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Orbital angular momentum problem

  1. Oct 18, 2009 #1
    1. The problem statement, all variables and given/known data

    Consider a particle that moves in three dimensions with wave function [tex] \varphi[/tex] . Use operator methods to show that if [tex]\varphi[/tex] has total angular momentm quantum number l=0 , then [tex]\varphi[/tex] satifies

    [tex] L\varphi=0[/tex]

    for all three components [tex]L_\alpha[/tex] of the total angular momentum L
    2. Relevant equations

    [tex] [L^2,L_\alpha][/tex]?

    [tex] L^2=L_x^2+L_y^2+L_z^2 [/tex]?

    [tex] L^2=\hbar*l(l+1)[/tex] , l=0,1/2,1,3/2,... ?

    [tex] L_z=m\hbar[/tex] , m=-l, -l+1,...,l-1,l. ?
    3. The attempt at a solution

    [tex] [L^2,L_\alpha]=[L_x^2+L_y^2+L_z^2,L_\alpha]=[L_x^2,L_\alpha]+[L_y^2,L_\alpha]+[L_z^2,L_\alpha][/tex]. [tex][AB,C]=A[B,C]+[A,C]B[/tex]; Therefore,[tex] [L_x^2,L_\alpha]+[L_y^2,L_\alpha]+[L_z^2,L_\alpha]=L_x[L_x,L_\alpha]+[L_x,L_\alpha]L_x+L_y[L_y,L_\alpha]+[L_y,L_\alpha]L_y+L_z[L_z,L_\alpha]+[L_z,L_\alpha]L_z[/tex]. Now what?
     
  2. jcsd
  3. Oct 19, 2009 #2
    People having a hard time reading the problem?
     
  4. Oct 19, 2009 #3

    gabbagabbahey

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    Do you not already know [itex][L^2,L_x][/itex], [itex][L^2,L_y][/itex] and [itex][L^2,L_z][/itex]?

    If so, just calculate [itex][L^2,L_x]\varphi[/itex], [itex][L^2,L_y]\varphi[/itex] and [itex][L^2,L_z]\varphi[/itex] and use the fact that [itex]L^2\varphi=l(l+1)\varphi=0[/itex] for [itex]l=0[/itex].
     
  5. Oct 19, 2009 #4

    Avodyne

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    I don't understand gabbagabbahey's method. I would (using bra-ket notation) define three states [itex]|\psi_i\rangle = L_i|\phi\rangle[/itex], [itex]i=x,y,z[/itex], and then compute [itex]\sum_{i}\langle\psi_i|\psi_i\rangle[/itex].
     
  6. Oct 19, 2009 #5
    Yes I calculated those commutators [tex][L^2,L_x]\varphi=(L_y*i\hbar*L_z)+(i*\hbar*L_z)(L_y)+L_z(i*\hbar*L_y)+(i*\hbar*L_y)L_z,[L^2,L_y]= L_x(i*\hbar*L_z)+(i*\hbar*L_z)L_y+L_z(i*\hbar*L_x)+(i*\hbar*L_x) +(i*\hbar*L_x)L_z, [L^2,L_z]=L_x(i*\hbar*L_y)+(i*\hbar*L_y)(L_x)+L_y(i*\hbar*L_x)+(i*\hbar*L_x)L_y [/tex], The comutators simply expand more and do not simplify . at [tex] l=0, l(l+1)\varphi=0[/tex]

    Please look at my latex for the expression [tex][L^2,L_z][/tex] PF is not completely showing the solution to this expression for some reason. Now is it readable
     
    Last edited: Oct 19, 2009
  7. Oct 19, 2009 #6

    gabbagabbahey

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    What you've just written is completely unreadable....How do you expect others to be able to help you, if you don't take the time to make sure they can read what you are posting?

    Begin by calculating the commutator [itex][L^2,L_x][/itex]....use the "go advanced" and "preview post" buttons to make sure whatever you write is actually readable, before submitting your post.
     
  8. Oct 19, 2009 #7
    [tex]
    [L^2,L_x]\varphi=(L_y*i\hbar*L_z)+(i*\hbar*L_z)(L_y)+L_z(i* \hbar*L_y)+(i*\hbar*L_y)L_z,[L^2,L_y]= L_x(i*\hbar*L_z)+(i*\hbar*L_z)L_y+L_z(i*\hbar*L_x) +(i*\hbar*L_x) +(i*\hbar*L_x)L_z, [L^2,L_z]=L_x(i*\hbar*L_y)+(i*\hbar*L_y)(L_x)+L_y(i*\hbar*L _x)+(i*\hbar*L_x)L_y
    [/tex] Now is it readable? [tex][L^2,L_z]=L_x(i*\hbar*L_y)+(i*\hbar*L_y)(L_x)+L_y(i*\hbar*L _x)+(i*\hbar*L_x)L_y[/tex]
     
  9. Oct 19, 2009 #8

    gabbagabbahey

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    I still can't make heads or tails of what you've written....When you read a textbook or article, or even someone else's posts here....do they lay things out in a jumbled mess like that?
     
  10. Oct 19, 2009 #9
    I will seperate my commutators into seperate latex code and aligned the commutators vertically. Maybe it should then be easier for you to read:

    [tex][L^2,L_x]\varphi=(L_y*i\hbar*L_z)+(i*\hbar*L_z)(L_y)+L_z(i* \hbar*L_y)+(i*\hbar*L_y)L_z[/tex]
    [tex][L^2,L_y]\varphi= L_x(i*\hbar*L_z)+(i*\hbar*L_z)L_y+L_z(i*\hbar*L_x) +(i*\hbar*L_x) +(i*\hbar*L_x)L_z[/tex]
    [tex][L^2,L_z]\varphi=L_x(i*\hbar*L_y)+(i*\hbar*L_y)(L_x)+L_y(i*\hbar*L _x)+(i*\hbar*L_x)L_y[/tex],
    at[tex]l=0, l(l+1)\varphi=0[/tex]
     
  11. Oct 19, 2009 #10

    gabbagabbahey

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    That's a little easier to read (Although I really wish you'd stop using that ugly [itex]*[/itex] symbol for multiplication). Why do you have a [itex]\varphi[/itex] on the LHS of each equation? And, you are missing some negative signs on the RHS....why don't you show me your steps for calculating [itex][L^2,L_x][/itex]?
     
  12. Oct 19, 2009 #11
    okay.[tex] [L^2,L_x]=[L_x^2+L_y^2+L_z^2,L_x]=[L_x^2,L_x]+[L_y^2,L_x]+[L_z^2,L_z][/tex]

    [tex][L_x^2,L_x]+[L_y^2,L_x]+[L_z^2,L_x]=[L_y^2,L_x]+[L_z^2,L_x][/tex]

    [tex][L_y^2,L_x]+[L_z^2,L_z]=L_y[L_y,L_x]+[L_y,L_x]L_y+L_z[L_z,L_x]+[L_z,L_x][L_z][/tex]

    [tex] [L_y,L_x]=i*\hbar*L_z, [L_z,L_x]=i*\hbar*L_y[/tex], therefore

    [tex]L_y[L_y,L_x]+[L_y,L_x]L_y+L_z[L_z,L_x]+[L_z,L_x][L_z]=i*L_y(\hbar*L_z),+i*\hbar*L_z*L_y+i*L_z*(\hbar*L_y)+i(\hbar*L_y)*L_z[/tex] .How did you get a negative term? (Sorry, if I don't used the [tex] * [/tex] symbol, latex will read the i's and L's as a subscript rather than a coeffcient
     
    Last edited: Oct 19, 2009
  13. Oct 20, 2009 #12

    gabbagabbahey

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    [itex][L_y,L_x]=-[L_x,L_y]=-i\hbar L_z[/itex]:wink:
     
  14. Oct 20, 2009 #13
    My calculations showed that [tex] [L^2,L_z]=0,[L^2,L_y]=0,[L^2,L_x]=0[/tex], now what?
     
  15. Oct 20, 2009 #14

    gabbagabbahey

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    Now, by definition what is [itex][L^2,L_x]\varphi[/itex]?...compare that to what you've just calculated....
     
  16. Oct 20, 2009 #15
    since[tex][L^2,L_x]=0[/tex],then[tex][L^2,L_x]\varphi=0\varphi=0[/tex]?
     
  17. Oct 20, 2009 #16

    gabbagabbahey

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    If I asked you the definition of [itex][A,B]\varphi[/itex], you would say [itex](AB-BA)\varphi[/itex], right?

    So, when I say "by definition what is [itex][L^2,L_x]\varphi[/itex]?", you say ____?
     
  18. Oct 20, 2009 #17
    yes thats right. [tex][L^2,L_x]=(L^2L_x-L_xL^2)\varphi[/tex], should I let [tex]L_x=(\hbar/i)*d/dx[/tex] and [tex] L=(\hbar/i)*d/dx+(\hbar/i)*d/dy+(\hbar/i)*d/dz[/tex] and let [tex] (-\hbar/i)*d/dx+(-\hbar/i)*d/dy+(-\hbar/i)*d/dz[/tex]
     
  19. Oct 20, 2009 #18

    gabbagabbahey

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    No need for all that, by definition you have [itex][L^2,L_x]\varphi=(L^2L_x-L_xL^2)\varphi[/itex], and you also just calculated that this was equal to zero, so

    [tex][L^2,L_x]\varphi=(L^2L_x-L_xL^2)\varphi=L^2L_x\varphi-L_xL^2\varphi=0[/tex]

    Now, express [itex]\varphi[/itex] as a linear combination of the eigenfunctions [itex]\psi_l[/itex] of [itex]L^2[/itex], and use the fact that [itex]L^2\psi_l=l(l+1)\psi_l=0[/itex] for [itex]l=0[/itex] to show that [itex]L_xL^2\varphi=0[/itex]
     
  20. Oct 20, 2009 #19

    Avodyne

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    I still don't think this works. We now have [itex]L^2L_x\varphi=0[/itex], but how do we go from there to prove that [itex]L_x\varphi=0[/itex]?
     
  21. Oct 20, 2009 #20
    [tex]L^2L_x\varphi=0[/tex] why not this expression? Can't I assume [tex]L^2=0[/tex] or [tex]L_x\varphi=0[/tex]
     
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