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What does this mean? (Intro QM)

  1. Jun 9, 2009 #1
    What does this mean?? (Intro QM)

    I am on the angular momentum unit in my introductory quantum mechanics course and we had to examine the following two commutators:

    1) [tex][Lx,Ly][/tex] it ends up equating to [tex]i\hbar Lz[/tex]

    [tex][Lx,Ly]= i\hbar Lz[/tex]
    [tex][Ly,Lz]= i\hbar Lx[/tex]
    [tex][Lz,Lx]= i\hbar Ly[/tex]

    Tt was concluded that no 2 components have similar eigenfunctions and no 2 components can be measured simultaneously.

    can anyone please explain how this is so? We were not required to take linear algebra for this course and even though I can get these answers, I do not understand what they physically mean.

    2) [tex][L^2,Lx]=0
    [L^2,Ly=0]
    [L^2,Lz]=0[/tex]

    And for the second commutator [tex][L^2,Lx]= 0[/tex] it was concluded that these CAN be measured simultaneously. You can know [tex]L^2[/tex] and one of its components but nothing more. Also [tex]L^2[/tex] and [tex]Lz[/tex] have similar eigenfunctions. How do you know this just from looking at commutators?


    Why is examining commutativity so important. What does looking at the results of the commutation relations give us physically??

    any insight towards this would be greatly appreciated thanks!!
     
  2. jcsd
  3. Jun 9, 2009 #2

    Cyosis

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    Homework Helper

    Re: What does this mean?? (Intro QM)

    For 1)
    Let a particle be in an eigenstate of [itex]L_z[/itex] with eigenvalue l. Why then can we not measure [itex]L_z, L_x[/itex] and [itex]L_y[/itex] simultaneously? We can simply calculate the expectation values and see if they equal 0. (Note that the operators are Hermitian and ignoring the ih term)

    [tex]
    \langle \psi|L_x|\psi \rangle=\langle \psi|[L_y,L_z]|\psi \rangle=\langle \psi|L_yL_z|\psi \rangle-\langle \psi|L_zL_y|\psi \rangle=l\langle \psi|L_y|\psi \rangle -l \langle \psi|L_y|\psi \rangle=0[/tex]

    Same for [itex]L_y[/itex].

    Now let the particle be in an eigenstate of [itex]L_x, L_y[/itex] with eigenvalue k. If our claim is correct all the expectation values should be 0 now.

    [tex]
    L_z |\psi \rangle=[L_y,L_x]|\psi \rangle=0=L_x|\psi \rangle=L_y|\psi \rangle
    [/tex]
     
    Last edited: Jun 9, 2009
  4. Jun 9, 2009 #3

    Pengwuino

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    Re: What does this mean?? (Intro QM)

    Well, do you remember what it means for two operators to commute? If they have a non-zero commutator, amongst other things, they have a non-zero uncertainty. Early in your class, you should have learned about the derivation of the uncertainty principle using the Schwartz Inequality. Long story short, [tex]\sigma _A ^2 \sigma _B ^2 \ge \frac{{|[A,B]|^2 }}{4}[/tex]. What it's basically telling you, as the uncertainty principle tells you in the first place, is that you can't measure the angular momentum along the z-axis simultaneously with the x or y axis without an in your measurement. We know this because [tex][L_i ,L_j ] = ih\varepsilon _{ijk} L_k [/tex] which is non-zero if you're looking at 2 different axis' angular momentum.

    As for your second question, it similarly follows from what you know about commutation relations. If the commutator is 0, the two operators share eigenfunctions and can be measured precisely simultaneously. There is a couple proofs showing that if a pair of operators have a commutator that is 0, they share eigenfunctions. Something that is also important is a proof that if you have 3 operators, A, B, and C and [A,C]=0 and [B,C]=0, that [A,B] is not necessarily 0 as well, that is A and B share eigenfunctions with C but that A and B don't necessarily share eigenfunctions with eachother.
     
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