega\bold{a}|U_n>

    = (E_n - \hbar\omega)\bold{a}|U_n>

    = H\bold{a}|U_n>

    And so we've shown \bold{a}|U_n> is also an eigenstate.

    So \bold{H}|U_{n-1}> = E_{n-1}|U_{n-1}.

    So..

    |U_n> \to E_n

    \bold{a}|U_n> \to E_n - \hbar\omega

    I believe \bold{a} is referred to as a step-down operator.

    \bold{H}\bold{a}|U_{n-1}> = (E_{n-1} - \hbar\omega)\bold{a}|U_{n-1}>

    This is equivalent to: \bold{H}\bold{a}\bold{a}|U_n> = (E_n - \hbar\omega - \hbar\omega)\bold{a}\bold{a}|U_n>

    \bold{H}\bold{a^2}|U_n> = (E_n - 2\hbar\omega)\bold{a^2}|U_n>

    And so |U_{n-2} = \bold{a^2}|U_n> \to E_n - 2\hbar\omega.

    We can generalize:

    \bold{H}\bold{a^n}|U_n> = (E_n - n\hbar\omega)\bold{a^n}|U_n>

    If \bold{H}|U_n> = E_n|U_n>, we consider

    \bold{a^\dagger}|U_n>

    \bold{H}\bold{a^\dagger}|U_n> = (E_n + \hbar\omega)\bold{a^\dagger}|U_n>

    I think then I have to prove for n+1, n+2, \ldots..

    U_{n+1}> = \bold{a^\dagger}|U_n> \to E_n + \hbar\omega

    .. well this is where I'm at. Still many things unanswered.
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  2. #2
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    Quote Originally Posted by DiscreteW View Post
    1.) These are all for a simple harmonic oscillator...

    a.) Prove the expression of \bold{a^\dagger} from \bold{a} in the equation below (1.1) using <\phi|\bold{A^\dagger}|\psi> = <\psi|\bold{A}|\phi>^*

    (1.1) \bold{a} = \sqrt{m\omega/2\hbar}\bold{x} + i\bold{p}/\sqrt{2m\hbar\omega}

    (1.1 continued) \bold{a^\dagger} = \sqrt{m\omega/2\hbar}\bold{x} - i\bold{p}/\sqrt{2m\hbar\omega}
    < \phi | \bold{a} | \psi >

    = < \phi | \sqrt{m\omega/2\hbar}\bold{x} + i\bold{p}/\sqrt{2m\hbar\omega} | \psi >

    = < \phi | \sqrt{m\omega/2\hbar}\bold{x} | \psi > + < \phi | i\bold{p}/\sqrt{2m\hbar\omega} | \psi >

    = \sqrt{\frac{m\omega}{2\hbar}} < \phi | \bold{x} | \psi > + \frac{i}{\sqrt{2m\hbar\omega}} < \phi | \bold{p} | \psi >

    So
    < \phi | \bold{a} | \psi > ^* = <br />
\sqrt{\frac{m\omega}{2\hbar}} < \phi | \bold{x} | \psi >^* - \frac{i}{\sqrt{2m\hbar\omega}} < \phi | \bold{p} | \psi >^*

    etc.

    -Dan
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by DiscreteW View Post
    b.) Write \bold{H} in terms of \bold{x} and \bold{p} and prove the expression of H in terms of \bold{a} and \bold{a^\dagger}.
    This one should be in your text somewhere.
    \bold{H} = \bold{T} + \bold{U}

    \bold{H} = \frac{1}{2m} \bold{p^2} + \frac{1}{2}m \omega ^2 \bold{x^2}

    Probably the best way to continue is to redefine this in terms of unitless operators:
    \bold{X} = \sqrt{\frac{m \omega}{\hbar}}\bold{x}
    and
    \bold{P} = \frac{1}{\sqrt{m \hbar \omega}}\bold{p}

    So that
    \bold{H} = \frac{\hbar \omega}{2}(\bold{X^2} + \bold{P^2})

    We may easily derive that
    [ \bold{X}, \bold{P} ] = i

    Now set
    \bold{a} = \frac{1}{\sqrt{2}}(\bold{X} + i \bold{P})
    and
    \bold{a}^{\dagger} = \frac{1}{\sqrt{2}}(\bold{X} - i \bold{P})

    Then we have that
    \bold{X} = \frac{1}{\sqrt{2}}(\bold{a}^{\dagger} + \bold{a})

    \bold{P} = \frac{1}{\sqrt{2}}(\bold{a}^{\dagger} - \bold{a})

    And using the commutation relation we can verify that
    [ \bold{a}, \bold{a}^{\dagger} ] = 1
    so that

    \bold{H} = \hbar \omega \left ( \bold{a} \bold{a}^{\dagger} - \frac{1}{2} \right )

    -Dan
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  4. #4
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    Quote Originally Posted by DiscreteW View Post
    d.) If |U_n> is the eigenstate of \bold{H} w/ eigenvalue E_n, ie., \bold{H}|U_n> = E_n|U_n>, prove |U_m> = \bold{a^\dagger} | U_n> is also an eigenstate of H w/ a different eigenvalue. Determine the eigenvalue that corresponds to |U_m>.
    We know that
    [ \bold{N}, \bold{a}^{\dagger} ] = \bold{a}^{\dagger}
    where \bold{N} = \bold{a}^{\dagger} \bold{a}

    Since [ \bold{H}, \bold{N} ] = 0, any eigenstate of H is an eigenstate of N. So basically we are to prove that \bold{a}^{\dagger}|U_n > is an eigenstate of N.

    So:
    [ \bold{N}, \bold{a}^{\dagger} ] |U_n> = \bold{a}^{\dagger} |U_n>

    So
    \bold{N}\bold{a}^{\dagger}|U_n> = \bold{a}^{\dagger}\bold{N}|U_n> + \bold{a}^{\dagger}|U_n>

    = n\bold{a}^{\dagger}|U_n> + \bold{a}^{\dagger}|U_n>
    (where n is the eigenvalue of |U_n> with respect to the operator N.)

    = (n + 1) \bold{a}^{\dagger}|U_n>

    In summation:
    \bold{N} \left ( \bold{a}^{\dagger}|U_n> \right ) = <br />
(n + 1) \left ( \bold{a}^{\dagger}|U_n> \right )

    Thus \bold{a}^{\dagger}|U_n> is an eigenvalue of N with eigenvalue n + 1. I leave it to you to translate this eigenvalue in terms of the corresponding energy eigenvalue.

    -Dan
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