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Basic question about quantum: Number operator usage

  1. Sep 25, 2006 #1
    So i was working on a homework problem to calculate energy, and i can calculate it directly. But i was thinking it should be easier to use the number operator on the wave function and find out what it's energy should be. The function involved is the initial wave state of a quantum harmonic oscillator. Anyways, it's getting late but i thought i'd ask to see if i'm thinking the right way, and try to solve it more tomorrow. so

    ANyways, Asub+ * Asub- *Psi(x) = n * Psi(x)
    where Psi is the wave function.

    by the definitions of Asub+ and Asub-, i will end up getting
    (2m *(Hamiltonian operator ) -m*omega* hbar) / (2m*omega*hbar)) * Psi = n *Psi

    i then insert the hamiltonian operator, which is
    - hbar^2/ 2m * d^2/dx^2 + V(x), where V(x) for a harmonic oscillator is
    kx^2/2

    if i do this and multiply by the wave function should it come out to N * the wave function so that i can learn what energy state the function is in?

    the wave function at issue is

    Psi (x,0) = A (1-2 *sqrt (mw/hbar)*x)^2 * e^ (-mwx^2/2hbar)

    from griffith 2.41

    i'm pretty sure i can solve it just by calculating the integral manually, but that seems inefficient..i thought since that in my notes it gave me a "number operator" and I have the energy for any given N, then that should be faster
     
  2. jcsd
  3. Sep 25, 2006 #2

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Could you write everything using LaTex ? I have troubles understanding what you wrote above.

    Daniel.
     
  4. Sep 25, 2006 #3
    i would love to, except i don't know what latex is or how to use it. Can you link me a tutorial on it?
     
  5. Sep 25, 2006 #4
    found it..working on it
     
  6. Sep 25, 2006 #5
    So i was working on a homework problem to calculate energy, and i can calculate it directly. But i was thinking it should be easier to use the number operator on the wave function and find out what it's energy should be. The function involved is the initial wave state of a quantum harmonic oscillator. Anyways, it's getting late but i thought i'd ask to see if i'm thinking the right way, and try to solve it more tomorrow. so

    ANyways, [tex] a_+ a+- /Psi (x) = n * /Psi (x) [/tex]
    where Psi is the wave function.

    by the definitions of [tex] a_+ and a_- [/tex], i will end up getting
    [tex] /frac{2m/hat(H) -m/omega */bar(h}{2m/omega/bar(h)} /Psi = n /Psi [tex]

    i then insert the hamiltonian operator, which is
    [tex] -/frac{/bar(h)^2} {2m} * /frac {d^2}{dx^2} + V(x), where V(x) for a harmonic oscillator is
    /frac{kx^2}{2}
    [/tex]

    if i do this and multiply by the wave function should it come out to N * the wave function so that i can learn what energy state the function is in?

    the wave function at issue is

    [tex] /Psi(x,0) = A (1-2x /sqrt(/frac{m/omega}{/bar(h)}))^2 * e^ -/frac{m/omega x^2}{2/bar(h)}
    [/tex]
    from griffith 2.41

    i'm pretty sure i can solve it just by calculating the integral manually, but that seems inefficient..i thought since that in my notes it gave me a "number operator" and I have the energy for any given N, then that should be faster
     
  7. Sep 25, 2006 #6
    I tried to do it for you - if something's wrong, perhaps you can quote from my posting and try to correct it yourself? Latex is very straightfroward, if you know some things. :smile:
     
    Last edited: Sep 25, 2006
  8. Sep 25, 2006 #7
    [itex] \Psi (x,0) = A (1-2 \sqrt {\frac{m \omega}{\hbar}} x)^2 e^ {-\frac{m \omega x^2}{2 \hbar}}[/itex]
     
    Last edited: Sep 25, 2006
  9. Sep 25, 2006 #8
    You have to do "\", not "/". :wink:
     
  10. Sep 25, 2006 #9
    lol ya i didn't notice it either..i just looked for other things..slashes are all the same to me:)
    and thanks for the help..i'll figure it out bit by bit
     
  11. Sep 25, 2006 #10
    I think this only works if [itex] \Psi [/itex] is an energy eigenstate, which it doesn't have to be. Consider for example the wavefunction [itex] \Psi(x) = \phi_1 + \phi_2 [/itex], where [itex] \phi_n[/itex] is the nth energy eigenstate, but [itex] \Psi [/itex] is no (energy) eigenstate at all.
     
    Last edited: Sep 25, 2006
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