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Dirac Notation: Am I doing this right?

  1. Apr 12, 2012 #1
    1. The problem statement, all variables and given/known data
    Find <P>. P = i√(mhw/2)(a†-a). Note a† and a are the ladder operators. P is the momentum operator of the harmonic oscillator.

    |ψ > = (1/sqrt(2))[ |1> - i |2>]

    The answer should be zero, can someone check my work?


    2. Relevant equations
    a† |n> = sqrt(n+1)|n+1>

    a |n> = sqrt(n)|n-1>


    3. The attempt at a solution
    I think my first step here might be wrong. I assumed that <ψ | = (1/sqrt(2))[ <1| + i <2|].

    <P> = <ψ | P | ψ > = (1/sqrt(2))[ <1| + i <2|] i√(mhw/2)(a†-a) (1/sqrt(2))[ |1> - i |2>]

    = (1/2)i√(mhw/2)[ <1| + i <2|] * (a†|1> - i*a†|2> - a|1> + i*a|2>]

    = (1/2)i√(mhw/2)[ <1| + i <2|] * [sqrt(2)|2> - i*sqrt(3)|3> - sqrt(1)|0> + i*sqrt(2)|1>]

    Since <x|y> = 1 when x = y but 0 otherwise, this reduces to

    = (1/2)i√(mhw/2)[ <1| + i <2|] * [sqrt(2)|2> - i*sqrt(3)|3> - sqrt(1)|0> + i*sqrt(2)|1>]

    =(1/2)i√(mhw/2) (isqrt(2)<1|1> + isqrt(2)<2|2>) = (1/2)i√(mhw/2) ( 2isqrt(2))

    = -sqrt(mhw)

    But I think the answer is supposed to be zero. What am I doing wrong here?

    I already checked for sign mistakes 3 times, and I can't find any.
     
  2. jcsd
  3. Apr 15, 2012 #2
    No matter how I work it out, your result is correct.

    The answer is not zero.

    What you'r thinking of is [itex]\left\langle n \right| P \left| n \right\rangle[/itex] = 0,

    which is a specific case of the general expression

    [itex]\left\langle n' \right| P \left| n \right\rangle= i\sqrt{\frac{mhω}{2}}(-\sqrt{n}δ_{n',n-1}+\sqrt{n+1}δ_{n',n+1})[/itex]

    Since your wavefunction is a superposition of states, it doesn't turn out to be zero. I can show my work if you like.

    I found this in a textbook I have. Quantum Mechanics Concepts and Applications 2nd, Zettili, pg 248.
     
    Last edited: Apr 15, 2012
  4. Apr 15, 2012 #3
    Thank you very much! I am a bit confused though, what are n and n prime? I don't really get how what I have is different from <n | P | n> = 0?

    Does this mean it only works for pure states?
     
  5. Apr 15, 2012 #4
    n and n prime are two different quantum states.

    This means [itex] \left\langle 1 \right| P \left| 1 \right\rangle = 0 [/itex] but [itex] \left\langle 2 \right| P \left| 1 \right\rangle[/itex] is not equal to zero.


    when I worked it out, it turns out that [itex]\left\langle 1 \right| P \left| 2 \right\rangle = -\left\langle 2 \right| P \left| 1 \right\rangle = -i\sqrt{mhω}[/itex]

    Here is a picture of my messy scratchwork. :redface:
    http://i.imgur.com/ekIjb.jpg
     
  6. Apr 15, 2012 #5
    Excellent, that was super helpful! Thanks again :) I understand now.
     
  7. Apr 15, 2012 #6
    You're welcome! I actually remembered not to forget this rule myself!
     
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