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Why can magnetic quantum no. only take values between -l and +l?

  1. May 10, 2014 #1
    Where l is the azimuthal quantum number!
     
  2. jcsd
  3. May 10, 2014 #2
    That's similar to asking why the cosine of an angle is only between +1 and -1.
     
  4. May 10, 2014 #3

    Nugatory

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    Intuitively, the ##m## quantum number is the value of the ##z## component of the angular momentum vector, and the total magnitude of that vector is ##l##. The magnitude of a single component of a vector has to be less than the magnitude of the vector; if I am moving in some direction at a speed of 10 m/sec, we don't know what my speed in a northwards direction is, but we know it has to be somewhere between -10 m/sec (I'm moving due south) and 10 m/sec (I'm moving due north).

    Mathematically, that's what falls out of the solution of the Schrodinger equation for the electron in the electric field of the nucleus. Google for "hydrogen atom Schodinger" to see how this works.
     
  5. May 10, 2014 #4

    Vanadium 50

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    It's actually [itex]\sqrt{l(l+1)}[/itex], but your argument still holds, since that is always less than [itex] l+ 1 [/itex].
     
  6. May 10, 2014 #5

    jtbell

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    More precisely (although it doesn't change the conclusion), the magnitude and z-component of the angular momentum are
    $$L = \sqrt{l(l+1)} \hbar\\
    L_z = m \hbar$$
    It's easy to see that if m were to equal ##l+1##, then we would have ##L_z > L##.
     
  7. May 10, 2014 #6

    I totally understand it now, great explanation!
     
    Last edited: May 10, 2014
  8. May 10, 2014 #7

    Nugatory

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    Glad it worked for you, but do note jtbell's and Vanadium50's correction/clarification - I was simplifying.

    I explained why the magnitude of ##m## cannot be greater than ##l##, but the fact that the magnitude of the angular momentum is actually ##\sqrt{l(l+1)}## not just ##l## is also important. That's what ensures that the the magnitude of ##L_z## cannot be exactly equal to the magnitude of ##L##, and therefore leaves room for ##L_x## and ##L_y## to not be determined even when ##L_z## is at its maximum value.
     
    Last edited by a moderator: May 10, 2014
  9. May 10, 2014 #8

    Lz = m [itex]\hbar[/itex]

    This means that the angular momentum is quantized right?

    Would a magnetic/electric field alter this relationship?
     
  10. May 10, 2014 #9

    jtbell

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    It means that one component of the angular momentum is quantized. By convention we usually call this the "z-component". The other two components (x and y) are indeterminate.

    What do you mean by "alter"?

    With a magnetic field, we use the direction of the field to define the "z-direction", and we get the Zeeman effect. The angular momentum values and quantum numbers don't change, but the energies associated with them do change:

    http://en.wikipedia.org/wiki/Zeeman_effect

    With an electric field, we have the Stark effect, which I don't know much about.

    http://en.wikipedia.org/wiki/Stark_effect
     
  11. May 10, 2014 #10

    WannabeNewton

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    No that is an inherent property of the angular momentum operator i.e. it is entirely independent of what system you consider. It is the Hamiltonian which depends on the system and accordingly it is the energy eigenvalues that will be shifted in the presence of electric or magnetic fields (c.f. Stark effect, Zeeman effect, fine structure of Hydrogen).
     
  12. May 11, 2014 #11

    Ok, I'm finding the whole spatial quantisation concept pretty difficult to grasp. Thanks for all your help. Sorry for the poorly phrased questions!
     
  13. May 11, 2014 #12

    WannabeNewton

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    I'm afraid I can't really help you there as I still have conceptual difficulties with the quantization of angular momentum as well. It's too bad typical QM classes completely brush aside concepts and focus so much on the rather straightforward calculations.
     
  14. May 11, 2014 #13
    The quantization of the z-component of the angular momentum is a result from the fact that the eigenfunction to ##\hat{L_z}##, ##Ce^{im \phi}##, must be periodic. I.e. ##Ce^{i m\phi}= Ce^{i m(\phi+2 \pi)}##.

    For Hydrogen( like) system(s):

    The quantization of angular momentum is a result gained from solving the Schroedinger equation with regards to the polar-angle function (let's call it ##f(\theta)##).

    If it's agreeable that the wavefunction can be written as ##\psi(r,\phi,\theta) = R(r) f(\theta) g(\phi)##, you can split up the schroedinger equation into a lhs dependant only on the radial distance, and a rhs one dependant on the angle variables.

    On the rhs, ##g(\phi)## will disappear after some algebra, and you'll end up with legendre differential equation http://en.wikipedia.org/wiki/Legendre_polynomials for ##f(\theta)##. From that you'll find that the eigenvalue of ##f(\theta)## must be quantized.

    The energy eigenvalue can be found on the lhs. After massaging the lhs with some algebra, it will yield another legendre differential equation. From this, it can be found that the energy too is quantized.

    Here's a good video explaining it all in full detail https://www.youtube.com/watch?v=i5Dcx2B8UCo
     
    Last edited: May 12, 2014
  15. May 11, 2014 #14

    WannabeNewton

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    That's not what I meant when I said I had conceptual difficulties with the quantization of angular momentum; that's just straightforward mathematics. However it may be what Aerozeppelin was referring to with regards to his own conceptual difficulties.
     
  16. May 11, 2014 #15

    ChrisVer

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    Also, apart from straightforward results obtained by solving the Schroedinger equation, this topic could also be addressed by the symmetries of your hamiltonian.
    However, the most straightforward way to see what you called "spatial quantization" is by looking at the eigenvalue problem of the angular momentum operators:
    [itex]L_{x},L_{y},L_{z}[/itex]
    Because they satisfy this commutation relation:
    [itex][L_{i},L_{j}]=ih\epsilon_{ijk}L_{k}[/itex] with [itex]i=1,2,3[/itex] same for j,k and [itex]\epsilon_{ijk}[/itex] being the form factor ([itex]\epsilon_{123}=+1[/itex] and then for odd permutations it's -1 (eg [itex]\epsilon_{132}=1[/itex]) for even +1 (eg [itex]\epsilon_{312}=+1[/itex]) and for any same index it's zero (eg [itex]\epsilon_{113}=0[/itex]...

    You can see that by choosing a basis to represent those operators as matrices, you can only diagonalize (diagonalization of a matrix is equivalent to finding its eigenvalues) one of them... that means that you can only choose one of them to have some certain eigenvalues (those [itex]m[/itex]). Now, whether you choose the z or the x or the y it doesn't matter (physics alone don't understand what a x,y,z is), but once you've chosen it- it will be the only one which will have known eigenvalues (it can solve [itex]L_{i}|l,m>= l_{i}|l,m>[/itex]. This is something coming from the [itex]so(3)[/itex] algebra which is isomorphic to the [itex]su(2)[/itex] one, and so it's associated with the symmetries of the problem in consideration- in particular rotations as the angular momenta are the generators of spatial rotations...that was addressed by Nikitin's post.

    The same algebra (after complexification=making the ladder operators) can give you the answer for why you can have the [itex]m[/itex] values between +l and -l (and with steps of 1). But I guess the explanations already given about this, are more than enough to be understood. I am just noting this, in order to emphasize how important the commutation relation above is.
     
    Last edited: May 11, 2014
  17. May 11, 2014 #16

    ChrisVer

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    Also if you'd like to break this, you would have to break the rotational symmetry of your problem. In that case different [itex]m[/itex] would give you different energy states...But you wouldn't be able to make |m|>l
     
  18. May 12, 2014 #17
    wannabenewton: Well, I just made a general reply really. But what exactly are your "conceptual understanding difficulties"? As far as I've learned, the behaviour of a system is ruled by the wave equation. Now, since the quantization of angular momentum is a direct result from that, the "straightforward math" is the conceptual explanation. Surely you don't expect to find any intuitive reason?

    Well, OK, probably there are some explanations/theories explaining all this degrees of freedom and quantization and so on in upper grad level physics. I duno. I'm nowhere near that level in my education yet.

    oops, I meant eigenvalue of the angular momentum squared operator. So ##\hat{L^2}[fg]=\hbar \sqrt{l(l+)} [fg]##. Excuse the brain fart, I just started cramming introductory Quantum physics for the exams like a few days ago.
     
  19. May 12, 2014 #18

    ChrisVer

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    The thing is that it's not in the upper grad- things get in a better position after doing some group theory of course, but in general, all these results you could get by working with the angular momenta commutator relation, you are also getting similar results working with the algebra of the raising and lowering operators in the Harmonic Oscillator problem (which is a common way of solving this problem in the undergrad)
     
  20. May 12, 2014 #19
    Nikitin: There are actually some technical issues that your explanation is glossing over. The first problem involves "internal angular momentum" or spin. This has nothing to do with the differential operator you mention, and using it to derive quantization conditions is not satisfactory. When one uses the purely algebraic method (which still applies), one gets half-integer values of [itex]l[/itex] as well. This is why ChrisVer stresses the commutation algebra as fundamental.

    The other issue is that we do not have any physical reason why the wave function needs to be the same at [itex]\theta[/itex] and [itex]\theta + 2\pi[/itex], since the wave function carries no physical information - only its magnitude does. So we can could conceivably create wave functions that pick up arbitrary phases under the shift, and we have the same physical theory (in fact, this is intimately tied with the existence of fermions). It turns out there's no consistent way to do this with spherical harmonics, even with half-integer m. Here's an old article by Merzbacher that discusses: http://scitation.aip.org/content/aapt/journal/ajp/30/4/10.1119/1.1941984
     
  21. May 13, 2014 #20
    Thanks for pointing out the faults, guys. Hope I'll be able to understand all this stuff in a few years :)
     
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