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Determining Eigenfunction of Operator

  1. Feb 20, 2010 #1
    1. The problem statement, all variables and given/known data

    Determing the constant c such that [tex]\psi_{c}(x,y,z) = x^{2}+cy^{2}[/tex] is an eigenfunction of [tex]\hat{L_{z}}[/tex]

    2. Relevant equations

    [tex]\hat{L_{z}} = -i \hbar (x\frac{\partial \psi}{\partial y} - y\frac{\partial \psi}{\partial x}[/tex]

    3. The attempt at a solution

    [tex]x\frac{\partial \psi}{\partial y} - y\frac{\partial \psi}{\partial x}) = 2x^{2}-2y^{2}c[/tex]

    Therefore:

    [tex]\hat{L_{z}} \psi = -i \hbar (2x^{2}-2y^{2}c) = -2i \hbar (x^{2}-y^{2}c)[/tex]

    .. and now I'm stuck. :|
     
  2. jcsd
  3. Feb 21, 2010 #2

    gabbagabbahey

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    You need to pay more attention to your notation. You can either say,

    [tex]\hat{L_{z}}\psi(x,y,z)= -i \hbar \left(x\frac{\partial \psi}{\partial y} - y\frac{\partial \psi}{\partial x}\right)[/tex]

    or

    [tex]\hat{L_{z}} \longrightarrow -i \hbar \left(x\frac{\partial}{\partial y} - y\frac{\partial}{\partial x}\right)[/tex]

    But you can't equate an abstract differential operator to a scalar function like you did above

    You need to double check this. :wink:
     
  4. Feb 21, 2010 #3
    I was meant to state this:

    [tex]\hat{L_{z}}\psi(x,y,z)= -i \hbar \left(x\frac{\partial \psi}{\partial y} - y\frac{\partial \psi}{\partial x}\right)= -i \hbar (2x^{2}-2y^{2}c) = -2i \hbar (x^{2}-y^{2}c)[/tex]

    [/tex]

    I don't know how to rearrange that the find the value of C.
     
  5. Feb 21, 2010 #4

    gabbagabbahey

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    [tex]\left(x\frac{\partial \psi}{\partial y} - y\frac{\partial \psi}{\partial x}\right)\neq (2x^{2}-2y^{2}c) [/tex]

    Recheck your calculation.
     
  6. Feb 21, 2010 #5
    I have:

    [tex]

    (x\frac{\partial \psi}{\partial y}) = x(2cy) = 2cxy

    [/tex]

    and:

    [tex]

    (y\frac{\partial \psi}{\partial x}) = y(2x + c) = 2cxy

    [/tex]

    so:

    [tex]

    \hat{L_{z}}\psi(x,y,z)

    = -i \hbar \left(x\frac{\partial \psi}{\partial y} - y\frac{\partial \psi}{\partial x}\right)

    = -i \hbar (2cxy - 2cxy)

    = 0

    [/tex]

    ?
     
  7. Feb 21, 2010 #6

    gabbagabbahey

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    Where's the [itex]c[/itex] in this part coming from?

    [tex]\frac{\partial\psi}{\partial x}=\frac{\partial}{\partial x}\left(x^2+cy^2\right)=2x[/itex]
     
  8. Feb 21, 2010 #7
    It should be:

    [tex](y\frac{\partial \psi}{\partial x}) = y(2x + 0) = 2xy[/tex]

    then.

    Ok, so:

    [tex]\hat{L_{z}}\psi(x,y,z)= -i \hbar \left(x\frac{\partial \psi}{\partial y} - y\frac{\partial \psi}{\partial x}\right)= -i \hbar (2cxy - 2xy)= -2i \hbar xy(c-1)[/tex] ??
     
  9. Feb 21, 2010 #8

    gabbagabbahey

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    Yup, so if [itex]\psi[/itex] is an eigenfunction of [tex]\hat{L}_z[/itex], what equation must be true?
     
  10. Feb 21, 2010 #9
    [tex]\lambda = \frac{-2i \hbar xy(c-1)}{x^{2}+cy^{2}}?[/tex]
     
  11. Feb 21, 2010 #10

    gabbagabbahey

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    Sure, but I'd write this as

    [tex]\lambda\left(x^2+cy^2\right)=-2i \hbar xy(c-1)[/itex]

    At first, it may not look like there are any values of [itex]c[/itex] that will make this true for all [itex]x[/itex] and [itex]y[/itex], but what if [itex]\lambda=0[/itex]?
     
  12. Feb 21, 2010 #11
    [tex]\lambda\left(\psi_{c}\right)=-2i \hbar xy(c-1)[/tex]

    It would be true if [tex]\lambda = 0[/tex]?
     
  13. Feb 21, 2010 #12

    gabbagabbahey

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    You tell me....if [itex]\lambda=0[/itex] what does that equation become? Are there any values of [itex]c[/itex] that make that equation true?
     
  14. Feb 21, 2010 #13
    [tex]\lambda\left(\psi_{c}\right)=-2i \hbar x y (c-1) = -2i \hbar x y c + 2i \hbar x y [/tex]

    [tex]0 = -2i \hbar x y c + 2i \hbar x y[/tex]

    [tex]2i \hbar x y c = 2i \hbar x y[/tex]

    [tex]c = 1

    [/tex]
     
  15. Feb 21, 2010 #14

    gabbagabbahey

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    Right, so for c=1, [itex]\psi[/itex] is an eigenfunction of [itex]\hat{L}_z[/itex] with corresponding eigenvalue [itex]\lambda=0[/itex].

    If [itex]\lambda\neq 0[/itex] are there any values of [itex]c[/itex] which satisfy the eigenvalue equation for all [itex]x[/itex] and [itex]y[/itex]?
     
  16. Feb 21, 2010 #15
    Um.. c can be any integer value? it's just a scaling value.
     
  17. Feb 26, 2010 #16

    gabbagabbahey

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    So, you are telling me that [itex]\lambda\left(x^2+cy^2\right)=-2i \hbar xy(c-1)[/itex] is satisfied for any integer value of [itex]c[/itex]???
     
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