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Problem about spherical angle operators

  1. Aug 11, 2011 #1
    Hi

    Here's the problem I am trying to do.

    a) Is the state [itex]\psi (\theta ,\phi)=e^{-3\imath \;\phi} \cos \theta [/itex]

    an eigenfunction of [itex]\hat{A_{\phi}}=\partial / \partial \phi[/itex] or of
    [itex]\hat{B_{\theta}}=\partial / \partial \theta [/itex] ?

    b) Are [itex]\hat{A_{\phi}} \;\mbox{and} \;\hat{B_{\theta}}[/itex] hermitian ?

    c)Evaluate the expressions [itex]\langle \psi \vert \hat{A_{\phi}} \vert \psi \rangle [/itex]
    and [itex]\langle \psi \vert \hat{B_{\theta}} \vert \psi \rangle [/itex]


    Now [itex] \hat{A_{\phi}}[/itex] has imaginary eigenvalues , so its not hermitian.
    I could show that [itex]\psi[/itex] is an eigenfunction of square of [itex] \hat{B_{\theta}}[/itex]. I have been able to show that the commutator of A and B is zero.
    So with this information, how do I check the hermiticity of B ?

    for part c) , since there are two state variables , I am little confused about how to go
    about it ? any guidance will be appreciated... thanks
     
  2. jcsd
  3. Aug 11, 2011 #2

    kuruman

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    Do you know how to represent the braket as an integral? If so, then to prove hermiticity you need to show that <ψ|Βψ> = <ψΒ|ψ>, i.e. that operator B gives the same result when it operates either to the left or to the right.
     
  4. Aug 11, 2011 #3
    but this wavefunction has two variables, so how do I evaluate these amplitudes ?
     
  5. Aug 11, 2011 #4

    kuruman

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    You do a double integral over two-dimensional spherical space.
     
  6. Aug 11, 2011 #5
    Commutator is irrelevant here. What is an eigenfunction of B? What is the corresponding eigenvalue? Is it real?
    What Kuruman said.
     
  7. Aug 11, 2011 #6
    [tex]\langle \psi \vert \hat{B_{\theta}} \vert \psi \rangle = \int \langle \psi \vert \theta ,\phi \rangle \langle \theta , \phi \vert \hat{B_{\theta}} \vert \psi \rangle d\theta \;d\phi [/tex]


    is this the correct way ? so what next ? these kets in [itex]\theta[/itex] and [itex]\phi[/itex] , sre those eigenfunctions of B ?
     
  8. Aug 11, 2011 #7

    vela

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    That's almost right. You need to throw a factor of sin θ in there because the infinitesimal element of solid angle is [itex]d\Omega = \sin\theta\,d\theta\,d\phi[/itex].
     
  9. Aug 12, 2011 #8
    ok vela

    so that would be

    [tex]\langle \psi \vert \hat{B_{\theta}} \vert \psi \rangle = \int \langle \psi \vert \theta ,\phi \rangle \langle \theta , \phi \vert \hat{B_{\theta}} \vert \psi \rangle \sin \theta \;d\theta \;d\phi [/tex]

    so now what ? how do I evaluate this ?
     
  10. Aug 12, 2011 #9

    kuruman

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    I guess you don't know how to represent a braket as an integral.

    [itex]<\psi|B|\psi>=\int \psi^*(B\psi) sin \theta \: d \theta \: d\phi[/itex]

    You operate on the wavefunction with B to get a new function then integrate as indicated.
     
  11. Aug 12, 2011 #10
    Hi kuruman

    to go from my equation to your equation , we need to assume the following

    [tex]\langle \theta , \phi \vert \hat{B_{\theta}} \vert \psi \rangle = \hat{B_{\theta}} \langle \theta , \phi \vert \psi \rangle [/tex]

    am i right ? if so , is the above equality always true , for all operators ?
     
    Last edited: Aug 12, 2011
  12. Aug 12, 2011 #11

    kuruman

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    You really need to beef up your understanding of Dirac bra-ket notation. When we write
    B|ψ> we mean "operate on function ψ with operator B", i.e. find a new function
    [itex]f(\theta,\phi)=\frac{\partial}{\partial \theta}\psi(\theta,\phi)[/itex]

    When we write <ψ|Β|ψ>,
    <ψ| stands for ψ*(θ,φ)
    Β|ψ> stands for f(θ,φ)
    and closing the braket stands for "do the integral" [itex]\int \psi^*(\theta,\phi)f(\theta,\phi) \: sin\theta \:d\theta\:d\phi[/itex]
     
  13. Aug 12, 2011 #12

    vela

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    Or to put it a little more formally,[itex]\langle \theta,\phi \vert \psi \rangle[/itex] is the representation of [itex]\vert \psi \rangle[/itex] in the [itex]\vert \theta,\phi \rangle[/itex] basis. That is, [itex]\langle \theta,\phi \vert \psi \rangle = \psi(\theta,\phi)[/itex]. Similarly, you have [tex]\langle \theta,\phi \lvert \hat{B}_\theta \rvert \psi \rangle = \frac{\partial}{\partial\theta} \psi(\theta,\phi)[/tex]
    Putting it all together, you get[tex]\langle \psi \vert \hat{B_{\theta}} \vert \psi \rangle
    = \int \langle \psi \vert \theta ,\phi \rangle \langle \theta , \phi \vert \hat{B_{\theta}} \vert \psi \rangle \sin \theta \,d\theta \,d\phi
    = \int \psi^*(\theta ,\phi) \left[\frac{\partial}{\partial\theta} \psi(\theta,\phi)\right] \sin \theta \,d\theta \,d\phi[/tex]
     
  14. Aug 12, 2011 #13
    kuruman , i was looking for more formal explanation. i am just using completeness property.



    which is what i was saying exactly in post # 10 .

    [tex]\langle \theta , \phi \vert \hat{B_{\theta}} \vert \psi \rangle = \hat{B_{\theta}} \langle \theta , \phi \vert \psi \rangle [/tex]

    is it not ?

    i remember studying somewhere about the derivation of such relationship. i will give an example about the momentum operator [itex]\hat{P}[/itex]

    [tex]\langle x \vert \hat{P} \vert \psi \rangle =\int \langle x \vert \hat{P} \vert p \rangle \langle p \vert \psi \rangle dp [/tex]

    [tex]= \int p \langle x \vert p \rangle \langle p \vert \psi \rangle dp [/tex]

    now we know

    [tex]\langle x \vert p \rangle = \frac{1}{(2\pi \hbar)^{1/2}}\; e^{ipx/ \hbar} [/tex]

    putting this in the integral ,

    [tex]= \int p \;\; \frac{1}{(2\pi \hbar)^{1/2}}\; e^{ipx/ \hbar} \langle p \vert \psi \rangle dp [/tex]

    [tex]= \frac{\partial}{\partial x}\int \frac{\hbar}{i} \frac{1}{(2\pi \hbar)^{1/2}}\; e^{ipx/ \hbar} \langle p \vert \psi \rangle dp [/tex]

    [tex]= \frac{\hbar}{i} \frac{\partial}{\partial x} \int \langle x \vert p \rangle \langle p \vert \psi \rangle dp [/tex]

    [tex] = \frac{\hbar}{i} \frac{\partial}{\partial x} \langle x \vert \psi \rangle [/tex]

    this is what i was talking about. is there any such process by which we can show that ,


    [tex]\langle \theta , \phi \vert \hat{B_{\theta}} \vert \psi \rangle = \hat{B_{\theta}} \langle \theta , \phi \vert \psi \rangle [/tex]

    thanks :cool:
     
  15. Aug 12, 2011 #14

    vela

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    To be a bit pedantic, sort of, but not exactly. I have [itex]\partial/\partial\theta[/itex] on the right-hand side, not [itex]\hat{B}_\theta[/itex]. Consider [itex]\hat{B}_\theta[/itex] as some abstract operator while [itex]\partial/\partial\theta[/itex] is its representation in the [itex]\vert\theta,\phi\rangle[/itex] basis. So you have[tex]\langle \theta , \phi \vert \hat{B}_\theta \vert \psi \rangle = \frac{\partial}{\partial \theta} \langle \theta , \phi \vert \psi \rangle [/tex]
    Though it's common, I see writing [itex]\hat{B}_\theta = \partial/\partial\theta[/itex] as kind of sloppy if you want to stick with the formalism of inserting complete sets, etc. It makes sense to apply [itex]\partial/\partial\theta[/itex] to the wave function of [itex]\theta[/itex], but not to an abstract ket.

    It's just like how [itex]\hat{p}[/itex] is the momentum operator in a general sense while [itex]p[/itex] and [itex](\hbar/i) \partial/\partial x[/itex] are its representations in, respectively, the momentum and position bases. Say you have some ket [itex]\vert \psi \rangle[/itex]. Then you have \begin{align*}
    \langle x \lvert \hat{p} \rvert \psi \rangle &= \frac{\hbar}{i}\frac{\partial}{\partial x}\langle x \vert \psi \rangle = \frac{\hbar}{i}\frac{\partial}{\partial x} \psi(x) \\
    \langle p \lvert \hat{p} \rvert \psi \rangle &= p\langle p \vert \psi \rangle = p\psi(p)
    \end{align*}
    It wouldn't make sense to use [itex](\hbar/i) \partial/\partial x[/itex] when you're working in the momentum basis and vice-versa. Once you choose a basis, you know how to represent the operator and you have also chosen which wave function to use.

    So getting back to your original question, I think saying [itex]\hat{B}_\theta = \partial/\partial\theta[/itex] pretty much tells you that you're working in a particular basis and that
    [tex]\langle \theta , \phi \vert \hat{B}_\theta \vert \psi \rangle = \frac{\partial}{\partial \theta} \langle \theta , \phi \vert \psi \rangle [/tex]
    holds.
     
  16. Aug 12, 2011 #15

    vela

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    You could also look at this way:
    [tex]\langle \theta , \phi \vert \hat{B}_\theta \vert \psi \rangle = \langle \theta , \phi \vert \hat{B}_\theta \psi \rangle = \frac{\partial}{\partial \theta} \psi(\theta,\phi)[/tex]
     
  17. Aug 13, 2011 #16
    so vela ,

    is this the definition of the the way operator acts on the ket ?

    so when an operator acts on the ket [itex]\vert \psi \rangle[/itex] one way to evaluate it
    is to take the inner product of this ket with some bra [itex]\langle x \vert[/itex] and then
    use the representation of the operator in the basis of this bra. i am using the book,
    "Quantum Mechanics: Concepts and Applications" by Nouredine Zettili and he uses
    the operators in a sloppy way. Can you suggest any book or online resource where the formalism of quantum mechanics is presented the way it should be.
     
  18. Aug 13, 2011 #17

    vela

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    I wouldn't say it's a definition, but yeah, what you're describing is essentially what it boils down to. When you multiply a ket by [itex]\langle x \vert[/itex] on the left, you're finding the representation of the ket in that basis. This is what I was getting at in post #15. When you act on [itex]\vert \psi \rangle[/itex] with [itex]\hat{B}_\theta[/itex], you get the state [itex]\vert \hat{B}_\theta \psi \rangle[/itex]. Then when you multiply by [itex]\langle \theta, \phi \vert[/itex], you're choosing a basis. In that basis, the state is represented by the derivative with respect to θ of the [itex]\psi(\theta,\phi)[/itex].

    You could also say
    \begin{align*}
    \hat{A} \vert \psi \rangle &= \int dx \vert x \rangle \langle x \vert \hat{A} \vert \psi \rangle \\
    &= \int dx \vert x \rangle \int dx'\langle x \vert \hat{A} \vert x'\rangle\langle x' \vert \psi \rangle
    \end{align*}
    which you can interpret as the representation of operator A in the x basis multiplied by the representation of the state ψ in the same basis.

    In any case, don't get too caught up in the mathematical formalism. What kuruman said back in post #11 is what you need in a practical sense.
     
    Last edited: Aug 13, 2011
  19. Aug 14, 2011 #18
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