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Homework Help: Checking if state is an eigenstate

  1. Nov 20, 2014 #1
    1. The problem statement, all variables and given/known data

    For the three-dimensional harmonic oscillator

    [tex]H_{xyz} = \frac{p_x^2}{2m}+\frac{p_y^2}{2m}+\frac{p_z^2}{2m}+\frac{1}{2}m \omega^2 x^2 + \frac{1}{2}m\omega^2 z^2 + \frac{1}{2}m\omega^2 z^2[/tex]


    [tex]| \alpha_1 > = \frac{1}{\sqrt{2}} (|n_x = 0, n_y = 0, n_z = 0> + |n_x = 0, n_y = 0, n_z = 1> )[/tex]


    [tex]| \alpha_2 > = \frac{1}{\sqrt{2}} (|n_x = 1, n_y = 0, n_z = 0> -i |n_x = 0, n_y = 1, n_z = 0> )[/tex]

    Does it correspond to:

    a) A stationary state
    b) an eigenstate of [tex]l^2[\tex]
    c) an eigenstate of [tex]l_z[\tex]

    2. Relevant equations

    a) [tex]H=(N_x +N_y + N_z +\frac{3}{2})\hbar \omega[/tex]

    b) [tex]L^2 = L_x^2 +L_y^2 +L_y^2[/tex]

    c) [tex]L_z=xp_y-yp_x[/tex]

    3. The attempt at a solution

    I think for a) I can just apply the operator and see whether it is a multiple of the original function of not.

    It seems like I should do c) before b) and I always have trouble with operator manipulation.

    What does [tex]L_z=xp_y-yp_x[/tex] applied to

    [tex]| \alpha_1 > = \frac{1}{\sqrt{2}} (|n_x = 0, n_y = 0, n_z = 0> + |n_x = 0, n_y = 0, n_z = 1> )[/tex]

    look like? How do you apply to position and momentum operators to alpha? What are the eigenvalues you are supposed to get out look like?
  2. jcsd
  3. Nov 21, 2014 #2

    Simon Bridge

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    The first thing to do is step back and see what you actually have there.
    i.e. how are these states constructed? i.e. what kinds of states are they made from?

    ##\renewcommand{\ket}[1]{\left| #1 \right\rangle}## It's shorter to write: using ##\ket{n_x,n_y,n_z}##
    ##\ket{\alpha_1} = \frac{1}{\sqrt{2}}\big( \ket{0,0,0} + \ket{0,0,1} \big)##
    ##\ket{\alpha_2} = \frac{1}{\sqrt{2}}\big( \ket{1,0,0} -i \ket{0,1,0} \big)##
    ... makes them easier to read.
  4. Nov 21, 2014 #3
    Alright, I have trouble with terminology: The states [tex]|\alpha_1>[/tex] and [tex]|\alpha_2>[/tex] are superpositions of three-dimensional harmonic oscillator eigenstates as expressed in the the [tex]|n_x,n_y,n_z>[/tex] basis (don't know what exactly you'd call this basis.) Is that correct? An alternative basis would be the [tex]|n,l,m_l>[/tex] ("angular momentum basis?") basis, correct?

    For part a) I did:

    [tex]H = (N+\frac{3}{2})\hbar\omega[/tex]


    [tex]H|\alpha_1>=\frac{1}{2}(\frac{3}{2}|0,0,0>+\frac{5}{2}|0,0,1>)[/tex] So not a stationary state since [tex]H|\alpha_1>\neq E|\alpha_1>[/tex]

    I found [itex]H|\alpha_2>=\frac{5}{2}|\alpha_2>[/itex]

    For part b) we are allowed to simply give an argument.In class we derived something like this picture: YbmdwBR.png

    (Source: http://arxiv.org/pdf/0808.2289v2.pdf, Page 5) where the y-axis is the HO energy levels. So, here is my attempt at an argument for whether they are eigenstates of [itex]l^2[/itex]...

    For [itex]|\alpha_2>[/itex] the state will always be in energy eigenstate with [itex]E=\frac{5}{2}\hbar\omega[/itex] which corresponds to an inaccessible hole in the above figure, so it is NOT an eigenstate of [itex]l^2[/itex]. It IS an eigen state of [itex]l^2[/itex] since both states correspond to l=1.
    For [itex]|\alpha_1>[/itex] it is a superposition of [itex]|0,0,0>[/itex] (accessible,defined [itex]l^2[/itex]) and [itex]|0,0,1>[/itex] (inaccessible). Since it is not strictly a superposition of accessible (and consistent) [itex]l^2[/itex] states, it is NOT an eigenstate.

    Is there any merit to this?

    EDIT: and for c) [itex][H,l_z]=0[/itex] so they are compatible observables...which means that if an H eigenstate exists, an [itex]l_z[/itex] must as well?
    Last edited: Nov 21, 2014
  5. Nov 21, 2014 #4

    Simon Bridge

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    The nx ny and nz label what? you have identifies |a2> as an energy eigenstate ...
    It would help if you could express the nx-z basis into angular momentum basis right?
    How do you change basis?

    Have you done some work on commutators and simultaneous eigenstates?
    [edit] off your edit: well done ... you can also do ##[H,L^2]## and check the other one as well.

    Careful - the commutation means that simultaneous eigenstates are possible, but a particular eigenstate may not be simultaneous.
    i.e. A linear combination of eigenstates for H may not be an eigenstate of a commuting operator.

    Something in there should sound familiar with something you've done in class recently.
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