Spin Angular Momentum and Operators

In summary: I was trying to use the expression for ##\hat{J}_+##, not ##\hat{J}_x##.In summary, the homework equations state that: - Angular momentum is conserved- The expression for angular momentum is given, but I'm unsure as to how to derive it- You can use the completeness relation for the space in question to derive the expression for angular momentum- Or, you can look at the problem from a more general Linear Algebra point of view
  • #1
samjohnny
84
1

Homework Statement



Attached
question.JPG


Homework Equations



##J_+|j,m⟩ = ћ\sqrt{j(j+1)-m(m+1)} |j,m+1⟩##
##J_-|j,m⟩ = ћ\sqrt{j(j+1)-m(m-1)} |j,m-1⟩##
##J_z|j,m⟩ = mћ |j,m⟩##
##J^2|j,m⟩ = ћ^2j(j+1) |j,m⟩##

The Attempt at a Solution



Hi there,

For part a, the expression we're looking for is given, but I'm unsure as to how to derive it from the equation for ##J_+## given above. Any help would be very much appreciated.
 
Physics news on Phys.org
  • #2
Try applying the expression you're given for [itex]J_+[/itex] in part a) to each of the eigenstates of [itex]J_z[/itex] individually.
 
  • #3
samjohnny said:
For part a
You can use the completeness relation for the space in question, the space of angular momentum ##j=3/2##, which is ##\sum_m |m\rangle \langle m| = \mathbf{1}## with ##m=-3/2,-1/2,1/2,3/2##. Since this relation is equal to an identity operator, you can place it wherever and how many times you want around an operator without altering the operator itself.
 
  • Like
Likes samjohnny and TSny
  • #4
blue_leaf77 said:
You can use the completeness relation for the space in question, the space of angular momentum ##j=3/2##, which is ##\sum_m |m\rangle \langle m| = \mathbf{1}## with ##m=-3/2,-1/2,1/2,3/2##. Since this relation is equal to an identity operator, you can place it wherever and how many times you want around an operator without altering the operator itself.
This is unnecessary. Simply apply [itex]J_+[/itex] to the [itex]J_z[/itex] eigenstates and use the fact that the eigenstates are orthogonal (i.e., [itex]\langle m | m' \rangle = \delta_{mm'}[/itex]).
 
  • #5
TeethWhitener said:
This is unnecessary.
It's just another and, in my opinion, more elegant way of doing the problem.
 
  • Like
Likes blue_leaf77 and DrClaude
  • #6
samjohnny said:
For part a, the expression we're looking for is given, but I'm unsure as to how to derive it from the equation for ##J_+## given above. Any help would be very much appreciated.

You can also look at this from a more general Linear Algebra point of view. In general, if you have a basis ##(e_i)## then ##|e_i> <e_j|## is an operator that picks out the ##j^{th}## component of a vector (##v_j##) and outputs a vector ##v_j e_i##. And, in fact, the set of all such operators is a basis for your operator space.

You can, therefore, express any operator as a linear combination of these ##|e_i> <e_j|## operators. If you think about what the raising and lowering opertors do, you may see that they are well suited to being expressed in this basis. In this case, the ##(e_i)## are the states with ##m = \pm \frac{1}{2}, \pm \frac{3}{2}##
 
  • Like
Likes blue_leaf77
  • #7
vela said:
It's just another and, in my opinion, more elegant way of doing the problem.
Ok I'm thoroughly confused. How the heck do you solve part a of this problem using the completeness relation? I guess it's a little too elegant for me.
 
  • #8
TeethWhitener said:
Ok I'm thoroughly confused. How the heck do you solve part a of this problem using the completeness relation? I guess it's a little too elegant for me.
The simplest way is to start with TeetWhitener's advice. First thing first: do you know how to apply the operator given in the answer to each of the ##J_z## states? For example, what do you get if you apply it to ##|1/2\rangle ## ?
 
  • #9
nrqed said:
The simplest way is to start with TeetWhitener's advice. First thing first: do you know how to apply the operator given in the answer to each of the ##J_z## states? For example, what do you get if you apply it to ##|1/2\rangle ## ?
I think TeethWhitener knows how he would solve it using his own advice. ;)
 
  • Like
Likes nrqed
  • #10
vela said:
I think TeethWhitener knows how he would solve it using his own advice. ;)
AH! My bad! Hehehe...
 
  • #11
TeethWhitener said:
Ok I'm thoroughly confused. How the heck do you solve part a of this problem using the completeness relation? I guess it's a little too elegant for me.
Fo any operator A, you may write
## A = \sum_{n,m} |n\rangle\langle n| A |m\rangle \langle m| ##
Knowing the effect of the operator on each basis state then yields the result.
 
  • #12
TeethWhitener said:
Ok I'm thoroughly confused. How the heck do you solve part a of this problem using the completeness relation?
##J_+ = J_+ \cdot \mathbb{1}## where ##\mathbb{1}## is the identity operator. Use blue_leaf's suggestion for the identity operator.
 
  • #13
Many thanks for all the replies. Hearing of both methods was very helpful.

Just on a final note, for the last part of the question, part d), I'm applying the expression given for ##J_+## to ##|ψ⟩## as it's given in d). However 'm ending up with:

##J_+|ψ⟩= \frac{h}{2\sqrt{2}} (\sqrt{3}|3/2⟩ - 2|1/2⟩ - 3|-1/2⟩ ) ##

Not sure if I'm approaching the question in the right way.
 
  • #14
##\hat{J}_x##, not ##\hat{J}_+##.
 
  • Like
Likes samjohnny
  • #15
vela said:
##\hat{J}_x##, not ##\hat{J}_+##.

D'oh!
 

Related to Spin Angular Momentum and Operators

1. What is spin angular momentum?

Spin angular momentum is a quantum mechanical property of particles that describes the intrinsic angular momentum of a particle. It is one of the two types of angular momentum in quantum mechanics, the other being orbital angular momentum. Spin angular momentum is a fundamental property of particles and cannot be explained by classical mechanics.

2. How is spin angular momentum measured?

Spin angular momentum is measured using spin operators, which are mathematical operators that act on the wavefunction of a particle. The eigenvalues of these operators give the possible values of spin angular momentum that can be measured. The unit of measurement for spin angular momentum is the reduced Planck constant, also known as h-bar.

3. What is the relationship between spin and magnetic moment?

Spin is directly related to the magnetic moment of a particle. This means that particles with nonzero spin have a magnetic moment, while particles with zero spin have no magnetic moment. The direction of the spin angular momentum is also related to the direction of the magnetic moment.

4. Can spin angular momentum be changed?

No, spin angular momentum is an intrinsic property of particles and cannot be changed. This means that the spin of a particle cannot be increased or decreased, it can only take on certain discrete values. However, the direction of the spin can be changed by applying a magnetic field.

5. Why is spin angular momentum important?

Spin angular momentum is important in many areas of physics, including quantum mechanics, particle physics, and condensed matter physics. It is crucial for understanding the behavior of subatomic particles and the structure of atoms. Spin also plays a role in many technological applications, such as magnetic storage devices and quantum computing.

Back
Top