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Spectrum of angular momentum operators

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paco_uk
#1
Sep22-10, 06:20 AM
P: 22
1. The problem statement, all variables and given/known data

I am trying to understand the allowed eigenvalues for the angular momentum operators J and L. In particular why, mj can take integer and half-integer values whereas ml can take only integer values.

2. Relevant equations

I have learned about angular momentum operators as generators of rotations.

So for complete rotation of the system:

[tex]

U_{rotation}( \bold{\alpha}) = \exp ( - i \bold{\alpha} \cdot \bold{J} )

[/tex]

where U is a rotation operator and J, the total angular momentum, is the generator of this rotation.

And for circular translation:

[tex]

U_{circular translation}( \bold{\alpha}) = \exp ( - i \bold{\alpha} \cdot \bold{L} )


[/tex]

where U generates circular translations and L, the orbital angular momentum, is the generator of this motion.


I also have:

[tex]

\bold{L} = \bold{r} \times \bold{p}

[/tex]

Which can be shown to be consistent with the previous equation. ( I am mostly using Binney: http://www-thphys.physics.ox.ac.uk/p...ney/QBhome.htm)

3. The attempt at a solution

I am happy with the proof that the eigenvalues of Lz and Jz must take half integer or integer values. I am confused about the argument that ml, the eigenvalues of Lz must take only integer values.

The argument is that moving the system all the way round in a circle ( so alpha is 2 pi radians) must leave the state unchanged. Plugging this into the equation involving L we find that ml must take only integer values unlike mj which can take half-integer values.

I don't understand why we can't apply the same argument to rotations through 2 pi and then argue that mj must take only integer values for the same reasons.
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Thaakisfox
#2
Sep22-10, 06:57 AM
P: 263
J is the total angular momentum. This means that it accounts for spin too. briefly it is the "sum" of the orbital angular momentum and spin. But this property (turn around by 2pi and dont flip sign) is not at all true for spin, hence it cant be a requirement for J either. It can only be a requirement for L.
paco_uk
#3
Sep22-10, 07:16 AM
P: 22
Thank you. I see what you mean that J can't be forced to obey a rule which spin doesn't obey but I still don't understand the difference between L and J which makes the original argument true for one and not the other.

In fact, the more I think about it the less I understand the argument for integer values of ml. Why does a rotation of 2 pi have to leave you with the same state? Common sense dictates that it has to describe the same physics but couldn't this be achieved by the same state with an arbitrary phase factor in front (as is the case for J)?

Thaakisfox
#4
Sep22-10, 07:27 AM
P: 263
Spectrum of angular momentum operators

Just as i answered, its not true for J because its not true for spin. Hence it cant be true for J.
L and J are not the same operators.
paco_uk
#5
Sep22-10, 07:40 AM
P: 22
Thank you, I understand there is a difference between L and J and you have made it clear why the argument can't hold for J. I'm afraid I still don't understand why the argument does hold for L.
Thaakisfox
#6
Sep22-10, 07:54 AM
P: 263
Right. So the z component of L: Lz is the generator of rotations around the z axis. explicitly written (take phi to be the angle of rotation):

Lz=-i*hbar*d/dphi

the eigenfunction of this operator is exp(i*m*phi). Now if we just rotate by 2pi, i.e. fully around, then the system cant change. For example you look at something in your room, then you turn around 360 degrees and look at it again, then nothing physical has changed. this is incorporated as a condition that m has to be an integer.
jambaugh
#7
Sep22-10, 08:06 AM
Sci Advisor
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Quote Quote by paco_uk View Post
Thank you. I see what you mean that J can't be forced to obey a rule which spin doesn't obey but I still don't understand the difference between L and J which makes the original argument true for one and not the other.

In fact, the more I think about it the less I understand the argument for integer values of ml. Why does a rotation of 2 pi have to leave you with the same state? Common sense dictates that it has to describe the same physics but couldn't this be achieved by the same state with an arbitrary phase factor in front (as is the case for J)?
Firstly a 2pi = 360deg rotation is by definition the identity transformation so by definition equivalent to doing nothing. The real question is how spinors can not be the same after a 2pi rotation if they represent a physical system. The answer of course is they do not represent the system itself but projectively represent information about the system hence the sign change for 2pi rotations of spinors has no direct physical meaning. psi and -psi correspond to the same physical mode.

As to why the orbital component of angular momentum should be integral valued, it is simply the only possibility for single valued wave-functions of an orbital particle. The angular momentum operators e.g. Sz take the form of differential operators on the wave-function psi(x,y,z).

[tex] L_z \propto [x\frac{\partial}{\partial y} - y\frac{\partial}{\partial x}][/tex]

If you then rewrite this in polar form you see that the eigen-functions are the spherical harmonics with integer eigen-values. You then see the spherical harmonics expressing the angular components of the wave-functions for a particle orbiting a central potential.

One then adds in the intrinsic angular momentum of the orbiting particle which we are surprised to discover may take on half integer values. Hence the total J is integral (orbital) plus half integral (intrinsic) which then may be half integral (3 +1/2 for example).

Ultimately though the answer to your question, "Why" is that that's how nature is observed to behave. There are possible exceptions, see for example anyons which may manifest as quasi-particles in a two dimensional material (such as thin films).
paco_uk
#8
Sep22-10, 08:19 AM
P: 22
Thank you both very much. It looks like I need to do some more reading about spin.


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