Orbital and Spin angular momentum

The physical idea is that this quantity is related to the cosine of the angle between the classical vectors {\vec L} and {\vec S}. But I prefere to keep my operators in a Hilbert space and not pretend they are classical vectors.In summary, for the states with l=4 and s=1/2, the largest possible values of j and mj are 9/2. To calculate the angle between L and S, we can use the operator L dot S which is equal to h-bar squared. The lengths of the vectors L and S are proportional to sqrt{l(l+1)} and sqrt{s(s+1)}, respectively. The angle between L and J can be found using the cosine law, but it should
  • #1
stunner5000pt
1,461
2

Homework Statement


Consider the states for which l=4 and s=1/2.For the state wit hteh largest j, and the largest mj, calculate
a) the nagle between L and S
the angle between [itex] \mu_{l}[/itex] and [itex] \mu_{s}[/itex]
c) the angle between J and the +z axis

Homework Equations


j=l+/-s

The Attempt at a Solution


the largest possible j is 4 +1/2 which is 9/2.
the largest mj is also 9/2
since mj ranges from -j to +j.

now my prof said that this could be done gemetrically that is
since l=4, then the value of value of L is [itex] \sqrt{4(4+1)}\hbar=3\hbar[/itex]
the projection on the z axis, that is ml may be -4,-3,...,3,4

the spin s=1/2, ms=1/2
the magnitude of the spin angular momentum vector is [itex] \sqrt{3}/2 \hbar[/itex]

usig the projection of both we can draw the vectors and we can find the nagles between them

i was jjust wondering if there was a way of doing this using [itex] \vec{L} \cdot \vec{S} [/tex]

[tex] L \cdot S = |L| |S| \cos \theta [/tex]

the magnitude of L is calcualted above
siilarly for S

but how would one go about calculating L dot S??
thanks for the help!
 
Last edited:
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  • #2
stunner5000pt said:

Homework Statement


Consider the states for which l=4 and s=1/2.For the state wit hteh largest j, and the largest mj, calculate
a) the nagle between L and S
the angle between [itex] \mu_{l}[/itex] and [itex] \mu_{s}[/itex]
c) the angle between J and the +z axis


Homework Equations


j=l+/-s

The Attempt at a Solution


the largest possible j is 4 +1/2 which is 9/2.
the largest mj is also 9/2
since mj ranges from -j to +j.

now my prof said that this could be done gemetrically that is
since l=4, then the value of value of L is [itex] \sqrt{4(4+1)}\hbar=3\hbar[/itex]
the projection on the z axis, that is ml may be -4,-3,...,3,4

the spin s=1/2, ms=1/2
the magnitude of the spin angular momentum vector is [itex] \sqrt{3}/2 \hbar[/itex]

usig the projection of both we can draw the vectors and we can find the nagles between them

i was jjust wondering if there was a way of doing this using [itex] \vec{L} \cdot \vec{S} [/tex]

[tex] L \cdot S = |L| |S| \cos \theta [/tex]

the magnitude of L is calcualted above
siilarly for S

but how would one go about calculating L dot S??
thanks for the help!


You can always write
[tex] \vec{L} \cdot \vec{S} = \frac{1}{2} \bigl( {\vec J}^2 - {\vec L^2} - {\vec S}^2 \bigr) [/tex]
So this operator sandwiched between states of definite j, l and s gives [itex] \frac{\hbar^2}{2} (j(j+1) - l(l+1) -s(s+1))[/itex]. This is used in the calculation of the spin orbit interaction in hydrogen, for example.
 
  • #3
Someone correct me if I am wrong but:
The lengths of the vectors (L, S, J) are proportional to sqrt{l(l+1)}, sqrt{s(s+1)}, sqrt{j(j+1)}
Then maybe the angle between L and J can be had by the cosine law (which to me looks similar to what nrqed has above).
BTW check your sqrt{4(4+1)}.
 
Last edited:
  • #4
here we are looking for the largest possible j so

j = 9/2
so
[tex] L \cdot S = \frac{\hbar^2}{2} (j(j+1) - l(l+1) -s(s+1)) = \frac{\hbar^2}{2} \left(\frac{9}{2}\left(\frac{9}{2} +1\right) - 4(4+1) - \frac{1}{2}\left(\frac{1}{2} +1\right) \right) = \hbar^2 [/tex]

[tex] |L| = \sqrt{20} \hbar [/tex]
[tex] |S| = \sqrt{3}/2 \hbar [/tex]

are u sure it should be cosine law or dot product??
 
  • #5
stunner5000pt said:
here we are looking for the largest possible j so

j = 9/2
so
[tex] L \cdot S = \frac{\hbar^2}{2} (j(j+1) - l(l+1) -s(s+1)) = \frac{\hbar^2}{2} \left(\frac{9}{2}\left(\frac{9}{2} +1\right) - 4(4+1) - \frac{1}{2}\left(\frac{1}{2} +1\right) \right) = \hbar^2 [/tex]

[tex] |L| = \sqrt{20} \hbar [/tex]
[tex] |S| = \sqrt{3}/2 \hbar [/tex]

are u sure it should be cosine law or dot product??

Well, since S, L and J are really quantum operators, one cannot really see them as ordinary vectors. So it does not really make sense to talk about the angle between two operators. But people define the angle between [itex] {\vec L} [/itex] and [itex] {\vec S} [/itex] in a certain state of definite L,S and J as the ratio of the expectation values [tex] \frac{ < {\vec L } \cdot {\vec S} >} { \sqrt{ < {\vec L}^2 >} \sqrt{ < {\vec S}^2>}} [/tex]
 

1. What is orbital angular momentum?

Orbital angular momentum is a physical quantity that describes the rotational motion of a particle or system of particles around a fixed point, axis, or center of rotation. It is a vector quantity and is dependent on the mass, velocity, and distance of the rotating object.

2. How is orbital angular momentum different from spin angular momentum?

Orbital angular momentum refers to the motion of a particle or system of particles around a fixed point, while spin angular momentum refers to the intrinsic angular momentum of a particle, which is an inherent property of the particle and is not due to its motion.

3. How is angular momentum conserved?

Angular momentum is conserved when there is no external torque acting on a system. This means that the total angular momentum of a closed system remains constant over time, even if individual components of the system may change.

4. What is the relationship between angular momentum and energy?

Angular momentum and energy are related through the conservation of energy and the conservation of angular momentum. Changes in angular momentum can lead to changes in energy, and vice versa. For example, when an object is in orbit, its angular momentum is conserved, but as it moves farther away from the center of rotation, its kinetic energy decreases.

5. How is angular momentum used in quantum mechanics?

In quantum mechanics, angular momentum plays a crucial role in describing the behavior of subatomic particles. The spin angular momentum of particles, which is quantized, can determine their quantum states and properties. The orbital angular momentum of particles also plays a role in determining their energy levels and behavior in certain physical systems.

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