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Dirac Notation!

  1. Aug 6, 2015 #1
    I've been working through some dirac notation and I'm stuck...

    Here's where I'm at:

    I understand that an expectation value: <x> = ∫ ψ* x ψ dx = <ψ|xψ> = <ψ|x|ψ>
    Also, we can say H|ψ> = E|ψ> where E is an eigenvalue of the operator H and |ψ> represents a state your acting on.
    I get that you can represent a vector 'in Hilbert space' instead of a wavefunction inside the ket and these can be operated on to transform your vector into another vector, Q|a> = |b>
    |a> represents a column vector, <a| represents a row vector

    It's taking some time to get used to using Dirac notation, and I've come across this in some angular momentum notes:

    Lz|m,l> = mħ|m,l> .. So Lz is the operator, mħ is the corresponding eigenvalue. But what does having two values in the ket mean?
    Does, L|m,l> = L|m> + L|l>

    Moreover, what if we have:


    Should this be written in more classical notation as ∫ (l,m)* L+L- (l,m)* dx

    Any guide or text book with a good section on Dirac formalism would be great, I've been looking but can't find anything to explain this trickier stuff. I've attached a page of my notes so that it's more clear with what I'm working with. I've read through Griffith's introduction to quantum mechanics chapter on Dirac formalism but I'm still getting undone by it!


    Attached Files:

  2. jcsd
  3. Aug 6, 2015 #2


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    The fact that you can write H|ψ> = E|ψ>, implies that the state |ψ> is an eigenstate of H, it cannot be any state.
    ##|l,m \rangle ## is eigenstate of both ##L^2## and ##L_z##, this can be so because ##L^2## and ##L_z## commute - ##[L^2,L_z] = 0##. If you have studied linear algbera you may remember that if two operators commute and no other operators can be added into this group of operators without violating the mutual commutativity, they must share the same set of eigenstates and it will be uniquely specified by two indices, in this case ##l## and ##m##. The former corresponds to the operator ##L^2## and the latter to ##L_z##.
    No it doesn't.
    You can do the calculation in position representation like you wrote there, but if you know the property of ##L_+## and ##L_-## and how they act on ##|l,m \rangle##, it saves a lot of time working in Dirac notation.
  4. Aug 6, 2015 #3


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    The ket ##|l,m\rangle## is a state that is both an eigenstate of ##L_z## with eigenvalue ##m## and an eigenstate of ##L^2## with eigenvalue ##l(l+1)##. This is possible only because ##L_z## and ##L^2## commute.
  5. Aug 6, 2015 #4
    When we have two commuting operators we can have two values inside the ket, both are eigenstates of my operators?
    Ok, I see that I can, I'm stuck on the how. Also, I'm confused at how eigenvalues seem to be plucked out of thin air in every example I look at. I can find how to find eigenvalues from matrices, but in examples like this is there some way to work out what the eigenvalue should be?

    E.g. Lz|l,m> = mħ|l,m> ... where has that mħ come from?!

    Is it easier to just learn all the eigenvalues for each operator? There's a way I can work out what they should be, my notes, other course notes and text books (that I've looked at) don't seem to show this. What if Lz|ψ> where ψ is some different eigenstate, are the eigenvalues the same or different?

    Another example from my notes is this:

    L2 |l, m >= l(l + 1)ħ2|l, m >


    L22 , m >= β2ħ22 , m >

    These are two different eigenstates and two different eigenvalues that are just stated in my notes, you can see in the images I attached above. How, why? ... in the same example changing the eigenstate Lz is acting on doesn't change the eigenvalue. Very confused :(
  6. Aug 6, 2015 #5


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    At some point in your textbook you should find relation like
    $$ L_+|l,m \rangle = C_+ |l,m+1 \rangle$$
    for ##m \neq l ##, and
    $$ L_-|l,m \rangle = C_- |l,m-1 \rangle$$
    for ##m \neq -l ##. The constants ##C_+## and ##C_-## are found using normalization condition.
    Working out the eigenvalues of ##L^2## and ##L_z## in Dirac notation does require some elaborate works, for this purpose I suggest Modern Quantum Mechanics by Sakurai sub-chapter 3.5. But honestly as I remember Griffith must also provide the same material, I think you just need to be persistent enough to read through that chapter in Griffith's book patiently.

    If |ψ> is not an eigenstate of ##L_z##, Lz|ψ> won't be eigenvalue equation anymore. In such a case, you can represent |ψ> as a linear combination of the eigenstates of ##L_z##.
  7. Aug 6, 2015 #6


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    Yes. You could start with the definition of the orbital momentum operator and plug in the linear momentum operator: [itex]\mathbf{L} = \mathbf{r} \times \mathbf{p} = -i \hbar(\mathbf{r} \times \nabla)[/itex]. Now write down the eigenvalue equation for the operators, for example [tex]L_z u(\vec r) = \mu u(\vec r)[/tex] This partial differential equation can be solved for [itex]u(\vec r)[/itex] if you go over to spherical coordinates. So in the continuous case, you don't calculate eigenvalues of matrices but solve differential equations.

    Now the case of angular momentum is special because there exists a simpler and more powerful technique to calculate the eigenvalues. The idea is to start with abstract equations and use only the commutation relations of the angular momentum operators to obtain the possible eigenvalues (it is quite remarkable that this works!).

    Since [itex]\mathbf{L^2}[/itex] and [itex]L_z[/itex] commute, we can find states which are eigenstates to both operators. So instead of two independent eigenvalue equations
    [tex]\mathbf{L^2} |\lambda \rangle = \lambda |\lambda \rangle\\
    L_z |\mu \rangle = \mu |\mu \rangle[/tex]
    we have
    [tex]\mathbf{L^2} |\lambda, \mu \rangle = \lambda |\lambda, \mu \rangle\\
    L_z |\lambda, \mu \rangle = \mu |\lambda, \mu \rangle[/tex]

    Now we could try to use the commutation relations in order to determine the possible values of [itex]\lambda[/itex] and [itex]\mu[/itex] but usually, the derivation uses additional information. First, we know that we are dealing with angular momentum in Quantum Mechanics, so it seems sensible to write the eigenvalues as multiples of [itex]\hbar[/itex] which has the correct unit and expected magnitude. Also some authors write the eigenvalues of [itex]\mathbf{L^2}[/itex] in the special form [itex]l(l+1) \hbar^2[/itex] from the beginning, because they already know that the result will involve an integer [itex]l[/itex]. I don't like the pedagogy of this but mathematically, it is of course valid.

    Now check if you understand the derivation of the eigenvalues in a textbook. It should be included in Griffiths. If not, check a couple of other textbooks, as blue_leaf77 wrote, Sakurai definitely has it. It is a really common derivation.
    Last edited: Aug 6, 2015
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